Example free response question from AP macroeconomics

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Language: en

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- [Instructor] In this video,
I want to tackle an entire

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AP macroeconomics free
response exercise with you.

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Assume that the economy of Country X has

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an actual unemployment rate of 7%,

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a natural rate of unemployment of 5%,

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and an inflation rate of 3%.

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Using the numerical values given above,

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draw a correctly labeled
graph of the short-run

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and long-run Phillips curves.

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Label the current short-run
equilibrium as point B.

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Plot the numerical values
above on the graph.

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So pause this video if
you are inspired to do so,

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but I will now work through it.

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So remember, Phillips
curves show the relationship

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or the theoretical relationship

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between the unemployment
rate and the inflation rate.

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So I'm gonna do the inflation rate

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in the vertical axis which is typical.

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And then on the horizontal axis,

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I am going to do my unemployment rate.

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So this is going to be my

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unemployment rate

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which is going to be a percentage.

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Now we want to graph the short-run

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and long-run Phillips curves.

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Let's do the long-run first
because we've seen before

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the long-run just sets
our unemployment rate

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at the natural rate of unemployment,

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and it isn't related
to our inflation rate.

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It'll just be a vertical line.

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And so it'll be a vertical
line at our natural rate

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of unemployment which is 5%.

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So maybe it looks just like this.

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So I could call that our
long-run Phillips curve,

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and it's going to be right there at 5%.

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And now I have to do the
short-run Phillips curve,

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and that will show a relationship

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between inflation rate and unemployment.

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And just think about what's going on.

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If you have low rate of unemployment,

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especially if it's below your
natural rate of unemployment,

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well then there's a lot
of demand for people.

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And so people say, hey,
if you want me to work,

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you gotta pay me a little bit more,

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and so that could just lead
to a higher inflation rate.

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And then if a lot of
people are unemployed,

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they might be willing to work for less

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or they might have less
money in their pocket

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with which to drive up the
prices, and so you will have

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this inverse relationship right over here.

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So this is the short-run Phillips curve,

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which is downward sloping.

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And then they say,

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label the short-run
equilibrium as point B.

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So let's say this is
point B right over here.

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And they say the short-run
equilibrium we have

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an unemployment rate of 7%
and an inflation rate of 3%.

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So our unemployment rate
right over here is 7%,

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and our inflation rate
right over here is 3%.

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All right, let's do the next section.

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Assume that the government of Country X

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takes no policy action
to reduce unemployment.

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In the long run, which of the
following shift to the right,

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shift to the left, or remain the same?

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So here they're saying short-run

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aggregate supply curve, explain.

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So here it's kinda tricky
'cause you might be thinking

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they're asking about what you just drew.

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But here they're talking
about aggregate supply.

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So let me draw a graph to
even help to visualize this.

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So if we're talking about aggregate demand

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and aggregate supply,

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our vertical axis is going
to be our price level,

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I'll just call that PL,

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and our horizontal axis that
is going to be our real GDP.

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So this is real GDP
right over here, G-D-P.

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Now you're just going to
have a long-run supply curve

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which is vertical.

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That's just the full employment
output for our country.

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So that's the long-run aggregate supply.

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But what about the short-run
aggregate supply curve?

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Well, that's going to be upward sloping.

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If price levels are low,
people might not be willing

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to output a lot, and if
price levels are high,

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people will output more.

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So our short-run aggregate supply

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would look like that.

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I'll call that sub one,
since we're gonna think

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about how it shifts,

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and then aggregate demand
would look something like this.

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Let me draw it like that.

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And notice, our equilibrium
point right over here,

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let me call that aggregate
demand right over here.

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Let's call that Y sub one,

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and we are at price level sub one.

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I drew it to the left

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of the long-run aggregate supply curve.

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I drew it to the left of
the full employment output

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because we are dealing
with a recession here.

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Our unemployment rate is higher

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than the natural level of unemployment.

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So if our actual
unemployment rate is higher

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than natural rate of
unemployment, what will happen

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to the short-run aggregate supply?

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So one way to think about
it, at a given price level,

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because there's people out
there looking for a job,

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you might be able to get more output.

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Or for a given amount of
output, it might cost less

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because there's just people out there

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competing for that work.

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And so you would have your
short-run aggregate supply curve

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shift to the right, short-run
aggregate supply sub two.

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And then your equilibrium
price level would go down,

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price level sub two would go down.

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So we could say because

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of high unemployment,

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that could apply wage pressure.

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We could say wages come down

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which would shift

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the short-run aggregate supply curve

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to the right.

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And there's a couple of
ways to think about that.

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You would have more output

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at a given price level.

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You could also think at
a given output level,

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you would have a lower price level,

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at a given price level.

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Part two, long-run Phillips curve,

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so that's this vertical
line right over here.

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And the thing to appreciate
is the long-run Phillips curve

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or the long-run aggregate supply curve,

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these don't change unless
something structurally changes

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in the economy, unless the economy changes

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in some very fundamental way,

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maybe a change in education
levels, change in population,

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or change in technology.

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And so here we would say
it just remains the same.

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All right, we have more parts here.

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Now let's go to part (c).

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Identify a fiscal policy
action that could be used

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to reduce the unemployment
rate in the short run.

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So you have to be very careful here.

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They're saying a fiscal policy action,

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not a monetary policy.

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If you said hey, we would
change the federal funds rate

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or we would increase the money supply

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or decrease the money supply,

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those would be monetary actions.

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We care about a fiscal policy action.

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And this would be in
relation to lowering taxes

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or raising taxes or increasing

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or decreasing government spending.

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Well, if we want to reduce
the unemployment rate,

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one way to do the that would be

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to shift aggregate demand to the right.

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And one way to do that,
would be to put more money

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in people's pockets,

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and one way to do that,
is to have a tax cut.

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Draw a correctly labeled
graph of aggregate demand

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and short-run aggregate
supply, and show the impact

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on the equilibrium
price level and real GDP

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of the fiscal policy action
identified in part (c).

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All right, let me draw that.

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So this is going to be so that we have

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our price level axis up here,

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and we just drew something
very similar to this, real GDP.

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And now let's draw our
short-run aggregate supply

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which we have seen before.

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That would be upward sloping,

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as the price level increases
or the economy might be willing

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to output more, so that's
short-run aggregate supply.

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And then let's draw an
aggregate demand curve.

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So let's call that AD sub one.

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And then you have the equilibrium output,

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let's call that Y sub one.

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And you have your equilibrium
price level, PL sub one.

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And now if you have a
tax cut, that would shift

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aggregate demand to the right.

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At any given price level,
people are gonna want more.

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They're gonna demand more

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'cause now they have more
money in their pockets,

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and so it's going to shift to the right.

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So I'll do a aggregate demand sub two.

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And now we have a different
equilibrium real GDP,

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so that is going to be Y sub two.

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And it happens, and then we
have price level sub two.

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So you see our price level goes up

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and our aggregate output, our GDP,

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our real GDP, goes up as well.

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Based on the change in real
GDP identified in part (d),

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will the supply of Country X's currency

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in the foreign exchange market increase,

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decrease, or remain the same, explain?

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The way I think about it is if
you have real GDP increasing,

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you're in a situation where you just have

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more economic activity, the
national income has gone up.

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And if national income has
gone up, people are gonna do

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a lot more of everything
including buying imports.

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And to buy imports, they would
have to increase the supply

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of their currency in exchange markets

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because they want to convert
it into foreign currencies

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to buy those imports, and
so this will increase.

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And we could say, because national income

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has gone up,

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people will buy more imports,

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so the supply

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of Country X's currency

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for exchange will go up.

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Currency X's currency

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for exchange

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will go up.

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All right, part (f).

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Based on your answer
to part (e) and assume

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a flexible exchange rate system,

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will Country X's currency
appreciate, depreciate,

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or remain the same in the
foreign exchange market?

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Well, if you hold all else equal,

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but you increase the supply of something,

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well, then the price of
it is going to go down.

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If the demand for it stays constant,

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but you increase the supply,

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and that's what we just
talked about in part (e),

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well, then the price is going to go down.

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And if we're talking about
the price of a currency

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and we say it's going
down, we would say that

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that currency is depreciating,

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so it would depreciate, and we're done.


Justification with the mean value theorem: equation

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Kind: captions
Language: en

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- [Tutor] Let g of x equal one over x.

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Can we use the mean value theorem

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to say that the equation g
prime of x is equal to one half

00:00:06.950 --> 00:00:08.580
has a solution where negative one

00:00:08.580 --> 00:00:10.170
is less than x is less than two,

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if so, write a justification.

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Alright, pause this video

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and see if you can figure that out.

00:00:15.510 --> 00:00:17.750
So the key to using
the mean value theorem,

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even before you even think about using it,

00:00:19.810 --> 00:00:20.770
you have to make sure

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that you are continuous
over the closed interval

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and differentiable over the open interval,

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so this is the open interval here

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and then the closed interval
would include the end points.

00:00:30.540 --> 00:00:32.370
But you might immediately realize

00:00:32.370 --> 00:00:35.210
that both of these intervals
contain x equals zero

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and if x equals zero,
the function is undefined

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and if it's undefined there,

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well it's not going to continuous

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or differentiable at that point

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and so no,

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not continuous

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or differentiable,

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differentiable

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over the interval,

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over the interval.

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Alright, let's do the second part.

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Can we use the mean value theorem

00:01:00.330 --> 00:01:03.010
to say that there is a value
c such that g prime of c

00:01:03.010 --> 00:01:04.530
is equal to negative one half

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and one is less than c is less than two

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if so, write a justification.

00:01:08.390 --> 00:01:10.490
So pause the video again.

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Alright, so in this
situation, between one and two

00:01:13.100 --> 00:01:14.836
on both the open and the closed intervals,

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well, this is a rational function

00:01:16.580 --> 00:01:18.360
and a rational function
is going to be continuous

00:01:18.360 --> 00:01:20.900
and differentiable at
every point in its domain

00:01:20.900 --> 00:01:22.980
and its domain completely contains

00:01:22.980 --> 00:01:24.847
this open and closed interval

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or another way to think about it,

00:01:26.260 --> 00:01:28.140
every point on this open interval

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and on the closed
interval is in the domain,

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so we can write g of x

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is a rational

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function,

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which lets us know that it is continuous

00:01:43.420 --> 00:01:48.420
and differentiable at
every point in its domain,

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at every point

00:01:53.560 --> 00:01:55.963
in its domain,

00:01:57.460 --> 00:01:59.830
the closed interval from one to two

00:01:59.830 --> 00:02:02.580
is in domain

00:02:03.623 --> 00:02:06.560
and so now, let's see what
the average rate of change is

00:02:06.560 --> 00:02:08.020
from one to two

00:02:08.020 --> 00:02:10.680
and so we get g of two

00:02:10.680 --> 00:02:12.380
minus g of one

00:02:12.380 --> 00:02:15.420
over two minus one is equal to one half

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minus one over one, which is
equal to negative one half,

00:02:20.110 --> 00:02:21.510
therefore,

00:02:21.510 --> 00:02:23.320
therefore

00:02:23.320 --> 00:02:25.540
by the

00:02:25.540 --> 00:02:27.930
mean value theorem,

00:02:27.930 --> 00:02:32.537
there must be a c

00:02:33.440 --> 00:02:37.540
where one is less than c is less than two

00:02:37.540 --> 00:02:42.540
and g prime of c is equal to

00:02:42.750 --> 00:02:46.390
the average rate of change
between the end points,

00:02:46.390 --> 00:02:48.637
negative one half

00:02:48.637 --> 00:02:51.463
and we're done, so we can put
a big yes right over there

00:02:51.463 --> 00:02:53.283
and then this is our justification.


Justification with the mean value theorem: table

WEBVTT
Kind: captions
Language: en

00:00:00.270 --> 00:00:01.940
- [Instructor] The table
gives selected values

00:00:01.940 --> 00:00:03.980
of the differentiable function f.

00:00:03.980 --> 00:00:07.060
All right, can we use
the mean value theorem

00:00:07.060 --> 00:00:10.300
to say that there is a value
c such that f prime of c

00:00:10.300 --> 00:00:13.030
is equal to five and c
is between four and six?

00:00:13.030 --> 00:00:15.300
If so, write a justification.

00:00:15.300 --> 00:00:19.000
Well it meets, to meet the,
to use the mean value theorem,

00:00:19.000 --> 00:00:21.250
you have to be differentiable
over the open interval

00:00:21.250 --> 00:00:23.630
and continuous over the closed interval,

00:00:23.630 --> 00:00:25.000
so it seems like we've met that.

00:00:25.000 --> 00:00:26.880
Because if you're
differentiable over an interval,

00:00:26.880 --> 00:00:28.670
you're definitely continuous
over that interval.

00:00:28.670 --> 00:00:31.820
It's saying that it's just
a generally differentiable

00:00:31.820 --> 00:00:33.500
function f I guess over any interval.

00:00:33.500 --> 00:00:35.080
But the next part is to say,

00:00:35.080 --> 00:00:38.060
all right, if that condition is met,

00:00:38.060 --> 00:00:39.680
then the slope of the secant line

00:00:39.680 --> 00:00:43.941
between four comma f of
four and six comma f of six,

00:00:43.941 --> 00:00:48.100
that some at least one point
in between four and six

00:00:48.100 --> 00:00:49.860
will have a derivative that is equal

00:00:49.860 --> 00:00:51.410
to the slope of the secant line.

00:00:51.410 --> 00:00:52.750
And so let's figure out what the slope

00:00:52.750 --> 00:00:55.890
of the secant line is
between four comma f of four

00:00:55.890 --> 00:00:57.290
and six comma f of six.

00:00:57.290 --> 00:00:59.090
And if it's equal to five,

00:00:59.090 --> 00:01:01.140
then we could use the mean value theorem.

00:01:02.076 --> 00:01:02.909
If it's not equal to five,

00:01:02.909 --> 00:01:04.040
then the mean value
theorem would not apply.

00:01:04.040 --> 00:01:05.410
And so let's do that.

00:01:05.410 --> 00:01:08.600
f of six minus f of four,

00:01:08.600 --> 00:01:12.380
all of that over six
minus four is equal to

00:01:12.380 --> 00:01:13.910
seven minus three

00:01:14.960 --> 00:01:17.130
over two, which is equal to two.

00:01:17.130 --> 00:01:20.270
So two not equal to five.

00:01:20.270 --> 00:01:24.853
So mean value theorem doesn't apply.

00:01:26.740 --> 00:01:29.237
All right let's, I'll
put an exclamation mark

00:01:29.237 --> 00:01:31.410
there for emphasis.

00:01:31.410 --> 00:01:32.650
All right let's do the next part.

00:01:32.650 --> 00:01:34.972
Can we use the mean value theorem to say

00:01:34.972 --> 00:01:36.650
that the equation f prime of
x is equal to negative one

00:01:36.650 --> 00:01:37.620
has a solution?

00:01:37.620 --> 00:01:39.300
And now the interval's from zero to two.

00:01:39.300 --> 00:01:41.140
If so, write a justification.

00:01:41.140 --> 00:01:42.650
All right so let's see this.

00:01:42.650 --> 00:01:45.940
So if we were to take the
slope of the secant line,

00:01:45.940 --> 00:01:49.160
so f of two minus f of zero,

00:01:49.160 --> 00:01:51.070
all that over two minus zero.

00:01:51.070 --> 00:01:53.993
This is equal to negative two minus zero,

00:01:54.830 --> 00:01:56.380
all of that over two,

00:01:56.380 --> 00:01:57.770
which is equal to negative two over two,

00:01:57.770 --> 00:02:00.400
which is equal to negative one.

00:02:00.400 --> 00:02:04.250
And so and we also know that we meet

00:02:04.250 --> 00:02:06.660
the continuity and
differentiability conditions,

00:02:06.660 --> 00:02:10.280
and so we could say and since

00:02:11.770 --> 00:02:15.800
f is generally differentiable,

00:02:15.800 --> 00:02:18.483
generally differentiable,

00:02:19.350 --> 00:02:20.593
differentiable,

00:02:21.830 --> 00:02:25.470
it will be

00:02:25.470 --> 00:02:27.767
differentiable, differentiable,

00:02:29.490 --> 00:02:32.620
and, and continuous

00:02:33.550 --> 00:02:36.680
over the interval from zero to two,

00:02:36.680 --> 00:02:37.620
and I'll say the closed interval.

00:02:37.620 --> 00:02:39.590
You just have to be differentiable
over the open interval,

00:02:39.590 --> 00:02:41.720
but it's even better I guess
if you're differentiable

00:02:41.720 --> 00:02:43.510
over the closed interval because
you have to be continuous

00:02:43.510 --> 00:02:44.920
over the closed interval.

00:02:44.920 --> 00:02:47.330
And so, and since f is
generally differentiable,

00:02:47.330 --> 00:02:51.200
it will be differentiable
and continuous over zero two.

00:02:51.200 --> 00:02:55.130
So the mean value theorem tells us,

00:02:55.130 --> 00:02:58.980
tells us, that there

00:02:58.980 --> 00:03:01.610
is an x

00:03:01.610 --> 00:03:04.960
in that interval from zero to two

00:03:04.960 --> 00:03:08.130
such that

00:03:08.130 --> 00:03:10.260
f prime of x

00:03:10.260 --> 00:03:13.730
is equal to that secant slope,

00:03:13.730 --> 00:03:15.560
or you could say that
average rate of change,

00:03:15.560 --> 00:03:18.700
is equal to negative one.

00:03:18.700 --> 00:03:19.970
And so I could write,

00:03:19.970 --> 00:03:23.510
yes, yes, and then this
would be my justification.

00:03:23.510 --> 00:03:25.150
This is the slope of the secant line

00:03:25.150 --> 00:03:26.430
or the average rate of change,

00:03:26.430 --> 00:03:28.490
and since f is generally differentiable,

00:03:28.490 --> 00:03:30.360
it will be differentiable and continuous

00:03:30.360 --> 00:03:31.560
over the closed interval.

00:03:31.560 --> 00:03:33.570
So the mean value theorem tells us

00:03:33.570 --> 00:03:35.930
that there is an x in this interval

00:03:35.930 --> 00:03:39.780
such that f prime of x
is equal to negative one.

00:03:39.780 --> 00:03:40.613
And we're done.


Justification with the intermediate value theorem: equation

WEBVTT
Kind: captions
Language: en

00:00:00.350 --> 00:00:03.110
- [Instructor] Let g
of x equal one over x.

00:00:03.110 --> 00:00:06.140
Can we use the intermediate value theorem

00:00:06.140 --> 00:00:08.028
to say that there is a value c,

00:00:08.028 --> 00:00:11.067
such that g of c is equal to zero,

00:00:11.067 --> 00:00:14.380
and negative one is
less than or equal to c,

00:00:14.380 --> 00:00:15.900
is less than or equal to one?

00:00:15.900 --> 00:00:18.204
If so, write a justification.

00:00:18.204 --> 00:00:21.890
So in order to even use the
intermediate value theorem,

00:00:21.890 --> 00:00:24.420
you have to be continuous
over the interval

00:00:24.420 --> 00:00:25.253
that you care about.

00:00:25.253 --> 00:00:27.450
And this interval that we
care about is from x equals

00:00:27.450 --> 00:00:29.350
negative one to one.

00:00:29.350 --> 00:00:32.953
And, one over x is not
continuous over that interval,

00:00:32.953 --> 00:00:36.240
it is not defined when x is equal to zero.

00:00:36.240 --> 00:00:40.693
And so, we could say, no, because,

00:00:43.551 --> 00:00:45.463
g of x not defined, or I
could say not continuous.

00:00:52.417 --> 00:00:55.290
It's also not defined on
every point of the interval,

00:00:55.290 --> 00:01:00.290
but let's say not continuous
over the closed interval

00:01:00.950 --> 00:01:03.290
from negative one to one.

00:01:03.290 --> 00:01:07.960
And we could even put in
parentheses not defined,

00:01:07.960 --> 00:01:11.270
at x is equal to zero.

00:01:11.270 --> 00:01:13.288
All right, now let's start
asking the second question.

00:01:13.288 --> 00:01:15.800
Can we use the intermediate
value theorem to say

00:01:15.800 --> 00:01:17.934
that the equation g of x is equal to 3/4

00:01:17.934 --> 00:01:21.510
has a solution where 1 is
less than or equal to x,

00:01:21.510 --> 00:01:23.270
is less than or equal to two?

00:01:23.270 --> 00:01:24.817
If so, write a justification.

00:01:24.817 --> 00:01:26.930
All right so first let's
look at the interval.

00:01:26.930 --> 00:01:29.444
If we're thinking about the
interval from one to two,

00:01:29.444 --> 00:01:31.900
well, yeah, our function
is going to be continuous

00:01:31.900 --> 00:01:36.900
over that interval, so we
could say g of x is continuous

00:01:40.849 --> 00:01:45.849
on the closed interval from one to two.

00:01:46.010 --> 00:01:48.200
And if you wanted to put
more justification there,

00:01:48.200 --> 00:01:52.373
you could say g defined
for all real numbers,

00:02:00.200 --> 00:02:03.373
such that x does not equal zero.

00:02:04.830 --> 00:02:08.380
I could write g of x
defined for all real numbers

00:02:08.380 --> 00:02:10.296
such that x does not equal to zero,

00:02:10.296 --> 00:02:14.740
and you could say rational
functions like one over x,

00:02:14.740 --> 00:02:19.337
are continuous at all
points in their domains.

00:02:27.520 --> 00:02:30.306
That's going really
establishing that g of x

00:02:30.306 --> 00:02:32.200
is continuous on that interval.

00:02:32.200 --> 00:02:35.473
And then we wanna see what
values does g take over,

00:02:36.790 --> 00:02:38.940
take on at the end point, or actually,

00:02:38.940 --> 00:02:41.060
these are the end points we're
looking at right over here.

00:02:41.060 --> 00:02:45.500
G of one is going to be
equal to one over one is one,

00:02:45.500 --> 00:02:50.500
and g of two is going to be one over two.

00:02:51.254 --> 00:02:56.254
So, 3/4 is between g of one and g of two,

00:03:02.910 --> 00:03:07.910
so by the intermediate value theorem,

00:03:08.015 --> 00:03:12.233
there must be an x that is in the interval

00:03:21.305 --> 00:03:24.993
from where it's talking about
the interval from one to two,

00:03:26.205 --> 00:03:31.205
such that g of x is equal to 3/4.

00:03:34.670 --> 00:03:39.155
And so, yes, we can use the
intermediate value theorem

00:03:39.155 --> 00:03:42.240
to say that the equation
g of x is equal to 3/4

00:03:42.240 --> 00:03:44.723
has a solution, and we are done.


Justification with the intermediate value theorem: table

WEBVTT
Kind: captions
Language: en

00:00:00.350 --> 00:00:02.240
- [Instructor] The table
gives selected values

00:00:02.240 --> 00:00:04.040
of the continuous function f.

00:00:04.040 --> 00:00:05.310
All right, fair enough.

00:00:05.310 --> 00:00:08.770
Can we use the intermediate value theorem

00:00:08.770 --> 00:00:11.930
to say that the equation f
of x equal is equal to zero

00:00:11.930 --> 00:00:13.320
has a solution where four

00:00:13.320 --> 00:00:15.710
is less than or equal to x
is less than or equal to six?

00:00:15.710 --> 00:00:17.860
If so, write a justification.

00:00:17.860 --> 00:00:20.030
So, pause this video and see
if you can think about this

00:00:20.030 --> 00:00:22.210
on your own before we do it together.

00:00:22.210 --> 00:00:23.683
Okay, well let's just
visualize what's going on

00:00:23.683 --> 00:00:27.230
and visually think about the
intermediate value theorem.

00:00:27.230 --> 00:00:30.350
So, if that's my y-axis there

00:00:30.350 --> 00:00:35.350
and then let's say that this is my x-axis

00:00:35.600 --> 00:00:37.170
right over here.

00:00:37.170 --> 00:00:39.470
We've been given some points over here.

00:00:39.470 --> 00:00:42.370
We know when x is equal to
zero, f of x is equal to zero.

00:00:42.370 --> 00:00:43.240
Let me draw those.

00:00:43.240 --> 00:00:44.910
So, we have that point.

00:00:44.910 --> 00:00:47.930
When x is equal to two,

00:00:47.930 --> 00:00:50.580
y or f of x, y equals f of x

00:00:50.580 --> 00:00:52.520
is gonna be equal to a negative two.

00:00:52.520 --> 00:00:55.600
So, we have a negative
two right over there.

00:00:55.600 --> 00:00:59.410
When x is equal to four, so, three, four,

00:00:59.410 --> 00:01:00.910
f of x is equal to three.

00:01:00.910 --> 00:01:02.550
One, two, three.

00:01:02.550 --> 00:01:04.110
I'm doing it on a slightly different scale

00:01:04.110 --> 00:01:05.890
so that I can show everything.

00:01:05.890 --> 00:01:09.440
And when x is equal to six, so, five, six,

00:01:09.440 --> 00:01:11.130
f of x is equal to seven.

00:01:11.130 --> 00:01:12.840
Three, four,

00:01:12.840 --> 00:01:14.720
five, six, seven.

00:01:14.720 --> 00:01:17.280
So, right over here.

00:01:17.280 --> 00:01:20.300
Now, they also tell us that
our function is continuous.

00:01:20.300 --> 00:01:22.930
So, one intuitive way of
thinking about continuity

00:01:22.930 --> 00:01:25.430
is I can connect all of these dots

00:01:25.430 --> 00:01:27.120
without lifting my pencil.

00:01:27.120 --> 00:01:28.470
So, the function might look,

00:01:28.470 --> 00:01:30.060
I'm just gonna make up some stuff,

00:01:30.060 --> 00:01:31.440
it might look something,

00:01:31.440 --> 00:01:33.880
anything like what I just drew just now.

00:01:33.880 --> 00:01:35.870
And it could have even wilder fluctuations

00:01:35.870 --> 00:01:38.390
but that is what my f looks like.

00:01:38.390 --> 00:01:40.100
Now, the intermediate value theorem

00:01:40.100 --> 00:01:42.560
says hey, pick a closed interval.

00:01:42.560 --> 00:01:43.930
And here, we're picking
the closed interval

00:01:43.930 --> 00:01:45.990
from four to six, so let me look at that.

00:01:45.990 --> 00:01:49.810
So, this is one, two, three, four here,

00:01:49.810 --> 00:01:51.040
this is six here,

00:01:51.040 --> 00:01:53.160
so we're gonna look at
this closed interval.

00:01:53.160 --> 00:01:55.620
And the intermediate
value theorem tells us

00:01:55.620 --> 00:01:58.710
that look, if we're continuous
over that closed interval,

00:01:58.710 --> 00:02:01.600
our function f is gonna
take on every value

00:02:01.600 --> 00:02:05.360
between f of four, which in this case,

00:02:05.360 --> 00:02:08.713
so, this is f of four, is equal to three,

00:02:10.730 --> 00:02:13.563
and f of six, which is equal to seven.

00:02:14.500 --> 00:02:17.130
f of six,

00:02:17.130 --> 00:02:19.280
which is equal to seven.

00:02:19.280 --> 00:02:22.170
And so, if someone said hey,
is there gonna be a solution

00:02:22.170 --> 00:02:26.330
to f of x is equal to, say,
five over this interval?

00:02:26.330 --> 00:02:27.420
Yes.

00:02:27.420 --> 00:02:30.210
Over this interval, for some x,

00:02:30.210 --> 00:02:32.870
you're going to have f
of x being equal to five.

00:02:32.870 --> 00:02:35.350
But they're not asking us for an f of x

00:02:35.350 --> 00:02:37.850
equaling something
between these two values.

00:02:37.850 --> 00:02:41.210
They're asking us for
an f of x equaling zero.

00:02:41.210 --> 00:02:44.880
Zero isn't between f of four and f of six,

00:02:44.880 --> 00:02:48.260
and so we cannot use the
intermediate value theorem here.

00:02:48.260 --> 00:02:49.910
And so, if we wanted to write it out,

00:02:49.910 --> 00:02:54.140
we could say f is continuous

00:02:57.083 --> 00:02:59.860
but zero is not

00:03:02.240 --> 00:03:05.090
between f of four

00:03:05.090 --> 00:03:08.350
and f of six.

00:03:08.350 --> 00:03:13.350
So, the intermediate value
theorem does not apply.

00:03:15.520 --> 00:03:17.350
All right, let's do the second one.

00:03:17.350 --> 00:03:20.410
So here they say, can we use
the intermediate value theorem

00:03:20.410 --> 00:03:25.410
to say that there is a value
c such that f of c equals zero

00:03:25.670 --> 00:03:27.590
and two is less than or equal to c

00:03:27.590 --> 00:03:28.850
is less than or equal to four?

00:03:28.850 --> 00:03:30.893
If so, write a justification.

00:03:31.870 --> 00:03:34.760
We are given that f is continuous,
so let write that down.

00:03:34.760 --> 00:03:37.570
We are given

00:03:38.600 --> 00:03:43.180
that f is continuous,

00:03:43.180 --> 00:03:44.860
and if you wanna be over that interval,

00:03:44.860 --> 00:03:47.970
but they're telling us
it's continuous in general.

00:03:47.970 --> 00:03:50.560
And then we can just
look at what is the value

00:03:50.560 --> 00:03:52.400
of the function at these end points?

00:03:52.400 --> 00:03:53.830
Our interval goes from two to four,

00:03:53.830 --> 00:03:57.430
so we're talking about this
closed interval right over here.

00:03:57.430 --> 00:03:59.210
We know that f of two

00:04:00.440 --> 00:04:02.070
is going to be equal to negative two.

00:04:02.070 --> 00:04:03.910
We see it on that table.

00:04:03.910 --> 00:04:06.163
And what's f of four?

00:04:09.192 --> 00:04:11.540
f of four is equal to three.

00:04:11.540 --> 00:04:13.720
So, zero

00:04:15.550 --> 00:04:17.430
is between

00:04:17.430 --> 00:04:21.530
f of two and f of four.

00:04:21.530 --> 00:04:23.390
And you can see it visually here.

00:04:23.390 --> 00:04:26.475
There's no way to draw between
this point and that point

00:04:26.475 --> 00:04:30.130
without picking up your pen,
without crossing the x-axis,

00:04:30.130 --> 00:04:31.070
without having to point

00:04:31.070 --> 00:04:33.080
where your function is equal to zero.

00:04:33.080 --> 00:04:34.410
And so, we can say

00:04:38.252 --> 00:04:42.990
according to the
intermediate value theorem,

00:04:42.990 --> 00:04:45.670
there is

00:04:45.670 --> 00:04:49.250
a value c

00:04:49.250 --> 00:04:51.320
such that

00:04:52.210 --> 00:04:55.120
f of c is equal to zero

00:04:55.120 --> 00:04:58.141
and two is less than or equal to c

00:04:58.141 --> 00:05:00.740
is less than or equal to four.

00:05:00.740 --> 00:05:03.820
So, all we're saying is hey,
there must be a value c,

00:05:03.820 --> 00:05:06.893
and the way I drew it here,
that c value is right over

00:05:06.893 --> 00:05:09.250
where c is between two and four,

00:05:09.250 --> 00:05:11.810
where f of c is equal to zero.

00:05:11.810 --> 00:05:12.856
And this seems all mathy

00:05:12.856 --> 00:05:14.990
and a little bit confusing sometimes

00:05:14.990 --> 00:05:16.910
but it's saying something very intuitive.

00:05:16.910 --> 00:05:18.970
If I had to go from
this point to that point

00:05:18.970 --> 00:05:23.000
without picking up my pen,
I am going to at least cross

00:05:23.000 --> 00:05:27.963
every value between f of two
and f of four at least once.


Horizontal tangent to implicit curve

WEBVTT
Kind: captions
Language: en

00:00:00.200 --> 00:00:01.033
- [Instructor] We're told to consider

00:00:01.033 --> 00:00:02.830
the curve given by the equation.

00:00:02.830 --> 00:00:04.070
They give this equation.

00:00:04.070 --> 00:00:07.210
It can be shown that the
derivative of y with respect to x

00:00:07.210 --> 00:00:09.610
is equal to this expression,
and you could figure that out

00:00:09.610 --> 00:00:11.180
with just some implicit differentiation

00:00:11.180 --> 00:00:14.160
and then solving for the
derivative of y with respect to x.

00:00:14.160 --> 00:00:15.900
We've done that in other videos.

00:00:15.900 --> 00:00:18.890
Write the equation of the horizontal line

00:00:18.890 --> 00:00:22.960
that is tangent to the curve
and is above the x-axis.

00:00:22.960 --> 00:00:25.960
Pause this video, and see
if you can have a go at it.

00:00:25.960 --> 00:00:28.300
So let's just make sure
we're visualizing this right.

00:00:28.300 --> 00:00:30.250
So let me just draw a
quick and dirty diagram.

00:00:30.250 --> 00:00:33.800
If that's my y-axis, this is my x-axis.

00:00:33.800 --> 00:00:35.860
I don't know exactly what
that curve looks like,

00:00:35.860 --> 00:00:38.350
but imagine you have some type of a curve

00:00:38.350 --> 00:00:41.630
that looks something like this.

00:00:41.630 --> 00:00:44.130
Well, there would be two tangent lines

00:00:44.130 --> 00:00:47.560
that are horizontal based
on how I've drawn it.

00:00:47.560 --> 00:00:49.280
One might be right over there,

00:00:49.280 --> 00:00:50.660
so it might be like there.

00:00:50.660 --> 00:00:54.120
And then another one might
be maybe right over here.

00:00:54.120 --> 00:00:56.010
And they want the equation
of the horizontal line

00:00:56.010 --> 00:00:59.340
that is tangent to the curve
and is above the x-axis.

00:00:59.340 --> 00:01:00.180
So what do we know?

00:01:00.180 --> 00:01:05.180
What is true if this
tangent line is horizontal?

00:01:05.570 --> 00:01:08.550
Well, that tells us that, at this point,

00:01:08.550 --> 00:01:11.500
dy/dx is equal to zero.

00:01:11.500 --> 00:01:13.890
In fact, that would be true
at both of these points.

00:01:13.890 --> 00:01:15.940
And we know what dy/dx is.

00:01:15.940 --> 00:01:19.190
We know that the derivative
of y with respect to x

00:01:19.190 --> 00:01:22.340
is equal to negative
two times x plus three

00:01:22.340 --> 00:01:26.280
over four y to the third
power for any x and y.

00:01:26.280 --> 00:01:28.860
And so when will this equal zero?

00:01:28.860 --> 00:01:30.970
Well, it's going to equal
zero when our numerator

00:01:30.970 --> 00:01:34.060
is equal to zero and
our denominator isn't.

00:01:34.060 --> 00:01:36.450
So when is our numerator going to be zero?

00:01:36.450 --> 00:01:38.520
When x is equal to negative three.

00:01:38.520 --> 00:01:42.730
So when x is equal to negative three,

00:01:42.730 --> 00:01:45.850
the derivative is equal to zero.

00:01:45.850 --> 00:01:48.870
So what is going to be
the corresponding y value

00:01:48.870 --> 00:01:50.780
when x is equal to negative three?

00:01:50.780 --> 00:01:52.650
And, if we know that, well, this equation

00:01:52.650 --> 00:01:54.600
is just going to be y
is equal to something.

00:01:54.600 --> 00:01:56.480
It's going to be that y value.

00:01:56.480 --> 00:01:57.550
Well, to figure that out,

00:01:57.550 --> 00:01:59.480
we just take this x equals negative three,

00:01:59.480 --> 00:02:01.600
substitute it back into
our original equation,

00:02:01.600 --> 00:02:03.230
and then solve for y.

00:02:03.230 --> 00:02:04.290
So let's do that.

00:02:04.290 --> 00:02:08.080
So it's going to be negative three squared

00:02:08.080 --> 00:02:10.230
plus y to the fourth

00:02:10.230 --> 00:02:14.180
plus six times negative
three is equal to seven.

00:02:14.180 --> 00:02:15.570
This is nine.

00:02:15.570 --> 00:02:17.750
This is negative 18.

00:02:17.750 --> 00:02:19.990
And so we're going to get y to the fourth

00:02:19.990 --> 00:02:23.470
minus nine is equal to seven,

00:02:23.470 --> 00:02:25.380
or, adding nine to both sides,

00:02:25.380 --> 00:02:28.910
we get y to the fourth
power is equal to 16.

00:02:28.910 --> 00:02:30.930
And this would tell us that y is going

00:02:30.930 --> 00:02:33.400
to be equal to plus or minus two.

00:02:33.400 --> 00:02:36.100
Well, there would be then
two horizontal lines.

00:02:36.100 --> 00:02:37.880
One would be y is equal to two.

00:02:37.880 --> 00:02:40.810
The other is y is equal to negative two.

00:02:40.810 --> 00:02:43.330
But they want us, the equation
of the horizontal line

00:02:43.330 --> 00:02:47.550
that is tangent to the curve
and is above the x-axis,

00:02:47.550 --> 00:02:51.130
so only this one is going
to be above the x-axis.

00:02:51.130 --> 00:02:52.000
And we're done.

00:02:52.000 --> 00:02:54.033
It's going to be y is equal to two.


Finding points with vertical tangents

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:02.400
- [Instructor] Consider the
closed curve in the xy-plane

00:00:02.400 --> 00:00:04.500
given by this expression here.

00:00:04.500 --> 00:00:07.520
Find the coordinates of
the two points on the curve

00:00:07.520 --> 00:00:11.540
where the line tangent
to the curve is vertical.

00:00:11.540 --> 00:00:14.510
So pause this video, and see
if you could have a go at it.

00:00:14.510 --> 00:00:16.270
So I don't know what the exact shape

00:00:16.270 --> 00:00:17.970
of this closed curve is.

00:00:17.970 --> 00:00:20.630
But if I were to draw some
type of a closed curve,

00:00:20.630 --> 00:00:22.560
maybe it looks something like this.

00:00:22.560 --> 00:00:24.730
This isn't the one that's right over here.

00:00:24.730 --> 00:00:26.320
This one also has two points

00:00:26.320 --> 00:00:28.510
where my tangent line is vertical.

00:00:28.510 --> 00:00:30.630
One point would be right over there.

00:00:30.630 --> 00:00:34.690
Another point would be right over there.

00:00:34.690 --> 00:00:36.360
Now, how do we figure this out?

00:00:36.360 --> 00:00:39.570
Well, what we could do is
use implicit differentiation

00:00:39.570 --> 00:00:42.330
to find the derivative
of y with respect to x,

00:00:42.330 --> 00:00:44.500
and think about the x and y values

00:00:44.500 --> 00:00:47.270
that would give us a situation
where that derivative

00:00:47.270 --> 00:00:51.080
is nonzero in the numerator
and zero in the denominator.

00:00:51.080 --> 00:00:52.290
So let's do that.

00:00:52.290 --> 00:00:54.460
So let's just, let me rewrite everything.

00:00:54.460 --> 00:00:57.400
I have x squared plus two x

00:00:57.400 --> 00:01:00.800
plus y to the fourth plus four y

00:01:00.800 --> 00:01:02.600
is equal to five.

00:01:02.600 --> 00:01:04.670
I want to take the
derivative with respect to x

00:01:04.670 --> 00:01:07.470
of both sides of this equation.

00:01:07.470 --> 00:01:09.380
I'm trying to find an expression

00:01:09.380 --> 00:01:11.590
for the derivative of y with respect to x.

00:01:11.590 --> 00:01:13.010
And so what am I going to get?

00:01:13.010 --> 00:01:17.140
This is going to be equal
to two x plus two plus,

00:01:17.140 --> 00:01:20.480
derivative of this with respect
to y is four y to the third

00:01:20.480 --> 00:01:23.490
and then times the derivative
of y with respect to x,

00:01:23.490 --> 00:01:25.400
that's just straight
out of the chain rule,

00:01:25.400 --> 00:01:28.600
plus derivative of this
with respect to y is four,

00:01:28.600 --> 00:01:31.590
times the derivative of y
with respect to x, once again,

00:01:31.590 --> 00:01:33.950
straight out of the chain
rule, is equal to, whoops,

00:01:33.950 --> 00:01:36.030
I want to take the derivative
with respect to x here,

00:01:36.030 --> 00:01:37.770
is equal to

00:01:37.770 --> 00:01:38.680
zero.

00:01:38.680 --> 00:01:41.100
And now we just have to solve for dy/dx.

00:01:41.100 --> 00:01:42.780
So there's a couple things we could do.

00:01:42.780 --> 00:01:45.380
We could take the two x plus two

00:01:45.380 --> 00:01:47.680
and subtract it from both sides,

00:01:47.680 --> 00:01:52.020
and we could also factor out a four dy/dx

00:01:52.020 --> 00:01:54.020
out of this stuff right over here.

00:01:54.020 --> 00:01:55.100
So let's do that.

00:01:55.100 --> 00:01:57.290
Let's subtract the two x
plus two from both sides

00:01:57.290 --> 00:01:59.670
and factor out the four times dy/dx.

00:01:59.670 --> 00:02:04.670
So we will get four times dy/dx

00:02:04.730 --> 00:02:05.930
times

00:02:05.930 --> 00:02:07.670
y to the third

00:02:07.670 --> 00:02:09.410
plus one

00:02:09.410 --> 00:02:11.010
is equal to,

00:02:11.010 --> 00:02:12.160
is equal to,

00:02:12.160 --> 00:02:14.990
I subtracted this from both
sides, so I'll get negative two.

00:02:14.990 --> 00:02:16.800
And actually, let me
factor out a negative two,

00:02:16.800 --> 00:02:20.240
so negative two times x plus one.

00:02:20.240 --> 00:02:22.380
And now I just have to divide both sides

00:02:22.380 --> 00:02:24.690
by four times all of this business.

00:02:24.690 --> 00:02:27.510
And I am going to get the derivative of y

00:02:27.510 --> 00:02:32.300
with respect to x is equal
to negative two times

00:02:32.300 --> 00:02:33.740
x plus one

00:02:33.740 --> 00:02:35.530
divided by,

00:02:35.530 --> 00:02:36.730
divided by

00:02:36.730 --> 00:02:39.510
four times y to the third

00:02:39.510 --> 00:02:40.610
plus one.

00:02:40.610 --> 00:02:42.230
And actually, this can be rewritten

00:02:42.230 --> 00:02:45.670
as being equal to negative of x plus one

00:02:45.670 --> 00:02:48.730
divided by two times y
to the third plus one.

00:02:48.730 --> 00:02:51.720
I just divided the numerator
and the denominator by two.

00:02:51.720 --> 00:02:53.160
Now, why is this useful?

00:02:53.160 --> 00:02:56.290
Well, we could think about
what y values, because y is

00:02:56.290 --> 00:02:58.240
the only variable we have
in the denominator here,

00:02:58.240 --> 00:03:00.440
would make the denominator equal zero,

00:03:00.440 --> 00:03:03.940
and then find the corresponding
x values for those y values

00:03:03.940 --> 00:03:06.130
by going to our original equation.

00:03:06.130 --> 00:03:08.330
Well, this is going to be zero,

00:03:08.330 --> 00:03:10.470
zero when

00:03:10.470 --> 00:03:12.740
y is equal to negative one.

00:03:12.740 --> 00:03:14.370
So when y is equal to negative one,

00:03:14.370 --> 00:03:15.960
let's figure out what x is.

00:03:15.960 --> 00:03:17.560
And to do that, we just
have to substitute y

00:03:17.560 --> 00:03:20.170
equals negative one back
in our original equation

00:03:20.170 --> 00:03:22.560
and then solve for x.

00:03:22.560 --> 00:03:24.410
So let's do that.

00:03:24.410 --> 00:03:27.760
Let me clear this out since
I need that real estate.

00:03:27.760 --> 00:03:29.210
So if we go back and we substitute

00:03:29.210 --> 00:03:30.600
y equals negative one up here,

00:03:30.600 --> 00:03:35.600
we're going to get x
squared plus two x plus one,

00:03:36.070 --> 00:03:37.270
plus one

00:03:37.270 --> 00:03:40.510
minus four is equal to five.

00:03:40.510 --> 00:03:42.090
This is going to be negative three.

00:03:42.090 --> 00:03:44.930
Subtract five from both
sides, you get x squared

00:03:44.930 --> 00:03:48.180
plus two x minus eight is equal to zero.

00:03:48.180 --> 00:03:49.590
This is just simple factoring.

00:03:49.590 --> 00:03:54.370
So it's going to be x plus
four times x minus two.

00:03:54.370 --> 00:03:55.870
What two numbers, when
I take their product,

00:03:55.870 --> 00:03:56.720
I get negative eight?

00:03:56.720 --> 00:03:57.553
Four, negative two.

00:03:57.553 --> 00:04:00.470
When I add four, negative two,
I get a positive two there,

00:04:00.470 --> 00:04:01.610
is equal to zero.

00:04:01.610 --> 00:04:03.570
So x is equal to negative four,

00:04:03.570 --> 00:04:06.130
or x is equal to positive two

00:04:06.130 --> 00:04:08.070
when y is equal to negative one.

00:04:08.070 --> 00:04:09.830
So to answer their question,

00:04:09.830 --> 00:04:11.900
find the coordinates of
the two points on the curve

00:04:11.900 --> 00:04:14.840
where the line tangent
to the curve is vertical,

00:04:14.840 --> 00:04:16.980
well, the answer here would be,

00:04:16.980 --> 00:04:18.430
get a little bit of a drum roll,

00:04:18.430 --> 00:04:21.010
it would be the points

00:04:21.010 --> 00:04:23.700
x equals negative four

00:04:23.700 --> 00:04:26.820
and y is equal to negative one and

00:04:26.820 --> 00:04:30.570
x equals two and y is
equal to negative one.

00:04:30.570 --> 00:04:31.403
And we're done.


Volume with cross sections: intro

WEBVTT
Kind: captions
Language: en

00:00:00.120 --> 00:00:01.820
- [Instructor] You are
likely already familiar

00:00:01.820 --> 00:00:04.040
with finding the area between curves.

00:00:04.040 --> 00:00:04.930
And, in fact, if you're not,

00:00:04.930 --> 00:00:07.070
I encourage you to review
that on Khan Academy.

00:00:07.070 --> 00:00:09.500
For example, we could
find this yellow area

00:00:09.500 --> 00:00:11.130
using a definite integral.

00:00:11.130 --> 00:00:12.300
But what we're going to do in this video

00:00:12.300 --> 00:00:14.390
is do something even more interesting.

00:00:14.390 --> 00:00:16.750
We're gonna find the volume of shapes

00:00:16.750 --> 00:00:19.260
where the base is defined in some way

00:00:19.260 --> 00:00:21.410
by the area between two curves.

00:00:21.410 --> 00:00:23.610
And, in this video, we're
gonna think about a shape,

00:00:23.610 --> 00:00:24.950
and I'm gonna draw it in three dimensions.

00:00:24.950 --> 00:00:26.790
So let me draw this over again,

00:00:26.790 --> 00:00:28.700
but with a little bit of perspective.

00:00:28.700 --> 00:00:29.860
So,

00:00:29.860 --> 00:00:32.633
let's make this the y-axis.

00:00:33.810 --> 00:00:35.540
So that's the y-axis.

00:00:35.540 --> 00:00:38.620
This is my x-axis.

00:00:38.620 --> 00:00:40.670
That is my x-axis.

00:00:40.670 --> 00:00:42.120
This is the

00:00:42.120 --> 00:00:44.800
line y is equal to six

00:00:44.800 --> 00:00:45.980
right over there.

00:00:45.980 --> 00:00:48.020
Y is equal to six.

00:00:48.020 --> 00:00:51.850
This dotted line, we could
just draw it like this,

00:00:51.850 --> 00:00:54.750
and so this would be
the point x equals two.

00:00:54.750 --> 00:00:57.000
And then the graph of y is equal to four

00:00:57.000 --> 00:00:59.010
times the natural log of three minus x

00:00:59.010 --> 00:01:01.380
would look something like this,

00:01:01.380 --> 00:01:02.610
look something like this.

00:01:02.610 --> 00:01:04.940
And so this region is this region,

00:01:04.940 --> 00:01:06.830
but it's going to be the base

00:01:06.830 --> 00:01:09.900
of a three-dimensional shape
where any cross section,

00:01:09.900 --> 00:01:13.010
if I were to take a cross
section right over here,

00:01:13.010 --> 00:01:14.490
is going to be a square.

00:01:14.490 --> 00:01:18.670
So whatever this length is,
we also go that much high,

00:01:18.670 --> 00:01:23.470
and so the cross section is
a square right over there.

00:01:23.470 --> 00:01:26.870
The cross section right over
here is going to be a square.

00:01:26.870 --> 00:01:29.700
Whatever the difference
between these two functions is,

00:01:29.700 --> 00:01:33.100
that's also how high we are going to go.

00:01:33.100 --> 00:01:36.720
This length, which is six at this point,

00:01:36.720 --> 00:01:39.670
this is also going to be the height.

00:01:39.670 --> 00:01:40.880
It is going to be a square.

00:01:40.880 --> 00:01:42.270
It's going to be quite big,

00:01:42.270 --> 00:01:44.970
might have to scroll down so
we can draw the whole thing,

00:01:44.970 --> 00:01:46.810
roughly at the right proportion.

00:01:46.810 --> 00:01:48.210
So it looks something like this.

00:01:48.210 --> 00:01:49.920
This should be a square.

00:01:49.920 --> 00:01:51.220
It's gonna look something like this.

00:01:51.220 --> 00:01:54.130
And so the whole shape would look,

00:01:54.130 --> 00:01:56.260
would look something

00:01:56.260 --> 00:01:58.300
like this,

00:01:58.300 --> 00:02:00.060
would look something like that,

00:02:00.060 --> 00:02:02.430
try to shade that in a
little bit so that you can

00:02:02.430 --> 00:02:04.700
appreciate it a little bit more,

00:02:04.700 --> 00:02:06.310
but hopefully you get the idea.

00:02:06.310 --> 00:02:07.670
And some of you might be excited,

00:02:07.670 --> 00:02:09.110
and some of you might
be a little intimidated.

00:02:09.110 --> 00:02:10.000
Well, hey, I've been dealing

00:02:10.000 --> 00:02:11.350
with the two dimensions for so long.

00:02:11.350 --> 00:02:13.350
What's going on with
these three dimensions?

00:02:13.350 --> 00:02:16.070
But you'll quickly appreciate
that you already have

00:02:16.070 --> 00:02:18.240
the powers of integration to solve this.

00:02:18.240 --> 00:02:21.100
And to do that, we just
have to break up the shape

00:02:21.100 --> 00:02:25.190
into a bunch of these, you
could view them as these little

00:02:25.190 --> 00:02:28.770
square tiles that have some depth to them.

00:02:28.770 --> 00:02:31.720
So let's make that into a
little tile, this one into it,

00:02:31.720 --> 00:02:34.380
that also has some depth to it.

00:02:34.380 --> 00:02:36.390
You could even, I could
draw it multiple places.

00:02:36.390 --> 00:02:37.223
You could view it as a,

00:02:37.223 --> 00:02:41.410
break it up into these things
that have a very small depth,

00:02:41.410 --> 00:02:43.310
that we could call dx.

00:02:43.310 --> 00:02:46.390
And we know how to figure
out what their volume is.

00:02:46.390 --> 00:02:48.630
What would be the volume
of one of these things?

00:02:48.630 --> 00:02:51.480
Well, it would be the
depth times the area,

00:02:51.480 --> 00:02:52.810
times the surface area

00:02:52.810 --> 00:02:55.150
of this cross section right over here.

00:02:55.150 --> 00:02:56.580
Let me do that in a different color.

00:02:56.580 --> 00:02:59.270
So what would be the area

00:02:59.270 --> 00:03:02.300
that I am shading in in
pink right over here?

00:03:02.300 --> 00:03:06.640
Well, that area is going to
be the base length squared.

00:03:06.640 --> 00:03:07.780
What's the base length?

00:03:07.780 --> 00:03:09.900
Well, it's the difference
between these two functions.

00:03:09.900 --> 00:03:12.560
It is going to be six minus,

00:03:12.560 --> 00:03:15.200
our bottom function is four

00:03:15.200 --> 00:03:19.220
times the natural log of three minus x,

00:03:19.220 --> 00:03:21.540
and so that would just
give us that length.

00:03:21.540 --> 00:03:24.500
But if we square it, we
get this entire area.

00:03:24.500 --> 00:03:26.730
We get that entire area, you square it.

00:03:26.730 --> 00:03:29.760
And then you multiply it times the depth.

00:03:29.760 --> 00:03:31.280
You multiply it times the depth.

00:03:31.280 --> 00:03:33.780
Now you have the volume
of just this little

00:03:33.780 --> 00:03:35.150
section right over here,

00:03:35.150 --> 00:03:37.610
and I think you might
see where this is going.

00:03:37.610 --> 00:03:39.930
Now, what if you were
to add up all of these

00:03:39.930 --> 00:03:42.810
from x equals zero to x equals two?

00:03:42.810 --> 00:03:45.150
Well, then you would have the
volume of the entire thing.

00:03:45.150 --> 00:03:47.450
This is the power of
the definite integral.

00:03:47.450 --> 00:03:49.380
So we could just integrate

00:03:49.380 --> 00:03:51.560
from x equals zero to x equals two,

00:03:51.560 --> 00:03:54.310
from x equals zero to x equals two.

00:03:54.310 --> 00:03:56.900
If you drew where these
intersect our base,

00:03:56.900 --> 00:03:59.770
you would say all right,
this thing right over here

00:03:59.770 --> 00:04:01.940
would be this thing

00:04:01.940 --> 00:04:04.750
right over here, where it's dx.

00:04:04.750 --> 00:04:06.540
And instead of just multiplying dx

00:04:06.540 --> 00:04:08.420
times the difference
between these functions,

00:04:08.420 --> 00:04:10.440
we're going to square the
difference of these functions

00:04:10.440 --> 00:04:13.130
'cause we're visualizing
this three-dimensional shape,

00:04:13.130 --> 00:04:14.830
the surface area of this
three-dimensional shape,

00:04:14.830 --> 00:04:19.060
as opposed to just the height
of this little rectangle.

00:04:19.060 --> 00:04:21.020
And if you were to evaluate this integral,

00:04:21.020 --> 00:04:24.840
you would indeed get the volume of this,

00:04:24.840 --> 00:04:28.130
this kind of pedestal horn-looking thing.

00:04:28.130 --> 00:04:31.210
This is not an easy definite
integral to evaluate by hand,

00:04:31.210 --> 00:04:33.700
but we can actually use
a calculator for that.

00:04:33.700 --> 00:04:34.850
And so,

00:04:34.850 --> 00:04:36.020
we can

00:04:36.020 --> 00:04:38.480
hit math

00:04:38.480 --> 00:04:41.670
and then hit choice number
nine for definite integral,

00:04:41.670 --> 00:04:44.070
and then we just have to input everything.

00:04:44.070 --> 00:04:46.940
We're going from zero till

00:04:46.940 --> 00:04:49.120
two of,

00:04:49.120 --> 00:04:51.580
and then we have, let me open parentheses

00:04:51.580 --> 00:04:52.710
'cause I'm gonna have
to square everything.

00:04:52.710 --> 00:04:56.860
Six minus four times the natural log

00:04:56.860 --> 00:04:58.570
of x,

00:04:58.570 --> 00:05:02.650
or actually the natural
log of three minus x.

00:05:02.650 --> 00:05:06.020
And so let me close the parentheses
on the natural log part.

00:05:06.020 --> 00:05:09.810
And then, if I close the
parentheses on this whole thing,

00:05:09.810 --> 00:05:11.620
I want to then square it.

00:05:11.620 --> 00:05:15.770
And then I'm integrating
with respect to x.

00:05:15.770 --> 00:05:20.410
Enter, I got approximately 26.27.

00:05:20.410 --> 00:05:24.580
So approximately 26.27,
and this is a volume here.

00:05:24.580 --> 00:05:25.850
So if we thought about units,

00:05:25.850 --> 00:05:28.743
it would be in our units
cubed or cubic units.


Integration using completing the square and the derivative of arctan(x)

WEBVTT
Kind: captions
Language: en

00:00:00.520 --> 00:00:01.353
- [Instructor] All right, let's see if

00:00:01.353 --> 00:00:04.010
we can find the indefinite integral of

00:00:04.010 --> 00:00:09.010
one over five x squared
minus 30x plus 65 dx.

00:00:14.010 --> 00:00:16.713
Pause this video and see
if you can figure it out.

00:00:17.620 --> 00:00:19.920
All right, so this is going
to be an interesting one.

00:00:19.920 --> 00:00:21.350
And it'll be a little bit hairy,

00:00:21.350 --> 00:00:23.630
but we're gonna work through it together.

00:00:23.630 --> 00:00:26.070
So, immediately you might try multiple

00:00:26.070 --> 00:00:28.800
integration techniques
and be hitting some walls.

00:00:28.800 --> 00:00:31.330
And what we're going to
do here is actually try

00:00:31.330 --> 00:00:35.170
to complete the square in this
denominator right over here.

00:00:35.170 --> 00:00:37.580
And then by completing the
square, we're gonna get it

00:00:37.580 --> 00:00:40.830
in the form that it looks
like the derivative of arctan.

00:00:40.830 --> 00:00:42.750
And if that's a big
hint to you, once again

00:00:42.750 --> 00:00:45.220
pause the video and try to move forward.

00:00:45.220 --> 00:00:47.050
All right, now let's do this together.

00:00:47.050 --> 00:00:49.650
So I'm just gonna try to
simplify this denominator

00:00:49.650 --> 00:00:53.300
so that my coefficient on
my x squared term is a one.

00:00:53.300 --> 00:00:55.820
And so I could just factor a
five out of the denominator.

00:00:55.820 --> 00:00:59.250
If I did that, then this
integral will become

00:00:59.250 --> 00:01:04.250
1/5 times the integral of one over,

00:01:05.020 --> 00:01:06.890
so I've factored a five
out of the denominator

00:01:06.890 --> 00:01:11.873
so it is x squared minus six x plus 13 dx.

00:01:13.990 --> 00:01:15.360
And then as I mentioned I'm gonna

00:01:15.360 --> 00:01:16.640
complete the square down here.

00:01:16.640 --> 00:01:19.540
So let me rewrite it,
so this is equal to 1/5

00:01:19.540 --> 00:01:24.480
times the integral of one over,

00:01:24.480 --> 00:01:27.570
and so x squared minus six x is clearly

00:01:27.570 --> 00:01:29.340
not a perfect square the way it's written.

00:01:29.340 --> 00:01:32.940
Let me write this plus 13 out here.

00:01:32.940 --> 00:01:35.460
Now what could I add and then I'm gonna

00:01:35.460 --> 00:01:36.610
to have to subtract if I don't want

00:01:36.610 --> 00:01:38.290
to change the value of the denominator.

00:01:38.290 --> 00:01:40.960
In order to make, In order to make

00:01:40.960 --> 00:01:44.000
this part right over
here a perfect square.

00:01:44.000 --> 00:01:45.720
Well we've done this before,

00:01:45.720 --> 00:01:47.410
you take half of your coefficient here,

00:01:47.410 --> 00:01:49.660
which is negative three
and you square that.

00:01:49.660 --> 00:01:51.420
So you want to add a nine here.

00:01:51.420 --> 00:01:52.640
But if you add a nine then you have

00:01:52.640 --> 00:01:54.970
to subtract a nine as well.

00:01:54.970 --> 00:01:57.820
And so this part is going to be

00:01:57.820 --> 00:02:01.990
x minus three squared and then this part

00:02:01.990 --> 00:02:03.380
right over here is going to be

00:02:03.380 --> 00:02:06.300
equal to a positive four and we of course

00:02:06.300 --> 00:02:08.730
don't want to forget our dx out here.

00:02:08.730 --> 00:02:11.130
And so let me write it in this form.

00:02:11.130 --> 00:02:12.404
So this is going to be equal to

00:02:12.404 --> 00:02:17.404
1/5 times the integral of one over,

00:02:17.620 --> 00:02:22.620
get myself some space,
x minus three squared

00:02:23.337 --> 00:02:27.700
plus four, which we could also
write as plus two squared.

00:02:27.700 --> 00:02:32.700
Actually let me do it that
way, plus two squared dx.

00:02:33.120 --> 00:02:34.890
Now many of y'all might already say

00:02:34.890 --> 00:02:36.740
hey this looks a lot like arc tangent,

00:02:36.740 --> 00:02:38.780
but I'm gonna to try to
simplify it even more

00:02:38.780 --> 00:02:40.350
so it becomes very clear that

00:02:40.350 --> 00:02:42.690
it looks like arc tangent
is going to be involved.

00:02:42.690 --> 00:02:43.523
I'm actually gonna do some

00:02:43.523 --> 00:02:45.880
u substitution in order to do it.

00:02:45.880 --> 00:02:47.242
So the first thing I'm going to do is

00:02:47.242 --> 00:02:52.242
let's factor a four out
of the denominator here.

00:02:53.010 --> 00:02:55.230
So if we do that, then this is going

00:02:55.230 --> 00:02:58.930
to become 1/5 times 1/4 which is going

00:02:58.930 --> 00:03:03.110
to be 1/20 times the integral of

00:03:03.110 --> 00:03:08.110
one over x minus three squared

00:03:09.180 --> 00:03:12.800
over two squared.

00:03:12.800 --> 00:03:15.293
And this is going to be a plus one,

00:03:15.293 --> 00:03:18.690
and of course we have our dx.

00:03:18.690 --> 00:03:21.460
And then we could write this as,

00:03:21.460 --> 00:03:23.270
and I'm trying to just do every step here.

00:03:23.270 --> 00:03:24.103
A lot of these you might have

00:03:24.103 --> 00:03:25.530
been able to do in your head.

00:03:25.530 --> 00:03:29.060
One over, and I'll just write this as,

00:03:29.060 --> 00:03:34.060
x minus three over two squared plus one.

00:03:34.620 --> 00:03:38.870
Plus one and then dx.

00:03:38.870 --> 00:03:41.630
And now the u substitution
is pretty clear.

00:03:41.630 --> 00:03:44.090
I'm just going to make the substitution

00:03:44.090 --> 00:03:48.170
that u is equal to x minus three over two.

00:03:48.170 --> 00:03:50.100
Or we could even say that's u is equal to

00:03:50.100 --> 00:03:54.130
one half x minus three halves,

00:03:54.130 --> 00:03:56.110
that's just x minus three over two.

00:03:56.110 --> 00:04:01.110
And du is going to be
equal to one half dx.

00:04:02.500 --> 00:04:04.600
And so what I can do here is, actually

00:04:04.600 --> 00:04:07.510
let me start to re-engineer
this integral a little bit

00:04:07.510 --> 00:04:09.680
so that we see a one half here.

00:04:09.680 --> 00:04:11.690
So if I make this a one half

00:04:11.690 --> 00:04:13.930
and then I multiply the outside by two.

00:04:13.930 --> 00:04:15.830
So I divide by two multiply by two.

00:04:15.830 --> 00:04:16.920
It's one way to think about it.

00:04:16.920 --> 00:04:20.380
This becomes 1/10 and so
doing my u substitution

00:04:20.380 --> 00:04:25.030
I get 1/10, that's the 1/10
there times the integral of.

00:04:25.030 --> 00:04:28.430
Well I have one half dx right over here

00:04:28.430 --> 00:04:30.600
which is the same thing as du.

00:04:30.600 --> 00:04:32.330
So I can put the du
either in the numerator

00:04:32.330 --> 00:04:33.980
or I could put it out here.

00:04:33.980 --> 00:04:37.000
And then I have one over,

00:04:37.000 --> 00:04:40.863
this is u squared, u squared plus one.

00:04:41.750 --> 00:04:43.300
Now you might immediately recognize,

00:04:43.300 --> 00:04:46.260
what's the derivative of arctan of u?

00:04:46.260 --> 00:04:48.200
Well that would be one
over u squared plus one.

00:04:48.200 --> 00:04:50.720
So this is going to be equal to 1/10

00:04:50.720 --> 00:04:55.440
times the arc tangent of u, and of course

00:04:55.440 --> 00:04:58.040
we can't forget our big constant C

00:04:58.040 --> 00:05:00.050
because we're taking
an indefinite integral.

00:05:00.050 --> 00:05:02.170
And now we just want to do
the reverse substitution.

00:05:02.170 --> 00:05:05.220
We know that u is equal to
this business right over here.

00:05:05.220 --> 00:05:07.070
So we deserve a little bit of a drum roll.

00:05:07.070 --> 00:05:09.280
This is going to be equal to 1/10

00:05:09.280 --> 00:05:12.950
times the arc tangent of u.

00:05:12.950 --> 00:05:15.190
Well u is just x minus three over two,

00:05:15.190 --> 00:05:16.970
which could also be written like this.

00:05:16.970 --> 00:05:21.970
So arctan of x minus three
over two and then plus C.

00:05:23.210 --> 00:05:26.103
And we are done.


Interpreting the meaning of the derivative in context

WEBVTT
Kind: captions
Language: en

00:00:00.320 --> 00:00:01.210
- [Instructor] We are told that Eddie

00:00:01.210 --> 00:00:03.470
drove from New York City to Philadelphia.

00:00:03.470 --> 00:00:04.640
The function d gives the

00:00:04.640 --> 00:00:07.120
total distance Eddie has driven

00:00:07.120 --> 00:00:10.430
in kilometers t hours after he left.

00:00:10.430 --> 00:00:14.890
What is the best interpretation
for the following statement?

00:00:14.890 --> 00:00:17.790
D prime of two is equal to 100.

00:00:17.790 --> 00:00:20.600
So pause this video, and I
encourage you to write it out.

00:00:20.600 --> 00:00:22.290
What do you think this means,

00:00:22.290 --> 00:00:26.070
and be sure to include
the appropriate units.

00:00:26.070 --> 00:00:27.440
And now let's do this together.

00:00:27.440 --> 00:00:31.110
If d is equal to the distance,

00:00:31.110 --> 00:00:35.440
the distance driven, then to get d prime,

00:00:35.440 --> 00:00:37.940
you're taking the derivative
with respect to time.

00:00:38.810 --> 00:00:40.460
So one way to think about it is,

00:00:40.460 --> 00:00:43.010
it is the rate of change of d.

00:00:43.010 --> 00:00:46.600
So we could view this as d prime is

00:00:46.600 --> 00:00:49.120
going to give you the instantaneous,

00:00:49.120 --> 00:00:54.120
instantaneous, instantaneous rate.

00:00:54.500 --> 00:00:56.740
And they are both functions of t.

00:00:56.740 --> 00:01:01.740
So one way to view d prime
of two is equal to 100,

00:01:02.580 --> 00:01:06.220
that would mean, well,
what is our time now?

00:01:06.220 --> 00:01:09.010
Well, that is our t, and that's in hours.

00:01:09.010 --> 00:01:11.840
So two hours, actually
let me color code it.

00:01:11.840 --> 00:01:16.840
So two hours after leaving,

00:01:17.000 --> 00:01:22.000
after leaving, Eddie,

00:01:23.280 --> 00:01:27.547
Eddie, drove, drove,

00:01:28.800 --> 00:01:31.530
and this means, so it'll
be grammatically correct.

00:01:31.530 --> 00:01:35.844
Drove at an instantaneous,

00:01:35.844 --> 00:01:40.844
instantaneous, instantaneous rate of,

00:01:43.380 --> 00:01:46.400
and let me use a different
color now for this part,

00:01:46.400 --> 00:01:49.700
of 100, and what are the units?

00:01:49.700 --> 00:01:53.680
Well, the distance was
given in kilometers,

00:01:53.680 --> 00:01:55.060
and now we're gonna be thinking

00:01:55.060 --> 00:01:57.010
about kilometers per unit time.

00:01:57.010 --> 00:02:02.010
Kilometers per hour, so
this is 100 kilometers,

00:02:02.130 --> 00:02:05.200
kilometers, per hour.

00:02:05.200 --> 00:02:06.740
So that's the interpretation there.

00:02:06.740 --> 00:02:07.993
Let's do another example.

00:02:09.130 --> 00:02:13.190
Here we are told a tank
is being drained of water.

00:02:13.190 --> 00:02:17.260
The function v gives the
volume of liquid in the tank,

00:02:17.260 --> 00:02:20.440
in liters, after t minutes.

00:02:20.440 --> 00:02:24.410
What is the best interpretation
for the following statement?

00:02:24.410 --> 00:02:27.600
The slope of the line tangent to the graph

00:02:27.600 --> 00:02:32.600
of v at t equals seven is
equal to negative three.

00:02:32.960 --> 00:02:34.230
So pause this video again

00:02:34.230 --> 00:02:36.610
and try to do what we just
did with the previous example.

00:02:36.610 --> 00:02:38.000
Write out that interpretation,

00:02:38.000 --> 00:02:39.873
make sure to get the units right.

00:02:40.810 --> 00:02:44.020
All right, so let's just remind
ourselves what's going on.

00:02:44.020 --> 00:02:49.020
V is going to give us the
volume, as a function of time.

00:02:49.610 --> 00:02:52.510
Volume is in liters,
and time is in minutes.

00:02:52.510 --> 00:02:54.290
And so they're talking about the slope

00:02:54.290 --> 00:02:56.200
of the tangent line to the graph,

00:02:56.200 --> 00:02:59.200
the slope of the tangent
line to the graph of v,

00:02:59.200 --> 00:03:00.800
that's just v prime.

00:03:00.800 --> 00:03:03.720
So if you take the derivative
with respect to time,

00:03:03.720 --> 00:03:05.330
that's going to give you v prime,

00:03:05.330 --> 00:03:07.550
and these are all functions of t.

00:03:07.550 --> 00:03:09.760
These are all functions of t.

00:03:09.760 --> 00:03:12.100
And they say at t equals seven,

00:03:12.100 --> 00:03:13.800
it's equal to negative three.

00:03:13.800 --> 00:03:16.270
So this, which is the
same thing as the slope

00:03:16.270 --> 00:03:20.580
of tangent line, slope
of tangent, tangent line.

00:03:22.220 --> 00:03:24.520
And they tell us that v prime

00:03:24.520 --> 00:03:27.930
of at time equals seven minutes,

00:03:27.930 --> 00:03:30.920
our rate of change of volume with respect

00:03:30.920 --> 00:03:33.450
to time is equal to negative three.

00:03:33.450 --> 00:03:36.330
And so you could say if
we were to write out,

00:03:36.330 --> 00:03:38.533
this means that after,

00:03:39.500 --> 00:03:44.500
after seven minutes, seven minutes,

00:03:44.540 --> 00:03:49.540
the tank is being drained
at an instantaneous,

00:03:56.120 --> 00:03:58.260
instantaneous, that's why we need

00:03:58.260 --> 00:04:00.750
that calculus for that instantaneous rate.

00:04:00.750 --> 00:04:05.150
An instantaneous rate of,

00:04:05.150 --> 00:04:06.530
now you might be tempted

00:04:06.530 --> 00:04:09.910
to say it's being drained
at an instantaneous rate

00:04:09.910 --> 00:04:13.540
of negative three liters per minute.

00:04:13.540 --> 00:04:15.820
But remember, the
negative three just shows

00:04:15.820 --> 00:04:18.210
that the volume is decreasing.

00:04:18.210 --> 00:04:19.390
So one way to think about it

00:04:19.390 --> 00:04:22.050
is this negative is
already being accounted

00:04:22.050 --> 00:04:24.530
for when you're saying it's being drained.

00:04:24.530 --> 00:04:27.490
If this was positive, that
means it is being filled.

00:04:27.490 --> 00:04:31.210
So it is being drained
at an instantaneous rate

00:04:31.210 --> 00:04:33.960
of three liters per minute.

00:04:33.960 --> 00:04:38.347
Three liters per minute.

00:04:39.370 --> 00:04:41.810
And how did I know the units
were liters per minute?

00:04:41.810 --> 00:04:44.430
Well, the volume function
is in terms of liters,

00:04:44.430 --> 00:04:46.290
and the time is in terms of minutes.

00:04:46.290 --> 00:04:48.780
And then I'm taking the
derivative with respect to time,

00:04:48.780 --> 00:04:52.033
so now it's gonna be liters
per minute, and we are done.


Second derivatives (implicit equations): evaluate derivative

WEBVTT
Kind: captions
Language: en

00:00:00.090 --> 00:00:01.540
- [Instructor] So we have a question here

00:00:01.540 --> 00:00:05.080
from the 2015 AP Calculus AB test,

00:00:05.080 --> 00:00:08.050
and it says consider the
curve given by the equation

00:00:08.050 --> 00:00:11.300
y to the third minus x y is equal to two.

00:00:11.300 --> 00:00:14.660
It can be shown that the
first derivative of y

00:00:14.660 --> 00:00:16.450
with respect to x is equal to that.

00:00:16.450 --> 00:00:19.020
So they solve that for us.

00:00:19.020 --> 00:00:21.660
And then part c of it,
I skipped parts a and b

00:00:21.660 --> 00:00:23.070
for the sake of this video.

00:00:23.070 --> 00:00:26.950
Evaluate the second derivative
of y with respect to x

00:00:26.950 --> 00:00:30.180
at the point on the curve
where x equal negative one

00:00:30.180 --> 00:00:32.470
and y is equal to one.

00:00:32.470 --> 00:00:35.630
So pause this video and
see if you can do that.

00:00:35.630 --> 00:00:37.090
All right now let's do it together.

00:00:37.090 --> 00:00:40.090
And so let me just first write
down the first derivative.

00:00:40.090 --> 00:00:42.570
So dy, derivative of y,

00:00:42.570 --> 00:00:45.310
with respect to x is equal to

00:00:45.310 --> 00:00:47.360
y over

00:00:47.360 --> 00:00:50.670
three y squared minus x.

00:00:50.670 --> 00:00:52.090
Well if we're concerning ourselves

00:00:52.090 --> 00:00:54.090
with the second derivative,

00:00:54.090 --> 00:00:56.220
well then we wanna take the
derivative with respect to x

00:00:56.220 --> 00:00:57.620
of both sides of this.

00:00:57.620 --> 00:00:59.250
So let's just do that.

00:00:59.250 --> 00:01:01.150
Do the derivative operator

00:01:01.150 --> 00:01:04.260
on both sides right over here.

00:01:04.260 --> 00:01:05.650
Now on the left hand side,

00:01:05.650 --> 00:01:08.590
we of course are going to get
the second derivative of y,

00:01:08.590 --> 00:01:11.030
with respect to x.

00:01:11.030 --> 00:01:12.610
But what do we get on the right hand side?

00:01:12.610 --> 00:01:14.730
And there's multiple
ways to approach this.

00:01:14.730 --> 00:01:15.760
But for something like this,

00:01:15.760 --> 00:01:18.360
the quotient rule probably
is the best way to tackle it.

00:01:18.360 --> 00:01:20.690
I sometimes complain
about the quotient rule

00:01:20.690 --> 00:01:23.440
saying hey, it's just a
variation of the product rule.

00:01:23.440 --> 00:01:25.470
But it's actually quite
useful in something like this.

00:01:25.470 --> 00:01:26.700
We just have to remind ourselves

00:01:26.700 --> 00:01:28.400
that this is going to be equal to

00:01:28.400 --> 00:01:30.430
the derivative of the numerator

00:01:30.430 --> 00:01:32.640
with respect to x.

00:01:32.640 --> 00:01:36.610
And so that's just going
to be derivative of y

00:01:36.610 --> 00:01:38.010
with respect to x

00:01:38.010 --> 00:01:39.690
times the denominator,

00:01:39.690 --> 00:01:43.110
three y squared minus x

00:01:43.110 --> 00:01:46.230
Minus the numerator, y,

00:01:46.230 --> 00:01:50.180
times the derivative of the
denominator with respect to x.

00:01:50.180 --> 00:01:51.080
Well, what's the derivative

00:01:51.080 --> 00:01:53.530
of this denominator with respect to x?

00:01:53.530 --> 00:01:56.760
Well the derivative of three
y squared with respect to x

00:01:56.760 --> 00:01:58.880
that's going to be the
derivative of three y squared

00:01:58.880 --> 00:02:01.270
with respect to y,

00:02:01.270 --> 00:02:04.390
which is just going to be six y.

00:02:04.390 --> 00:02:06.380
I'm just using the power rule there.

00:02:06.380 --> 00:02:10.290
Times the derivative
of y with respect to x.

00:02:10.290 --> 00:02:12.430
All I did just now is I
took the derivative of that

00:02:12.430 --> 00:02:14.050
with respect to x.

00:02:14.050 --> 00:02:16.680
Which is derivative of
that with respect to y

00:02:16.680 --> 00:02:18.540
times the derivative
of y with respect to x.

00:02:18.540 --> 00:02:20.250
Comes straight out of the chain rule.

00:02:20.250 --> 00:02:22.720
Minus the derivative of
this with respect to x,

00:02:22.720 --> 00:02:25.100
which is just going to be equal to one.

00:02:25.100 --> 00:02:26.210
All of that over, remember,

00:02:26.210 --> 00:02:28.280
we're in the middle of the
quotient rule right over here.

00:02:28.280 --> 00:02:29.980
All of that over

00:02:29.980 --> 00:02:31.830
the denominator squared.

00:02:31.830 --> 00:02:36.830
All of that over three y
squared minus x, squared.

00:02:37.010 --> 00:02:38.190
Now lucky for us,

00:02:38.190 --> 00:02:40.090
they want us to evaluate this at a point,

00:02:40.090 --> 00:02:41.510
as opposed to have to do a bunch

00:02:41.510 --> 00:02:43.570
of algebraic simplification here.

00:02:43.570 --> 00:02:45.650
So we can say,

00:02:45.650 --> 00:02:46.800
when,

00:02:46.800 --> 00:02:48.270
let me do it over here.

00:02:48.270 --> 00:02:50.490
So when, I'll it right here.

00:02:50.490 --> 00:02:51.323
When

00:02:52.350 --> 00:02:54.700
x is equal to negative one,

00:02:54.700 --> 00:02:58.070
and y is equal to one,

00:02:58.070 --> 00:02:59.530
well first of all, what's the,

00:02:59.530 --> 00:03:01.640
what's dy dx going to be?

00:03:01.640 --> 00:03:03.760
The derivative of y with respect to x,

00:03:03.760 --> 00:03:04.970
let me scroll down a little bit,

00:03:04.970 --> 00:03:07.130
so we have a little bit more space.

00:03:07.130 --> 00:03:08.770
The derivative of y with respect to x

00:03:08.770 --> 00:03:13.770
is going to be equal to one
over three times one squared,

00:03:14.100 --> 00:03:15.770
which is just three,

00:03:15.770 --> 00:03:17.750
minus negative one.

00:03:17.750 --> 00:03:19.580
So that's just going be plus one.

00:03:19.580 --> 00:03:21.680
It's going to be equal to 1/4.

00:03:21.680 --> 00:03:24.930
And so this whole expression over here,

00:03:24.930 --> 00:03:29.340
so I can write the second
derivative of y with respect to x

00:03:29.340 --> 00:03:30.910
is going to be equal to,

00:03:30.910 --> 00:03:33.520
well we know that that's
going to be equal to 1/4.

00:03:33.520 --> 00:03:37.960
1/4 times three times one squared,

00:03:37.960 --> 00:03:39.640
which is just three,

00:03:39.640 --> 00:03:41.610
minus negative one,

00:03:41.610 --> 00:03:43.440
so plus one,

00:03:43.440 --> 00:03:45.490
minus one,

00:03:45.490 --> 00:03:48.960
so I'll just leave that minus out there,

00:03:48.960 --> 00:03:51.710
times six times one,

00:03:51.710 --> 00:03:53.320
times 1/4.

00:03:53.320 --> 00:03:54.153
Let me just write it out.

00:03:54.153 --> 00:03:57.280
Six, six times one

00:03:57.280 --> 00:03:59.700
times 1/4

00:03:59.700 --> 00:04:01.770
minus one

00:04:01.770 --> 00:04:03.340
all of that over

00:04:04.230 --> 00:04:05.670
let's see this is going to be

00:04:05.670 --> 00:04:07.180
three times y squared,

00:04:07.180 --> 00:04:10.000
y is one, so is just gonna be three,

00:04:10.000 --> 00:04:12.980
three minus negative one,

00:04:12.980 --> 00:04:14.253
so plus one,

00:04:15.100 --> 00:04:16.030
squared.

00:04:16.030 --> 00:04:17.230
Now what is this going to be?

00:04:17.230 --> 00:04:19.360
And this is just
simplifying something here.

00:04:19.360 --> 00:04:21.670
1/4 times four

00:04:21.670 --> 00:04:23.490
that's going to simplify to one.

00:04:23.490 --> 00:04:28.180
And let's see, this is
going to be 1 1/2 minus one

00:04:28.180 --> 00:04:30.280
so that's going to be 1/2.

00:04:30.280 --> 00:04:33.520
And then we're going to be
have all of that over 16.

00:04:33.520 --> 00:04:35.660
And so this is going to be equal to,

00:04:35.660 --> 00:04:37.890
we get a mini drum roll here,

00:04:37.890 --> 00:04:40.700
this is going to be
equal to one minus 1/2,

00:04:40.700 --> 00:04:42.400
which is equal to 1/2.

00:04:42.400 --> 00:04:43.790
Over 16,

00:04:43.790 --> 00:04:47.210
which is the same thing as one over 32.

00:04:47.210 --> 00:04:48.603
And we are done.


Second derivatives (implicit equations): find expression

WEBVTT
Kind: captions
Language: en

00:00:00.260 --> 00:00:01.940
- [Instructor] Let's say
that we're given the equation

00:00:01.940 --> 00:00:05.060
that y squared minus x squared

00:00:05.060 --> 00:00:07.430
is equal to four.

00:00:07.430 --> 00:00:11.890
And our goal is to find
the second derivative of y

00:00:11.890 --> 00:00:14.930
with respect to x, and we want to

00:00:14.930 --> 00:00:19.130
find an expression for it
in terms of x's and y's.

00:00:19.130 --> 00:00:22.750
So pause this video, and see
if you can work through this.

00:00:22.750 --> 00:00:24.860
All right, now let's do it together.

00:00:24.860 --> 00:00:28.550
Now, some of you might
have wanted to solve for y

00:00:28.550 --> 00:00:30.510
and then use some traditional techniques.

00:00:30.510 --> 00:00:32.200
But here, we have a y squared,

00:00:32.200 --> 00:00:34.450
and so it might involve a
plus or a minus square root.

00:00:34.450 --> 00:00:36.410
And so some of y'all
might have realized, hey,

00:00:36.410 --> 00:00:38.810
we can do a little bit of
implicit differentiation,

00:00:38.810 --> 00:00:41.310
which is really just an
application of the chain rule.

00:00:41.310 --> 00:00:42.180
So let's do that.

00:00:42.180 --> 00:00:44.070
Let's first find the first derivative

00:00:44.070 --> 00:00:45.470
of y with respect to x.

00:00:45.470 --> 00:00:47.540
And to do that, I'll
just take the derivative

00:00:47.540 --> 00:00:51.343
with respect to x of both
sides of this equation.

00:00:52.290 --> 00:00:55.020
And then what do we get?

00:00:55.020 --> 00:00:57.590
Well, the derivative with
respect to x of y squared,

00:00:57.590 --> 00:00:59.080
we're gonna use the chain rule here.

00:00:59.080 --> 00:01:00.270
First, we can take the derivative

00:01:00.270 --> 00:01:02.280
of y squared with respect to y,

00:01:02.280 --> 00:01:04.260
which is going to be equal to two y,

00:01:04.260 --> 00:01:08.000
and then that times the
derivative of y with respect to x.

00:01:08.000 --> 00:01:10.850
Once again, this comes
straight out of the chain rule.

00:01:10.850 --> 00:01:14.020
And then, from that, we will subtract,

00:01:14.020 --> 00:01:16.710
what's the derivative of x
squared with respect to x?

00:01:16.710 --> 00:01:18.670
Well, that's just going to be two x.

00:01:18.670 --> 00:01:21.040
And then last, but not least,

00:01:21.040 --> 00:01:25.020
what is the derivative of a
constant with respect to x?

00:01:25.020 --> 00:01:25.853
Well, it doesn't change,

00:01:25.853 --> 00:01:28.770
so it's just going to be equal to zero.

00:01:28.770 --> 00:01:30.870
All right, now we can solve for

00:01:30.870 --> 00:01:33.610
our first derivative
of y with respect to x.

00:01:33.610 --> 00:01:34.443
Let's do that.

00:01:34.443 --> 00:01:36.430
We can add two x to both sides,

00:01:36.430 --> 00:01:40.410
and we would get two y
times the derivative of y

00:01:40.410 --> 00:01:43.860
with respect to x is equal to

00:01:43.860 --> 00:01:45.160
two x.

00:01:45.160 --> 00:01:50.160
And now I can divide both
sides by two y, and I am going

00:01:50.480 --> 00:01:51.850
to get

00:01:51.850 --> 00:01:54.280
that the derivative of y

00:01:54.280 --> 00:01:59.110
with respect to x is equal to x,

00:01:59.110 --> 00:02:01.803
x over y.

00:02:02.770 --> 00:02:04.630
Now, the next step is
let's take the derivative

00:02:04.630 --> 00:02:06.930
of both sides of this with respect to x,

00:02:06.930 --> 00:02:09.770
and then we can hopefully
find our second derivative

00:02:09.770 --> 00:02:11.540
of y with respect to x.

00:02:11.540 --> 00:02:13.520
And to help us there,
actually let me rewrite this.

00:02:13.520 --> 00:02:15.800
And I always forget the quotient rule,

00:02:15.800 --> 00:02:18.400
although it might be a useful
thing for you to remember.

00:02:18.400 --> 00:02:20.240
But I could rewrite this as a product,

00:02:20.240 --> 00:02:22.010
which will help me at least.

00:02:22.010 --> 00:02:24.740
So I'm going to rewrite
this as the derivative of y

00:02:24.740 --> 00:02:28.100
with respect to x is equal to

00:02:28.100 --> 00:02:31.620
x times y to the negative one power,

00:02:31.620 --> 00:02:34.670
y to the negative one power.

00:02:34.670 --> 00:02:37.380
And now, if we want to
find the second derivative,

00:02:37.380 --> 00:02:40.100
we apply the derivative operator

00:02:40.100 --> 00:02:42.710
on both sides of this equation,

00:02:42.710 --> 00:02:45.150
derivative with respect to x.

00:02:45.150 --> 00:02:47.350
And our left-hand side is exactly

00:02:47.350 --> 00:02:49.480
what we eventually wanted to get,

00:02:49.480 --> 00:02:53.520
so the second derivative
of y with respect to x.

00:02:53.520 --> 00:02:56.900
And what do we get here
on the right-hand side?

00:02:56.900 --> 00:02:59.480
Well, we can apply the product rule.

00:02:59.480 --> 00:03:04.320
So first, we can say the
derivative of x with respect to x,

00:03:04.320 --> 00:03:07.840
well, that is just going to
be one times the other thing,

00:03:07.840 --> 00:03:11.460
so times y to the negative one power,

00:03:11.460 --> 00:03:13.630
y to the negative one power.

00:03:13.630 --> 00:03:15.890
And then we have plus

00:03:15.890 --> 00:03:19.470
x times the derivative
of y to the negative one.

00:03:19.470 --> 00:03:21.180
So plus x,

00:03:21.180 --> 00:03:22.170
what's the, times,

00:03:22.170 --> 00:03:24.880
what's the derivative of y
to the negative one power?

00:03:24.880 --> 00:03:26.540
Well, first, we can find the derivative

00:03:26.540 --> 00:03:29.660
of y to the negative one
power with respect to y.

00:03:29.660 --> 00:03:31.150
We'll just leverage the power rule there.

00:03:31.150 --> 00:03:33.410
So that's going to be negative one

00:03:33.410 --> 00:03:36.340
times y to the negative two power.

00:03:36.340 --> 00:03:38.560
And then we would multiply that

00:03:38.560 --> 00:03:41.200
times the derivative
of y with respect to x,

00:03:41.200 --> 00:03:43.340
just an application of the chain rule,

00:03:43.340 --> 00:03:46.180
times dy/dx.

00:03:46.180 --> 00:03:48.090
And remember, we know what the derivative

00:03:48.090 --> 00:03:49.740
of y with respect to x is.

00:03:49.740 --> 00:03:51.000
We already solved for that.

00:03:51.000 --> 00:03:52.900
It is x over y.

00:03:52.900 --> 00:03:56.620
So this over here is going to be x over y.

00:03:56.620 --> 00:03:59.490
And so now we just have to
simplify this expression.

00:03:59.490 --> 00:04:02.210
This is going to be equal to,

00:04:02.210 --> 00:04:03.720
and I'll try to do it part by part,

00:04:03.720 --> 00:04:07.520
that part right over there is
just going to be a one over y.

00:04:07.520 --> 00:04:10.210
And then all of this business,

00:04:10.210 --> 00:04:12.140
let's see if I can simplify that.

00:04:12.140 --> 00:04:15.030
This negative is going to
go out front, so minus,

00:04:15.030 --> 00:04:18.770
and then I'm going to have
x times x in the numerator.

00:04:18.770 --> 00:04:22.010
And then it's going to
be divided by y squared

00:04:22.010 --> 00:04:24.250
and then divided by another y.

00:04:24.250 --> 00:04:27.810
So it's going to be minus x squared

00:04:27.810 --> 00:04:30.390
over y to the third,

00:04:30.390 --> 00:04:32.840
over y to the third, or
another way to think about it,

00:04:32.840 --> 00:04:35.830
x squared times y to the negative three.

00:04:35.830 --> 00:04:37.050
And we are done.

00:04:37.050 --> 00:04:41.080
We have just figured out
the second derivative of y

00:04:41.080 --> 00:04:43.730
with respect to x

00:04:43.730 --> 00:04:46.303
in terms of x's and y's.