In this video, we will answer
the question, "Why is the derivative
"of the volume of a sphere equal
"to the surface area of a sphere?"
The volume equals four-thirds pi r-cubed,
and the surface area
equals four pi r-squared.
Given the volume of a sphere,
V equals four-thirds pi r-cubed,
let's begin by determining dV / dr
or V-prime of r.
To do this, we apply the
power rule of differentiation.
And therefore, the derivative is equal
to four-thirds pi times
the derivative of r-cubed
with respect to r
which gives us four-thirds
pi times three r-squared.
Simplifying, we have dV / dr
or V-prime of r equals four pi r-squared.
And four pi r-squared is equal
to the surface area of a sphere.
So the question is, "Why is the derivative
"of the volume equal to the surface area?"
To begin, let's review some notation.
Where differential V, or dV, describes
the change in the volume,
differential r, or dr, describes
the change in the radius.
And therefore, dV / dr, or V-prime of r,
describes the change in the volume
with respect to the radius.
And now beginning with
a sphere with radius r,
let's consider a small
increase in the radius.
And let's let that small increase
be differential r as shown here.
Notice how differential r
or the change in the radius
produces a small increase in the volume
which would be the volume of the shell
or extra layer created when the radius
is increased by differential r.
And therefore, if we
can determine the volume
of this layer of our
shell, we can determine
differential V, the change in the volume,
given a small change in the radius.
So if we flatten this shell or layer
created by the increase in the radius,
it'll produce a right circular
cylinder as shown here.
Let's look at some additional images.
Let's let a ball filled with air
represent the shell or layer
we want to find the volume of.
As we begin to squish the ball,
it might look something like this.
Once it's completely flattened,
it will look like this
which will be a right circular cylinder
where the top is a circle,
the bottom is a circle, and the height
or thickness is going to
be two times differential r
where again, differential r represents
the change in the radius.
So we need to find the volume
of this right circular
cylinder to find the volume.
We need to find the area of the base
and multiply it by the height.
And now to find the area of the base,
we know it's a circle but the area
is going to be equal to
half of the surface area
of the sphere because if we take a look
at this picture here, we can see
as we begin to flatten the ball,
the top and the bottom will be a circle.
And both circles are formed
by the total surface area.
And therefore, half the
surface area will give us
the area of the base.
And therefore, the area
of the base is equal
to four pi r-squared divided by two
where four pi r-squared
is the surface area.
And therefore, the area of
the base is two pi r-squared
and the height is two differential r.
So going back to our previous slide,
the volume of this right circular cylinder
which represents the change in the volume
is equal to two pi r-squared
times two differential r
which again represents
the change in the volume.
And therefore, we can say
that differential V, or dV,
equals four pi r-squared
times differential r.
Now if we divide both
sides by differential r,
we have dV / dr, or V-prime of r,
equals four pi r-squared
which is the surface area of a sphere.
So the derivative of
the volume of a sphere
is the surface area.
Increasing the radius will result
in an increase in volume proportional
to the surface area.
I hope you have found this
informal discussion helpful
to explain why the derivative
of the volume of a sphere equals
the surface area of a sphere.
Thank you for watching.
