# United States Calculus 2 - Calculus in Polar Coordinates

1. Dec 6, 2011

### GreenPrint

1. The problem statement, all variables and given/known data

Find the slope of the line tangent to the polar curve at the given point. At the point where the curve intersects the origin, find the equation of the tangent line in polar coordinates.

r = 6 sinθ; (-3 7∏/6)

2. Relevant equations

3. The attempt at a solution

Select the correct choice below and, if necessary, fill in the answer box within your choice.

A. The slope of the curve at the point (-3, 7∏/6) is _
B. The slope of the curve at the point (-3, 7∏/6) is undefined.

I selected A. and put sqrt(3) and it told me I was correct.

The equation of the tangent line when the curve intersects the origin is _
(Type an equation. Type any angles in radians between 0 and 2∏.)

I'm not sure how to do this part. I believe I put in the correct equation in Cartesian. coordinate system in terms of y and x and it told me I was wrong. I wasn't exactly sure why and then I realized that it wanted me to input a polar coordinate equation in the instructions, which I'm not exactly sure how to do.
When I put the equation it told me I was wrong and this came up

If the graph of the polar curve passes through the origin of some angles θ_0, then f(θ_0)=0 and dy/dx=tan θ_0, which means that the equation of a tangent line through the origin is θ=θ_0. Thus, to determine the equation of the tangent line(s) when the curve intersects the origin, find the value(S) of θ when r=0. Note that this hold for f'(θ_0)=/=0.

I'm not exactly sure what it wants me to do.

I got
r= 6sinθ
x=6sinθcosθ
y=6(sinθ)^2
dy/dx= ( 2cosθsinθ )/( (cosθ)^2 - (sinθ)^2 )

I'm not exactly sure why they tell me dy/dx = tanθ_0 or why the equation of the line that it wants me to enter is just an angle θ=θ_0 if that is what it wants me to do.

I do not see what setting r=0 and solving for θ does

r = 6sinθ = 0
θ = 0,∏,2∏

Is this the answer it wants?

Thanks for any help.