Posted by Jessica on .
Find the equation (in terms of x and y) of the tangent line to the curve r=3sin3(theta) at theta=pi/4.
I don't know if I'm taking the right approach for this question but her goes what I did:
I found the slope
I got (9(1/2)(sqrt3/2)+3(sqrt3/2)(1/2))/(9(1/2)(1/2)3(sqrt3/2)(sqrt3/2))
Is that right? what else do I have to do?

PLEASE I really need help!!! 
Jessica,
I don't know what to do for this problem.

Calculus 
Reiny,
Since you want the equation in terms of x and y, let's change what looks a polar coordinate equation to the xy form
r = 3sin(3Ø)
let k=3Ø
so r = 3sin k
r = 3(y/r)
r^2 = 3y
x^2 + y^2 = 3y
then
2x + 2y dy/dx = 3dy/dx
2x/(32y) = dy/dx
when Ø = π/4
k = 3(π/4) = 3π/4
so r = 3sin(3π/4) = 3√2/2
but r^2 = 3y
9/2 = 3y
y = 3/2
sub that into x^2 + y^2 = 3y
x^2 + 9/4 = 9/2
x^2 = 9/4
x = 3/2
so the point of contact is (3/2 , 3/2)
dy/dx = 2(3/2) / (32(3/2) = 3/(0) > undefined !
aaahhhh!, so the tangent line is a vertical line
so its equation is simply x = 3/2
check my arithmetic. 
Calculus 
Jessica,
Webwork asks me to enter an answer for y=

Calculus 
Reiny,
A vertical line cannot be written in the form y = ...
Will it let you enter x + 0y = 3/2 ??
Did you look at my solution and follow the steps?
Do you see any errors? 
Calculus 
Jessica,
I don't see any error, but webwork just says y= , I can't enter anything besides what y is supposed to be. The slope I took was a method I found in my book but it only says that, how to take the slope, and I don't even know if I had that right. But I cant enter anything besides the answer for y=.

Calculus 
Steve,
Leaving things in polar coordinates is a lot simpler here.
if r = f(t)
dy/dx = [f'(t)*sin(t) + f(t)*cos(t)] / [f'(t)*cos(t)  f(t)*sin(t)]
t = π/4
f = 3sin(3t) = 3/√2
f' = 9cos(3t) = 9/√2
sin(t) = 1/√2
cos(t) = 1/√2
dy/dx = (9/√2 * 1/√2 + 3/√2 * 1/√2)/(9/√2 * 1/√2  3/√2 * 1/√2)
= (6/2 +3/2)/(9/2  3/2) = 6/12 = 1/2
google polar coordinates graphing and the first hit is a nice simple grapher for polar coordinates. Looks like slope = 1/2
So, now we have a point (π/4,3/√2) and a slope: 1/2
y3/√2 = 1/2(xπ/4) 
Calculus 
MathMate,
I am not sure I'm right, but it seems to me that:
r=3sin3(theta)
is meant to be interpreted as
r(t)=3sin³(θ),
in which case dy/dx=2.
I do not know why, but when I started looking at it, I did the same interpretation of 3θ.