Posted by Jessica on Thursday, October 13, 2011 at 9:27pm.
I don't know what to do for this problem.
Since you want the equation in terms of x and y, let's change what looks a polar coordinate equation to the x-y 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/(3-2y) = 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) / (3-2(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.
Webwork asks me to enter an answer for y=
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?
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=.
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
y-3/√2 = 1/2(x-π/4)
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.