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April 30, 2016
Posted by **Jessica** on Thursday, October 13, 2011 at 9:27pm.

I dont 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**, Thursday, October 13, 2011 at 9:59pmI dont know what to do for this problem.

- Calculus -
**Reiny**, Thursday, October 13, 2011 at 10:29pmSince 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. - Calculus -
**Jessica**, Thursday, October 13, 2011 at 10:37pmWebwork asks me to enter an answer for y=

- Calculus -
**Reiny**, Thursday, October 13, 2011 at 10:52pmA 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**, Thursday, October 13, 2011 at 11:03pmI 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**, Friday, October 14, 2011 at 10:37amLeaving 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) - Calculus -
**MathMate**, Friday, October 14, 2011 at 6:24pmI 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.