Extrema polar equation

**jones123** · December 21st 2012, 09:13 AM

Given r² = 4cos(2Ɵ) , 0 < Ɵ < π/4

Calculate the point on this curve with the highest y-coordinate.

How would you calculate this? I do not even know how to start this

Thanks already!

**chiro** · December 21st 2012, 05:54 PM

Hey jones123.

Hint: How do you represent the (x,y) co-ordinates given the polar representation?

**skeeter** · December 21st 2012, 06:02 PM

you're going to have to determine where $\frac{dy}{d\theta} = 0$ in the given interval ... this will require doing the grunt work of finding both $\frac{dy}{d\theta}$ and $\frac{dr}{d\theta}$ .

Derivatives of polar equations are a P-I-T-A.

**jones123** · December 22nd 2012, 04:30 AM

Originally Posted by chiro

Hey jones123.

Hint: How do you represent the (x,y) co-ordinates given the polar representation?

So I know that the conversion to cartesian form gives: x^4 + 2(x^2)(y^2) + y^4 - 4x^2 + 4y^2 = 0

**HallsofIvy** · December 22nd 2012, 04:53 AM

Good. Now, differentiate that with respect to x and set y' equal to 0.

Another way to do this, without converting to Cartesian coordinates, is to maximise $y= rsin(\theta)$ subject to the constraint $r^2- 4cos(2\theta)= 0$ , perhaps using Lagrange multipliers.

**skeeter** · December 22nd 2012, 07:24 AM

$y = r\sin{\theta}$

$\frac{dy}{d\theta} = r\cos{\theta} + \sin{\theta} \cdot \frac{dr}{d\theta}$

$r^2 = 4\cos(2\theta)$

$\frac{dr}{d\theta} = -\frac{4\sin(2\theta)}{r}$

$\frac{dy}{d\theta} = r\cos{\theta} - \frac{8\sin^2{\theta}\cos{\theta}}{r}$

$0 = r\cos{\theta} - \frac{8\sin^2{\theta}\cos{\theta}}{r}$

$r\cos{\theta} = \frac{8\sin^2{\theta}\cos{\theta}}{r}$

$r^2\cos{\theta} = 8\sin^2{\theta}\cos{\theta}$

$4\cos(2\theta) = 8\sin^2{\theta}$

$4(1-2\sin^2{\theta}) = 8\sin^2{\theta}$

$\sin^2{\theta} = \frac{1}{4}$

$\sin{\theta} = \frac{1}{2}$

$\theta = \frac{\pi}{6}$

**jones123** · December 23rd 2012, 05:38 AM

Originally Posted by skeeter

$r^2 = 4\cos(2\theta)$

$\frac{dr}{d\theta} = -\frac{4\sin(2\theta)}{r}$

$\frac{dy}{d\theta} = r\cos{\theta} - \frac{8\sin^2{\theta}\cos{\theta}}{r}$

Thanks a lot! Just one more question: I do not understand the part of your calculation i have quoted...

If r² = 4cos(2Ɵ) than r = sqrt(4cos(2Ɵ)), shouldn't the derivative dr/dƟ be - 2sin(2Ɵ) / sqrt(cos(2Ɵ)) ?

**skeeter** · December 23rd 2012, 06:11 AM

$\frac{d}{d\theta}\left[r^2 = 4\cos(2\theta)\right]$

$2r \cdot \frac{dr}{d\theta} = -8\sin(2\theta)$

$\frac{dr}{d\theta} = \frac{-8\sin(2\theta)}{2r} = - \frac{-4\sin(2\theta)}{r}$

If r² = 4cos(2Ɵ) than r = sqrt(4cos(2Ɵ)), shouldn't the derivative dr/dƟ be - 8sin(2Ɵ) / 2sqrt(cos(2Ɵ)) ?

note that sqrt(cos(2Ɵ)) = r

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