Given rē = 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!

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- Dec 21st 2012, 09:13 AMjones123Extrema polar equation
Given rē = 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! - Dec 21st 2012, 05:54 PMchiroRe: Extrema polar equation
Hey jones123.

Hint: How do you represent the (x,y) co-ordinates given the polar representation? - Dec 21st 2012, 06:02 PMskeeterRe: Extrema polar equation
you're going to have to determine where $\displaystyle \frac{dy}{d\theta} = 0$ in the given interval ... this will require doing the grunt work of finding both $\displaystyle \frac{dy}{d\theta}$ and $\displaystyle \frac{dr}{d\theta}$.

Derivatives of polar equations are a P-I-T-A. - Dec 22nd 2012, 04:30 AMjones123Re: Extrema polar equation
- Dec 22nd 2012, 04:53 AMHallsofIvyRe: Extrema polar equation
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 $\displaystyle y= rsin(\theta)$ subject to the constraint $\displaystyle r^2- 4cos(2\theta)= 0$, perhaps using Lagrange multipliers. - Dec 22nd 2012, 07:24 AMskeeterRe: Extrema polar equation
$\displaystyle y = r\sin{\theta}$

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \theta = \frac{\pi}{6}$ - Dec 23rd 2012, 05:38 AMjones123Re: Extrema polar equation
- Dec 23rd 2012, 06:11 AMskeeterRe: Extrema polar equation
$\displaystyle \frac{d}{d\theta}\left[r^2 = 4\cos(2\theta)\right]$

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

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

Quote:

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