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!
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.
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.
$\displaystyle y = r\sin{\theta}$
$\displaystyle \frac{dy}{d\theta} = r\cos{\theta} + \sin{\theta} \cdot \frac{dr}{d\theta}$
$\displaystyle r^2 = 4\cos(2\theta)$
$\displaystyle \frac{dr}{d\theta} = -\frac{4\sin(2\theta)}{r}$
$\displaystyle \frac{dy}{d\theta} = r\cos{\theta} - \frac{8\sin^2{\theta}\cos{\theta}}{r}$
$\displaystyle 0 = r\cos{\theta} - \frac{8\sin^2{\theta}\cos{\theta}}{r}$
$\displaystyle r\cos{\theta} = \frac{8\sin^2{\theta}\cos{\theta}}{r}$
$\displaystyle r^2\cos{\theta} = 8\sin^2{\theta}\cos{\theta}$
$\displaystyle 4\cos(2\theta) = 8\sin^2{\theta}$
$\displaystyle 4(1-2\sin^2{\theta}) = 8\sin^2{\theta}$
$\displaystyle \sin^2{\theta} = \frac{1}{4}$
$\displaystyle \sin{\theta} = \frac{1}{2}$
$\displaystyle \theta = \frac{\pi}{6}$
$\displaystyle \frac{d}{d\theta}\left[r^2 = 4\cos(2\theta)\right]$
$\displaystyle 2r \cdot \frac{dr}{d\theta} = -8\sin(2\theta)$
$\displaystyle \frac{dr}{d\theta} = \frac{-8\sin(2\theta)}{2r} = - \frac{-4\sin(2\theta)}{r}$
note that sqrt(cos(2Ɵ)) = rIf r² = 4cos(2Ɵ) than r = sqrt(4cos(2Ɵ)), shouldn't the derivative dr/dƟ be - 8sin(2Ɵ) / 2sqrt(cos(2Ɵ)) ?