Find the slope dy/dx of the polar curve r = 3 / (2 - cos(theta)) at the point (r, theta) = (3/2, pi/2)

Mathematics

1 answer:

sashaice [31]3 years ago

6 0

Given polar equation is

$r=\frac{3}{2-\cos\left(\theta\right)}$

for polar equation we use

$x=r\cos\left(\theta\right)$

and $y=r\sin\left(\theta\right)$

plug the given value of r into these equations we get:

$x=r\cos\left(\theta\right)=\frac{3\cos\left(\theta\right)}{2-\cos\left(\theta\right)}$

$y=r\sin\left(\theta\right)=\frac{3\sin\left(\theta\right)}{2-\cos\left(\theta\right)}$

find derivative with respect to theta

$\frac{dx}{d\theta}=-\frac{6\sin\left(\theta\right)}{\left(2-\cos\left(\theta\right)\right)^2}$

$\frac{dy}{d\theta}=-\frac{3\left(\sin^2\left(\theta\right)+\cos^2\left(\theta\right)-2\cos\left(\theta\right)\right)}{\left(2-\cos\left(\theta\right)\right)^2}$

now slope is given by formula

$m=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

plug the above values into slope formula and the given angle theta = pi/2

$m=\frac{-\frac{3\left(\sin^2\left(\theta\right)+\cos^2\left(\theta\right)-2\cos\left(\theta\right)\right)}{\left(2-\cos\left(\theta\right)\right)^2}}{-\frac{6\sin\left(\theta\right)}{\left(2-\cos\left(\theta\right)\right)^2}}$