I was given the following problem:
For which of the following points on the circle $r=\cos\theta$ is the slope equal to $2$? $\\
a. \displaystyle \theta=\frac{1}{2} \arctan \Big(-\frac{1}{2} \Big)\\
b. \displaystyle \theta=\frac{1}{2} \arctan (2)\\
c. \displaystyle \theta=\frac{1}{2} \arctan \Big(\frac{1}{2} \Big)$
The way I attempted the problem was by converting the polar equation to a parametric (cartesian) equation and using the formula $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}$ – and setting that equal to two. This did not seem to get me the right answer.
I looked in the answer key, and the first step given was to:
Set the equation for the slope equal to 2 and solve for $\theta$.
$$\frac{dy}{dx}=\frac{f(\theta)\cos(\theta)+f'(\theta)\sin(\theta)}{-f(\theta)\sin(\theta)+f'(\theta)\cos(\theta)}=2$$
I understand why they are setting the slope equation equal to two, but how did they get this equation as the slope equation?
Best Answer
$y=f(\theta) \sin\theta$ and $x=f(\theta) \cos\theta$ and so (by product rule) $$\frac{d y}{d \theta} = f(\theta) \cos \theta + f'(\theta)\sin \theta$$ and $$\frac{d x}{d \theta} = -f(\theta) \sin \theta + f'(\theta)\cos \theta$$ You said you understand that $\frac{dy}{dx} = \frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$ so I'll let you do that.