Answer
a) $\frac{dF}{d\theta}=\frac{-\mu^2 W\cos\theta+\mu W\sin\theta}{(\mu sin\theta+cos\theta)^2}$
b) $\theta=\arctan(\mu)$
c) $\arctan(0.6)=0.5404195$, consistent
Work Step by Step
a) The rate of change of $F$ with respect to $\theta$ is $\frac{dF}{d\theta}$. Since $\mu$ and $\mu W$ are constants, their derivatives with respect to $\theta$ will be 0.
$\frac{dF}{d\theta}=\frac{-(\mu sin\theta+cos\theta)'\mu W}{(\mu sin\theta+cos\theta)^2}=\frac{-\mu^2 W\cos\theta+\mu W\sin\theta}{(\mu sin\theta+cos\theta)^2}$
b) $-\mu W(\mu\cos\theta-\sin\theta)=0$
$\mu\cos\theta=\sin\theta$
$\mu=\tan\theta$
$\theta=\arctan(\mu)$
c) For $\mu=0.6$, the value of $\theta$ is $\arctan(0.6)=0.5404195$
From the graph, we can see this is consistent with our answer.
