Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 38

Suppose that an object with weight $w$ is dragged along a horizontal plane by a force acting along a rope attracted to the objet. If the rope makes an angle $\theta$ with the plane then the magnitude of the force;
$\displaystyle F = \frac{\mu\omega}{\mu \sin \theta + \cos \theta}$

where $\mu$ is a constant called the coefficient of friction.
a.) Find the rate of change of $F$ with respect to $\theta$.
b.) When is this rate of change equal to 0?
c.) If $ w= 50 lb$ and $\mu = 0.6$, draw the graph of $F$ as a function of $\theta$ and use it to locate the value of $\theta$ for which $\displaystyle \frac{d_F}{d\theta} = 0$. Is the value consistent with your answer to part(b)?



$
\begin{equation}
\begin{aligned}
\text{a.) } \frac{d_F}{d\theta} &= \frac{[\mu\sin\theta+\cos\theta] \frac{d}{d\theta}(\mu \omega) - \mu \omega \frac{d}{d\theta}[\mu \sin \theta + \cos \theta]}{[\mu \sin \theta + \cos \theta]^2}\\
\\
\frac{d_F}{d\theta} &= \frac{0 - \mu \omega [ \mu \cos \theta + (-\sin\theta)]}{[\mu \sin \theta + cos \theta]^2}\\
\\
\frac{d_F}{d\theta} &= \frac{\mu \omega [\sin \theta - \mu \cos \theta]}{[\mu \sin \theta + \cos \theta]^2}\\
\\

\end{aligned}
\end{equation}
$


$\displaystyle \text{b.) } \frac{d_F}{d\theta} = 0 = \frac{\mu \omega [\sin \theta - \mu \cos \theta]}{[\mu \sin \theta + \cos \theta]^2}$

$
\begin{equation}
\begin{aligned}
\\
\mu \omega [\sin \theta - \mu \cos \theta] &= 0 \\
\\
\sin \theta - \mu \cos \theta &= 0\\
\\
\sin \theta &= \mu \cos \theta\\
\\
\frac{\sin \theta}{\cos \theta} &= \mu\\
\\
\tan \theta &= \mu\\
\\
\theta & = \tan^{-1} \mu
\end{aligned}
\end{equation}
$


The rate of change is equal to 0 when $\theta = \tan^{-1} \mu$


$
\begin{equation}
\begin{aligned}
\text{c.) } F &= \frac{\mu \omega}{\mu \sin \theta + \cos \theta} && ; \omega = 50^h, \, \mu = 0.6\\
\\
F &= \frac{50(0.6)}{0.6 \sin \theta+ \cos \theta}\\
\\
F &= \frac{30}{0.6 \sin \theta + \cos \theta}
\end{aligned}
\end{equation}
$




We know that $\displaystyle \frac{d_F}{d\theta} = 0$ when $\theta = \tan^{-1}\mu$ so

$
\begin{equation}
\begin{aligned}
\theta &= \tan^{-1} (0.6) = 0.5404\\
\\
F &= \frac{30}{0.6 \sin \theta + \cos \theta} && ; \text{where } \theta = 0.5404\\
\\
F(\theta) &= \frac{30}{0.6\sin(0.5404)+\cos(0.5404)} = 25.725

\end{aligned}
\end{equation}
$


Based from the graph, our answer in part(b) matches our answer in part(c) since the value of tangent line of that point is zero.