I'm stuck at the following related rates problem: As shown in the diagram the apex angle of a cone is $2\theta$ and the slope length is 10cm. The rate of increase of $\theta$ is 0.01 radians per second. The cone starts with $\theta=\frac \pi 6$. Find the initial rate of change of the volume of the cone.
What I've tried:
The volume $V$ of a cone is given by $V = \frac 1 3 \pi r^2h$. We can replace $r$ and $h$ with $\theta$ using the law of sines if we look at the cross section of the cone:
$\frac {10}{\sin \frac \pi2} = \frac r {\sin \theta}$ => $r = 10\sin \theta$
Similarly
$\frac {10}{\sin \frac \pi2} = \frac h {\sin (\frac \pi 2 -\theta)}$ => $ h = 10 \sin (\frac \pi 2 -\theta) = 10 \cos \theta$
There's probably a better way since this is a right angled triangle but this should work.
From this follows:
$V=\frac 1 3 \pi r^2h = \frac 1 3 \pi (10 \sin \theta)^2 (10 \cos \theta) = \frac {1000} 3 \pi \sin^2 \theta \cos\theta $
So: $\frac {dV} {dt} = \frac {d}{dt} (\frac {1000} 3 \pi \sin^2 \theta \cos\theta)$
$\frac {dV} {dt} = \frac {1000} 3 \pi \frac {d}{dt} (\sin^2 \theta \cos\theta)$ $\frac {dV} {dt} = \frac {1000} 3 \pi ( \cos\theta \frac {d}{dt} \sin^2 \theta + \sin^2 \theta \frac {d}{dt} \cos\theta)$
$\frac {dV} {dt} = \frac {1000} 3 \pi ( 2\cos\theta \sin\theta * \frac {d}{dt} \sin \theta- \sin^2 \theta \sin\theta\frac {d \theta}{dt})$
$\frac {dV} {dt} = \frac {1000} 3 \pi ( 2\cos\theta \sin\theta * \cos\theta\frac {d \theta}{dt}- \sin^2 \theta \sin\theta\frac {d \theta}{dt})$
$\frac {dV} {dt} = \frac {1000} 3 \pi ( 2\cos^2\theta \sin \theta \frac {d \theta}{dt} - \sin^3 \theta \frac {d \theta}{dt})$
Then I plugged in the given values: $\frac {dV} {dt} = \frac {1000} 3 \pi ( 2 \cos^2\frac \pi 6 \sin \frac \pi 6 (0.01) - \sin^3 \frac \pi 6 (0.01)) = 6.545 \frac {cm^3}{s}$.
However my textbook says that it should be $0.654 \frac {cm^3}s$ which seems pretty close to what I have found. Did I make a blunder somewhere or is my reasoning incorrect?
Your answer is correct. The book is incorrect by a factor of $10$.
To simplify your calculation, we have $$V(t) = \frac{1000\pi}{3} \sin^2 \theta(t) \cos \theta(t) = \frac{1000\pi}{3} (\cos \theta(t) - \cos^3 \theta(t)).$$ At time $t = 0$, $\theta(0) = \pi/6$ and $\theta'(0) = \frac{1}{100}$, so $$\begin{align} V'(0) &= \frac{1000\pi}{3} \left(-\theta'(0) \sin \theta(0) + 3 \theta'(0) \cos^2 \theta(0) \sin \theta(0) \right) \\ &= \frac{1000\pi}{3} \theta'(0) \sin \theta(0)(3 \cos^2 \theta(0) - 1) \\ &= \frac{5\pi}{3} \left(3 \cos^2 \frac{\pi}{6} - 1 \right) \\ &= \frac{5\pi}{3} \left(\frac{9}{4} - 1\right) \\ &= \frac{25\pi}{12}. \end{align}$$