Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 37

a.) At what rate is the distance from the television camera to the rocket changing at that instant?
b.) At what rate is the camera's angle of elevation changing at the same moment, suppose that the television camera is always kept aimed at the rocket.







a.) By Pythagorean theorem,
$z^2 = y^2 + 4000^2$; when $y = 3000$, $z = \sqrt{3000^2 + 4000^2} = 5000$ ft

Taking the derivative with respect to time,

$
\begin{equation}
\begin{aligned}
\cancel{2} z \frac{dz}{dt} &= \cancel{2}y \frac{dy}{dt}\\
\\
\frac{dz}{dt} =&= \frac{y}{z} \frac{dy}{dt} && \text{; where } \frac{dy}{dt} = 600 \frac{\text{ft}}{s}
\end{aligned}
\end{equation}
$


Plugging all the values we get,

$
\begin{equation}
\begin{aligned}
\frac{dz}{dt} &= \frac{3000}{5000} (600)\\
\\
\frac{dz}{dt} &= 360 \frac{\text{ft}}{s}
\end{aligned}
\end{equation}
$


b.) By applying tangent function,
$\displaystyle \tan \theta = \frac{y}{4000}$; when $\displaystyle y = 3000, \theta = \tan^{-1} \left[ \frac{3000}{4000} \right]= 36.8699^{\circ}$

Taking its derivative with respect to time we have,

$
\begin{equation}
\begin{aligned}
\sec^2 \theta \frac{d \theta}{d t} &= \frac{\frac{dy}{dt}}{4000} \quad ; \sec \theta = \frac{1}{\cos \theta}\\
\\
\frac{d \theta}{d t} &= \frac{\cos^2 \theta \left( \frac{dy}{dt} \right)}{4000}\\
\\
\frac{d \theta}{dt} &= \frac{\cos^2(36.8699)(600)}{4000}\\
\\
\frac{d \theta}{dt} &= 0.096\frac{^\circ}{s}
\end{aligned}
\end{equation}
$

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