Question

a television camera is positioned 4,000 ft from the base of a rocket launching pad. the angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. also the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. lets assume the rocket rises vertically and its speed is 500 ft/s when it has risen 3,000 ft. (a) how fast (in ft/s) is the distance from the television camera to the rocket changing at that moment? (b) if the television camera is always kept aimed at the rocket, how fast (in rad/s) is the cameras angle of elevation changing at that same moment?

Explanation:

Step1: Set up variables and find initial distance

Let the height of the rocket be $y$, the distance from the camera to the rocket be $z$, and the horizontal - distance from the camera to the launch - pad be $x = 4000$ ft. By the Pythagorean theorem, $z^{2}=x^{2}+y^{2}$. When $y = 3000$ ft, $z=\sqrt{4000^{2}+3000^{2}}=\sqrt{16000000 + 9000000}=\sqrt{25000000}=5000$ ft.

Step2: Differentiate the Pythagorean equation with respect to time $t$

Differentiating $z^{2}=x^{2}+y^{2}$ with respect to $t$ gives $2z\frac{dz}{dt}=2y\frac{dy}{dt}$ (since $x$ is constant, $\frac{dx}{dt}=0$).

Step3: Solve for $\frac{dz}{dt}$

We know that $y = 3000$ ft, $z = 5000$ ft, and $\frac{dy}{dt}=500$ ft/s. Substitute these values into $z\frac{dz}{dt}=y\frac{dy}{dt}$:
$\frac{dz}{dt}=\frac{y}{z}\cdot\frac{dy}{dt}=\frac{3000}{5000}\times500 = 300$ ft/s.

Step4: Set up the tangent function for the angle of elevation

Let $\theta$ be the angle of elevation. Then $\tan\theta=\frac{y}{x}$.

Step5: Differentiate the tangent - function equation with respect to time $t$

Differentiating $\tan\theta=\frac{y}{x}$ with respect to $t$ gives $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{x}\frac{dy}{dt}$.
Since $\tan\theta=\frac{y}{x}=\frac{3000}{4000}=\frac{3}{4}$, then $\sec^{2}\theta = 1+\tan^{2}\theta=1 + (\frac{3}{4})^{2}=1+\frac{9}{16}=\frac{25}{16}$.

Step6: Solve for $\frac{d\theta}{dt}$

We know that $x = 4000$ ft and $\frac{dy}{dt}=500$ ft/s. Substitute into $\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{x}\frac{dy}{dt}$:
$\frac{25}{16}\frac{d\theta}{dt}=\frac{1}{4000}\times500$.
$\frac{d\theta}{dt}=\frac{1}{4000}\times500\times\frac{16}{25}=\frac{4}{100}=0.04$ rad/s.

Answer:

(a) 300 ft/s
(b) 0.04 rad/s