A television camera is 4000 ft from the base of a rocket launching pad. The angle of which the camera elecates changes such that it keeps the rocket in sight. The mechanism for the camera also must account for the distance the rocket reaches after launch-off. Assume the rocket rises vertically at a speed of 600 ft per second when it has risen 3000 ft.

a) How fast 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 is the angle of elevation at that same moment?


Let $h=$ hypotenuse of the triangle, i.e., the distance between the camera and the rocket. Let $x=$ base, i.e., the distance between the camera and the base of the rocket launcher. This doesn't change, therefore $dx/dt=0$. Let $y=$ vertical length, i.e., the distance the rocket as travelled. I know this can be expressed in terms of speed at the moment I'm interested in, $y=600t$ where $t$ is the number of seconds per feet.

Therefore, \begin{align*} h^2&=x^2+y^2 \\ \frac{d}{dt}[h^2&=x^2+y^2] \\ 2h\frac{dh}{dt}&=2x\frac{dx}{dt}+2y\frac{dy}{dt} \\ \frac{dh}{dt}&=\cfrac{1200t}{2\sqrt{4000^2+(600t)^2}}\end{align*}

Now, I know that in order to find the moment height is $3000$, I solve $600t=3000$ to get $t=5$. Therefore, I plug that into my $dh/dt$ to get $$\cfrac{dh}{dt}=360 \text{ feet/sec}$$

This was marked correct, so this part of the problem is complete. However, I'm struggling with the next part.


I know radius $h$ is $\sqrt{4000^2+(600t)^2}$, so that means \begin{align*}(\sqrt{4000^2+(600t)^2})\sin\theta&=\cfrac{600t}{\sqrt{4000^2+(600t)^2}} \\ \sin\theta&=600t \\ \frac{d}{dt}[\sin\theta]&=\frac{d}{dt}[600t] \\ \frac{d\theta}{dt}\cos\theta &= 600 \\ \frac{d\theta}{dt}&= \frac{600}{4000} = \frac{3}{20} \text{ rad per sec}\end{align*}

However, the answer should be $d\theta/dt=\cfrac{12}{125}$ rad per sec. How do I get this answer?


1 Answer 1


Consider $y=4000.tan\theta$



$\Rightarrow \frac{d\theta}{dt}=\frac{12}{125} rad/s$

  • $\begingroup$ Where does $\frac54$ come from? $\endgroup$
    – Lex_i
    Commented Apr 10, 2020 at 20:53
  • $\begingroup$ At y=3000 and x=4000, hypotenuse of right triangle is 5000 $\endgroup$ Commented Apr 10, 2020 at 20:54
  • $\begingroup$ Oh okay I see that now, but why is $dh/dh=600$? I thought it was established from part a that its 360? $\endgroup$
    – Lex_i
    Commented Apr 10, 2020 at 21:05
  • 1
    $\begingroup$ I took h for height. Changed it to y now $\endgroup$ Commented Apr 10, 2020 at 21:07

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