Related Rated in Calc 1

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. Let's assume the rocket rises vertically and its speed is 800 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 camera's angle of elevation changing at that same moment?

Answer 1

(a) Let L is distance from camera to rocket. L² = 4000² + h²; 2LdL/dt = 2hdh/dt; dL/dt = (hdh/dt)/L

L = 5000ft; h = 3000ft; dh/dt = 800ft/s; dL/dt = 3000·800/5000 = 480 ft/s

(b) tanθ = h/4000; sec²θdθ/dt = (dh/dt)/4000; dθ/dt = (dh/dt)cos²θ/4000 = (dh/dt)·(4000/5000)²/4000;

dθ/dt = 480·0.8²/4000 = 0.0768 rad/s

Answer 2

Make a sketch in the form of a right triangle with the vertical axis = h (height) and the horizontal axis as the distance from the launch to the camera (4000f). D becomes the hypotenuse of a right triangle and = sqrt(4000^2+h^2). Take the derivative of D to find dD/dt =dh/dt /(2sqrt(4000^2+h^2)). Note that dh/dt = 800 f/s and h=3000f. Substitute to find dD/dt= 0.08 f/s a)

Note that the tan(angle)= h/4000 where angle is the camera angle x. Take the derivative to find sec^2xdx/dt=dh/dt/4000 thus dx/dt= dh/dt/4000/sec^x. Again, dh/dt= 800 f/x. Secx=1/cosx=D/4000. Note that D=sqrt(4000^2+3000^2)=5000f, so secx=1.2. Substitute to find dx/dt= .139 rad/sec.