Answer to Question #159633 in Calculus for MaryJane Copeland

Question

A camera is mounted 3.000 feet from a rocket launching pad. The camera needs to swivel as the ro launched to keep it in focus, a) If the rocket is rising vertically at 800 ft/sec when it is 4,000 feet hi fast is the camera-to-rocket distance changing?

Answer 1

The camera-to-rocket distance as a function of time t is

$s(t)=\sqrt{3000^2+(800t)^2}=\sqrt{3000^2+800^2t^2}$

$s'(t)=\frac{1}{2}\frac{800^2*2t}{\sqrt{3000^2+800^2t^2}}=$

$=\frac{800^2t}{\sqrt{3000^2+800^2t^2}}$

when rocket is 4,000 feet hi $t=\frac{4000}{800}=5$

$s'(5)=\frac{800^2*5}{\sqrt{3000^2+800^2*5^2}}=$

$=\frac{640000*5}{\sqrt{9000000+16000000}}=$

$=\frac{640000*5}{\sqrt{25000000}}=\frac{640000*5}{5000}=640$

Answer: the camera-to-rocket distance is rising with velocity 640 feet/sec.