Question #159633

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?


1
Expert's answer
2021-02-01T07:25:35-0500

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

s(t)=30002+(800t)2=30002+8002t2s(t)=\sqrt{3000^2+(800t)^2}=\sqrt{3000^2+800^2t^2}


s(t)=1280022t30002+8002t2=s'(t)=\frac{1}{2}\frac{800^2*2t}{\sqrt{3000^2+800^2t^2}}=

=8002t30002+8002t2=\frac{800^2t}{\sqrt{3000^2+800^2t^2}}


when rocket is 4,000 feet hi t=4000800=5t=\frac{4000}{800}=5


s(5)=8002530002+800252=s'(5)=\frac{800^2*5}{\sqrt{3000^2+800^2*5^2}}=

=64000059000000+16000000==\frac{640000*5}{\sqrt{9000000+16000000}}=

=640000525000000=64000055000=640=\frac{640000*5}{\sqrt{25000000}}=\frac{640000*5}{5000}=640


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



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