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# After blast off, a space shuttle climbs vertically and a radar tracking dish, located 1000 yd from the launch pad, follows the shuttle. How

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After blast off, a space shuttle climbs vertically and a radar tracking dish, located 1000 yd from the launch pad, follows the shuttle. How fast is the radar dish revolving 10 sec after blast-off if at the time the velocity of the shuttle is 100 yd/sec and the shuttle is 500 yd above the ground?

Guest Dec 14, 2014

#1
+94120
+10

This is my best guess

When t=10sec  y=500yards   and dy/dt=100 yards/sec   and    $$\theta=atan (\frac{500}{1000})=atan (0.5)$$

 $$\\tan\theta=\frac{y}{1000}\\\\ y=1000tan\theta\\\\ \frac{dy}{d\theta}=1000sec^2\theta\\\\ \frac{d\theta}{dy}=\frac{cos^2\theta}{1000}\\\\ \frac{d\theta}{dt}=\frac{d\theta}{dy}\times \frac{dy}{dt}\\\\ When \;\;t=10 \\\\ \frac{d\theta}{dt}=\frac{cos^2\theta}{1000}\times 100\\\\ \frac{d\theta}{dt}=\frac{cos^2(atan(0.5))}{10}\\\\ \frac{d\theta}{dt}=\frac{4}{5}\div 10\\\\ \frac{d\theta}{dt}=\frac{4}{50}\\\\ \frac{d\theta}{dt}=\frac{2}{25} \;\;radians/sec\\\\$$

I think that this is correct.

Melody  Dec 14, 2014
#1
+94120
+10

This is my best guess

When t=10sec  y=500yards   and dy/dt=100 yards/sec   and    $$\theta=atan (\frac{500}{1000})=atan (0.5)$$

 $$\\tan\theta=\frac{y}{1000}\\\\ y=1000tan\theta\\\\ \frac{dy}{d\theta}=1000sec^2\theta\\\\ \frac{d\theta}{dy}=\frac{cos^2\theta}{1000}\\\\ \frac{d\theta}{dt}=\frac{d\theta}{dy}\times \frac{dy}{dt}\\\\ When \;\;t=10 \\\\ \frac{d\theta}{dt}=\frac{cos^2\theta}{1000}\times 100\\\\ \frac{d\theta}{dt}=\frac{cos^2(atan(0.5))}{10}\\\\ \frac{d\theta}{dt}=\frac{4}{5}\div 10\\\\ \frac{d\theta}{dt}=\frac{4}{50}\\\\ \frac{d\theta}{dt}=\frac{2}{25} \;\;radians/sec\\\\$$

I think that this is correct.

Melody  Dec 14, 2014
#2
+92808
+5

Yep....nice job !!!

CPhill  Dec 14, 2014
#3
+94120
0

Thanks Chris :)

Melody  Dec 14, 2014