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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

Best Answer 

 #1
avatar+93289 
+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
avatar+93289 
+10
Best Answer

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
avatar+88775 
+5

Yep....nice job !!!

 

CPhill  Dec 14, 2014
 #3
avatar+93289 
0

Thanks Chris :)

Melody  Dec 14, 2014

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