The dreaded mid terms....

[Daniel_Feldman]

Hey guys.


I'm just doing some reviewing for my calc mid term, and can't quite figure out this related rates problem. Maybe I'm just blanking out, but it seems like I need to convert from .125rev/min to radians per minute, then use pythagoras and differentiate? Any help would be much appreciated.


John leaves his house and quickly moves down the street. 30' across the street from John's house, a policer officer puts a turning spotlight on him. When john is 40' away from his house, the spotlight is turning at .125 rev/min. How fast is John moving at that moment?


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[galactus]

Hey Daniel: I think you're on the right track with your thinking. Here's my take. I hope someone confirms or denies.

Let x=distance from John's house to where the light hits John

We want \(\displaystyle \frac{dx}{dt}\) when x=40.

We know \(\displaystyle \frac{d{\theta}}{dt}=(.125\frac{rev}{min})(\frac{2{\pi}rad}{1rev})=\frac{\pi}{4} rad/min\)

\(\displaystyle x=30tan({\theta})\)

\(\displaystyle \frac{dx}{dt}=30sec^{2}({\theta})\frac{d{\theta}}{dt}\)

When \(\displaystyle x=40, tan({\theta})=\frac{4}{3}\) and \(\displaystyle sec{\theta}=\frac{5}{3}\),

so \(\displaystyle \frac{dx}{dt}=30(\frac{5}{3})^{2}(\frac{{\pi}}{4})=\frac{125{\pi}}{6}=65.45 \frac{ft}{min}\)

 

[Daniel_Feldman]

It looks right...I keep trying to make these problems more complicated than they are. I've got to think more creatively.

Thanks galactus.

 

[soroban]

Hey, galactus!

I agree totally . . . nice job!