You are videotaping a race from a stand 132 ft from the track, following a car that is moving at 180 mph (264ft/sec). About how fast will your camera angle theta be changing when the car is right in front of you? a half second later?

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- Oct 30th 2006, 09:35 PMturtleanother related rates problem!
You are videotaping a race from a stand 132 ft from the track, following a car that is moving at 180 mph (264ft/sec). About how fast will your camera angle theta be changing when the car is right in front of you? a half second later?

- Oct 31st 2006, 02:27 AMCaptainBlack
Take your position to be (0,0), and the point where the car is closest to you

(that is right infront of you) to be (132,0), and take this to occur at time

t=0.

Then the position of the car atime t is (132, 264t). Then the camera angle

theta (measured from the position of the car at t=0) satisfies:

tan(theta) = 264t/132=2t.

so:

d/dt(tan(theta)=2

Therefore:

sec^2(theta) dtheta/dt = 2,

or:

dtheta/dt = 2 cos^2(theta).

Now at t=0, theta=0, so dtheat/dt=2 radians/s = 2*180/pi degrees/s

When t=0.5 theta=pi/4, so dtheat/dt=2 cos^2(pi/4)= 1 radian/s = 180/pi degrees/s.

RonL