Find the rate of change in the angle of elevation of the camera at 10 seconds after lift-off

Thanks to both Mufasa and Ron Gordon, I now understand when I difficultly came from, misunderstanding of the term vertically.

Problem:

A rocket takes off vertically from the ground. 2000 ft. away, a camera captures it image. The rocket lift-off vertically to the position equation $s=50t^2$. Find the rate of change of the angle of the camera at 10 seconds after lift-off.

I am having trouble solving this problem. Also, a bit unsure if I fully understand the language used in the problem.

I know that the rocket lift off vertically, and I know the graph of the position function $s$.

Find the rate of change in the angle of elevation of the camera at 10 seconds after lift-off

I know to use the right half of the graph, otherwise time would being going backwards.

However, how to give the angle is puzzling me, from the camera's reference point. I know how to use the arc length to find the angle, but only for circles.

Now, I understand that vertically means a straight line, but how does one solve it if it doesn't follow the strict meaning of vertically, and insteads follows a parabola?


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Find the rate of change in the angle of elevation of the camera at 10 seconds after lift-off

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Updated On: 27-06-2022

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Find the rate of change in the angle of elevation of the camera at 10 seconds after lift-off

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How do you find the rate of change of a triangle?

Let the area of the triangle be xy/2 and dA/dt becomes dA/dt= (xy' +yx')/2. To find y' use the Pythagorean theorem x2 +y2 =z2 where z is constant. So xx' +yy' =0 and y'=-xx'/y and substituting this into dA/dt gives dA/dt =(-x2x'/y +x'y)/2 = x'(y2 -x2)/2y.