Question 575008: A camera is filming the progress as a daredevil attempts to scale the wall of a skyscraper. The climber is moving vertically at a constant rate of 16 feet per minute, and the camera is 400 feet from the base of skyscraper. Through how many radians per minute is the camera angle changing when the climber is 300 feet up the building?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Maybe you are taking calculus and know everything about derivatives.
Otherwise, (for those in geometry class) I'll give a good approximation solution.
The height of the climber in feet (h), as a function of t, the time since he/she left the ground is
If the angle between the direction the camera is pointing and the horizontal is A, its tangent is a linear function of t
Angle A is also a function of t, but its a more complicated function and I do not feel like writing it out.
WITH CALCULUS
If we say that y is a function of A and A is a function of t,
Two of those derivatives are easy
 and  , so
 --> 
When the climber is at 300 ft up the building
 and
So at that point 
JUST GEOMETRY
The rate of change of the camera angle with time would be the slope of the graph of A as a function of t. It would be the slope of the line if A was a linear function of t. As A does not vary linearly with t, the best we can do is to approximate the slope of the curve by calculating the slope of the line connecting two points on the curve that are very close to either side of the point where h=300.
For h=301, 301=16t and t=301/16
For h=299, 299=16t and t=299/16
The time difference is 
For h=301, tan(A)=301/400 and A = approx. 0.64509919
For h=299, tan(A)=301/400 and A = approx. 0.64189919
The angle difference is  (approximate, but a very good approximation).
The slope of the line connecting those two points in the graph of A vs t is approximately
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