Posted by Amy on Thursday, May 5, 2011 at 4:54am.
Make a diagram showing the lamppost and the man's position any time.
Join the top of the lamppost to the man's top and extend it to the ground.
let the distance between the lamppost and the man along the ground be x ft, let the length of his shadow on the ground be y ft.
Since we have similar triangles, use the ratio
(x+y)/15 = y/6
6x + 6y = 15y
6x = 9y
2x = 3y
then 2dx/dt = 3dy/dt
2(5) = 3dy/dt
dy/dt = 10/3
So it appears that the position of the man is irrelevant and the shadow is changing at 10/3 ft/sec
Had it asked "how fast is his shadow moving ?" you would have added the 5ft/sec to the above answer.
2nd problem:
Again, make a diagram showing the plane moving in a line parallel to the ground.
Let the distance of the plane's path be x miles and the angle of elevation be Ø
I see that
tan Ø = 5/x
x tanØ = 5
x sec^2 Ø dØ/dt + tanØ dx/dt = 0
so when Ø = 60° , x = 5/√3
(5/√3)(1/3) dØ/dt + √3(600) = 0
dØ/dt = -1080 radians/hour
or 3/10 radians/sec , (I divided by 3600 to get to seconds)
(since derivatives in trig only work with radians, the answer obtained would be in radians)
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