SOLUTION: A man 2m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 25m tall. When the man is 10m from the tower, how fast is the angle of elevation changing if

Algebra ->  Trigonometry-basics -> SOLUTION: A man 2m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 25m tall. When the man is 10m from the tower, how fast is the angle of elevation changing if      Log On


   



Question 728358: A man 2m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 25m tall. When the man is 10m from the tower, how fast is the angle of elevation changing if that angle is measured from the horizontal to the line joining the top of the man's head to the top of the tower?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
To solve this problem in a strictly accurate way requires calculus, because the rate of change of the angle is not constant.
We could an expression for the angle as a function of time, calculate the derivative of that function, and then calculate the value of that derivative at the time the man is 10m from the tower.

Alternatively we could obtain an approximate value by calculating the average rate of changeover short period(s) of time at around that time.

APPROXIMATE SOLUTION:
If theta= angle of elevation
when the man is 10 meters from the base of the tower,
tan%28theta%29=23%2F10 because the horizontal from the top of the 2 meter tall man's head passes by a point 2 meters above the base of the tower and
25meters-2meters=23meters below the top of the tower
(assuming the ground is horizontal between the man and the tower).
When the man is 11 meters from the base of the tower,
tan%28theta%29=23%2F11 , and
when the man is 9 meters from the base of the tower,
tan%28theta%29=23%2F9
Approximate calculations of tangents and the corresponding angles in radians:
tan%28theta%29=23%2F9=2.555556 --> theta=1.197809
tan%28theta%29=23%2F10=2.3 --> theta=1.160669
The average rate of change of that angle of elevation over the second the man shortens his distance from the foot of the tower from 10m to 9m , in radians per second, is
%281.197809-1.160669%29%2F1=0.037140
In degrees per second, it would be approximately 2.13.
tan%28theta%29=23%2F11=2.090909 --> theta=1.124691
The average rate of change of that angle of elevation over the second the man shortens his distance from the foot of the tower from 11m to 10m , in radians per second, is
%281.124691-1.197809%29%2F1=0.035978
In degrees per second, it would be approximately 2.06.
The instantaneous rate of change at thye precise time when the man is 10 meters from the base of the tower must be somewhere in between, but maybe highlight%282%5Eo%29 per second is accurate enough for an answer.
Otherwise, we could average the results an come up with
%280.037140%2B0.035978%29%2F2=about0.03656 radians per second = about 2.09%5Eo per second

WITH CALCULUS:
t= time until the man reaches the bottom of the tower, in seconds.
so t= horizontal distance between the man and the tower in meters
tan%28theta%29=23%2Ft
theta=tan%5E%28-1%29%28theta%29=tan%5E-1%2823%2Ft%29
Since the derivative of f%28x%29=tan%5E-1%28x%29 is df%2Fdx=1%2F%281%2Bx%5E2%29 ,
the rate of change of theta+as+a+function+of+%7B%7B%7Bt is

That rate of change is negative, of course, because as the time needed to reach the foot of the tower increases, the angle decreases.
We are interested in the opposite because we are decreasing the time needed to reach the foot of the tower.
The rate of increase of the angle as the man approaches the tower, as a function
of time is
23%2F%28t%5E2%2B23%5E2%29=23%2F%28t%5E2%2B529%29
When t=10 that rate is exactly 23%2F%2810%5E2%2B529%29=23%2F100%2B529=23%2F629 radians per second, with units of s%5E-1 .
In the more familiar degrees per second, the numerical value would be exactly
%2823%2F629%29%28180%2Fpi%29%7D%7D+=+approx.+%7B%7B%7B2.09508 .