Related Rates Problem?

Question

At a distance of 45ft from the pad, a man observes a helicopter taking off from a heliport. If the helicopter lifts off vertically and is rising at a speed of 49ft/sec when it is at an altitude of 112 ft, how fast is the distance between the helicopter and the man changing at that instant?

Answer 1

see the website cited in source, i uploaded a file in Microsoft Paint, entitled Problem A

Answer 2

The distance between the man and the helicopter is the length of the hypotenuse of a right triangle, with legs of length 45 ft and 49t ft, where t is the time after takeoff in seconds.

Let D be this distance. Then, by the Pythagorean theorem,

D = sqrt(45^2 + (49t)^2).

Differentiate with respect to t:

dD/dt = (1/2)(1/sqrt(45^2 + (49t)^2))(2*49t)
= 49t / sqrt(45^2 + (49t)^2).

The helicopter needs 16/7 seconds to reach 112 ft, so plug in t=16/7 and get

dD/dt = 112 / sqrt(45^2 + 112^2) = 0.928 ft/s.