MATH SOLVE

2 months ago

Q:
# A rocket is launched at the rate of 11 feet per second from a point on the ground 15 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the rocket is 30 feet above the ground.

Accepted Solution

A:

let the angle of elevation is x and the height of the rocket from the ground is y

tanx = y/15

by differentiating both sides with respect to T

sec²x·dx/dt = (dy/dt)/15

at y = 30 , the hypotenuse of the triangle = 15√5

sec²x=(15√5/15)²=5

5 dx/dt = 11/15

dx/dt = 11/75 rad/sec

tanx = y/15

by differentiating both sides with respect to T

sec²x·dx/dt = (dy/dt)/15

at y = 30 , the hypotenuse of the triangle = 15√5

sec²x=(15√5/15)²=5

5 dx/dt = 11/15

dx/dt = 11/75 rad/sec