Question

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.

Answer 1

Answer:

0.15 rad/s

Step-by-step explanation:

A rocket is launched at the rate of 11 feet per second from a point on the ground 15 feet from an observer.

To find rate of change of angle when the rocket is 30 feet above the ground

Height of rocket, y = 30 feet

Speed of rocket, $\dfrac{dy}{dt}= 11 ft/s$

Distance of observer from rocket, d = 15 feet

Angle between observer and rocket at height 30 feet is Ф

$\tan\theta=\dfrac{30}{15}=2$

$\tan\theta=\dfrac{\text{Perpendicular}}{\text{base}}$

Distance between observer and rocket launch doesn't change.

So, d=15 will remain constant.

$\tan\theta=\dfrac{y}{d}$

$15\tan\theta=y$

differentiate w.r.t t

$15\sec^2\theta \dfrac{d\theta}{dt}=\dfrac{dy}{dt}$

$15(1+\tan^2\theta)\cdot \dfrac{d\theta}{dt}=11$

$\dfrac{d\theta}{dt}=\dfrac{11}{15(1+2^2)}$

$\dfrac{d\theta}{dt}=\dfrac{11}{75}\ rad/s\approx 0.15\ rad/s$

Hence, The rate of change of the angle of elevation is 0.15 rad/s

Answer 2

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