Mathematics: Arocket 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.

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clinton1790

15.03.2022

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

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