Question

You stand 20 ft from a bottle rocket on the ground and watch it as it takes off vertically into the air at a rate of 19 ft/sec. Find the rate at which the angle of elevation from the point on the ground at you feet and the rocket changes when the rocket is 30 ft in the air.

Answer 1

Answer:

0.292 rad/s

Step-by-step explanation:

$b=\text{Distance of rocket from your feet}=20\ \text{ft}$

$\dfrac{dp}{dt}=\text{Rate of change of height}=19\ \text{ft/s}$

$p=\text{Distance of the rocket from the ground}=30\ \text{ft}$

Finding distance between the feet and when the rocket is 30 ft in the air

$d=\sqrt{b^2+p^2}\\\Rightarrow d=\sqrt{20^2+30^2}\\\Rightarrow d=36.1\ \text{ft}$

$\tan\theta=\dfrac{p}{b}=\dfrac{p}{20}$

Differentiating with respect to time

$\sec^2\theta\dfrac{d\theta}{dt}=\dfrac{1}{20}\dfrac{dp}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\cos^2\theta}{20}\dfrac{dp}{dt}$

$\cos\theta=\dfrac{b}{h}\\\Rightarrow \cos\theta=\dfrac{20}{36.1}\\\Rightarrow \cos^2\theta=\dfrac{20^2}{36.1^2}$

$\dfrac{d\theta}{dt}=\dfrac{\dfrac{20^2}{36.1^2}}{20}\times 19\\\Rightarrow \dfrac{d\theta}{dt}=0.292\ \text{rad/s}$

Rate of change of the angle is 0.292 rad/s.