Question

You are standing on the ground, watching a bird fly away horizontally at a rate of 9 meters per second. The bird is initially located 25 meters above your head. Let θ be the angle measured from the vertical direction to the direction from you to the bird. How fast (in radians per second) is θ changing when the horizontal distance between you and the bird is 12 meters

Answer 1

Answer:

$\dfrac{d\theta}{dt}= 0.293\ rad/s$

Explanation:

horizontal distance traveled by the bird is equal to be x

given,

height of the bird above his head = 25 m

angle between them is equal to θ

$\dfrac{dx}{dt} = 9 m/s$

d² = x² + y²

$\theta = tan^{-1}{\dfrac{x}{25}}$

$\dfrac{d\theta}{dx}= \dfrac{1}{1+(\dfrac{x}{25})^2}\times(\dfrac{1}{25})$

$\dfrac{d\theta}{dx}= \dfrac{25}{x^2+625}$

now,

$\dfrac{d\theta}{dt}= \dfrac{d\theta}{dx}\times \dfrac{dx}{dt}$

$\dfrac{d\theta}{dt}= \dfrac{25}{x^2+625}\times 9$

$\dfrac{d\theta}{dt}= \dfrac{225}{x^2+625}$

at x = 12 m

$\dfrac{d\theta}{dt}= \dfrac{225}{12^2+625}$

$\dfrac{d\theta}{dt}= 0.293\ rad/s$