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

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

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

Answer 2

The speed of the bird when the horizontal distance between you and the bird is 12 meters is 0.2926 rad/sec.

What is Velocity?

Velocity is the rate of change of position of the object with respect to time.

$v=\dfrac{dx}{dt}$

We know that the vertical distance between you the bird is 25 m, while the speed of the bird is 9 m/s. Therefore, the change θ in the horizontal distance between you and the bird can be given as,

$\dfrac{d \theta}{dt}$

Further, the change can be written as

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

We know that the angle between the bird and you can be written as θ,

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

Differentiating the angle with respect to x,

$\dfrac{d\theta}{dx} = \dfrac{1}{1+(\dfrac{x}{25})^2} \times \dfrac{1}{25}\\\\\dfrac{d\theta}{dx}=\dfrac{25}{x^2+625}$

As it is given that the velocity of the bird is 9 m/s, we know that the velocity can be written as the rate of change of position of the object with respect to time. therefore,

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

Now, the change θ in the horizontal distance between you and the bird can be written as,

$\dfrac{d \theta}{dt} = \dfrac{d \theta}{dx} \times \dfrac{d x}{dt}\\\\\dfrac{d \theta}{dt} = \dfrac{25}{x^2+625} \times 9\\\\\dfrac{d \theta}{dt} = \dfrac{225}{x^2+625}$

As we need to know the change θ in the horizontal distance between you and the bird when the horizontal distance between you and the bird is 12 meters. therefore,

$\dfrac{d \theta}{dt} = \dfrac{225}{x^2+625}\\\\\dfrac{d \theta}{dt} = \dfrac{225}{(12)^2+625}\\\\\dfrac{d \theta}{dt} = \dfrac{225}{769}\\\\\dfrac{d \theta}{dt} =0.2926\rm\ rad/sec$

Hence, the speed of the bird when the horizontal distance between you and the bird is 12 meters is 0.2926 rad/sec.

Learn more about Velocity:

https://brainly.com/question/862972