Question

You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the ball drops at a constant rate of 2 m/s, what is the rate of change of the
angel between you and the ball when the angle is ( ?

Answer 1

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

$\mathbf{x = 100\ m}$

$\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}$ ---- the rate

$\mathbf{\theta = \frac{\pi}{4}}$

First, we calculate the vertical distance (y) using tangent ratio

$\mathbf{\tan(\theta) = \frac{y}{x}}$

Substitute 100 for x

$\mathbf{y = 100\tan(\theta) }$

$\mathbf{\tan(\theta) = \frac{y}{100}}$

Differentiate both sides with respect to time (t)

$\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}$

Substitute values for the rates and $\mathbf{\theta }$

$\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

This gives

$\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

$\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

Divide both sides by 2

$\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }$

$\mathbf{ \frac{d\theta}{dt} = 0.01 }$

Hence, the rate of change between you and the ball is 0.01 rad per second

Read more about related rates at:

brainly.com/question/16981791