Question

A road runs east to west, and a camera is mounted 10 ft south of the road. A dragster races along the road at 440 ft/s to the west. The camera is programmed to look directly at the car at all times. Let 0 degrees represent east, and angles be measured positively counterclockwise, as in trigonometry. Let Θ be the angle of the line of sight of the camera in this way, and let x be the horizontal distance between the camera and the car. As the car races past, Θ increases from near 0 to near π.
Keep in mind: East is the positive x direction, west is the negative x direction. Use the unit s throughout to represent seconds, to keep it from being confused with the secant function.

-Find an equation relating Θ and x
-Derive this equation with respect to time to create a related-rates equation
-At the instant when the car is 10 ft east of the camera, determine the rate of change of the camera angle Θ, in radians per second.

Answer 1

Answer:

$\textsf{A)}\quad \theta=\arctan\left(\dfrac{10}{x}\right)$

$\textsf{B)}\quad \dfrac{\text{d}\theta}{\text{d}t}=\dfrac{4400}{x^2+100}$

C)  2.2 radians per second

Step-by-step explanation:

Part A

The line of sight from the camera to the dragster creates a right triangle with the road.

To create an equation relating θ and x, we can use the tangent trigonometric ratio, which states that the tangent of an angle is equal to the ratio of the length of the side opposite the angle to the length of the adjacent side.

In this case, the vertical distance between the camera and the car is the side opposite the angle, and the horizontal distance between the camera and the car is the side adjacent to the angle. Therefore:

$\tan\theta=\dfrac{10}{x}$

$\theta=\arctan\left(\dfrac{10}{x}\right)$

Note that, even though the camera is mounted south of the road, angles are to be measured positively counterclockwise, making the vertical distance positive in this context.

$\hrulefill$

Part B

To create a related-rates equation, we want to find an equation for dθ/dt in terms of x. To do this, we need to use the chain rule:

$\dfrac{\text{d}\theta}{\text{d}t}=\dfrac{\text{d}x}{\text{d}t} \times \dfrac{\text{d}\theta}{\text{d}x}$

We know that the car is racing along the road at 440 ft/s to the west. Since west is the negative x direction, then:

$\dfrac{\text{d}x}{\text{d}t} =-440$

Differentiate the equation from part A:

$\dfrac{\text{d}\theta}{\text{d}x}=\dfrac{\text{d}}{\text{d}x}\left(\arctan\left(\dfrac{10}{x}\right)\right)\\\\\\\\\dfrac{\text{d}\theta}{\text{d}x}=\dfrac{1}{1+\left(\dfrac{10}{x}\right)^2} \cdot \dfrac{\text{d}}{\text{d}x}\left(\dfrac{10}{x}\right)\\\\\\\\\dfrac{\text{d}\theta}{\text{d}x}=\dfrac{1}{1+\dfrac{100}{x^2}} \cdot -\dfrac{10}{x^2}\\\\\\\\\dfrac{\text{d}\theta}{\text{d}x}=-\dfrac{10}{\left(1+\dfrac{100}{x^2}\right)x^2}\\\\\\\\\dfrac{\text{d}\theta}{\text{d}x}=-\dfrac{10}{x^2+100}$

Substitute the expressions into the chain rule:

$\dfrac{\text{d}\theta}{\text{d}t}=-440 \times -\dfrac{10}{x^2+100}\\\\\\\\\dfrac{\text{d}\theta}{\text{d}t}=\dfrac{4400}{x^2+100}$

$\hrulefill$

Part C

At the instant when the car is 10 ft east of the camera, x = 10. Therefore, to determine the rate of change of the camera angle θ, in radians per second, substitute x = 10 into dθ/dt:

$\dfrac{\text{d}\theta}{\text{d}t}=\dfrac{4400}{(10)^2+100}\\\\\\\\\dfrac{\text{d}\theta}{\text{d}t}=\dfrac{4400}{100+100}\\\\\\\\\dfrac{\text{d}\theta}{\text{d}t}=\dfrac{4400}{200}\\\\\\\\\dfrac{\text{d}\theta}{\text{d}t}=2.2\; \sf rad/s$

Therefore, at the instant when the car is 10 ft east of the camera, the rate of change of the camera angle θ is 2.2 radians per second.