Question

A rotating light is located 15 feet from a wall. The light completes one rotation every 2 seconds. Find the rate at which the light projected onto the wall is moving along the wall when the light's angle is 20 degrees from perpendicular to the wall.

Answer 1

Answer:

  53.4 ft/s

Step-by-step explanation:

The distance from the point closest to the light is ...

  d = (15 ft)tan(α)

where α is the angle from perpendicular. Since the light travels through an angle of 2π radians in 2 seconds, the angle can be represented by ...

  α = πt . . . . radians

and the rate of change of α is ...

  dα/dt = π . . . . radians/second

The rate of change of distance is ...

  dd/dt = (15 ft)sec(α)²(dα/dt) = (15π)(sec(20°)²) ft/s ≈ 53.4 ft/s

Answer 2

This is about differential calculus.

Rate = dy/dt = 41.61 ft/s

We are told that the light is located 15 ft from the wall and completes rotation every 2 seconds. Thus; we can say that the distance of the light's projection onto the given wall is expressed as;

y = 15 tan θ

where;

θ is the angle of light perpendicular to the wall.

One rotation is completed every 2 seconds and this means that;

Angular speed; ω = 2π/2 = π

• Angular speed can also be expressed as;

ω = θ/t

Thus;

θ = wt = πt

dθ/dt = π

• Since y = 15 tan θ

Then

dy/dt = 15 sec² θ (dθ/dt)

(dy/dt) = 15π sec² θ

 

• when θ = 20°

dy/dt = 15π sec²(20°)

dy/dt = 41.61 ft/s

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