Arotating 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.

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

ω = θ/t

Thus;

θ = wt = πt

dθ/dt = π

Then

dy/dt = 15 sec² θ (dθ/dt)

(dy/dt) = 15π sec² θ

dy/dt = 15π sec²(20°)

dy/dt = 41.61 ft/s

Read more at; link

  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: my brother can figure it out, imma go get him so i can answer : ) edit: my brother is sleeping, sorry man

step-by-step explanation:

try soting them out first

step-by-step explanation: