Calculus help?

Question

1. A patrol car is parked 50 feet from a long warehouse. The revolving light on to ofhte car turns at a rate of 30 revolutions per minute. how fast is the light beam moving along the wall when the beam makes an angle of 30 degrees and an angle of 60 degrees with the line perpendicular from the light to the wall?

Answer 1

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You must know that,
Linear velocity = Radius x Angular Velocity => v = r.ω

From the question we know that Angular velocity, ω is 30 revolutions per minute.

Our job is now to find the radius at 30 degrees and at 60 degrees.

Let the point at which the patrol car headlight is revolving be P
Let the perpendicular from the headlight to the wall meet the wall at Q.

(i) Now, let the point at which the beam makes an angle of 30º be T

In the right angled triangle PQT

angle(PQT) = 90 degrees
angle(QPT) = 30 degrees

cos (QPT) = PQ/PT

cos (QPT) = Perpendicular Distance / Length of the beam (Radius)

cos (30º) = (50 feet) / Radius

Radius = (50) / (√3/2) feet
Radius = 57.7350 feet

Linear velocity, v = r.ω

v = 57.7350 x 30 feet/min

Linear velocity, v = 1732.0508 feet/min //


(ii) Now, let the point at which the beam makes an angle of 60º be S

In the right angled triangle PQS

angle(PQS) = 90 degrees
angle(QPS) = 60 degrees

cos (QPS) = PQ/PS

cos (QPS) = Perpendicular Distance / Length of the beam (Radius)

cos (60º) = (50 feet) / Radius

Radius = (50) / (1/2) feet

Radius = 100 feet

Linear velocity, v = r.ω

v = 100 x 30 feet/min

Linear velocity, v = 3000 feet/min //

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Answer 2

It goes like this:

Draw the diagram as indicated in the question, then add in the 30° and 60° angles. The 'right angle' should be at the top left. Sorry...can't do diagrams here.

You know the vertical length (50'), that will be 'x', and the shorter horizontal length at the top will be 'y'.

With this diagram you can say:

y = rsin(θ)

Then:

dy/dt = (dr/dt)sin(θ) + rcos(θ)(dθ/dt)

Well...dr/dt = 0, that's the vertical length that is a constant 50'. That doesn't change.

So you have: dy/dt = rcos(θ)(dθ/dt)

dy/dt is what you want. This is the velocity of this length of the triangle.

dθ/dt = 30 rpm
r = 50'

Therefore you have: dy/dt = 50*cos(30)(30)

dy/dt = 1299 ft/min


Now plug in for θ = 60°

((done))

Hope this helps.

Answer 3

I used this source with question 53.2:

http://polo.k12.mo.us/homework/hholt_web/PDFLsns/Algebra3/Lsn53.pdf

You need to first find the distance of the hypotenuse at 30 and 60 degrees. I get 57.74 feet at 30 degrees and 100 feet at 60 degrees. Now take each of those values and plug in as r in the formula shown at the website. Convert this the same way they converted it. You can skip the in to ft conversion as your result is already in ft. At 0 degrees I get 107.1 MPH, 30 degrees is 123.7 MPH, and 60 degrees is 214.2 MPH using this formula. Hope this helps.