Calculus help for related helps please!!! :)

Question

A lighthouse is stationed at a point P some distance away from the nearest point Q on a straight shoreline. It's light is making 3 revolutions per minute. It is found that the beam of light is moving along the shoreline at a rate of 200pi m/s when it passes point R on the shoreline 1km away from Q. How far away is the lighthouse from the shoreline? 

 

I just need to know how to start , as I keep cancelling out the variable I'm looking to solve when I differentiate, but a full solution/explanation would be appreciated so much!! Thank you in advance!

Answer 1

I cannot draw the diagram here. However, picture a right triangle PQR.

Let  Q be at the origin, P at the point (d,0), R at the point (0,1).

Consider a point along QR that is y away from Q. Let this point be r away from P.

Then r^2 = d^2 + y^2            

Differentiate implicitly with respect to time.

2r dr/dt = 2y dydt              dy/dt = -200pi m/s = 0.2 pi km/s when y = 1 km and r = √(d^2 + 1)

 dr/dt = 3 rev/min = 3 (2pi)/60 rad/s

 

r dr/dt = y dydt

 

Substitute,

√(d^2 +1) x 3 (2pi)/60 rad/s = 1 km x (- 0.2 pi km/s)

 

Solve d^2 + 1 = 4         d^2 = 3               d = √3 km

Answer 2

Hi Emma

 

I could be wrong, but it doesn't look like you actually need calculus to solve this problem. Basically, you draw a picture with P, Q, and R so that they form a right triangle. PQ and QR are the legs, and the segment QR (I'll call this distance r) is the hypotenuse. The key to this problem is the formula

 

v = ωr

 

where ω is the angular velocity. In this case, the lighthouse rotates three times per minute, so its angular velocity is 3·2π radians per minute. And we know v = 200π m/s. So we have

 

200π m/s = 6πr /min.

 

We have to change the left side to m/min instead of m/sec.

 

60·200π m/min = 6πr /min           then cancel 6π from both sides and end up with

 

2000 m/min = r /min.

 

Therefore r is 2000 m, or 2 kilometers. Then just use the Pythagorean theorem: PQ is √3 kilometers.

 

(I assumed that the light's speed "moving along the shoreline" just means its speed tangent to the circle, not straight along the shore, as otherwise I'm pretty sure you can't solve the problem.)