How fast is the light sweeping along the 2 km shoreline?

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In summary, the light is revolving at 10\pi radians/minute and is 1 km away from the shore. The shore is 2 km away from the lighthouse and the light is moving at a rate of tan(theta) km/minute, where theta is the angle formed by the light, the shore, and the lighthouse.
  • #1
hatelove
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A lighthouse 1 km from a beach shore revolves at 10[tex]\pi[/tex] radians/minute. How fast is the light sweeping across the shore 2 km from the lighthouse?

I drew a diagram here to try and help:

KokY2.png


I think it will have something to do with a triangle. So just to find the other side, I'll use the Pythagorean theorem and get [tex]\sqrt{3}[/tex]. I already knew this because I recognized the 30-60-90 triangle, so I know the angles too, but I don't know if we need them yet.

What I don't understand is what the question is asking. I thought the light was going around at a constant speed of 10[tex]\pi[/tex] radians/minute? What speed do they want?
 
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  • #2
A lighthouse 1 km from a beach shore revolves at 10[tex]\pi[/tex] radians/minute. How fast is the light sweeping across the shore 2 km from the lighthouse?

1. Don't put the quetion in a quote, it makes it more difficult to construct a coherent reply.

2. \(v=\dot{\theta}r\), \(\dot{\theta}\) in units of radians per unit time, is probably what they are looking for, but there is really insufficient information in what you posted to be sure.

CB
 
  • #3
daigo said:
I drew a diagram here to try and help:

KokY2.png


I think it will have something to do with a triangle. So just to find the other side, I'll use the Pythagorean theorem and get [tex]\sqrt{3}[/tex]. I already knew this because I recognized the 30-60-90 triangle, so I know the angles too, but I don't know if we need them yet.

What I don't understand is what the question is asking. I thought the light was going around at a constant speed of 10[tex]\pi[/tex] radians/minute? What speed do they want?
You are asked how fast the light is moving along the shoreline- that would be a speed in km per min or km per hr. What you are given is how fast the light is rotating. Yes, at the moment shown, the section along the shortline that the light reaches is 2 times [tex]\sqrt{3}[/tex]. But the light is moving so that right triangle is changing. If you call the angle theta, do you see that opposite leg of the triangle is given by [itex]tan(theta)[/itex]? What is the rate of change of that length relative to the rate of change of the angle.
 

Related to How fast is the light sweeping along the 2 km shoreline?

What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is the slope of the tangent line to the function at that point.

What is the purpose of finding the derivative of a function?

The derivative of a function provides information about the behavior of the function, such as its slope, concavity, and extrema. This information is useful in solving real-world problems involving rates of change.

How do you find the derivative of a function?

The derivative of a function can be found using various methods, such as the power rule, product rule, quotient rule, and chain rule. It involves taking the limit of a difference quotient as the change in the independent variable approaches zero.

What is the difference between average rate of change and instantaneous rate of change?

The average rate of change of a function over an interval is the slope of the secant line connecting two points on the function. The instantaneous rate of change, or simply the derivative, is the slope of the tangent line to the function at a specific point.

Can the derivative of a function be negative?

Yes, the derivative of a function can be negative. It indicates that the function is decreasing at that point. The sign of the derivative also provides information about the concavity of the function - a negative derivative corresponds to a concave down function.

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