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Related rates - word problem

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Homework Statement


A man walks along a straight path at 5km/hr. A search light is located on the ground
30 metres form the path and is kept focused on the man. At what rate is the searchlight
rotating when the man is 20 metres from the point on the path closest to the searchlight?


Homework Equations


[tex]\frac{dx}{dt}[/tex] = 5km/h

[tex]\frac{d\theta}{dt}[/tex] = ???

The Attempt at a Solution


Reading the question, it is obvious that the situation above describes a right triangle:

calculus.jpg


I need to find the rate of change of [tex]\theta[/tex] with respect to x, or time. The rate of change of x with respect to time is 5km/h. First I believe I need to relate [tex]\theta[/tex] with x (time). This is probably obvious, but I'm not sure how to do it. I think I need to set up theta as a function of x as follows:

[tex]\theta[/tex] = tan[tex]_{-1}[/tex][tex]\frac{20}{30}[/tex]

Thus,


[tex]\theta[/tex] = tan[tex]_{-1}[/tex][tex]\frac{x}{30}[/tex]
 

Answers and Replies

  • #2
CompuChip
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Can you now express [itex]\frac{d\theta}{dt}[/itex] in terms of [itex]\frac{dx}{dt}[/itex] ?
 
  • #3
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Can you now express [itex]\frac{d\theta}{dt}[/itex] in terms of [itex]\frac{dx}{dt}[/itex] ?
Would I do this by differentiating tan[tex]_{-1}[/tex][tex]\frac{x}{30}[/tex]?
 
  • #4
CompuChip
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Presuming you know what the derivative of arctan u is, yes.
 
  • #5
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Ok I think I have it:

To find [tex]\frac{d\theta}{dt}[/tex], apply chain rule:


[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{d\theta}{dx}[/tex] [tex]\frac{dx}{dt}[/tex]


[tex]\frac{dx}{dt}[/tex] = 5km/h


Therefore,


[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{d}{dx}[/tex] (tan[tex]_{-1}[/tex][tex]\frac{x}{30}[/tex])(5)

Just don't know how to differentiate tan[tex]_{-1}[/tex][tex]\frac{x}{30}[/tex]
 
  • #6
CompuChip
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The derivative of arctan(u) is 1/(1 + u^2)
 
  • #7
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The derivative of arctan(u) is 1/(1 + u^2)
So [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{1}{1 + (x/30)^2}[/tex] (5)

= [tex]\frac{5}{1 + (x/30)^2}[/tex]

Assuming this is correct (please do correct me if I'm wrong), all that's left now is to enter 20m into the equation for x?:

[tex]\frac{5}{1 + (20/30)^2}[/tex] = 3.46 degrees per hour?

I feel like I'm off.
 
  • #8
CompuChip
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You don't have the derivative of arctan(u), but of arctan(x / 30). So you might want to use the chain rule :smile:
But don't worry, that just means you are off by a constant factor.
Also, if you're getting a very small answer, note that you probably get something in radians (2 pi radians corresponding to 360 degrees).

If you've never seen the derivative of arctan and chain rule etc before, please tell us - then there's probably another way to solve the problem. Mathematically inclined as I am, this is the first that came to mind though.
 
  • #9
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You don't have the derivative of arctan(u), but of arctan(x / 30). So you might want to use the chain rule :smile:
But don't worry, that just means you are off by a constant factor.
Also, if you're getting a very small answer, note that you probably get something in radians (2 pi radians corresponding to 360 degrees).

If you've never seen the derivative of arctan and chain rule etc before, please tell us - then there's probably another way to solve the problem. Mathematically inclined as I am, this is the first that came to mind though.
Yeah I'm only just getting into differentiation now so I'm still finding my feet.

So, applying the chain rule I got something like this:

[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{d}{dx}[/tex]arctan(u) (5), where u = (x/30)

= (5) [tex]\frac{d}{dx}[/tex]arctan(x/30) [tex]\frac{d}{dx}[/tex](x/30)

= [tex]\frac{150}{1 + (x/30)^2}[/tex]

Therefore,

[tex]\frac{150}{1 + (20/30)^2}[/tex] = 103.85 degrees per hour.

Are you getting something similar?
 
  • #10
CompuChip
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I didn't do the calculation but it looks precisely like what I had in mind.
 
  • #11
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eh.. im not sure if this is right but try

tan(theta)=x/30 ; thus theta = tan^-1(x/30)=tan^-1(20/30)=tan^-1(2/3) aprox = 33.6 degrees

then differentiating from tan(theta)=x/30 ...
sec^2(theta)theta'=(1/30)x' = sec^2(33.6)theta'=(1/30)5
theta'=5/(30sec^2(33.6)) aprox = .115 deg/hour seems small...but all the steps seem correct
 
  • #12
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I didn't do the calculation but it looks precisely like what I had in mind.
Thanks so much. I'll compare my answer with peers and see if they match.
 
  • #13
Cyosis
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The first mistake I can see is the differentiation,[tex]\frac{d}{dx} \frac{x}{30} \neq 30[/tex]. Secondly we have to be very careful with the units. The velocity is given in km/h, the distances in meters.
 
Last edited:
  • #14
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The first mistake I can see is the differentiation,[tex]\frac{d}{dx} \frac{x}{30} \neq 30[/tex]. Secondly we have to be very careful with the units. The velocity is given in km/h, the distances in meters.
Yeah I was wondering about the units, should have used 5000m instead.

Try again:

Chain rule:

[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{d\theta}{dx}[/tex] [tex]\frac{dx}{dt}[/tex] = [tex]\frac{d}{dx}[/tex]arctan(u) (5000), where u = [tex]\frac{x}{30}[/tex]

Applying chain rule to arctan(u):

[tex]\frac{d}{dx}[/tex]arctan(u) = [tex]\frac{1}{1+(u)^2}[/tex] [tex]\frac{30}{900}[/tex]

Therefore,

[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{30}{900(1 + (x/30)^2)}[/tex] (5000)

[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{150000}{900(1 + (x/30)^2)}[/tex]

Substituting 20m for x yields:

[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{150000}{900(1 + (20/30)^2)}[/tex] = 115.38 degrees per hour.
 

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