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Related rates - word problem

  1. Apr 14, 2009 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations
    [tex]\frac{dx}{dt}[/tex] = 5km/h

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

    3. 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]
     
  2. jcsd
  3. Apr 14, 2009 #2

    CompuChip

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    Can you now express [itex]\frac{d\theta}{dt}[/itex] in terms of [itex]\frac{dx}{dt}[/itex] ?
     
  4. Apr 14, 2009 #3
    Would I do this by differentiating tan[tex]_{-1}[/tex][tex]\frac{x}{30}[/tex]?
     
  5. Apr 14, 2009 #4

    CompuChip

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    Presuming you know what the derivative of arctan u is, yes.
     
  6. Apr 14, 2009 #5
    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]
     
  7. Apr 14, 2009 #6

    CompuChip

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    The derivative of arctan(u) is 1/(1 + u^2)
     
  8. Apr 14, 2009 #7
    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.
     
  9. Apr 14, 2009 #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.
     
  10. Apr 14, 2009 #9
    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?
     
  11. Apr 14, 2009 #10

    CompuChip

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    I didn't do the calculation but it looks precisely like what I had in mind.
     
  12. Apr 14, 2009 #11
    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
     
  13. Apr 14, 2009 #12
    Thanks so much. I'll compare my answer with peers and see if they match.
     
  14. Apr 14, 2009 #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: Apr 14, 2009
  15. Apr 14, 2009 #14
    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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