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Related Rates - Not getting answer in book

  1. Mar 27, 2006 #1
    The problem given in my book is:

    I set up and solved the problem this way:
    dx/dt = 500 mi/h
    x = 2mi
    y = 1mi (constant)

    Distance:
    s2 = y2 + x2

    s2 = 1 + x2

    s = (1 + x2)1/2

    d/dt = d/dt[sqrt(1 + x2)1/2]

    ds/dt = (1/2)(1 + x2)-1/2 * (2x(dx/dt))

    ds/dt = (1 + x2)-1/2 * (x(dx/dt))

    ds/dt = [tex]\frac{x(dx/dt)}{(1 + x^2)^{1/2}}[/tex]

    Substiting:
    ds/dt = [tex]\frac{(2)(500)}{(1 + (2)^2)^{1/2}}[/tex]

    ds/dt = [tex]\frac{1000}{\sqrt{5}}[/tex]

    The final answer:
    ds/dt = [tex]200\sqrt{5}[/tex] mi/h

    However the back of my book has: [tex]250\sqrt{3}[/tex] mi/h

    What am I doing wrong?
    Thank you,
    -GM-
     
    Last edited: Mar 27, 2006
  2. jcsd
  3. Mar 27, 2006 #2

    benorin

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    don't solve for s, just leave it at [tex]s^2=1+x^2[/tex] and differentiate implicitly to get

    [tex]2s\frac{ds}{dt}=2x\frac{dx}{dt}[/tex]

    solve for ds/dt

    [tex]\frac{ds}{dt}=\frac{x}{s}\frac{dx}{dt}[/tex]
     
  4. Mar 27, 2006 #3

    benorin

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    You work was correct that far, however, since [tex]s=\sqrt{x^2+1}[/tex] anyhow. The error is the sqrt(5) should be sqrt(3): 2 mi away is straight line distance.
     
  5. Mar 27, 2006 #4

    arildno

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    Gold Member
    Dearly Missed

    You have misinterpreted the meaning of "2 mi away from the station".

    It is s that equals 2, not x!

    Thus, you have:
    [tex]2^{2}=y^{2}+x^{2}\to{x}=\sqrt{3}[/tex]
     
  6. Mar 27, 2006 #5
    AH! Thank you!
    I understand now, I was taking 2mi as horizontal distance.

    Thanks!
    -GM-
     
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