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Homework Help: Rate of change

  1. Feb 19, 2005 #1
    Imagine a trough filled with water (I can't put up a picture).
    The water in the tank at time t seconds is given by [tex]V = 12x^2[/tex]. Given that water is flowing into the trough at the rate of [tex]60 cm^3/s[/tex], find the rate at which x is increasing when x = 10.

    [tex]\frac{dV}{dx} = 24x[/tex]
    [tex]\frac{d?}{ds} = 60 cm^3/s[/tex]
    Then what do I do?
     
  2. jcsd
  3. Feb 19, 2005 #2

    dextercioby

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    Who's "x" and what does your second equation stand for...?

    Daniel.
     
  4. Feb 19, 2005 #3
    If it shouldn't be x, can you tell me what it should be? I don't know what to do for the second equation. I just know its [tex]\frac{d\text{ something}}{dx} = 60 cm^3/s[/tex]. Again, if I'm wrong please tell me what it should be.
     
  5. Feb 19, 2005 #4

    dextercioby

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    I don't know about "x",what is its point...But:
    [tex] \frac{dV}{dt}=24x \frac{dx}{dt} [/tex]

    ,uding the chain rule...Now plug in the values and solve for the rate of "x"...

    Daniel.
     
  6. Feb 19, 2005 #5
    Oh now I know why you asked me about the second equation. I always get mixed up and put ds instead of dt :grumpy: .
    So I got 1440x. I know thats not the final answer because theres still the x = 10 part. And I'm confused on that part.
     
  7. Feb 19, 2005 #6

    dextercioby

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    Well,[itex] \frac{dV}{dt}=60cm^{3}s^{-1} [/itex]...x=10,you must find [itex] \frac{dx}{dt} [/itex]...

    Daniel.
     
  8. Feb 19, 2005 #7
    I thought [tex]\frac{dx}{dt} = 60 cm^3/s[/tex] and I've to find [tex]\frac{dV}{dt}[/tex] which is 1440x?
     
  9. Feb 19, 2005 #8

    dextercioby

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    Nope,it's the other way aroung,the volume is increasing at the rate given in the problem...:wink:

    Daniel.
     
  10. Feb 19, 2005 #9

    HallsofIvy

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    The point is that
    [tex]\frac{dV}{dt}= \frac{dV}{dx}\frac{dx}{dt}[/tex]
    (the chain rule).

    You are given [itex]\frac{dV}{dt}[/itex] and you can (and did) calculate [itex]\frac{dV}{dx}[/itex]. Put them into that equation and solve for [itex]\frac{dx}{dt}[/itex].
     
  11. Feb 19, 2005 #10
    Oh... Got it! Thank you!
     
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