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Math Help - Volume change

  1. #1
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    Volume change

    The edge of a cube expands 3 centimeters per second. In the moment in which each edge has a longitude of 10 cm, the volume changes with a velocity of:

    a) 300 cm^3/s
    b) 900 cm^3/s
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  2. #2
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    We want \frac{dV}{dt} when x=10. We know \frac{dx}{dt}=+3

    V=x^{3}

    \frac{dV}{dt}=3x^{2}\frac{dx}{dt}

    3(10)^{2}(3)
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  3. #3
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    whats the difference between dv/dt and dx/dt?
    Why multiply 3cm/s times the first derivative of the volume of a cube?
    Last edited by bret80; August 4th 2006 at 10:17 AM.
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  4. #4
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    What do you mean?. I took the derivative of x^{3}, which is 3x^2.

    dx/dt=3
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  5. #5
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    Quote Originally Posted by bret80
    The edge of a cube expands 3 centimeters per second. In the moment in which each edge has a longitude of 10 cm, the volume changes with a velocity of:

    a) 300 cm^3/s
    b) 900 cm^3/s
    Let me try to explain it diffrently. You need to find what \frac{dV}{dt} is at when the side is 10 centimeters because that is the instantenous rate of change in volume. But you know that the volume of a cube is:
    V=s^3
    Take, the derivative to respect to time, (implictly),
    \frac{dV}{dt}=3s^2\frac{ds}{dt}
    Now, the problem says the side at s=10 and the rate which the sides change is \frac{ds}{dt}=3 thus,
    \frac{dV}{dt}=3\cdot (10)^2 \cdot 3=900
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