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angelarios

  • 2 years ago

all edges of a cube are expanding at a rate of 3 centimeters per second. how fast is the volume changing when each edge is (a) 1 cm. (b) 10 cm.

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  1. anonymous
    • 2 years ago
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    looks like a related rate problem fortunately the formula for the volume of a cube is easy, it is \(V(x)=x^3\)

  2. anonymous
    • 2 years ago
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    take the derivative with respect to time and get \[V'=3x^2x'\]

  3. anonymous
    • 2 years ago
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    you are told \(x'=3\) so this is really \[V'=9x^2\] replace \(x\) by the various values to get \(V'\)

  4. angelarios
    • 2 years ago
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    dV/dt=9x^2dx/dt ?:)

  5. anonymous
    • 2 years ago
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    no

  6. anonymous
    • 2 years ago
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    i used \(V'\) and \(x'\) instead of \(\frac{dV}{dt}\) and \(\frac{dx}{dt}\) because it was easier for me to write also for you it is easier

  7. anonymous
    • 2 years ago
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    but in your notation, you have \[V=x^3\] so \[\frac{dV}{dt}=3x^2\frac{dx}{dt}\]

  8. anonymous
    • 2 years ago
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    this line all edges of a cube are expanding at a rate of 3 centimeters per second tells you that \(\frac{dx}{dt}=3\)

  9. anonymous
    • 2 years ago
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    that means \(\frac{dV}{dt}=3x^2\times 3=9x^2\)

  10. anonymous
    • 2 years ago
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    let me know if i lost you there

  11. angelarios
    • 2 years ago
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    Ok got it(:

  12. angelarios
    • 2 years ago
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    So dats da answer?:)

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