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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.

Calculus1
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looks like a related rate problem fortunately the formula for the volume of a cube is easy, it is \(V(x)=x^3\)
take the derivative with respect to time and get \[V'=3x^2x'\]
you are told \(x'=3\) so this is really \[V'=9x^2\] replace \(x\) by the various values to get \(V'\)

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Other answers:

dV/dt=9x^2dx/dt ?:)
no
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
but in your notation, you have \[V=x^3\] so \[\frac{dV}{dt}=3x^2\frac{dx}{dt}\]
this line all edges of a cube are expanding at a rate of 3 centimeters per second tells you that \(\frac{dx}{dt}=3\)
that means \(\frac{dV}{dt}=3x^2\times 3=9x^2\)
let me know if i lost you there
Ok got it(:
So dats da answer?:)

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