Question

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.

Answer 1

looks like a related rate problem
fortunately the formula for the volume of a cube is easy, it is \(V(x)=x^3\)

Answer 2

take the derivative with respect to time and get
\[V'=3x^2x'\]

Answer 3

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

Answer 4

dV/dt=9x^2dx/dt ?:)

Answer 5

no

Answer 6

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

Answer 7

but in your notation, you have
\[V=x^3\] so
\[\frac{dV}{dt}=3x^2\frac{dx}{dt}\]

Answer 8

this line
all edges of a cube are expanding at a rate of 3 centimeters per second
tells you that \(\frac{dx}{dt}=3\)

Answer 9

that means \(\frac{dV}{dt}=3x^2\times 3=9x^2\)

Answer 10

let me know if i lost you there

Answer 11

Ok got it(:

Answer 12

So dats da answer?:)