Question

I don't need the answer, I need to know how to set the word problem up.........Changing Volume. If each edge of a cube is increasing at the constant rate of 3 centimeters per second, how fast is the volume increasing when x, the length of an edge, is 10 centimeters long?

Answer 1

Volume = (3x)^3 maybe? assuming that, at time 0, your cube is non-existent I guess? then, for the rate, Rate = 27x^2?

Answer 2

no no no, wait, I screwed it up, I put x as the time here

Answer 3

x = initial x + 3t
Volume = x^3 = (initial x + 3t)^3 = (initial x)^3 + 9t^3
For the rate of increase of the volume, since initial x is a constant (well, I think), you can scratch it and you'll be left with 27t^2

Answer 4

I know that the volume is increasing at a rate of 900 cubic centimeters per second. this is the answer

Answer 5

Sorry, I'm at a loss here

Answer 6

its ok

Answer 7

900 looks a lot like 10^2 * 9... formula could be something like 3x^3, which would give you 9x^2

Answer 8

but that coefficient, as far as I would've thought, would've also been cubed

Answer 9

oh, yeah, I'm dumb... 3^3 = 27, not 9... So, we have the volume which is :
V = x^3 = (3t)^3 = 27t^3
V' = 81t^2
when x = 10, t = 3,33 seconds, right (x = 3t)
81 (3.33)^2 = 900

Answer 10

And, if you kept x before differentiating the equation (is it differentiating? English isn't my first language, so I struggle with the vocabulary at times), you'll get this :
V = x^3
V' = 3x^2 dx
x = 3t
dx = 3dt
V' = 3(3t)^2 * dx * (3dt/dx) = 81t^2 dt

Something like that. Not too sure how dx and dt move around, but I have a vague souvenir that it might be how this works. It would insure that the equation gives you the same results whichever way you pick

Answer 11

thank you