Question

If the length of one a side of a cube is increasing at a rate of 4m/s how fast is the surface area of the cube increasing when the length is 16m.

Answer 1

This is related rates, right?

Answer 2

yea

Answer 3

768

Answer 4

I get 768 also. Hint. Relate the information, take a derivative, plug in what you know.

Answer 5

Oh thanks.. opps.

Answer 6

Its surface area not volume:
\[A = 6\cdot l^2\]\[\frac{dl}{dt} = 4\]\[\frac{dA}{dl} = 12 \cdot l \cdot \frac{dl}{dt}\]\[12 \cdot 16 \cdot 4 = 768\]

Answer 7

so if I want to know the rate of increase of volume it's increasing much faster than the surface area so ...
(I'm a bit confused as to taking the derivative of V here...)
I know
dL/dt = 4m/s

so for volume...when the length is 16m

V = L^3
dV/dt =(3L^2)dL/dt = (3)(16)^2)(4) = 3072m/s

I assume that's correct. But I'm not used to seeing an extra dL/dt. I guess the explanation for that is that it's a dependent variable.