Change in Volume: vol. decr. @ 60 in^3/hr; find change in si

[matsh3]

Heres the problem:
A block of ice is in the shape of a cube ( all 3 dimensions are the same). The block is melting uniformly so that it is always a cube. If the volume is decreasing at a rate of 60 in^3 per hour, at what rate is the length of a side of the cube changing when the volume is 1000 in^3?

Now since I know that the volume of a cube= side^3
I know that the side length is 10 inches.

my question is how do i apply the decreasing rate of the volume (60 in^3 per hour) into the problem in order to determine the decreasing rate of the length of the side.
Can anyone help me??

Thanks,
Matsh3

 

[stapel]

A block of ice is in the shape of a cube ( all 3 dimensions are the same). The block is melting uniformly so that it is always a cube. If the volume is decreasing at a rate of 60 in^3 per hour, at what rate is the length of a side of the cube changing when the volume is 1000 in^3?

The volume V of a cube with side-length s is V(s) = s[sup:9su8est9]3[/sup:9su8est9]. You are given that dV/dt = -60, and are asked to find the value of ds/dt when s = 10.

Differentiate the volume equation implicitly, with respect to time t. Then plug in the known values and solve for the requested value.

If you get stuck, please reply showing all of your work and reasoning. Thank you!

Eliz.