Volume of a Changing Rectangular Prism: How Fast is it Changing?

[mikee]

Homework Statement

A rectangular prism has its length increasing by 12 cm/min, its width increasing by 4 cm/min and its height increasing by 2 cm/min. How fast is it's volume changing when the dimensions are 200 cm in length, 50 cm in width and 30 cm in height?

Homework Equations

The Attempt at a Solution

I wasnt positive on how to do this but the way i attempted it was Since the V=LxWxH i tryed to put them all into one variable x, therefore 200(0.25x)(0.15x)=V in this case x=200 and then i took the derivative with respect to time to find the Rate of the volume is this the right approach?

 

[HallsofIvy]

Not quite. If you assert that "x is the length, 0.25x is the width, and 0.15x is the height" for all x, then you are asserting that the length is changing at rate 0.25 dx/dt and the height is changing at 0.15 dx/dt. But the problem tells you that dx/dt= 12 cm/min, the width is changing at 4 cm/min which is not 0.25(12), and the height is changing at increasing by 3 cm/min which is not 0.15(12).

The width is 1/4 the length and the height is 3/20 the length only at that one instant, not in general.

The simplest thing to do is write V(t)= x(t)y(t)z(t), were x(t) is the length at time t, y(t) is the width at time t, and z(t) is the height at time t. Use the product rule and the chain rule to determind dV/dt in terms of x, y, z, dx/dt, dy/dt, and dz/dt.

Another way to do this, although I think it requires more computation, is to use that fact that constant rate of change implies a linear function. If you take t= 0 to be the time at which "the dimensions are 200 cm in length, 50 cm in width and 30 cm in height" then the length at time t is x(t)= 200+ 12t, y(t)= 50+ 4t, and z(t)= 30+ 3t. Form V from those and differentiate. In fact do it both ways and you should see that those are basically the same. The "more computation" is in the finding the linear functions which I did for you!