Expert: Paul Klarreich - 4/19/2007

Question
at the start, a cube's edge measures 6 inches and is increasing at a rate of 2in/min for 5 mins. at what rate is the volume of the cube changing after the five mins. have passed? (V=edge^3)    i know the answer is 1536in^3/min, but i cant derive at the answer...please help me!

Answer
Questioner:   crystal
Category:  Calculus
 
Subject:  related rates of cube
Question:  at the start, a cube's edge measures 6 inches and is increasing at a rate of 2in/min for 5 mins. at what rate is the volume of the cube changing after the five mins. have passed? (V=edge^3)    i know the answer is 1536in^3/min, but i can't derive at the answer...please help me!
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Hi, Crystal,

Is this your first attempt at R-R problems?  If so, the scheme is something like this:

1. Identify the variables in the problem -- the things that change.  Give them names.

2. Write their rates of change as derivatives with respect to time.  Note which are known and which is to be found.

3. Determine a relationship (yes, it is called 'related rates' for a reason) between the variables.  Use a diagram, use your life experience, use your general knowledge and brilliance, do whatever you have to.  This is the hard part, because it isn't always mechanical.

4. Now differentiate implicitly, then substitute the known quantities and rates, and solve for the unknown rate.

Think I just made that up?  Yours is only about the 439th related rates problem people have sent me.

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In your example:

Let  e = the length of one edge.
    V = volume of the cube.

de/dt = the rate of increase of the edge, GIVEN as 2 in/min.

dV/dt = the rate of increase of the volume, TO BE FOUND.

Relation:  V = e^3  [You knew that.]

Differentiate:

dV/dt = 3e^2 de/dt

Substitute known quantities:

de/dt = 2.

t = 5 minutes, at which time we have e = 6 + 2(5) = 16 inches.

dV/dt = 3(16)^2 (2) = 3(256)(2) = 1536.

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I think we can also do the problem along these lines.  Since the edge starts (t=0) at 6 inches and increases by 2 in/min, we have:

e = 6 + 2t

V = (6 + 2t)^3

Then dV/dt = 3(6 + 2t)^2(2), using the chain rule.
dV/dt = 6(6 + 2t)^2
At t = 5:

dV/dt = 6(6 + 2(5))^2 = 6(6 + 10)^2 = 6(16)^2 = 6(256) = 1536.
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Take your pick.  I like the first one.