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Do you mean that the length of the each edge is growing at a rate of 5cm / sec?
They key to this problem is to express the surface area of the cube at any point in time in terms of the length of an edge. Differentiate with respect to t and then substitute edge length = 4.5cm into that.
Yes, 5 cm/s. And ... would that be SA'(t)=6(s(t))^2*ds/dt? I'm so bad at these.
I googled "Related Rates Cube Expanding" this is the most helpful reply, even though it doesn't have a pic:
The answer does not mention UNITS, which are SO IMPORTANT. So the answer for the problem I linked should have units cm^2/sec.
Not the answer you are looking for? Search for more explanations.
Thank you, but I'm still having trouble figuring everything out. Sigh.
This is a class of problems called "related Rates"
http://patrickjmt.com/related-rates-using-cones/ there is a nice example. Additionally, I would look into
Paul's Online Math Notes:
SOme more great videos are here: http://www.mathtv.com/
Go to Calculus > Application of Derivatives > Related Rates
You're right, Surface Area = 6*(side)^2.
We also know that side length is a function of time, since
the side is increasing with respect to time.
So let’s denote that s(t).
Then SA = 6*[s(t)]^2.
Now we want to know how fast the surface area is increasing.
So we want to find [SA(t)]’.
So take the derivative with respect to t on both sides.
On the right side, remembering that s(t) is a function of t, we
know that we’ll need to use the chain rule.
So the derivative of the right side is
2 * 6[s(t)] * s’(t).
(The derivative of 6x^2 is 2 * 6x. But here s(t) is a function of t, so
the derivative has a factor of s’(t).)
So now we have that [SA(t)]’ = 2 * 6[s(t)] * s’(t).
Plug in your numbers for s(t) and s’(t) to find how fast the surface area is increasing at that particular moment.