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anonymous
 5 years ago
The edges of a cube are expanding at a rate of 5 centimeters per second. How fast is the surface area changing when each edge is 4.5 centimeters?
anonymous
 5 years ago
The edges of a cube are expanding at a rate of 5 centimeters per second. How fast is the surface area changing when each edge is 4.5 centimeters?

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mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0Do 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 5 cm/s. And ... would that be SA'(t)=6(s(t))^2*ds/dt? I'm so bad at these.

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0I googled "Related Rates Cube Expanding" this is the most helpful reply, even though it doesn't have a pic: http://en.allexperts.com/q/Calculus2063/2009/3/relatedrates15.htm The answer does not mention UNITS, which are SO IMPORTANT. So the answer for the problem I linked should have units cm^2/sec.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you, but I'm still having trouble figuring everything out. Sigh.

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0This is a class of problems called "related Rates" http://patrickjmt.com/relatedratesusingcones/ there is a nice example. Additionally, I would look into Paul's Online Math Notes: http://tutorial.math.lamar.edu/Problems/CalcI/RelatedRates.aspx SOme more great videos are here: http://www.mathtv.com/ Go to Calculus > Application of Derivatives > Related Rates

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You'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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02*6*(4.5 cm)*(5 cm/s) = 270 cm^2 / s. So yep.
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