increasing surface area

**TGS** · December 7th 2009, 12:27 PM

The volume of a cube is increasing at a rate of 10cm^3/min. How fast is the surface area increasing when the length of an edge is 10cm?

I don't understand this question at all..

**alexmahone** · December 7th 2009, 07:53 PM

Originally Posted by TGS

The volume of a cube is increasing at a rate of 10cm^3/min. How fast is the surface area increasing when the length of an edge is 10cm?

I don't understand this question at all..

$V=a^3$

$S=6a^2$

$\frac{dV}{dt}=10cm^3/min$

$3a^2\frac{da}{dt}=10$

$3*10^2\frac{da}{dt}=10$

$\frac{da}{dt}=\frac{1}{30}$

$\frac{dS}{dt}=6*2a\frac{da}{dt}=12*10*\frac{1}{30} =4cm^2/min$

So, the surface area is increasing at a rate of $4cm^2/min$ .

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