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karush
Well-known member
- Jan 31, 2012
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The volume of a cube is increasing at a rate of $$\displaystyle\frac{10 cm^3}{\text {min}}$$
How fast is the surface area increasing when the lenght of an edge is $$\displaystyle 30 \text{ cm}$$
The answer to this is $$\displaystyle \frac{4}{3} \text{ cm}^2$$
$$V=x^3$$ and $$SA=6x^2$$ and $$\displaystyle\frac {dV}{dt} = \displaystyle\frac{10 cm^3}{\text {min}}$$
actually I don't know how to set this up relating $SA$ to $V$
How fast is the surface area increasing when the lenght of an edge is $$\displaystyle 30 \text{ cm}$$
The answer to this is $$\displaystyle \frac{4}{3} \text{ cm}^2$$
$$V=x^3$$ and $$SA=6x^2$$ and $$\displaystyle\frac {dV}{dt} = \displaystyle\frac{10 cm^3}{\text {min}}$$
actually I don't know how to set this up relating $SA$ to $V$
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