Calculating Volume Increase for Cube w/ Expanding Surface Area

[Tom McCurdy]

If a cube is expanding so the surface area is increasing at a constant rate of 7 in^2/sec How fast is the volume increasing at the instant the surface area is 204 in^2 ?

How would I go about starting this problem to set it up?

surface area = 6x^2
volume = x^3

 

[Galileo]

Use the chain rule.

You're given:
$$\frac{d}{dt}S(x(t))=\frac{dS}{dx}\frac{dx}{dt}=7$$

You have to find:
$$\frac{dV}{dt}=\frac{dV}{dx}\frac{dx}{dt}$$

 

[Tom McCurdy]

$$\frac{dS}{dx}\frac{dx}{dt}$$ = 7


$$12 x \frac{dx}{dt}=7$$

$$\frac{dx}{dt}=\frac{7}{12x}$$

$$\frac{dV}{dt}=\frac{dV}{dx} \frac{dx}{dt}$$

$$\frac{dV}{dt}= 3x^2 \frac{7}{12x} = \frac {7x}{4}$$

$$SA = 204 = 6x^2$$

$$SA = \sqrt{34}$$

$$rate = \frac{7\sqrt{34}}{4}$$

$$rate= 10.2 in^3/sec$$

 

[Tom McCurdy]

Does my work seem correct?

 

[HallsofIvy]

$$\frac{dS}{dx}\frac{dx}{dt}$$ = 7


$$12 x \frac{dx}{dt}=7$$

$$\frac{dx}{dt}=\frac{7}{12x}$$

$$\frac{dV}{dt}=\frac{dV}{dx} \frac{dx}{dt}$$

$$\frac{dV}{dt}= 3x^2 \frac{7}{12x} = \frac {7x}{4}$$

$$SA = 204 = 6x^2$$

$$SA = \sqrt{34}$$

you mean
$$x= \sqrt{34}$$

$$rate = \frac{7\sqrt{34}}{4}$$

$$rate= 10.2 in^3/sec$$

Yes.

 

[Tom McCurdy]

Thank you for your help

So the general formula for something like this would simply be
SA rate= X
Volume rate wanted at Surface Area = Y

$$\frac{X \sqrt{\frac{Y}{6}}}{4} = Volume Rate$$

 

[dextercioby]

Hold on,this is a cube we're talking about.So if your formula works for the cube,i can guantee u it doesn't work for any other closed surface.

Daniel.

 

[Tom McCurdy]

Yes I figured it was for the cube only, thanks for the clafication and the help though