# [SOLVED]How fast is the surface area increasing

#### karush

##### Well-known member
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 $V$

Last edited:

#### topsquark

##### Well-known member
MHB Math Helper
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 $V$
Can you solve the SA equation for x? Then you can put that value of x into the V equation...

-Dan

#### karush

##### Well-known member
Can you solve the SA equation for x? Then you can put that value of x into the V equation...

-Dan
like thIs?

 $\displaystyle SA=6x^2 \Rightarrow x=\sqrt{\frac{(SA)}{6}}$
Then since

$V=x^3$
$\displaystyle V=\left(\sqrt{\frac{(SA)}{6}}\right)^3=\frac{\sqrt{6}}{36}\cdot(SA)^{\frac{3}{2}}$

#### MarkFL

Staff member
like thIs?

 $\displaystyle SA=6x^2 \Rightarrow x=\sqrt{\frac{(SA)}{6}}$
Then since

$V=x^3$
$\displaystyle V=\left(\sqrt{\frac{(SA)}{6}}\right)^3=\frac{\sqrt{6}}{36}\cdot(SA)^{\frac{3}{2}}$
Another approach would be to use related rates:

$$\displaystyle V=x^3\implies\frac{dV}{dt}=3x^2\frac{dx}{dt}$$

$$\displaystyle S=6x^2\implies\frac{dS}{dt}=12x\frac{dx}{dt}$$

From the derivative of the first equation, we have:

$$\displaystyle \frac{dx}{dt}=\frac{1}{3x^2}\frac{dV}{dt}$$

Now, substitute for $$\displaystyle \frac{dx}{dt}$$ into the derivative of the second equation, and you will have $$\displaystyle \frac{dS}{dt}$$ in terms of $x$ and $$\displaystyle \frac{dV}{dt}$$ both of which are given.

#### karush

##### Well-known member
Another approach would be to use related rates:

$$\displaystyle V=x^3\implies\frac{dV}{dt}=3x^2\frac{dx}{dt}$$

$$\displaystyle S=6x^2\implies\frac{dS}{dt}=12x\frac{dx}{dt}$$

From the derivative of the first equation, we have:

$$\displaystyle \frac{dx}{dt}=\frac{1}{3x^2}\frac{dV}{dt}$$

Now, substitute for $$\displaystyle \frac{dx}{dt}$$ into the derivative of the second equation, and you will have $$\displaystyle \frac{dS}{dt}$$ in terms of $x$ and $$\displaystyle \frac{dV}{dt}$$ both of which are given.
$$\displaystyle \frac{dS}{dt} =12x\frac{1}{3x^2}\frac{dV}{dt} \Rightarrow 12(30)\frac{1}{3(30)^2}\cdot 10 =\frac{4}{3}\frac{m^2}{min}$$