[SOLVED] How fast is the surface area increasing

[karush]

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 $V$

 

[topsquark]

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]

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]

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]

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.

\(\displaystyle \frac{dS}{dt}
=12x\frac{1}{3x^2}\frac{dV}{dt}
\Rightarrow
12(30)\frac{1}{3(30)^2}\cdot 10
=\frac{4}{3}\frac{m^2}{min}
\)