# Math Help - A volume/SA problem

1. ## A volume/SA problem

The volume of a cube is decreasing 10cm^3/min. How fast is the Surface Area decreasing when the length of one edge is 30cm?

2. Originally Posted by nirva
The volume of a cube is decreasing 10cm^3/min. How fast is the Surface Area decreasing when the length of one edge is 30cm?
If $s$ is the side of the cube then $V=s^3$ thus taking the derivative to respect to time from both time we have $\frac{dV}{ds}=3s^2\frac{ds}{dx}$, conditions of the problem says the edge is 30 thus, $s=30$ and the rate of change is 10, thus, $\frac{dV}{ds}=10$. Thus, placing in these values,
$10=3(10)^2\frac{ds}{dt}$, thus,
$\frac{ds}{dt}=\frac{1}{30}$. But this tells you the instantenous rate of change of the side, the problem asks for the instantenous rate of change of the surface area. Since $A=6s^2$ for the cube then taking the derivate to respect to time we have,
$\frac{dA}{dt}=12s\frac{ds}{dt}$ know we know that $s=30$ for those are the conditions of this problem and in the other problem you found that $\frac{ds}{dt}=\frac{1}{30}$ thus, substituting these values,
$\frac{dA}{dt}=12(30)\frac{1}{30}=12$.
Just placing in the proper unit values to get,
$\frac{dA}{dt}=12\frac{\mbox{cm}^2}{\mbox{min}}$

3. Originally Posted by nirva
The volume of a cube is decreasing 10cm^3/min. How fast is the Surface Area decreasing when the length of one edge is 30cm?
Here is one way.

Cube of side s cm.
Volume, V = s^3 -----------------------(1)
Surface area, A = 6(s^2) = 6s^2 -------(2)

From (1),
s = V^(1/3)
Substitute that into (2),
A = 6(V^(1/3))^2
A = 6V^(2/3)
Differentiate both sides with respect to time t,
dA/dt = 6(2/3)*V^(-1/3)*(dV/dt)
Plug into that the given dV/dt = -10 cu.cm/min,
dA/dt = 4*V^(-1/3) *(-10)
dA/dt = (-40)/[cubrt(V)] ---------------(3)
So when s=30cm,
V = (30)^3 cu.cm. Plug that into (3),
dA/dt = (-40)/[cubrt(30^3)]
dA/dt = -40/[30] = -4/3 sq.cm/min

Therefore, at that time, the surface area is decreasing by 4/3 sq.cm/min. ------------answer.

--------------------------
Another way.

V = s^3
dV/dt = 3s^2 *(ds/dt) --------(i)

A = 6s^2
dA/dt = 12s*(ds/dt) ----------(ii)

When s=30cm, and that dV/dt is always -10cu.cm/min, then, plug those into (i),
-10 = 3(30^2)*(ds/dt)
-10 = 2700(ds/dt)
ds/dt = -10/2700 = -1/270 cm/min.
Substitute that, and s=30cm, into (ii),
dA/dt = 12(30)(-1/270)
dA/dt = -4(10)(1/30)
dA/dt = -40/30 = -4/3 sq.cm/min. ---------same as above.

4. Originally Posted by ThePerfectHacker
If $s$ is the side of the cube then $V=s^3$ thus taking the derivative to respect to time from both time we have $\frac{dV}{ds}=3s^2\frac{ds}{dx}$, conditions of the problem says the edge is 30 thus, $s=30$ and the rate of change is 10, thus, $\frac{dV}{ds}=10$. Thus, placing in these values,
$10=3(10)^2\frac{ds}{dt}$, thus,
That should be $30^2$ not $10^2$

5. Originally Posted by nirva
The volume of a cube is decreasing 10cm^3/min. How fast is the Surface Area decreasing when the length of one edge is 30cm?
Hello,

here is a more practical way: There are 4 squares involved, when the surface area decreases. I've attached a drawing to demonstrate, what I would like to explain.
So the part of the surface, which is "lost" per minute, is a rectangle with 4*30 cm length and $\frac{10\ cm^3}{900\ cm^2}$ height. That means, it's an area of $\frac{1}{90} cm \cdot 120 cm=\frac{4}{3} cm^2$ per min.

Greetings

EB

6. Originally Posted by CaptainBlack
That should be $30^2$ not $10^2$
Sorry, it hate making mistakes like that. Now the guy is going to take the test an get everything wrong because of me.