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?

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- Mar 1st 2006, 04:52 PMnirvaA 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?

- Mar 1st 2006, 06:21 PMThePerfectHackerQuote:

Originally Posted by**nirva**

$\displaystyle 10=3(10)^2\frac{ds}{dt}$, thus,

$\displaystyle \frac{ds}{dt}=\frac{1}{30}$. 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 $\displaystyle A=6s^2$ for the cube then taking the derivate to respect to time we have,

$\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}$ know we know that $\displaystyle s=30$ for those are the conditions of this problem and in the other problem you found that $\displaystyle \frac{ds}{dt}=\frac{1}{30}$ thus, substituting these values,

$\displaystyle \frac{dA}{dt}=12(30)\frac{1}{30}=12$.

Just placing in the proper unit values to get,

$\displaystyle \frac{dA}{dt}=12\frac{\mbox{cm}^2}{\mbox{min}}$ - Mar 1st 2006, 11:18 PMticbolQuote:

Originally Posted by**nirva**

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. - Mar 2nd 2006, 03:17 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

- Mar 2nd 2006, 07:06 AMearbothQuote:

Originally Posted by**nirva**

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 $\displaystyle \frac{10\ cm^3}{900\ cm^2}$ height. That means, it's an area of $\displaystyle \frac{1}{90} cm \cdot 120 cm=\frac{4}{3} cm^2$ per min.

Greetings

EB - Mar 2nd 2006, 12:29 PMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**