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, 05:52 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, 07:21 PMThePerfectHackerQuote:

Originally Posted by**nirva**

, thus,

. 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 for the cube then taking the derivate to respect to time we have,

know we know that for those are the conditions of this problem and in the other problem you found that thus, substituting these values,

.

Just placing in the proper unit values to get,

- Mar 2nd 2006, 12:18 AMticbolQuote:

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, 04:17 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

- Mar 2nd 2006, 08:06 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 height. That means, it's an area of per min.

Greetings

EB - Mar 2nd 2006, 01:29 PMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**