Volume and surface are of a cube??

[aff]

The side of a cube is increasing at a constant rate of 0.2 centimeter per second. In terms of surface are S, what is the rate of change of the volume of the cube, in cubic meters per second?

i know that:
S=6s^2
DA/dt=12s ds/dt
^ and ds/dt equals 0.2
DA/dt=2.4s

V=s^3
DV=3s^2

but i don't know how to put it together?

-Thanks in advance

 

[tkhunny]

\(\displaystyle V = s^{3}\)

\(\displaystyle dV = 3s^{2}\cdot ds\)


\(\displaystyle S = 6s^{2}\)

\(\displaystyle dS = 12s\cdot ds\)

Thus...

\(\displaystyle dV = 3s^{2}\cdot \left(\frac{dS}{12s}\right)\)

Ponder it for a little while. Do we need that "ds"? What does it mean to be "in terms of the Surface Area"?

\(\displaystyle dV = 6s^{3}\cdot \left(\frac{dS}{24s^{2}}\right) = S\cdot \left(\frac{dS}{24s^{2}}\right)\)

Rather guessing at what is wanted. Did you include the ENTIRE problem?

 

[aff]

the answer is suppose to be 0.1S,
S is the surface area

 

[lookagain]

The > > > side of a cube < < <
is increasing at a constant rate of 0.2 centimeter per second.

The side of a cube is one of the 6 square faces of the cube, not one of the edges of the cube.



\(\displaystyle Volume \ = \ the\ cube\ of \ the \ length\ of \ an \ edge \ \ or \ \ V \ = \ e^3.\)


But as "e" is already used as a constant, you may want to type:

\(\displaystyle Let \ s \ = \ the \ \ length \ \ of \ \ an \ \ edge \ \ of \ \ a \ \ cube.\)

\(\displaystyle Then \ V \ = \ s^3.\)



In terms of surface are S, what is the rate of change of the volume of the cube,
in cubic meters per second?


i know that:
S=6s^2

DA/dt=12s ds/dt
When you use S for the surface area of the cube, then you must
stay with that variable. The equation could be dS/dt = (12s)dt.

** But, we don't need dS/dt for this solution. We are not concerned
with the change of the surface area of the cube with respect to time,
only with the change of volume of the cube with respect to time,
in terms of the expression for the total surface area of the cube.


^ and ds/dt equals 0.2
DA/dt=2.4s

V=s^3
DV=3s^2

but i don't know how to put it together?

\(\displaystyle V \ = \ s^3.\)

\(\displaystyle \dfrac{dV}{dt} \ = \ (3s^2)\dfrac{ds}{dt}.\)


Note: If S = the surface area of the whole cube, then

\(\displaystyle S \ = \ 6s^2.\)


\(\displaystyle \dfrac{dV}{dt} \ = \ 3s^2(0.2) \ = \ 0.6s^2.\)


\(\displaystyle But \ \ 0.6s^2 \ = \)


\(\displaystyle 0.1(6s^2) \ = \)


\(\displaystyle \boxed{ 0.1S}\)