The edge of a cube expands 3 centimeters per second. In the moment in which each edge has a longitude of 10 cm, the volume changes with a velocity of:

a) 300 cm^3/s

b) 900 cm^3/s

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- Aug 4th 2006, 09:07 AMbret80Volume change
The edge of a cube expands 3 centimeters per second. In the moment in which each edge has a longitude of 10 cm, the volume changes with a velocity of:

a) 300 cm^3/s

b) 900 cm^3/s - Aug 4th 2006, 09:42 AMgalactus
We want $\displaystyle \frac{dV}{dt}$ when x=10. We know $\displaystyle \frac{dx}{dt}=+3$

$\displaystyle V=x^{3}$

$\displaystyle \frac{dV}{dt}=3x^{2}\frac{dx}{dt}$

$\displaystyle 3(10)^{2}(3)$ - Aug 4th 2006, 10:05 AMbret80
whats the difference between dv/dt and dx/dt?

Why multiply 3cm/s times the first derivative of the volume of a cube? - Aug 4th 2006, 10:11 AMgalactus
What do you mean?. I took the derivative of x^{3}, which is 3x^2.

dx/dt=3 - Aug 4th 2006, 11:26 AMThePerfectHackerQuote:

Originally Posted by**bret80**

$\displaystyle V=s^3$

Take, the*derivative to respect to time*, (implictly),

$\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, the problem says the side at $\displaystyle s=10$ and the rate which the sides change is $\displaystyle \frac{ds}{dt}=3$ thus,

$\displaystyle \frac{dV}{dt}=3\cdot (10)^2 \cdot 3=900$