# Volume change

• Aug 4th 2006, 09:07 AM
bret80
Volume 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 AM
galactus
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 AM
bret80
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 AM
galactus
What do you mean?. I took the derivative of x^{3}, which is 3x^2.

dx/dt=3
• Aug 4th 2006, 11:26 AM
ThePerfectHacker
Quote:

Originally Posted by bret80
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

Let me try to explain it diffrently. You need to find what $\displaystyle \frac{dV}{dt}$ is at when the side is 10 centimeters because that is the instantenous rate of change in volume. But you know that the volume of a cube is:
$\displaystyle V=s^3$
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,
$\displaystyle \frac{dV}{dt}=3\cdot (10)^2 \cdot 3=900$