The volume of a cone is calculated using the formula:
V = \frac{1}{3} \pi r^2 h
Where:
- V = Volume of the cone
- r = Radius of the base of the cone
- h = Height of the cone
In this scenario, both the radius (r) and the height (h) are increasing at a constant rate of 12 centimeters per second. We need to find the rate of change of volume (\frac{dV}{dt}) with respect to time.
Using the chain rule of differentiation, we can express \frac{dV}{dt} as follows:
\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} + \frac{dV}{dh} \cdot \frac{dh}{dt}
First, we calculate \frac{dV}{dr} and \frac{dV}{dh}:
- Finding \frac{dV}{dr}:
- Finding \frac{dV}{dh}:
V = \frac{1}{3} \pi r^2 h
\frac{dV}{dr} = \frac{d}{dr} \left( \frac{1}{3} \pi r^2 h \right) = \frac{2}{3} \pi r h
\frac{dV}{dh} = \frac{d}{dh} \left( \frac{1}{3} \pi r^2 h \right) = \frac{1}{3} \pi r^2
Now, we can plug in values for \frac{dV}{dr}, \frac{dV}{dh}, \frac{dr}{dt}, and \frac{dh}{dt}:
Given:
- \frac{dr}{dt} = 12 cm/s
- \frac{dh}{dt} = 12 cm/s
Thus, the rate of change of volume becomes:
\frac{dV}{dt} = \left(\frac{2}{3} \pi r h \cdot 12 \right) + \left( \frac{1}{3} \pi r^2 \cdot 12 \right)
Factoring out the common terms:
\frac{dV}{dt} = 12 \cdot \frac{\pi}{3} \left(2rh + r^2 \right)
Therefore, the rate of change of the volume of the cone, where radius and height both increase at a rate of 12 centimeters per second, can be represented by:
\frac{dV}{dt} = 4\pi \left(2rh + r^2 \right) cm^3/s
This formula efficiently describes how the volume of the cone changes as both the height and radius increase, offering a clear understanding of the dynamics involved.