A cone has a volume of V cubic units. Another cone has the same height and 3 times the radius of the original cone. Write an expression for its volume.

Steps to Solve

1. Recall the formula for the volume of a cone

The volume $V$ of a cone is given by the formula:

$$ V = \frac{1}{3} \pi r^2 h $$

where $r$ is the radius and $h$ is the height of the cone.

2. Define the original cone's volume

Let $r_1$ be the radius and $h_1$ be the height of the original cone. Its volume $V_1$ is:

$$ V_1 = \frac{1}{3} \pi r_1^2 h_1 $$

Since the original cone's volume is given as $V$, we have $V_1 = V$.

3. Define the new cone's volume

Let $r_2$ be the radius and $h_2$ be the height of the new cone. We are given that the new cone has the same height as the original, so $h_2 = h_1$. We are also given that the radius of the new cone is 3 times the radius of the original, so $r_2 = 3r_1$. The volume $V_2$ of the new cone is:

$$ V_2 = \frac{1}{3} \pi r_2^2 h_2 $$

4. Express the new cone's volume in terms of the original cone's radius and height

Substitute $r_2 = 3r_1$ and $h_2 = h_1$ into the equation for $V_2$:

$$ V_2 = \frac{1}{3} \pi (3r_1)^2 h_1 = \frac{1}{3} \pi (9r_1^2) h_1 = 9 \left( \frac{1}{3} \pi r_1^2 h_1 \right) $$

5. Relate the new cone's volume to the original cone's volume

Since $V_1 = \frac{1}{3} \pi r_1^2 h_1$, we can substitute $V_1$ into the equation for $V_2$:

$$ V_2 = 9 V_1 $$

Since $V_1 = V$, the new cone's volume is:

$$ V_2 = 9V $$