If the radius of a right circular cone is increased by p% without increasing its height, then what is the percentage increase in the volume of the cone?
p(2 + p/100)
Calculating Percentage Increase in Cone Volume
This problem asks us to find the percentage increase in the volume of a right circular cone when its radius is increased by a certain percentage, specifically p%, while its height remains unchanged. We need to use the formula for the volume of a cone and then calculate the percentage change.
Understanding the Cone Volume Formula
The volume (V) of a right circular cone is given by the formula:
\(V = \frac{1}{3} \pi r^2 h\)
where:
- \(r\) is the radius of the base of the cone.
- \(h\) is the height of the cone.
- \(\pi\) is a mathematical constant approximately equal to 3.14159.
Setting up the Initial and New Dimensions
Let's denote the initial dimensions and volume as follows:
- Initial radius = \(r_1\)
- Initial height = \(h_1\)
- Initial volume = \(V_1\)
So, the initial volume is \(V_1 = \frac{1}{3} \pi r_1^2 h_1\).
Now, the problem states that the radius is increased by p% without increasing the height. So, the new dimensions are:
- New radius = \(r_2\)
- New height = \(h_2\)
- New volume = \(V_2\)
According to the problem:
- \(h_2 = h_1\) (height remains the same)
- \(r_2\) is \(r_1\) increased by p%. This can be written as:
\(r_2 = r_1 + r_1 \times \frac{p}{100} = r_1 \left(1 + \frac{p}{100}\right)\)
Calculating the New Volume
Using the formula for the volume of a cone with the new radius \(r_2\) and new height \(h_2\):
\(V_2 = \frac{1}{3} \pi r_2^2 h_2\)
Substitute \(r_2 = r_1 \left(1 + \frac{p}{100}\right)\) and \(h_2 = h_1\):
\(V_2 = \frac{1}{3} \pi \left(r_1 \left(1 + \frac{p}{100}\right)\right)^2 h_1\)
\(V_2 = \frac{1}{3} \pi r_1^2 \left(1 + \frac{p}{100}\right)^2 h_1\)
Determining the Increase in Volume
The increase in volume is the difference between the new volume and the initial volume:
\(\text{Increase in Volume} = V_2 - V_1\)
\(V_2 - V_1 = \frac{1}{3} \pi r_1^2 \left(1 + \frac{p}{100}\right)^2 h_1 - \frac{1}{3} \pi r_1^2 h_1\)
Factor out the common term \(\frac{1}{3} \pi r_1^2 h_1\) (which is \(V_1\)):
\(V_2 - V_1 = \frac{1}{3} \pi r_1^2 h_1 \left[ \left(1 + \frac{p}{100}\right)^2 - 1 \right]\)
\(V_2 - V_1 = V_1 \left[ \left(1 + \frac{p}{100}\right)^2 - 1 \right]\)
Calculating the Percentage Increase in Volume
The percentage increase in volume is given by the formula:
\(\text{Percentage Increase} = \left(\frac{\text{Increase in Volume}}{\text{Initial Volume}}\right) \times 100\)
\(\text{Percentage Increase} = \left(\frac{V_2 - V_1}{V_1}\right) \times 100\)
Substitute the expression for \(V_2 - V_1\):
\(\text{Percentage Increase} = \left(\frac{V_1 \left[ \left(1 + \frac{p}{100}\right)^2 - 1 \right]}{V_1}\right) \times 100\)
\(\text{Percentage Increase} = \left[ \left(1 + \frac{p}{100}\right)^2 - 1 \right] \times 100\)
Now, let's expand the term \(\left(1 + \frac{p}{100}\right)^2\) using the algebraic identity \((a+b)^2 = a^2 + 2ab + b^2\):
\(\left(1 + \frac{p}{100}\right)^2 = 1^2 + 2 \times 1 \times \frac{p}{100} + \left(\frac{p}{100}\right)^2\)
\(\left(1 + \frac{p}{100}\right)^2 = 1 + \frac{2p}{100} + \frac{p^2}{10000}\)
Substitute this back into the percentage increase formula:
\(\text{Percentage Increase} = \left[ \left(1 + \frac{2p}{100} + \frac{p^2}{10000}\right) - 1 \right] \times 100\)
\(\text{Percentage Increase} = \left[ \frac{2p}{100} + \frac{p^2}{10000} \right] \times 100\)
Distribute the multiplication by 100:
\(\text{Percentage Increase} = \left(\frac{2p}{100} \times 100\right) + \left(\frac{p^2}{10000} \times 100\right)\)
\(\text{Percentage Increase} = 2p + \frac{p^2}{100}\)
This expression can be factored by taking 'p' common:
\(\text{Percentage Increase} = p \left(2 + \frac{p}{100}\right)\)
This is the percentage increase in the volume of the cone.
Comparing with Options
Let's compare our result \(p \left(2 + \frac{p}{100}\right)\) with the given options:
| Option | Expression |
|---|---|
| 1 | \(p^2\) |
| 2 | \(2p^2\) |
| 3 | \(\frac{p^2}{100}\) |
| 4 | \(p \left(2 + \frac{p}{100}\right)\) |
Our calculated percentage increase matches the expression in Option 4.
Revision Table: Cone Volume Percentage Change
| Concept | Formula/Calculation |
|---|---|
| Initial Volume \(V_1\) | \(\frac{1}{3} \pi r_1^2 h_1\) |
| New Radius \(r_2\) (after p% increase) | \(r_1 \left(1 + \frac{p}{100}\right)\) |
| New Height \(h_2\) | \(h_1\) (no change) |
| New Volume \(V_2\) | \(\frac{1}{3} \pi r_2^2 h_2 = \frac{1}{3} \pi \left(r_1 \left(1 + \frac{p}{100}\right)\right)^2 h_1\) |
| Percentage Increase in Volume | \(\left(\frac{V_2 - V_1}{V_1}\right) \times 100\) |
| Simplified Percentage Increase | \(p \left(2 + \frac{p}{100}\right)\) |
Additional Information: Geometric Volume Changes
When dimensions of a geometric shape change, its volume changes based on how the dimension affects the volume formula. For a cone, volume is proportional to the square of the radius (\(r^2\)) and linearly proportional to the height (\(h\)).
- If only height changes by a percentage, the volume changes by the same percentage.
- If only radius changes by a percentage, the volume change is more complex because the radius is squared in the formula. An increase of p% in radius results in a change factor of \(\left(1 + \frac{p}{100}\right)^2\) in the \(r^2\) term.
- If both dimensions change, the calculation involves multiplying the respective change factors.
In this problem, increasing the radius by p% means the \(r^2\) term increases by a factor of \(\left(1 + \frac{p}{100}\right)^2\). Since the height is constant, the volume changes by this same factor relative to its dependence on \(r^2\), leading to the percentage increase we calculated.
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