Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Question 15: Volume Change in a Right Circular Cone
**Question:**
If we double the radius of a right circular cone, what effect does that have on the volume? Explain your answer.
#### Explanation:
When considering the volume of a right circular cone, it is important to remember the formula for the volume:
\[ V = \frac{1}{3} \pi r^2 h \]
Where:
- \( V \) is the volume,
- \( r \) is the radius of the base,
- \( h \) is the height of the cone.
If we double the radius of the cone, the new radius \( r_{\text{new}} \) becomes \( 2r \).
Substituting \( 2r \) into the volume formula:
\[ V_{\text{new}} = \frac{1}{3} \pi (2r)^2 h \]
This simplifies to:
\[ V_{\text{new}} = \frac{1}{3} \pi (4r^2) h \]
\[ V_{\text{new}} = 4 \left( \frac{1}{3} \pi r^2 h \right) \]
\[ V_{\text{new}} = 4V \]
Therefore, if we double the radius of a right circular cone, the volume increases by a factor of four. This means the volume of the cone becomes four times larger.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8311114-a7cb-49e3-b44e-1e05de4bc29a%2F1d4ab042-adea-4d74-adf6-80f22ec531db%2Frl4syvs.png&w=3840&q=75)
Transcribed Image Text:### Question 15: Volume Change in a Right Circular Cone
**Question:**
If we double the radius of a right circular cone, what effect does that have on the volume? Explain your answer.
#### Explanation:
When considering the volume of a right circular cone, it is important to remember the formula for the volume:
\[ V = \frac{1}{3} \pi r^2 h \]
Where:
- \( V \) is the volume,
- \( r \) is the radius of the base,
- \( h \) is the height of the cone.
If we double the radius of the cone, the new radius \( r_{\text{new}} \) becomes \( 2r \).
Substituting \( 2r \) into the volume formula:
\[ V_{\text{new}} = \frac{1}{3} \pi (2r)^2 h \]
This simplifies to:
\[ V_{\text{new}} = \frac{1}{3} \pi (4r^2) h \]
\[ V_{\text{new}} = 4 \left( \frac{1}{3} \pi r^2 h \right) \]
\[ V_{\text{new}} = 4V \]
Therefore, if we double the radius of a right circular cone, the volume increases by a factor of four. This means the volume of the cone becomes four times larger.
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