The radius of a right circular cone of fixed height 20 ft is increasing at a rate of 4 in/s. How fast is the volume increasing when the radius is 8 ft? I
The radius of a right circular cone of fixed height 20 ft is increasing at a rate of 4 in/s. How fast is the volume increasing when the radius is 8 ft? I
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![Certainly! Below is the transcribed text intended for an educational website. Note that for this particular image, there are no diagrams or graphs to explain.
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### Calculus Problem: Rate of Change in Volume of a Cone
**Problem Statement:**
The radius of a right circular cone of fixed height 20 ft is increasing at a rate of 4 in/s. How fast is the volume increasing when the radius is 8 ft?
---
To understand how to solve this problem, we can use principles from calculus, specifically related rates. Let's break it down step by step:
1. **Identify the given information:**
- The height of the cone (\(h\)) is constant at 20 feet.
- The radius of the cone (\(r\)) is increasing at a rate of 4 inches per second.
- We need to find the rate at which the volume (\(V\)) of the cone is increasing when the radius is 8 feet.
2. **Convert units if necessary:**
- Since the radius change rate is given in inches per second, convert it to feet per second.
\[
4 \text{ inches/second} = \frac{4}{12} \text{ feet/second} = \frac{1}{3} \text{ feet/second}
\]
3. **Volume of a cone formula:**
\[
V = \frac{1}{3} \pi r^2 h
\]
4. **Differentiate both sides with respect to time (\(t\)):**
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left(2r \frac{dr}{dt} \right) h
\]
Since \(h\) is constant, it can be factored out as a constant in the differentiation.
5. **Substitute the known values:**
- \( r = 8 \text{ feet} \)
- \( \frac{dr}{dt} = \frac{1}{3} \text{ feet/second} \)
- \( h = 20 \text{ feet} \)
Now, plug these values into the differentiated equation:
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left(2(8) \frac{1}{3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa589a0b1-5940-47b0-b798-8fa24b0ff23c%2Fac64a13e-7c2c-49e6-b752-08077a4ea1fa%2Ft2dx42h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Certainly! Below is the transcribed text intended for an educational website. Note that for this particular image, there are no diagrams or graphs to explain.
---
### Calculus Problem: Rate of Change in Volume of a Cone
**Problem Statement:**
The radius of a right circular cone of fixed height 20 ft is increasing at a rate of 4 in/s. How fast is the volume increasing when the radius is 8 ft?
---
To understand how to solve this problem, we can use principles from calculus, specifically related rates. Let's break it down step by step:
1. **Identify the given information:**
- The height of the cone (\(h\)) is constant at 20 feet.
- The radius of the cone (\(r\)) is increasing at a rate of 4 inches per second.
- We need to find the rate at which the volume (\(V\)) of the cone is increasing when the radius is 8 feet.
2. **Convert units if necessary:**
- Since the radius change rate is given in inches per second, convert it to feet per second.
\[
4 \text{ inches/second} = \frac{4}{12} \text{ feet/second} = \frac{1}{3} \text{ feet/second}
\]
3. **Volume of a cone formula:**
\[
V = \frac{1}{3} \pi r^2 h
\]
4. **Differentiate both sides with respect to time (\(t\)):**
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left(2r \frac{dr}{dt} \right) h
\]
Since \(h\) is constant, it can be factored out as a constant in the differentiation.
5. **Substitute the known values:**
- \( r = 8 \text{ feet} \)
- \( \frac{dr}{dt} = \frac{1}{3} \text{ feet/second} \)
- \( h = 20 \text{ feet} \)
Now, plug these values into the differentiated equation:
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left(2(8) \frac{1}{3
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