Approximate the change in the volume of a sphere when its radius changes from r = 40 ft to r=40.1 ft. 1² (V(₁) = 235 ²³). 4 When r changes from 40 ft to 40.1 ft, AV ft³ (Type an integer or a decimal. Round to the nearest hundredth as needed.) (...) Use a linear approximation. 4
Approximate the change in the volume of a sphere when its radius changes from r = 40 ft to r=40.1 ft. 1² (V(₁) = 235 ²³). 4 When r changes from 40 ft to 40.1 ft, AV ft³ (Type an integer or a decimal. Round to the nearest hundredth as needed.) (...) Use a linear approximation. 4
Chapter3: Polynomial Functions
Section3.5: Mathematical Modeling And Variation
Problem 7ECP: The kinetic energy E of an object varies jointly with the object’s mass m and the square of the...
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![### Linear Approximation of Volume Change in a Sphere
**Objective:**
Approximate the change in the volume of a sphere when its radius changes from \( r = 40 \, \text{ft} \) to \( r = 40.1 \, \text{ft} \). Use a linear approximation.
---
**Instructions:**
1. Start with the formula for the volume \( V \) of a sphere:
\[
V(r) = \frac{4}{3} \pi r^3
\]
2. Use linear approximation to estimate the change in volume \( \Delta V \).
**Given:**
When \( r \) changes from \( 40 \, \text{ft} \) to \( 40.1 \, \text{ft} \), calculate \( \Delta V \) using:
\[
\Delta V \approx f'(r) \Delta r
\]
where \( f(r) = \frac{4}{3} \pi r^3 \).
**Calculation:**
1. Find \( f'(r) \):
\[
f'(r) = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2
\]
2. Evaluate \( f'(r) \) at \( r = 40 \, \text{ft} \):
\[
f'(40) = 4 \pi (40)^2 = 4 \pi (1600) = 6400 \pi
\]
3. The change in \( r \), \( \Delta r \), is \( 0.1 \, \text{ft} \).
4. Therefore, the change in volume \( \Delta V \approx 6400 \pi \times 0.1 \).
5. Calculate the approximate change:
\[
\Delta V \approx 6400 \pi \times 0.1 = 640 \pi \, \text{cubic feet}
\]
Thus, the approximate change in volume \( \Delta V \) is \( 640 \pi \, \text{ft}^3 \).
---
**Question:**
When \( r \) changes from \( 40 \, \text{ft} \) to \( 40.1 \, \text{ft} \), \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa204fb01-1593-427c-b210-f132b42420bd%2F17ebb4f2-906e-4423-8bf0-076f84592fc1%2Fc13h4w9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Linear Approximation of Volume Change in a Sphere
**Objective:**
Approximate the change in the volume of a sphere when its radius changes from \( r = 40 \, \text{ft} \) to \( r = 40.1 \, \text{ft} \). Use a linear approximation.
---
**Instructions:**
1. Start with the formula for the volume \( V \) of a sphere:
\[
V(r) = \frac{4}{3} \pi r^3
\]
2. Use linear approximation to estimate the change in volume \( \Delta V \).
**Given:**
When \( r \) changes from \( 40 \, \text{ft} \) to \( 40.1 \, \text{ft} \), calculate \( \Delta V \) using:
\[
\Delta V \approx f'(r) \Delta r
\]
where \( f(r) = \frac{4}{3} \pi r^3 \).
**Calculation:**
1. Find \( f'(r) \):
\[
f'(r) = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2
\]
2. Evaluate \( f'(r) \) at \( r = 40 \, \text{ft} \):
\[
f'(40) = 4 \pi (40)^2 = 4 \pi (1600) = 6400 \pi
\]
3. The change in \( r \), \( \Delta r \), is \( 0.1 \, \text{ft} \).
4. Therefore, the change in volume \( \Delta V \approx 6400 \pi \times 0.1 \).
5. Calculate the approximate change:
\[
\Delta V \approx 6400 \pi \times 0.1 = 640 \pi \, \text{cubic feet}
\]
Thus, the approximate change in volume \( \Delta V \) is \( 640 \pi \, \text{ft}^3 \).
---
**Question:**
When \( r \) changes from \( 40 \, \text{ft} \) to \( 40.1 \, \text{ft} \), \(
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