The volume of oil in a cylindrical container is increasing at a rate of 150 cubic inches per second. The height of the cylinder is approximately 10 times the radius. At what rate is the height of the oil changing when the oil is 35 inches high? The formula for the volume of a cylinder is V = tr h. Draw a sketch of the situation and show all work. %3D
The volume of oil in a cylindrical container is increasing at a rate of 150 cubic inches per second. The height of the cylinder is approximately 10 times the radius. At what rate is the height of the oil changing when the oil is 35 inches high? The formula for the volume of a cylinder is V = tr h. Draw a sketch of the situation and show all work. %3D
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Concept explainers
Cylinders
A cylinder is a three-dimensional solid shape with two parallel and congruent circular bases, joined by a curved surface at a fixed distance. A cylinder has an infinite curvilinear surface.
Cones
A cone is a three-dimensional solid shape having a flat base and a pointed edge at the top. The flat base of the cone tapers smoothly to form the pointed edge known as the apex. The flat base of the cone can either be circular or elliptical. A cone is drawn by joining the apex to all points on the base, using segments, lines, or half-lines, provided that the apex and the base both are in different planes.
Question
![**Problem Statement:**
The volume of oil in a cylindrical container is increasing at a rate of 150 cubic inches per second. The height of the cylinder is approximately 10 times the radius. At what rate is the height of the oil changing when the oil is 35 inches high? The formula for the volume of a cylinder is \( V = \pi r^2 h \).
**Solution:**
1. **Given Information:**
- Rate of change of volume \(\frac{dV}{dt} = 150\) cubic inches per second
- Height (\(h\)) is approximately 10 times the radius (\(r\)) \( \Rightarrow h = 10r \)
- To find: Rate of change of height (\(\frac{dh}{dt}\)) when \( h = 35 \) inches.
2. **Formula for Volume:**
\[
V = \pi r^2 h
\]
3. **Substitute \(h = 10r\) into the volume formula:**
\[
V = \pi r^2 (10r) = 10\pi r^3
\]
4. **Differentiate both sides with respect to time (\(t\)):**
\[
\frac{dV}{dt} = 30\pi r^2 \frac{dr}{dt}
\]
5. **Substitute \(\frac{dV}{dt} = 150\):**
\[
150 = 30\pi r^2 \frac{dr}{dt}
\]
6. **Solve for \(\frac{dr}{dt}\):**
\[
\frac{dr}{dt} = \frac{150}{30\pi r^2} = \frac{5}{\pi r^2}
\]
7. **Find \(r\) when \( h = 35 \):**
\[
h = 10r \Rightarrow 35 = 10r \Rightarrow r = 3.5 \text{ inches}
\]
8. **Substitute \(r = 3.5\) into \(\frac{dr}{dt}\):**
\[
\frac{dr}{dt} = \frac{5}{\pi (3.5)^2} = \frac{5}{\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F880b5248-7c0b-499c-91d5-d57492b6ff41%2F483243ce-3f20-43bd-847c-b30eb4592ffa%2Fgb6b2bj.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
The volume of oil in a cylindrical container is increasing at a rate of 150 cubic inches per second. The height of the cylinder is approximately 10 times the radius. At what rate is the height of the oil changing when the oil is 35 inches high? The formula for the volume of a cylinder is \( V = \pi r^2 h \).
**Solution:**
1. **Given Information:**
- Rate of change of volume \(\frac{dV}{dt} = 150\) cubic inches per second
- Height (\(h\)) is approximately 10 times the radius (\(r\)) \( \Rightarrow h = 10r \)
- To find: Rate of change of height (\(\frac{dh}{dt}\)) when \( h = 35 \) inches.
2. **Formula for Volume:**
\[
V = \pi r^2 h
\]
3. **Substitute \(h = 10r\) into the volume formula:**
\[
V = \pi r^2 (10r) = 10\pi r^3
\]
4. **Differentiate both sides with respect to time (\(t\)):**
\[
\frac{dV}{dt} = 30\pi r^2 \frac{dr}{dt}
\]
5. **Substitute \(\frac{dV}{dt} = 150\):**
\[
150 = 30\pi r^2 \frac{dr}{dt}
\]
6. **Solve for \(\frac{dr}{dt}\):**
\[
\frac{dr}{dt} = \frac{150}{30\pi r^2} = \frac{5}{\pi r^2}
\]
7. **Find \(r\) when \( h = 35 \):**
\[
h = 10r \Rightarrow 35 = 10r \Rightarrow r = 3.5 \text{ inches}
\]
8. **Substitute \(r = 3.5\) into \(\frac{dr}{dt}\):**
\[
\frac{dr}{dt} = \frac{5}{\pi (3.5)^2} = \frac{5}{\
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