Problem 4
Question
Volume The radius \(r\) and height \(h\) of a right circular cone are related to the cone's volume \(V\) by the equation \(V=(1 / 3) \pi r^{2} h .\) a. How is \(d V / d t\) related to \(d h / d t\) if \(r\) is constant? b. How is \(d V / d t\) related to \(d r / d t\) if \(h\) is constant? c. How is \(d V / d t\) related to \(d r / d t\) and \(d h / d t\) if neither \(r\) nor \(h\) is constant?
Step-by-Step Solution
Verified Answer
a. \( \frac{dV}{dt} = \frac{1}{3} \pi r^2 \frac{dh}{dt} \). b. \( \frac{dV}{dt} = \frac{2}{3} \pi r h \frac{dr}{dt} \). c. \( \frac{dV}{dt} = \frac{1}{3} \pi (2r h \frac{dr}{dt} + r^2 \frac{dh}{dt}) \).
1Step 1: Understand the Given Equation
The volume of a right circular cone is given by the equation \( V = \frac{1}{3} \pi r^2 h \). In this equation, \( r \) is the radius of the base of the cone, \( h \) is the height of the cone, and \( V \) is the volume.
2Step 2: Calculate \( \frac{dV}{dt} \) with Constant \( r \)
Given that \( r \) is constant, we can treat \( \pi r^2 \) as a constant factor. Differentiating \( V = \frac{1}{3} \pi r^2 h \) with respect to time \( t \) gives \( \frac{dV}{dt} = \frac{1}{3} \pi r^2 \frac{dh}{dt} \). Thus, the relationship is direct: \( \frac{dV}{dt} \) is proportional to \( \frac{dh}{dt} \) with \( r \) constant.
3Step 3: Calculate \( \frac{dV}{dt} \) with Constant \( h \)
Given that \( h \) is constant, treat \( \frac{1}{3} \pi h \) as a constant. Differentiating \( V = \frac{1}{3} \pi r^2 h \) with respect to time \( t \) gives \( \frac{dV}{dt} = \frac{2}{3} \pi r h \frac{dr}{dt} \). Thus, \( \frac{dV}{dt} \) is proportional to \( \frac{dr}{dt} \) with \( h \) constant.
4Step 4: Calculate \( \frac{dV}{dt} \) with Changing \( r \) and \( h \)
If both \( r \) and \( h \) change over time, use the product rule to differentiate: \( \frac{dV}{dt} = \frac{1}{3} \pi (2r \cdot h \cdot \frac{dr}{dt} + r^2 \cdot \frac{dh}{dt}) \). This equation shows how \( \frac{dV}{dt} \) is related to both \( \frac{dr}{dt} \) and \( \frac{dh}{dt} \).
Key Concepts
cone volumedifferentiationvolume equation
cone volume
Understanding the volume of a cone hinges on the equation \( V = \frac{1}{3} \pi r^2 h \). This represents the volume \( V \) of a right circular cone with a base radius \( r \) and a height \( h \).
- The term \( \pi r^2 \) calculates the area of the circular base. Multiplied by the height \( h \), it finds the volume if the shape were a perfect cylinder. - However, because a cone comes to a point, not a flat top, it's only one-third of that cylinder volume.
Thus, the equation incorporates a factor of \( \frac{1}{3} \), making the volume effectively one-third of a cylinder with the same radius and height. This foundational concept allows us to delve into understanding how changes in \( r \) or \( h \) — or both — influence the cone's volume dynamically.
- The term \( \pi r^2 \) calculates the area of the circular base. Multiplied by the height \( h \), it finds the volume if the shape were a perfect cylinder. - However, because a cone comes to a point, not a flat top, it's only one-third of that cylinder volume.
Thus, the equation incorporates a factor of \( \frac{1}{3} \), making the volume effectively one-third of a cylinder with the same radius and height. This foundational concept allows us to delve into understanding how changes in \( r \) or \( h \) — or both — influence the cone's volume dynamically.
differentiation
Differentiation is a key tool in calculus that allows us to study how a function changes over time or across different values. When we talk about differentiating the volume equation of the cone, we are exploring how the volume \( V \) changes as time \( t \) passes and as the radius \( r \) or height \( h \) vary.
- If \( r \) is constant, differentiating with respect to \( t \) involves applying the derivative to \( h \) only: \( \frac{dV}{dt} = \frac{1}{3} \pi r^2 \frac{dh}{dt} \).
- If \( h \) remains constant, the differentiation focuses on \( r \): \( \frac{dV}{dt} = \frac{2}{3} \pi r h \frac{dr}{dt} \).
- When both \( r \) and \( h \) change, the product rule is used: \( \frac{dV}{dt} = \frac{1}{3} \pi (2r \cdot h \cdot \frac{dr}{dt} + r^2 \cdot \frac{dh}{dt}) \).
These formulas ensure clarity on how the volume's rate of change is influenced by changes in its defining parameters.
- If \( r \) is constant, differentiating with respect to \( t \) involves applying the derivative to \( h \) only: \( \frac{dV}{dt} = \frac{1}{3} \pi r^2 \frac{dh}{dt} \).
- If \( h \) remains constant, the differentiation focuses on \( r \): \( \frac{dV}{dt} = \frac{2}{3} \pi r h \frac{dr}{dt} \).
- When both \( r \) and \( h \) change, the product rule is used: \( \frac{dV}{dt} = \frac{1}{3} \pi (2r \cdot h \cdot \frac{dr}{dt} + r^2 \cdot \frac{dh}{dt}) \).
These formulas ensure clarity on how the volume's rate of change is influenced by changes in its defining parameters.
volume equation
The volume equation for a cone, \( V = \frac{1}{3} \pi r^2 h \), is more than just a static formula. It becomes dynamic when we understand how it behaves under changing conditions. By focusing on the related rates, we find out how shifts in the variables affect the volume.
- **Constant radius**: When \( r \) is fixed, the equation simplifies to study the effect of changing height alone on the volume. Here, \( \frac{dV}{dt} = \frac{1}{3} \pi r^2 \frac{dh}{dt} \) shows direct proportionality between the cone's height change rate and the volume change rate.
- **Constant height**: With a fixed \( h \), the change in volume is dependent solely on the radius change \( \frac{dV}{dt} = \frac{2}{3} \pi r h \frac{dr}{dt} \). This highlights the sphere of influence the radius has when height is constant.
- **Dynamic radius and height**: When both \( r \) and \( h \) change, the equation uses the product rule to integrate both changes: \( \frac{dV}{dt} = \frac{1}{3} \pi (2r \cdot h \cdot \frac{dr}{dt} + r^2 \cdot \frac{dh}{dt}) \).
Each of these scenarios offers a distinct perspective of the intricate dance between geometry and calculus, captured through the cone's volume equation.
- **Constant radius**: When \( r \) is fixed, the equation simplifies to study the effect of changing height alone on the volume. Here, \( \frac{dV}{dt} = \frac{1}{3} \pi r^2 \frac{dh}{dt} \) shows direct proportionality between the cone's height change rate and the volume change rate.
- **Constant height**: With a fixed \( h \), the change in volume is dependent solely on the radius change \( \frac{dV}{dt} = \frac{2}{3} \pi r h \frac{dr}{dt} \). This highlights the sphere of influence the radius has when height is constant.
- **Dynamic radius and height**: When both \( r \) and \( h \) change, the equation uses the product rule to integrate both changes: \( \frac{dV}{dt} = \frac{1}{3} \pi (2r \cdot h \cdot \frac{dr}{dt} + r^2 \cdot \frac{dh}{dt}) \).
Each of these scenarios offers a distinct perspective of the intricate dance between geometry and calculus, captured through the cone's volume equation.
Other exercises in this chapter
Problem 4
In Exercises \(1-4,\) find the linearization \(L(x)\) of \(f(x)\) at \(x=a\) $$ f(x)=\sqrt[3]{x}, \quad a=-8 $$
View solution Problem 4
Find \(d y / d x\) in Exercises \(1-10\) $$ y=\sqrt[4]{5 x} $$
View solution Problem 4
In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(f^{\prime}(g(x)) g^{\prime}(x)\). $$ y=\cos u, \quad u=-x / 3 $$
View solution Problem 4
In Exercises \(1-12,\) find \(d y / d x\) $$ y=x^{2} \cot x-\frac{1}{x^{2}} $$
View solution