Derivatives are powerful tools for understanding how quantities change. These tools appear in many real-world settings, from physics to population growth. However, geometric applications also play a vital role. In AP® Calculus, “4.3 rates of change other than motion” includes problems involving shapes like cones and cylinders, such as finding the derivative of volume of a cone.
This article highlights the derivative of volume of a cone, the derivative of the volume of a cylinder, and derivative word problems. These areas demonstrate how calculus can link geometry and real-life scenarios. By understanding how dimensions change, it becomes easier to solve a variety of related rates problems on exams and beyond.
What We Review
Connecting Derivatives and Volumes
Why Volumes Matter in Rate-of-Change Problems
Volumes often depend on more than one variable. For example, a cone’s volume depends on both its radius and height. When these variables change over time, a derivative can measure how fast the volume increases or decreases. Therefore, volume-based derivative problems make an excellent case study in understanding “4.3 rates of change other than motion.”
Reviewing Essential Derivative Principles
A derivative, denoted \frac{dy}{dx} or \frac{d}{dt}, represents the instantaneous rate of change of a function. In geometry-based problems, the function is the volume, and the variables might include time, height, and radius. According to “CHA-3.C.1: The derivative can be used to solve problems involving rates of change in applied contexts,” it is crucial to connect the derivative concept to physical or geometric changes.
Derivative of Volume of a Cone
Formula for the Volume of a Cone
The standard formula for the volume of a cone is: V = \frac{1}{3}\pi r^2 h.
Here,
- r is the radius of the cone’s base,
- h is the cone’s height.
In many related rates problems, both r and h can change with respect to time t. Therefore, applying derivatives offers insight into how fast the volume is changing.
Step-by-Step Derivation
- Start with V = \frac{1}{3}\pi r^2 h.
- Differentiate both sides with respect to t. Use the product rule if r and h depend on t.
- The product rule states \frac{d}{dt}[u \cdot v] = u \frac{dv}{dt} + v \frac{du}{dt}. Here, treat u = \frac{1}{3}\pi r^2 and v = h.
- Also apply the chain rule when differentiating r^2, because r depends on t.
- Simplify and interpret the derivative result, which is often written as \frac{dV}{dt}.
Example Problem with Step-by-Step Solution
A water tank shaped like a cone has water flowing in so that the radius of the water’s surface increases at \frac{dr}{dt} = 0.2 meters per minute. The tank’s height increases at \frac{dh}{dt} = 0.3 meters per minute. At a particular instant, r = 2 meters and h = 3 meters. Find \frac{dV}{dt}, the rate at which the volume is increasing.
Step-by-Step:
- Write the volume formula: V = \frac{1}{3}\pi r^2 h.
- Differentiate with respect to t: \frac{dV}{dt} = \frac{1}{3}\pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right). This follows from applying the product rule to r^2 \cdot h.
- Plug in known values:
- r = 2,
- h = 3,
- \frac{dr}{dt} = 0.2,
- \frac{dh}{dt} = 0.3.
- Therefore,\frac{dV}{dt} = \frac{1}{3}\pi \left( 2 \cdot 2 \cdot 0.2 \cdot 3 + 2^2 \cdot 0.3 \right).
- Simplify step by step:
- First part: 2 \cdot 2 \cdot 0.2 \cdot 3 = 2.4.
- Second part: 2^2 \cdot 0.3 = 4 \cdot 0.3 = 1.2.
- Sum: 2.4 + 1.2 = 3.6.
- Multiply by \frac{1}{3}\pi: \frac{dV}{dt} = \frac{1}{3} \pi \cdot 3.6 = 1.2\pi.
- Conclude that the volume is increasing at a rate of 1.2\pi cubic meters per minute.
Derivative of the Volume of a Cylinder
Formula for the Volume of a Cylinder
A cylinder’s volume formula is V = \pi r^2 h. This formula closely relates to “4.3 rates of change other than motion” as it involves geometric changes in height and radius. In many derivative word problems, both r and h may vary over time.
Step-by-Step Derivation
- Begin with V = \pi r^2 h.
- Differentiate both sides with respect to t. Remember that r and h can both change: \frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right).
- Use the product rule as before, and properly handle constants. In this case, \pi is a constant that can remain outside the differentiation step.
Example Problem with Step-by-Step Solution
A metal cylinder has its radius increasing at 0.1 cm/s, while its height decreases at 0.05 cm/s. At a given moment, r = 5 cm and h = 10 cm. Determine \frac{dV}{dt}, the rate of change of the volume.
Step-by-Step:
- Use V = \pi r^2 h.
- Differentiate: \frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right).
- Substitute:
- r = 5,
- h = 10,
- \frac{dr}{dt} = 0.1,
- \frac{dh}{dt} = -0.05 (negative sign for decreasing height).
- Calculate carefully:
- First term: 2 \cdot 5 \cdot 0.1 \cdot 10 = 2 \cdot 0.1 \cdot 50 = 10.
- Second term: 5^2 \cdot (-0.05) = 25 \cdot (-0.05) = -1.25.
- Add both: 10 + (-1.25) = 8.75.
- Multiply by \pi: 8.75\pi.
- Conclude that the volume is changing at 8.75\pi cubic centimeters per second.
Other Common Derivative Word Problems
Overview of Related Rates
Related rates problems focus on how multiple variables change together. These problems extend beyond motion to shapes that grow or shrink over time. The main strategy is to:
- Express variables in a single equation (often geometric).
- Differentiate that equation with respect to time or another variable.
- Substitute known rates and values.
Example: Rates of Change in a Sphere
A sphere has volume V = \frac{4}{3}\pi r^3. If r changes with respect to time, the derivative word problems often look similar to those of cones and cylinders. The principle remains the same: use the chain rule to find \frac{dV}{dt}, then substitute known rates.
Putting It All Together
Derivatives involving volume formulas help solve “4.3 rates of change other than motion.” Whether it is the derivative of volume of a cone, derivative of the volume of a cylinder, or another shape, the process is straightforward:
- Identify the formula.
- Differentiate using product or chain rules.
- Plug in the known rates and values to solve for the unknown rate of change.
This approach applies to any shape where multiple dimensions vary over time. Mastering these steps also supports a deeper understanding of geometry and calculus connections.
Quick Reference Chart
| Term or Formula | Definition or Explanation |
| Derivative \frac{dy}{dx} | The instantaneous rate of change of a function with respect to x |
| Volume of a Cone V = \frac{1}{3}\pi r^2 h | Formula for the volume of a cone (depends on r and h) |
| Volume of a Cylinder V = \pi r^2 h | Formula for the volume of a cylinder (depends on r and h) |
| Related Rates | Problems where multiple variables change with respect to time |
| Product Rule \frac{d}{dt}[uv] = u\frac{dv}{dt} + v \frac{du}{dt} | Used to differentiate products of two functions |
| Chain Rule \frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t) | Used to differentiate composite functions |
Conclusion
Rates of change in cones, cylinders, and other shapes connect geometry to calculus. By using product and chain rules, it becomes possible to see how each dimension affects volume. These derivative word problems are core examples in AP® Calculus, reflecting the broader concept of “4.3 rates of change other than motion.” Students are encouraged to practice with different shapes and dimensions. Consistent practice helps build competence in recognizing and solving derivative problems in both exams and real-life contexts.
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