Introduction to Related Rates - AP Calculus BC Study Guide

The Core Idea: Introduction to Related Rates

Related rates problems involve situations where multiple quantities are changing over time. The core idea is to determine the rate of change of one of these quantities by using the known rate of change of another. The fundamental challenge is that we are not given a direct relationship between the rates themselves, but rather an equation that relates the underlying quantities (e.g., a geometric formula for area or volume, or the Pythagorean theorem).

To connect the rates of change, we differentiate the entire equation that relates the quantities with respect to a common variable, almost always time ($t$). This process of implicit differentiation relies heavily on the chain rule. By applying the chain rule, we generate a new equation that explicitly links the rates of change of the variables, allowing us to solve for an unknown rate using the given information.

Key Rule: The Chain Rule in Context

The foundation of every related rates problem is the chain rule. In this context, we treat all variables as implicit functions of time, $t$. When we differentiate an expression with respect to time, the chain rule requires us to multiply by the derivative of that variable with respect to time.

For a variable $x$ that is a function of time, its rate of change is $\frac{dx}{dt}$. The chain rule, applied to an expression involving $x$, is as follows:

$$\frac{d}{dt}[f(x)]=f^{\prime }(x)\cdot \frac{dx}{dt}$$

For example, if an equation involves the term $r^2$, where $r$ is a function of time, its derivative with respect to time is not simply $2r$. Applying the chain rule gives:

$$\frac{d}{dt}[r^2]=2r\cdot \frac{dr}{dt}$$

Similarly, for a volume formula $V=\frac{4}{3}\pi r^3$, differentiating with respect to time $t$ yields:

$$\frac{dV}{dt}=\frac{d}{dt}\left[\frac{4}{3}\pi r^3\right]=\frac{4}{3}\pi \cdot (3r^2)\cdot \frac{dr}{dt}=4\pi r^2\frac{dr}{dt}$$

This application of the chain rule is the essential mechanical step that transforms an equation relating quantities into an equation relating their rates of change.

Understanding the Procedure

The Essential Knowledge for this topic outlines a clear, systematic procedure for solving related rates problems. The goal is to find an unknown rate of change by leveraging a known rate of change.

The Method:

1. Identify Quantities and Rates: Read the problem carefully to identify all quantities that are changing and all quantities that are constant. Assign variables to the changing quantities and note their rates of change. Identify which rate is given and which rate you need to find. Rates of change are derivatives with respect to time (e.g., $\frac{dr}{dt}$, $\frac{dV}{dt}$, $\frac{dx}{dt}$).

2. Find a Relating Equation: Determine an equation that connects the variables from Step 1. This equation should hold true for any moment in time. Often, this is a geometric formula (e.g., area, volume, Pythagorean theorem) or a trigonometric relationship.

3. Differentiate with Respect to Time: Differentiate both sides of the equation with respect to time ($t$). This is the crucial step where the chain rule is applied to every variable that is a function of time. Any variable that represents a constant will have a derivative of zero.

4. Substitute and Solve: After differentiating, substitute all known values for the variables and their rates of change into the resulting derivative equation. Solve algebraically for the desired unknown rate of change. This step should only be performed after the differentiation is complete.

Core Concepts & Rules

• The Chain Rule is Foundational: The chain rule is the primary mathematical tool used to solve related rates problems. It allows us to differentiate an equation of related quantities with respect to time.

• The Goal is to Find a Rate: The objective in these problems is always to calculate the rate of change of one quantity by using the known rate of change of a different, but related, quantity.

• An Equation is the Starting Point: The first step in the procedure is to establish an equation that relates the quantities involved in the problem (e.g., $A=\pi r^2$, $x^2+y^2=z^2$).

• Differentiate with Respect to Time: The core of the process is to take the derivative of the entire relating equation with respect to time ($t$), creating a new equation that relates the rates of change (e.g., $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$).

Step-by-Step Example 1: Basic Application

Problem: The radius of a sphere is increasing at a constant rate of 3 centimeters per minute. At the instant when the radius of the sphere is 5 centimeters, what is the rate of change of the volume of the sphere?

Step 1: Identify Quantities and Rates

• The radius $r$ is changing. The given rate is $\frac{dr}{dt}=3\text{ cm/min}$.

• The volume $V$ is changing. We need to find $\frac{dV}{dt}$.

• We are interested in the specific instant when $r=5\text{ cm}$.

Step 2: Find a Relating Equation

The equation that relates the volume and radius of a sphere is:

$$V=\frac{4}{3}\pi r^3$$

Step 3: Differentiate with Respect to Time

Differentiate both sides of the equation with respect to $t$. Remember to use the chain rule on the $r^3$ term.

$$\frac{d}{dt}[V]=\frac{d}{dt}\left[\frac{4}{3}\pi r^3\right]$$

$$\frac{dV}{dt}=\frac{4}{3}\pi \cdot (3r^2)\cdot \frac{dr}{dt}$$

$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$

Step 4: Substitute and Solve

Now, substitute the known values from Step 1 into the differentiated equation.

• $r=5$

• $\frac{dr}{dt}=3$

$$\frac{dV}{dt}=4\pi (5{)}^2(3)$$

$$\frac{dV}{dt}=4\pi (25)(3)$$

$$\frac{dV}{dt}=300\pi$$

The rate of change of the volume is $300\pi$ cubic centimeters per minute at the instant when the radius is 5 centimeters.

Step-by-Step Example 2: Exam-Style Application

Problem: A 25-foot ladder is leaning against a vertical wall. The bottom of the ladder is pulled horizontally away from the wall at a rate of 2 feet per second. At the instant when the bottom of the ladder is 7 feet from the wall, how fast is the top of the ladder sliding down the wall?

Step 1: Identify Quantities and Rates

• Let $x$ be the horizontal distance from the wall to the bottom of the ladder. We are given $\frac{dx}{dt}=2\text{ ft/s}$.

• Let $y$ be the vertical distance from the ground to the top of the ladder. We need to find $\frac{dy}{dt}$.

• The length of the ladder is constant: $L=25\text{ ft}$.

• The specific instant is when $x=7\text{ ft}$.

Step 2: Find a Relating Equation

The wall, the ground, and the ladder form a right triangle. The Pythagorean theorem relates the quantities:

$$x^2+y^2=L^2$$

$$x^2+y^2=25^2$$

Step 3: Differentiate with Respect to Time

Differentiate both sides of the equation with respect to $t$. Apply the chain rule to both $x^2$ and $y^2$. The derivative of the constant $25^2$ is 0.

$$\frac{d}{dt}[x^2+y^2]=\frac{d}{dt}[25^2]$$

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$

Step 4: Substitute and Solve

We need to solve for $\frac{dy}{dt}$. We know $x=7$ and $\frac{dx}{dt}=2$. However, we also need the value of $y$ at this instant. We can find it using the original relating equation:

$$x^2+y^2=25^2$$

$$7^2+y^2=25^2$$

$$49+y^2=625$$

$$y^2=576⟹y=24\text{ ft}$$

Now substitute all known values into the differentiated equation:

$$2(7)(2)+2(24)\frac{dy}{dt}=0$$

$$28+48\frac{dy}{dt}=0$$

$$48\frac{dy}{dt}=-28$$

$$\frac{dy}{dt}=-\frac{28}{48}=-\frac{7}{12}$$

The top of the ladder is sliding down the wall at a rate of $\frac{7}{12}$ feet per second. The negative sign indicates that the distance $y$ is decreasing.

Using Your Calculator

Related rates problems are primarily analytical and must be solved by hand. The procedure requires finding a relating equation and differentiating it symbolically using the chain rule, which a calculator cannot perform.

The calculator's role is limited to performing the final arithmetic calculation after all calculus steps are complete.

• Use for Calculation: In Example 2, if the numbers were less convenient (e.g., $\frac{dy}{dt}=-\frac{2(7.1)(2.3)}{2\sqrt{25^2-7.1^2}}$), you would use your calculator to compute the final numerical value.

• Not for Differentiation: You cannot use calculator functions like nDeriv or $d/dx$ to solve the problem, as these functions compute the derivative at a single point and cannot be used to find the symbolic derivative equation that relates the rates.

AP Exam Quick Hit

Common Question Types

• Geometric Contexts: You are given a geometric shape (circle, sphere, cone, cylinder) and asked to find the rate of change of one of its properties (area, volume, radius, height) given the rate of change of another.

  • Example: Water is draining from a conical tank at a rate of 3 m^3/s. Find the rate at which the water level is falling when the water is 2 m deep.

• Right Triangle Contexts: You are given a scenario that can be modeled by a right triangle where the side lengths are changing. This often involves ladders, shadows, or moving objects (cars, planes, boats).

  • Example: A person 6 ft tall walks away from a 15 ft lamppost at 5 ft/s. At what rate is the tip of their shadow moving when they are 10 ft from the post?

Common Mistakes

• Substituting Before Differentiating: A critical error is to substitute a value that is changing (like $r=5$ in Example 1) into the relating equation before differentiating. This treats a variable as a constant, making its derivative zero and leading to an incorrect result. Only values that are constant for the entire problem (like the ladder's length) can be substituted before differentiation.

• Forgetting the Chain Rule: A common mistake is to differentiate an expression like $x^2$ with respect to $t$ and write $2x$ instead of the correct $2x\frac{dx}{dt}$. Every variable that is a function of time requires the chain rule.

• Omitting Units or Using Incorrect Signs: Forgetting to include units in the final answer (e.g., cm/s, ft^3/min). Also, failing to recognize that a decreasing quantity corresponds to a negative rate of change. If a ladder is sliding down a wall, its $\frac{dy}{dt}$ must be negative.

• Using an Incorrect Relating Equation: The entire problem depends on starting with the correct formula (e.g., volume of a cone vs. a cylinder, Pythagorean theorem). An error in this initial setup makes the rest of the work incorrect.