Solving Related Rates Problems - AP Calculus BC Study Guide

The Core Idea: Solving Related Rates Problems

A "related rates" problem involves finding the rate of change of one quantity by using the known rate of change of one or more other quantities. The core idea is that multiple variables in a given situation are changing over time, but they are connected or "related" to each other through an underlying equation, often a geometric formula. For example, as the radius of a sphere increases, its volume also increases, but not at the same rate. The rates of change of the radius ($\frac{dr}{dt}$) and the volume ($\frac{dV}{dt}$) are related.

This topic uses the principles of implicit differentiation to find these unknown rates. Since all the variables involved are considered to be functions of time ($t$), we differentiate an equation that relates the variables with respect to time. This process, which relies heavily on the chain rule, creates a new equation that relates the rates of change of those variables, allowing us to solve for an unknown rate.

Key Concepts: The Chain Rule and Implicit Differentiation

The foundational principles for solving related rates problems are the chain rule and implicit differentiation. Since the variables in the problem are functions of time, we must differentiate with respect to the variable $t$.

Essential Knowledge CHA-3.F.1 & CHA-3.F.4: The chain rule is the basis for solving related rates problems, and the derivative of each variable with respect to time is found using implicit differentiation.

When we differentiate an expression with respect to time, any variable that is not $t$ itself is treated as a function of $t$. Applying the chain rule results in a derivative factor for each variable.

For a variable $x$ which is a function of time, $x(t)$:

• The derivative of $x$ with respect to $t$ is $\frac{dx}{dt}$.

• The derivative of $x^n$ with respect to $t$ is $n{x}^{n-1}\cdot \frac{dx}{dt}$.

Consider the Pythagorean theorem relating variables $x$, $y$, and $z$, where each is a function of time:

$$x^2+y^2=z^2$$

To find the relationship between their rates of change, we differentiate the entire equation with respect to $t$:

$$\frac{d}{dt}(x^2+y^2)=\frac{d}{dt}(z^2)$$

Applying the power rule and chain rule to each term implicitly gives:

$$2x\cdot \frac{dx}{dt}+2y\cdot \frac{dy}{dt}=2z\cdot \frac{dz}{dt}$$

This new equation relates the variables ($x,y,z$) and their rates of change ($\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}$).

Understanding Variables as Functions of Time

Essential Knowledge CHA-3.F.2: In a related rates problem, the variables are functions of time.

This is the most critical conceptual point in related rates. A variable like $r$ for radius is not just a number; it is a function $r(t)$ whose value changes as time $t$ changes. Similarly, volume $V$ is a function $V(t)$. This is why the chain rule is necessary.

When we see an equation like the volume of a sphere, $V=\frac{4}{3}\pi r^3$, we must understand it as a relationship between two functions of time: $V(t)=\frac{4}{3}\pi [r(t){]}^3$.

When we differentiate with respect to $t$, we are asking, "How does $V$ change as $t$ changes?"

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

The derivative of the left side is simply $\frac{dV}{dt}$. For the right side, $\frac{4}{3}\pi$ is a constant. We must use the chain rule on $[r(t){]}^3$. The "outside" function is $(\cdot {)}^3$ and the "inside" function is $r(t)$.

• The derivative of the outside is $3(\cdot {)}^2$, so we have $3[r(t){]}^2$.

• The derivative of the inside, $r(t)$, with respect to $t$ is $\frac{dr}{dt}$.

Combining these gives the final differentiated equation:

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

Forgetting that $r$ is a function of $t$ is the most common source of error and leads to omitting the $\frac{dr}{dt}$ factor.

Core Concepts & Rules

• The Goal: The objective of a related rates problem is to determine an unknown rate of change from one or more given rates of change.

• Variables are Functions of Time: All quantities that are changing in the problem must be treated as functions of time, $t$.

• Relating Equation: The first step is to establish an equation that relates the variables. This equation often comes from a geometric formula (e.g., area, volume, Pythagorean theorem).

• Implicit Differentiation: The relating equation is differentiated implicitly with respect to time, $t$.

• The Chain Rule is Essential: Every time a variable (that is a function of $t$) is differentiated, the chain rule requires you to multiply by its derivative with respect to $t$ (e.g., $\frac{d}{dt}(x^2)=2x\frac{dx}{dt}$).

• Substitute and Solve: After differentiating, substitute the known instantaneous values for the variables and their rates of change into the derivative equation to solve for the unknown rate.

Step-by-Step Example 1: Expanding Sphere

Problem: Air is being pumped into a spherical balloon at a rate of 100 cubic centimeters per second. How fast is the radius of the balloon increasing at the instant when the radius is 5 cm?

Step 1: Identify Given Information and the Goal

• The rate of change of volume is given: $\frac{dV}{dt}=100{\text{ cm}}^3/\text{s}$.

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

• The goal is to find the rate of change of the radius, $\frac{dr}{dt}$.

Step 2: Write the Relating Equation

The variables are volume ($V$) and radius ($r$). The geometric formula that relates them for a sphere is:

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

Step 3: Differentiate with Respect to Time ($t$)

Differentiate both sides of the equation with respect to $t$, remembering that both $V$ and $r$ are functions of $t$.

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

Applying the chain rule to the $r^3$ term:

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

Simplify the expression:

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

Step 4: Substitute Known Values and Solve

Now, substitute the given values ($\frac{dV}{dt}=100$ and $r=5$) into the differentiated equation.

$$100=4\pi (5{)}^2\frac{dr}{dt}$$

$$100=4\pi (25)\frac{dr}{dt}$$

$$100=100\pi \frac{dr}{dt}$$

Solve for $\frac{dr}{dt}$:

$$\frac{dr}{dt}=\frac{100}{100\pi }=\frac{1}{\pi }$$

Conclusion: The radius is increasing at a rate of $\frac{1}{\pi }$ centimeters per second at the instant the radius is 5 cm.

Step-by-Step Example 2: Sliding Ladder

Problem: A 25-foot ladder is leaning against a vertical wall. The bottom of the ladder is sliding away from the wall at a constant rate of 3 feet per second. At the instant 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 Given Information and the Goal

• Let $x$ be the distance from the base of the wall to the bottom of the ladder.

• Let $y$ be the height of the top of the ladder on the wall.

• The ladder length is constant at 25 feet.

• The rate at which the bottom is sliding away is $\frac{dx}{dt}=3\text{ ft/s}$.

• We are interested in the instant when $x=7\text{ ft}$.

• The goal is to find the rate at which the top is sliding down, $\frac{dy}{dt}$.

Step 2: Write the Relating Equation

The wall, the ground, and the ladder form a right triangle. The Pythagorean theorem relates $x$, $y$, and the ladder's length.

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

Step 3: Differentiate with Respect to Time ($t$)

Differentiate the equation implicitly with respect to $t$.

$$\frac{d}{dt}(x^2+y^2)=\frac{d}{dt}(25^2)$$

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

Note that the derivative of the constant $25^2$ is zero.

Step 4: Find Missing Variable Values at the Specific Instant

The derivative equation contains $x$, $y$, $\frac{dx}{dt}$, and $\frac{dy}{dt}$. We know $x=7$ and $\frac{dx}{dt}=3$, but we need the value of $y$ at that same instant. Use the original relating equation to find it.

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

$$(7{)}^2+y^2=(25{)}^2$$

$$49+y^2=625$$

$$y^2=576⟹y=24\text{ ft}$$

Step 5: Substitute Known Values and Solve

Substitute $x=7$, $y=24$, and $\frac{dx}{dt}=3$ into the differentiated equation.

$$2(7)(3)+2(24)\frac{dy}{dt}=0$$

$$42+48\frac{dy}{dt}=0$$

$$48\frac{dy}{dt}=-42$$

$$\frac{dy}{dt}=-\frac{42}{48}=-\frac{7}{8}$$

Conclusion: The top of the ladder is sliding down the wall at a rate of $\frac{7}{8}$ feet per second. The negative sign indicates that the height $y$ is decreasing.

Using Your Calculator

Related rates problems are fundamentally analytical. The process of identifying variables, setting up the relating equation, and differentiating with respect to time must be done by hand. A calculator is not used to find the derivative in this context.

The primary use for a calculator in these problems is for numerical computation at the final step.

• Arithmetic: After substituting all known values into the differentiated equation, a calculator can be used to solve for the unknown rate, especially if the numbers are complex. For example, in the ladder problem, you could compute $-42/48$.

• Finding Instantaneous Values: A calculator can be used to solve for a missing variable value if the algebra is difficult (e.g., solving $y=\sqrt{625-49}$).

You cannot use a calculator's numerical derivative functions (like nDeriv) to solve the core part of the problem, as the process requires symbolic differentiation (implicit differentiation with the chain rule) to establish the relationship between the rates.

AP Exam Quick Hit

Common Question Types

• Geometric Contexts: A question will describe a situation involving a standard geometric shape (sphere, cone, cylinder, rectangle) and ask for a rate of change.

  • Example: "The radius of a cone is increasing at 2 in/s while its height is decreasing at 1 in/s. Find the rate of change of the volume when the radius is 4 inches and the height is 6 inches."

• Right Triangle Contexts: A problem based on the Pythagorean theorem, often involving moving objects, ladders, or shadows.

  • Example: "Two cars leave an intersection at the same time, one traveling north and the other east. Given their speeds, find the rate at which the distance between them is increasing after 2 hours."

• Angle Contexts: A problem involving trigonometry where an angle is changing.

  • Example: "A camera on the ground is tracking a rising rocket. Given the rocket's vertical speed, find the rate of change of the camera's angle of elevation at a specific height."

Common Mistakes

• Substituting Before Differentiating: Plugging in an instantaneous value (like $r=5$) into the geometric formula before taking the derivative with respect to $t$. This incorrectly treats a variable as a constant, making its derivative zero.

• Forgetting the Chain Rule: The most frequent error is omitting the $\frac{d(\text{variable})}{dt}$ factor after differentiating. For example, writing $\frac{d}{dt}(V)=4\pi r^2$ instead of $\frac{d}{dt}(V)=4\pi r^2\frac{dr}{dt}$.

• Incorrect Sign: Forgetting that a decreasing quantity (like a shrinking radius or a sliding-down ladder) must have a negative rate of change. If the problem states a quantity is decreasing, you must assign a negative sign to its rate.

• Product/Quotient Rule Errors: For formulas involving multiple variables (like the volume of a cone, $V=\frac{1}{3}\pi r^{2}h$), students often forget to use the product rule when differentiating the $r^{2}h$ term.

• Ignoring Constants: Mistakenly differentiating a known constant in the problem. For example, in the ladder problem, the ladder's length (25 ft) is constant, so its rate of change is zero.