Solving Related Rates Problems - AP Calculus AB Study Guide

The Core Idea: Solving Related Rates Problems

Related rates problems involve scenarios where multiple quantities are changing with respect to time. The core idea is to find the rate at which one of these quantities is changing by using a known rate of change of another quantity. This is possible because the quantities are connected or "related" by an underlying equation, often from geometry. The derivative is the fundamental tool used to solve these problems. By differentiating the equation that relates the quantities with respect to time, we create a new equation that relates their rates of change. This allows us to solve for an unknown rate using the information we are given.

The process hinges on the understanding that if two or more variables are related by an equation, and each variable is a function of time, then their rates of change are also related. The chain rule is the essential mechanism that allows us to find this relationship between the rates.

The Key Technique: Implicit Differentiation with Respect to Time

The foundation for solving related rates problems is the chain rule, applied implicitly with respect to the variable of time, $t$. When an equation relates several variables (e.g., $x$, $y$, $V$, $r$), and we know these variables are changing over time, we differentiate the entire equation with respect to $t$.

For any variable $y$ in the equation that is a function of time, its derivative with respect to time $t$ is $\frac{dy}{dt}$. The chain rule must be used for any function of that variable.

For example, consider an equation relating $V$ and $r$:

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

To find the relationship between their rates, we differentiate both sides with respect to $t$:

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

Applying the power rule in conjunction with the chain rule on the right side gives:

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

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

This new equation relates the rate of change of volume, $\frac{dV}{dt}$, to the rate of change of the radius, $\frac{dr}{dt}$.

Understanding the Problem-Solving Process

Solving a related rates problem involves a systematic process that moves from a physical description to a mathematical solution. The key is to translate the word problem into a set of equations and known values, and then apply calculus.

1. Identify Variables 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. List all known rates of change (e.g., $\frac{dx}{dt}$) and the specific instant in time (e.g., when $x=5$). Identify the rate of change you are asked to find (e.g., $\frac{dy}{dt}$). Pay close attention to units.

2. Find a Relating Equation: Determine an equation that connects the variables from Step 1. This equation often comes from a geometric formula (e.g., area, volume, Pythagorean theorem) or a given relationship in the problem.

3. Differentiate with Respect to Time: Differentiate both sides of the equation with respect to time ($t$). This is the crucial calculus step. Remember that every variable is a function of $t$, so the chain rule must be applied for each one. For example, the derivative of $x^2$ with respect to $t$ is $2x\frac{dx}{dt}$.

4. Substitute and Solve: After differentiating, substitute all the known values for the variables and their rates of change into the resulting equation. The only remaining unknown should be the rate you are asked to find. Solve algebraically for this unknown rate. Ensure your final answer includes the correct units.

Core Concepts & Rules

• The derivative is the tool used to solve related rates problems by relating a quantity's rate of change to the known rates of other quantities.

• The chain rule is the basis for the differentiation process in all related rates problems.

• When differentiating an equation with respect to time $t$, every variable is treated as a function of $t$.

• The derivative of a variable $x$ with respect to time $t$ is written as $\frac{dx}{dt}$.

• The differentiation step must be performed before substituting any instantaneous values for the variables.

• Rates of change for quantities that are decreasing must be represented by negative numbers.

Step-by-Step Example 1: Area of a Circle

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

Step 1: Identify Variables and Rates

• Known rate: The radius is increasing, so $\frac{dr}{dt}=3$ in/min.

• Known value at a specific instant: $r=5$ inches.

• Unknown rate: We need to find the rate of change of the area, $\frac{dA}{dt}$.

Step 2: Find a Relating Equation

The area of a circle is related to its radius by the formula:

$$A=\pi r^2$$

Step 3: Differentiate with Respect to Time

Differentiate both sides of the area formula with respect to time, $t$. Remember to use the chain rule on the $r^2$ term.

$$\frac{d}{dt}(A)=\frac{d}{dt}(\pi r^2)$$

$$\frac{dA}{dt}=\pi \cdot 2r\cdot \frac{dr}{dt}$$

$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$$

Step 4: Substitute and Solve

Now, substitute the known values ($r=5$ and $\frac{dr}{dt}=3$) into the differentiated equation.

$$\frac{dA}{dt}=2\pi (5)(3)$$

$$\frac{dA}{dt}=30\pi$$

The rate of change of the area is $30\pi$ square inches per minute.

Step-by-Step Example 2: Conical Tank

Problem: A water tank has the shape of an inverted circular cone with a base radius of 6 feet and a height of 12 feet. Water is being pumped into the tank at a rate of 10 cubic feet per minute. Find the rate at which the water level is rising when the water is 4 feet deep.

Step 1: Identify Variables and Rates

• Known rate: Water is being pumped in, so the volume is increasing. $\frac{dV}{dt}=10$ ft^3/min.

• Known value at a specific instant: The water depth is $h=4$ feet.

• Unknown rate: We need to find the rate at which the water level ($h$) is rising, $\frac{dh}{dt}$.

• Variables: The volume of water ($V$), the radius of the water's surface ($r$), and the height of the water ($h$) are all changing over time.

Step 2: Find a Relating Equation

The volume of a cone is given by:

$$V=\frac{1}{3}\pi r^{2}h$$

This equation has two changing variables, $r$ and $h$. We need to express one in terms of the other. We can use similar triangles from the cone's cross-section. The ratio of the radius to the height is constant for the cone.

$$\frac{r}{h}=\frac{\text{Total Radius}}{\text{Total Height}}=\frac{6}{12}=\frac{1}{2}$$

Solving for $r$, we get $r=\frac{1}{2}h$. Now substitute this into the volume formula to eliminate $r$.

$$V=\frac{1}{3}\pi {\left(\frac{1}{2}h\right)}^{2}h$$

$$V=\frac{1}{3}\pi \left(\frac{1}{4}{h}^2\right)h$$

$$V=\frac{1}{12}\pi h^3$$

Step 3: Differentiate with Respect to Time

Differentiate the simplified volume formula with respect to time, $t$.

$$\frac{d}{dt}(V)=\frac{d}{dt}\left(\frac{1}{12}\pi h^3\right)$$

$$\frac{dV}{dt}=\frac{1}{12}\pi \cdot (3h^2)\cdot \frac{dh}{dt}$$

$$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$$

Step 4: Substitute and Solve

Substitute the known values ($\frac{dV}{dt}=10$ and $h=4$) into the differentiated equation.

$$10=\frac{1}{4}\pi (4{)}^2\frac{dh}{dt}$$

$$10=\frac{1}{4}\pi (16)\frac{dh}{dt}$$

$$10=4\pi \frac{dh}{dt}$$

Now, solve for $\frac{dh}{dt}$:

$$\frac{dh}{dt}=\frac{10}{4\pi }=\frac{5}{2\pi }$$

The water level is rising at a rate of $\frac{5}{2\pi }$ feet per minute.

Using Your Calculator

Related rates problems are solved analytically. The primary steps of setting up the equation, differentiating using the chain rule, and performing algebraic substitution must be done by hand. A calculator is not used to find the derivative in these problems.

The calculator's role is limited to performing the final arithmetic calculation if a decimal approximation is required. For example, in the conical tank problem, if the question asked for a decimal answer, you would use your calculator to compute $\frac{5}{2\pi }\approx 0.796$. All supporting calculus work must be shown to receive credit on the AP Exam.

AP Exam Quick Hit

Common Question Types

• Geometric Shapes: A problem involving a simple geometric shape where one dimension is changing, affecting another property like area or volume.

  • Example: Air is pumped into a spherical balloon at a rate of 5 cm^3/min. How fast is the radius of the balloon increasing when the diameter is 20 cm?

• Pythagorean Theorem Contexts: A problem based on a right triangle where the lengths of one or more sides are changing with time.

  • Example: A 13-foot ladder is leaning against a vertical wall. The bottom of the ladder is sliding away from the wall at a rate of 2 ft/sec. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 5 feet from the wall?

• Similar Triangles (Shadows/Cones): Problems where the geometric setup involves similar triangles, such as a person walking away from a lamppost or liquid filling/draining from a conical tank.

  • Example: A 6-foot-tall person walks away from a 15-foot-tall lamppost at a rate of 4 ft/sec. At what rate is the tip of their shadow moving when they are 10 feet from the lamppost?

Common Mistakes

• Substituting Before Differentiating: Plugging in an instantaneous value (like $r=5$) into the primary equation ($A=\pi r^2$) 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 common calculus error is forgetting to multiply by the rate of change of the variable. For instance, differentiating $r^2$ with respect to $t$ and writing $2r$ instead of the correct $2r\frac{dr}{dt}$.

• Ignoring Negative Rates: Failing to assign a negative sign to a rate of change for a quantity that is decreasing. If a ladder is sliding down a wall, its height $y$ is decreasing, so $\frac{dy}{dt}$ must be negative. If a tank is draining, $\frac{dV}{dt}$ is negative.

• Forgetting Units: Not including the correct units in the final answer. Rates should have units like distance/time, area/time, or volume/time (e.g., ft/sec, in^2/min, m^3/hr).

• Mishandling Product/Quotient Rules: In problems where variables are not eliminated before differentiating (e.g., finding the rate of change of a rectangle's area $A=lw$), students may forget to use the product rule: $\frac{dA}{dt}=l\frac{dw}{dt}+w\frac{dl}{dt}$.