Introduction

This topic introduces the concept of related rates, where multiple variables change with respect to the same independent variable (usually time). The essential knowledge tells us that the chain rule is the basis for differentiating variables in a related rates problem with respect to the same independent variable and other differentiation rules, such as the product rule and the quotient rule, may also be necessary to differentiate all variables with respect to the same independent variable. This foundational understanding prepares us for solving complex related rates problems.

Understanding Related Rates

The Basic Concept

Related rates problems involve two or more variables that are changing with respect to time (or another common variable). These variables are related by an equation, and we use differentiation to find how their rates of change are connected.

Key Idea: If variables are related by an equation, their rates of change are also related.

For variables $x$ and $y$ related by an equation, we differentiate both sides with respect to time $t$ to find the relationship between $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

The Chain Rule Foundation

The chain rule is essential because we're differentiating with respect to time, but our variables are functions of time.

Chain Rule Application: If $y=f(x)$ and $x=g(t)$, then $\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}$

Example 1:

If $x^2+y^2=25$ and both $x$ and $y$ are functions of time:

Differentiating both sides with respect to $t$: $\frac{d}{dt}(x^2+y^2)=\frac{d}{dt}(25)$

Using the chain rule: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

This gives us the relationship: $x\frac{dx}{dt}+y\frac{dy}{dt}=0$

Implicit Differentiation with Respect to Time

Using the Chain Rule

When differentiating equations implicitly with respect to time, we must use the chain rule for each term containing a variable.

Example 2:

For the equation $xy=12$, find the relationship between $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

Differentiating both sides with respect to $t$: $\frac{d}{dt}(xy)=\frac{d}{dt}(12)$

Using the product rule: $x\frac{dy}{dt}+y\frac{dx}{dt}=0$

More Complex Relationships

Example 3:

If $x^{2}y+y^3=8$, find the relationship between the rates.

Differentiating with respect to $t$: $\frac{d}{dt}(x^{2}y+y^3)=\frac{d}{dt}(8)$

Using product rule for $x^{2}y$ and chain rule for $y^3$: $2x\frac{dx}{dt}\cdot y+x^2\frac{dy}{dt}+3y^2\frac{dy}{dt}=0$

Factoring: $2xy\frac{dx}{dt}+(x^2+3y^2)\frac{dy}{dt}=0$

Applying Different Differentiation Rules

Product Rule in Related Rates

When an equation contains products of variables, use the product rule.

Example 4:

For $x^2{y}^2=16$, find the rate relationship.

Differentiating: $\frac{d}{dt}(x^2{y}^2)=\frac{d}{dt}(16)$

Using the product rule: $2x\frac{dx}{dt}\cdot y^2+x^2\cdot 2y\frac{dy}{dt}=0$

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

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

$xy(y\frac{dx}{dt}+x\frac{dy}{dt})=0$

Quotient Rule in Related Rates

When equations involve quotients, the quotient rule may be necessary.

Example 5:

If $\frac{x}{y}=4$, find the rate relationship.

Method 1 - Rewrite as $x=4y$: $\frac{dx}{dt}=4\frac{dy}{dt}$

Method 2 - Use quotient rule directly: $\frac{d}{dt}\left(\frac{x}{y}\right)=\frac{d}{dt}(4)$

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

This gives: $y\frac{dx}{dt}-x\frac{dy}{dt}=0$

Since $x=4y$: $y\frac{dx}{dt}-4y\frac{dy}{dt}=0$

$\frac{dx}{dt}=4\frac{dy}{dt}$

Setting Up Related Rates Equations

Identifying Variables and Relationships

Step 1: Identify all variables that change with time

Step 2: Write an equation relating these variables

Step 3: Differentiate both sides with respect to time

Step 4: Apply chain rule, product rule, or quotient rule as needed

Common Geometric Relationships

Circle: $x^2+y^2=r^2$

Rectangle: Area $A=xy$, Perimeter $P=2x+2y$

Triangle: Area $A=\frac{1}{2}bh$

Volume of cube: $V=s^3$

Volume of sphere: $V=\frac{4}{3}\pi r^3$

Example 6:

A square is expanding. If the side length is $s$, relate the rate of change of area to the rate of change of side length.

Area: $A=s^2$

Differentiating: $\frac{dA}{dt}=2s\frac{ds}{dt}$

This shows that the rate of area change depends on both the current side length and the rate of side length change.

Working with Trigonometric Functions

Applying Chain Rule to Trigonometric Functions

When angles are changing with time, use the chain rule with trigonometric functions.

Example 7:

If $\sin (\theta )=\frac{x}{10}$ where both $\theta$ and $x$ change with time, find their rate relationship.

Differentiating both sides: $\frac{d}{dt}[\sin (\theta )]=\frac{d}{dt}\left(\frac{x}{10}\right)$

$\cos (\theta )\frac{d\theta }{dt}=\frac{1}{10}\frac{dx}{dt}$

Problem-Solving Framework

General Strategy

Recognizing When to Use Different Rules

Use Chain Rule: When differentiating composite functions like $f(g(t))$

Use Product Rule: When the equation contains products of variables

Use Quotient Rule: When the equation contains quotients of variables

Combine Rules: Many problems require multiple differentiation rules

Practice Section