AP Calculus AB/BC · Unit 4 · Topic 4.4
Introduction to Related Rates
Connect rates of variables constrained by one geometric or physical relationship.
1. Topic Focus
Interpret derivatives as rates in context, connect motion quantities, solve related-rate models, linearize, and evaluate indeterminate limits.
This topic: Connect rates of variables constrained by one geometric or physical relationship.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
Differentiate x²+y²=25 with respect to time to connect dx/dt and dy/dt.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
1. What Makes Rates “Related”?
In a related-rates problem, two or more quantities change with respect to the same independent variable, usually time, and an equation constrains those quantities. Differentiating the constraint creates an equation connecting their instantaneous rates.
The values \(x\) and \(y\) describe the state at one instant; \(dx/dt\) and \(dy/dt\) describe how that state is changing.
2. Every Changing Quantity Is a Function of Time
Even when notation is shortened to \(x\), \(r\), or \(V\), remember that changing variables mean \(x(t)\), \(r(t)\), and \(V(t)\). This hidden time dependence is why the chain rule produces rate factors:
Forgetting the factor \(dr/dt\) treats the variable as though it depended directly on itself rather than on time.
3. Define Variables and Directions First
Assign a symbol, meaning, and unit to every changing quantity. When geometry is involved, draw and label a diagram. Choose positive directions before assigning signs: increasing radius, rising height, or distance away from a wall is commonly positive; shrinking, falling, or moving toward a reference point is commonly negative.
4. Separate Values from Rates
| Information | Example | Role |
|---|---|---|
| Instantaneous value | \(r=5\text{ cm}\) | Describes size at the requested instant |
| Known rate | \(dr/dt=2\text{ cm/s}\) | Describes how a given quantity changes |
| Requested rate | \(dV/dt\) | The derivative to solve for |
| Constant | A fixed ladder length \(L=13\text{ ft}\) | Its time derivative is zero |
A length value and a length rate have different units and cannot replace one another.
5. Write One Relationship among the Quantities
Choose an equation containing the requested quantity and quantities whose values or rates are known. Common relationships include:
| Situation | Relationship |
|---|---|
| Circle | \(A=\pi r^2\) |
| Sphere | \(V=\frac43\pi r^3\) |
| Rectangle | \(A=\ell w\) |
| Right triangle | \(x^2+y^2=z^2\) |
| Similar triangles | A proportion between corresponding sides |
Do not add unnecessary variables if a constant already describes a fixed quantity.
6. Differentiate with Respect to Time
Apply the chain, product, or quotient rule to the entire relationship. Typical patterns are
Constants differentiate to zero. If \(x^2+y^2=L^2\) and \(L\) is fixed, then
7. Differentiate Before Substituting Instantaneous Values
A value such as \(r=5\) is true only at one instant. Replacing \(r(t)\) by 5 before differentiation turns a changing radius into a constant and incorrectly removes \(dr/dt\). Preserve variable quantities through differentiation, then insert their values into the rate equation.
8. Use the Original Relationship to Find Missing Values
The differentiated equation may require an instantaneous value not stated directly. Find it from the original geometric or physical relationship. For example, if a fixed 13-foot ladder has base distance \(x=5\), then \(y=12\) follows from \(x^2+y^2=169\). Use physically appropriate roots, such as positive lengths.
9. Interpret Signs Rather Than Erasing Them
A negative rate communicates direction. If \(dy/dt=-5/6\) feet per second for a ladder's height, the ladder top is moving downward at speed \(5/6\) feet per second. If the prompt gives “decreasing at 3,” encode the derivative as \(-3\), unless it explicitly requests only a nonnegative speed.
10. Check Units through the Rate Equation
Dimensional consistency provides a strong error check. In
length multiplied by length per time gives area per time. For sphere volume, \(r^2\,dr/dt\) gives length cubed per time. A requested area rate should never finish with only length-per-time units.
11. Rates Depend on the Current State
The same input rate can produce different output rates at different sizes. For a circle, \(dA/dt=2\pi r\,dr/dt\), so a fixed radial rate creates a larger area rate when the current radius is larger. Related rates are instantaneous relationships, not necessarily constant proportionalities.
12. A Reliable Introductory Workflow
- Draw and label the situation when useful.
- Define all changing variables as functions of time.
- Record given values, signed rates, and the requested rate.
- Write a relationship among the variables.
- Differentiate every term with respect to time.
- Use the original equation for any missing instantaneous values.
- Substitute values and rates only now.
- Solve, attach units, and interpret the sign.
13. Common Introductory Errors
- Substituting an instantaneous value before differentiating.
- Omitting \(dx/dt\), \(dr/dt\), or another chain-rule factor.
- Treating a fixed constant as a changing variable, or vice versa.
- Using an unsigned rate when the quantity is decreasing.
- Choosing an equation that does not include the requested variable.
- Using the differentiated equation to find missing lengths instead of the original relationship.
- Reporting a number without units or contextual direction.
6. Detailed Worked Example and Error Check
Example 1: Expanding circle. Let \(A(t)=\pi[r(t)]^2\). Differentiating with respect to time gives
If \(r=4\) centimeters and \(dr/dt=0.5\) centimeter per second, then
The circle's area is increasing at \(4\pi\) square centimeters per second.
Example 2: Shrinking square. A square has side \(s(t)\) and area \(A(t)=s(t)^2\). If the side is decreasing at 3 centimeters per second, then \(ds/dt=-3\). At \(s=5\),
The negative rate means the area is decreasing at 30 square centimeters per second.
Example 3: Fixed hypotenuse. Suppose \(x(t)^2+y(t)^2=100\). Differentiating gives
When \(x=6\), \(y=8\), and \(dx/dt=2\),
As one leg grows, the other must shrink to keep the hypotenuse fixed.
Example 4: Both rectangle dimensions change. For \(A=\ell w\), the product rule gives
When \(\ell=8\) centimeters, \(w=5\) centimeters, \(d\ell/dt=2\) centimeters per second, and \(dw/dt=-1\) centimeter per second,
The increasing length contributes more area than the shrinking width removes, so net area increases.
Example 5: Recover a radius rate from volume rate. A spherical balloon satisfies
If \(dV/dt=36\pi\) cubic centimeters per second when \(r=3\) centimeters, then
7. AP Reasoning Routine
Name variables and units, write the relationship before differentiating, substitute values at the correct time, and interpret the sign in context.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Set up and solve each introductory related-rates question.
(a) A square has side \(s(t)\) and area \(A(t)\). Derive the equation relating \(dA/dt\) and \(ds/dt\).
(b) Use that equation when \(s=5\) centimeters and the side increases at 3 centimeters per second.
(c) A sphere satisfies \(V=\frac43\pi r^3\). Derive the relationship between \(dV/dt\) and \(dr/dt\).
(d) Find \(dr/dt\) when \(r=2\) centimeters and \(dV/dt=32\pi\) cubic centimeters per second.
(e) A 13-foot ladder has base distance \(x\) and height \(y\). If \(x=5\), \(dx/dt=2\) feet per second, and the ladder length is fixed, find and interpret \(dy/dt\).
(f) A rectangle has \(\ell=7\) centimeters, \(w=4\) centimeters, \(d\ell/dt=1.5\) centimeters per second, and \(dw/dt=-0.5\) centimeter per second. Find \(dA/dt\).
(g) Explain precisely why replacing \(r(t)\) with \(r=2\) before differentiating \(V=\frac43\pi r^3\) produces an invalid related-rates equation.
(h) Explain how signs and units can be used to check the answer to a related-rates problem.
Check the solution
In part (a), differentiating \(A=s^2\) with respect to time gives \(dA/dt=2s\,ds/dt\). In part (b), \(dA/dt=2(5)(3)=30\) square centimeters per second, so area is increasing. In part (c), \(dV/dt=4\pi r^2\,dr/dt\). In part (d), \(32\pi=4\pi(2)^2\,dr/dt\), so \(dr/dt=2\) centimeters per second. In part (e), \(x^2+y^2=169\) gives \(y=12\) when \(x=5\). Differentiating gives \(x\,dx/dt+y\,dy/dt=0\), so \(dy/dt=-5/6\) foot per second; the ladder top moves downward. In part (f), the product rule gives \(dA/dt=\ell\,dw/dt+w\,d\ell/dt=7(-0.5)+4(1.5)=2.5\) square centimeters per second. In part (g), \(r=2\) is true only at the requested instant. Substituting it early treats radius as a constant, makes its derivative zero, and erases the needed factor \(dr/dt\). In part (h), a decreasing quantity should have a negative derivative under the chosen positive direction, while dimensions on both sides of the differentiated equation must agree; for example, a volume rate must have cubic-length-per-time units.