Related rates problems ask, “If one quantity changes over time, how quickly does a connected quantity change at the same instant?” Therefore, the topic sits at the heart of differential calculus, appearing in AP® Calculus AB (CHA-3.E.1) and BC alike. Section 4.5 solving related rates problems builds on earlier derivative skills—especially the chain rule and implicit differentiation.

This article first presents a five-step roadmap, then walks through several related rates sample problems. Each idea is paired with a clear example, so every reader can move from confusion to confidence.

The Core Strategy for 4.5 Solving Related Rates Problems

Five-Step Roadmap

1. Read carefully and assign variables. Label every changing quantity.
2. Sketch a diagram. A quick picture often reveals hidden relationships.
3. Write an equation that links the variables. Use geometry, trigonometry, or science formulas.
4. Differentiate implicitly with respect to time t. Remember that every variable is a function of time.
5. Substitute known values, solve, and attach correct units.

Tip: Numbers should enter the work only after differentiating.

Worked Example #1 – “Sliding Ladder”

A 10-ft ladder leans against a vertical wall. The bottom slides away at \dfrac{dx}{dt}=3\text{ ft/s}. How fast is the top descending when the bottom is 6 ft from the wall?

Step 1. Variables

Let

x(t) = distance from wall to ladder’s foot

y(t) = height of ladder’s top

Step 2. Diagram

Right triangle with legs x and y, hypotenuse 10.

Step 3. Equation

x^{2}+y^{2}=10^{2}

Step 4. Differentiate

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

Divide by 2:

x\dfrac{dx}{dt}+y\dfrac{dy}{dt}=0

Step 5. Substitute and solve

When x=6, find y using Pythagorean theorem:

6^{2}+y^{2}=100;\Rightarrow;y=\sqrt{64}=8\text{ ft}

Plug in:

6(3)+8\dfrac{dy}{dt}=0

18+8\dfrac{dy}{dt}=0

\dfrac{dy}{dt}=-\dfrac{18}{8}=-2.25\text{ ft/s}

The ladder’s top descends at 2.25 ft/s. The negative sign confirms downward motion.

Must-Know Related Rates Formulas

Many questions rely on geometry classics. Memorizing them speeds work, yet understanding when each variable depends on time is even more valuable.

• Circle area: A=\pi r^{2}
• Sphere volume: V=\dfrac{4}{3}\pi r^{3}
• Cone volume: V=\dfrac{1}{3}\pi r^{2}h
• Pythagorean theorem: x^{2}+y^{2}=z^{2}

Notice that r, h, x, and y usually change with time, so their derivatives appear after differentiation.

Worked Example #2 – “Inflating Balloon”

Air is pumped into a spherical balloon at \dfrac{dV}{dt}=100\text{ cm}^{3}/\text{s}. How fast is the radius increasing when r=5\text{ cm}?

Equation: V=\dfrac{4}{3}\pi r^{3}

Differentiate:

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

Solve for \dfrac{dr}{dt}:

\dfrac{dr}{dt}=\dfrac{1}{4\pi r^{2}}\dfrac{dV}{dt}=\dfrac{100}{4\pi(5)^{2}}=\dfrac{100}{100\pi}= \dfrac{1}{\pi}\text{ cm/s}

Therefore, the radius grows at approximately 0.318 cm/s when it is 5 cm wide.

Recognizing When to Use Related Rates (Graph Sense Included)

Verbal Clues

Phrases such as “at what rate,” “how fast,” or “find \dfrac{dy}{dt} when…” immediately signal a related rates question.

Visual Clues

Graphs showing changing slopes or tangent lines also hint that quantities vary with time. For instance, the slope of a height-versus-time curve equals \dfrac{dh}{dt}.

Worked Example #3 – “Filling a Conical Tank”

A water tank is an inverted cone with radius 3 m and height 9 m. Water pours in at \dfrac{dV}{dt}=2\text{ m}^{3}/\text{min}. Find \dfrac{dh}{dt} when the water depth is 4 m.

1. Variables

h(t) = water depth

r(t) = surface radius

2. Diagram

A smaller cone of water inside the large cone.

3. Relationship between r and h

Similar triangles give \dfrac{r}{h}=\dfrac{3}{9}=\dfrac{1}{3}\Rightarrow r=\dfrac{h}{3}.

4. Volume formula

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

Substitute r: V=\dfrac{1}{3}\pi\left(\dfrac{h}{3}\right)^{2}h=\dfrac{\pi h^{3}}{27}

5. Differentiate

\dfrac{dV}{dt}=\dfrac{\pi}{9}h^{2}\dfrac{dh}{dt}

6. Solve when h = 4 m

2=\dfrac{\pi}{9}(4)^{2}\dfrac{dh}{dt};\Longrightarrow;2=\dfrac{16\pi}{9}\dfrac{dh}{dt}

\dfrac{dh}{dt}=\dfrac{18}{16\pi}=\dfrac{9}{8\pi}\text{ m/min}\approx0.36\text{ m/min}

Tips, Traps, and Unit Checks

• Always track units. Mixing centimeters and meters can triple errors.
• Differentiate first, substitute later. Therefore constants stay constant.
• Express linked variables before differentiating. Example: in the conical tank, write r=\dfrac{h}{3} early.
• Finally, confirm the sign of each answer. A negative rate often signals a quantity is shrinking.

Quick Reference Chart: Key Vocabulary & Definitions

Term	Meaning in Related Rates Problems
Related Rates	Technique of finding an unknown rate by linking it to known rates through an equation.
Implicit Differentiation	Differentiating an equation with respect to time when variables are functions of time.
\dfrac{dA}{dt}, \dfrac{dV}{dt}	Notation for rate of change of area, volume, etc., with respect to time.
Instantaneous Rate	The exact rate at a specific moment, found using derivatives.
Chain Rule	The derivative rule that underpins every related-rates calculation: \dfrac{d}{dt}f(g(t))=f'(g(t))\cdot g'(t)

Conclusion

Mastering 4.5 solving related rates problems boils down to one repeatable method:

1. Translate words into variables and a diagram.
2. Connect those variables with a solid equation.
3. Apply the chain rule through implicit differentiation.
4. Substitute, solve, and label units.

With practice, the provided related rates formulas and the worked examples above will turn intimidating word problems into routine exercises. Keep the five-step plan nearby, double-check units, and success on the AP® exam will follow.

Sharpen Your Skills for AP® Calculus AB-BC

Are you preparing for the AP® Calculus exam? We’ve got you covered! Try our review articles designed to help you confidently tackle real-world math problems. You’ll find everything you need to succeed, from quick tips to detailed strategies. Start exploring now!

• 4.4 Introduction to Related Rates
• 4.6 Approximating Values of a Function Using Local Linearity and Linearization

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