Related Rates Problems

Why This Matters

Related rates problems are where calculus meets the real world—and they're a favorite on AP exams precisely because they test whether you can apply derivative rules, not just compute them. These problems require you to connect multiple changing quantities through an equation, then use implicit differentiation and the chain rule to find how fast one quantity changes given information about another. You're being tested on your ability to translate a physical scenario into mathematical relationships, identify which variables depend on time, and execute the differentiation correctly.

The core skills here—implicit differentiation, the chain rule, and geometric reasoning—appear throughout calculus. Whether it's a ladder sliding down a wall or water draining from a tank, the underlying principle is the same: if two quantities are related by an equation, their rates of change are also related. Don't just memorize problem types—understand why you set up each equation and how the chain rule connects $\frac{dV}{dt}$ to $\frac{dr}{dt}$ or $\frac{dh}{dt}$. That conceptual understanding is what separates a 3 from a 5.

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Pythagorean Relationship Problems

These problems involve right triangles where two sides change over time while satisfying $a^2 + b^2 = c^2$. The key insight is that differentiating this constraint links the rates of change of perpendicular distances.

Ladder Sliding Down a Wall

• The Pythagorean theorem $x^2 + y^2 = L^2$ relates the distance from the wall ($x$), height on the wall ($y$), and ladder length ($L$)
• Implicit differentiation gives $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ since $L$ is constant—note the rates have opposite signs
• The ladder accelerates as it falls; $\frac{dy}{dt}$ approaches negative infinity as $y \to 0$, a common exam trick question

Distance Between Two Moving Objects

• Set up coordinates so each object's position is a function of time: $(x_1(t), y_1(t))$ and $(x_2(t), y_2(t))$
• The distance formula $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ must be differentiated using the chain rule
• Simplify before differentiating when possible—sometimes squaring both sides ($D^2 = ...$) avoids messy square root derivatives

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Similar Triangle Problems

When objects cast shadows or light creates geometric relationships, similar triangles provide the equation you need. The ratios of corresponding sides remain equal, giving you a proportion to differentiate.

Shadow Length Problems

• Similar triangles relate the height of the light source, object height, and shadow length through proportional ratios
• Set up the proportion carefully: $\frac{H}{x+s} = \frac{h}{s}$ where $H$ is light height, $h$ is object height, $s$ is shadow length, and $x$ is distance from light
• Cross-multiply before differentiating—this eliminates fractions and makes implicit differentiation cleaner

Angle Change in a Triangle with Moving Sides

• The Law of Cosines $c^2 = a^2 + b^2 - 2ab\cos(\theta)$ relates sides and angles when you don't have a right triangle
• Differentiate implicitly to connect $\frac{d\theta}{dt}$ to the rates of change of the sides
• Trigonometric derivatives appear here: remember that $\frac{d}{dt}[\cos(\theta)] = -\sin(\theta)\frac{d\theta}{dt}$

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Spherical Volume Problems

Spheres appear constantly because $V = \frac{4}{3}\pi r^3$ creates a clean relationship between volume and radius. The chain rule shows that $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$—notice the surface area appears naturally.

Changing Radius of a Sphere or Circle

• For spheres, $V = \frac{4}{3}\pi r^3$ differentiates to $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$; for circles, $A = \pi r^2$ gives $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$
• The coefficient $4\pi r^2$ is the surface area—this isn't coincidence; volume changes occur through the surface
• Watch your units: if $r$ is in cm and $t$ is in seconds, $\frac{dV}{dt}$ is in $\text{cm}^3/\text{s}$

Expanding or Contracting Balloon

• Same formula as sphere problems: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$, but context matters for sign
• Air pumped in means $\frac{dV}{dt} > 0$; air escaping means $\frac{dV}{dt} < 0$
• Rate of radius change decreases as the balloon grows—even with constant $\frac{dV}{dt}$, larger $r$ means smaller $\frac{dr}{dt}$

Growing Snowball

• Identical setup to balloon problems, but snow accumulation is typically given as a volume rate
• Surface area matters if the problem describes snow sticking to the surface at a rate per unit area
• Common variation: "radius grows at constant rate"—then you're finding $\frac{dV}{dt}$, which increases over time

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Conical Tank Problems

Cones are trickier than cylinders because radius and height are related by the tank's geometry. You must use similar triangles to eliminate one variable before differentiating.

Filling or Draining Tanks (Cylindrical, Conical, or Spherical)

• For cylinders, $V = \pi r^2 h$—if the radius is constant, this simplifies to $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$
• For cones, the ratio $\frac{r}{h}$ is fixed by the tank's shape; substitute $r = kh$ to get $V = \frac{1}{3}\pi k^2 h^3$
• Differentiate the simplified formula: $\frac{dV}{dt} = \pi k^2 h^2 \frac{dh}{dt}$, then solve for the unknown rate

Water Leaking from an Inverted Conical Tank

• "Inverted cone" means vertex at bottom—water forms a smaller cone similar to the tank
• Find the ratio $\frac{r}{h} = \frac{R}{H}$ where $R$ and $H$ are the tank's dimensions
• Height changes faster when water is low (small $h$) because the cross-sectional area is smaller—this is a frequent conceptual question

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Cylindrical Volume Problems

Cylinders with changing dimensions require the product rule since $V = \pi r^2 h$ involves two variables. When both $r$ and $h$ change, you get $\frac{dV}{dt} = \pi(2rh\frac{dr}{dt} + r^2\frac{dh}{dt})$.

Volume Change in a Cylinder with Varying Dimensions

• Product rule is essential: $\frac{dV}{dt} = \pi \left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right)$ when both radius and height vary
• If only height changes (fixed radius), this simplifies to $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$
• Check for constraints: sometimes $r$ and $h$ are related (e.g., $r = 2h$), requiring substitution before differentiation

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Quick Reference Table

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Self-Check Questions

1. Both the ladder problem and the distance-between-objects problem use the Pythagorean theorem. What's the key difference in how you set up the differentiation for each?

2. In a conical tank problem, why must you express $r$ in terms of $h$ (or vice versa) before differentiating? What geometric principle allows you to do this?

3. Compare a balloon being inflated at a constant volume rate to a snowball whose radius grows at a constant rate. In which scenario does $\frac{dV}{dt}$ increase over time, and why?

4. An FRQ asks: "Explain why the water level in a conical tank drops faster when the tank is nearly empty than when it's nearly full, even if water drains at a constant rate." How would you answer this using your related rates setup?

5. For a cylinder where both radius and height change over time, write out $\frac{dV}{dt}$ and identify which differentiation rule(s) you used. How does this differ from the sphere case?