Related Rates Problems

Why This Matters

Related rates problems test whether you can apply derivative rules to real-world scenarios, not just compute them mechanically. You start with a physical situation where multiple quantities change over time, connect those quantities through an equation, then use implicit differentiation and the chain rule to find how fast one quantity changes given information about another.

The core skills here are implicit differentiation, the chain rule, and geometric reasoning. 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 $\frac{dh}{dt}$.

General Strategy for Any Related Rates Problem

1. Draw a diagram and label all quantities that change with time as variables (not numbers).
2. Write an equation relating the variables. This comes from geometry, physics, or whatever the problem describes.
3. Eliminate extra variables if possible. Use a constraint (like similar triangles) to reduce to fewer variables before differentiating.
4. Differentiate both sides with respect to $t$, applying the chain rule to every variable that depends on time.
5. Substitute known values (the specific measurements and rates given in the problem) after differentiating.
6. Solve for the unknown rate.

A very common mistake is plugging in numbers before differentiating. Once you substitute a constant for a variable, its derivative becomes zero, and you lose information.

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Pythagorean Relationship Problems

These problems involve right triangles where two sides change over time while satisfying $a^2 + b^2 = c^2$. 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 ($L$). Since $L$ is constant, implicit differentiation gives:

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

Notice the rates have opposite signs. As the base slides out ($\frac{dx}{dt} > 0$), the top slides down ($\frac{dy}{dt} < 0$). The ladder also accelerates as it falls: $\frac{dy}{dt}$ approaches negative infinity as $y \to 0$. Exam problems sometimes ask you to explain this behavior.

Distance Between Two Moving Objects

Set up coordinates so each object's position is a function of time. The distance formula $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ must be differentiated using the chain rule.

A useful trick: square both sides first to get $D^2 = (x_2-x_1)^2 + (y_2-y_1)^2$. This avoids messy square root derivatives. Then differentiate:

$2D\frac{dD}{dt} = 2(x_2-x_1)\left(\frac{dx_2}{dt}-\frac{dx_1}{dt}\right) + 2(y_2-y_1)\left(\frac{dy_2}{dt}-\frac{dy_1}{dt}\right)$

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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. A typical setup:

$\frac{H}{x+s} = \frac{h}{s}$

where $H$ is the light height, $h$ is the object height, $s$ is the shadow length, and $x$ is the distance from the light to the object. Cross-multiply before differentiating to eliminate fractions, which makes the implicit differentiation much cleaner.

Angle Change in a Triangle with Moving Sides

When a problem involves a changing angle in a non-right triangle, the Law of Cosines provides the relationship:

$c^2 = a^2 + b^2 - 2ab\cos(\theta)$

Differentiating implicitly connects $\frac{d\theta}{dt}$ to the rates of change of the sides. Remember that $\frac{d}{dt}[\cos(\theta)] = -\sin(\theta)\frac{d\theta}{dt}$, so trigonometric derivatives will appear.

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Spherical Volume Problems

Spheres appear frequently because $V = \frac{4}{3}\pi r^3$ creates a clean relationship between volume and radius. Applying the chain rule:

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

The coefficient $4\pi r^2$ is the surface area of the sphere. This isn't a coincidence: volume changes occur through the surface.

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

The formula is the same as any sphere problem, but context determines the sign. Air pumped in means $\frac{dV}{dt} > 0$; air escaping means $\frac{dV}{dt} < 0$.

An important conceptual point: the rate of radius change decreases as the balloon grows. Even with constant $\frac{dV}{dt}$, a larger $r$ means a larger $4\pi r^2$ in the denominator when you solve for $\frac{dr}{dt}$, so $\frac{dr}{dt}$ gets smaller.

Growing Snowball

The setup is identical to balloon problems, but snow accumulation is typically given as a volume rate. If the problem describes snow sticking to the surface at a rate per unit area, you'll need to multiply by the surface area $4\pi r^2$ to get $\frac{dV}{dt}$.

A common variation: "the radius grows at a constant rate." Here you're finding $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$, which increases over time because $r$ is growing while $\frac{dr}{dt}$ stays fixed.

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Conical Tank Problems

Cones are trickier than cylinders because radius and height are both changing. You must use similar triangles to eliminate one variable before differentiating.

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

• For cylinders with fixed radius: $V = \pi r^2 h$ 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. If the tank has radius $R$ at height $H$, then $r = \frac{R}{H}h$. Substituting gives $V = \frac{1}{3}\pi \left(\frac{R}{H}\right)^2 h^3$
• Differentiate the simplified formula: $\frac{dV}{dt} = \pi \left(\frac{R}{H}\right)^2 h^2 \frac{dh}{dt}$, then solve for the unknown rate

Water Leaking from an Inverted Conical Tank

"Inverted cone" means vertex at the bottom, so water forms a smaller cone that is similar to the tank. Find the ratio $\frac{r}{h} = \frac{R}{H}$ where $R$ and $H$ are the tank's full dimensions, then substitute to eliminate $r$.

Height changes faster when water is low (small $h$) because the cross-sectional area $\pi r^2$ is smaller there. The same volume leaving the tank causes a bigger drop in height when the cone is narrow near the bottom. 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:

$\frac{dV}{dt} = \pi\left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right)$

Volume Change in a Cylinder with Varying Dimensions

• Product rule is essential when both radius and height vary. The formula above comes from differentiating $r^2 \cdot h$ as a product (where $r^2$ itself requires the chain rule).
• 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, just like in cone problems

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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?