Can I use Related Rates if I only know some of the rates?

Yes—that's the whole point. Related Rates connects several rates through one equation, so if you know all but one of them, you can solve for the missing one. Make sure you have enough information: all variables' values at the moment in question, all but one rate, and the geometric or physical relationship that connects the variables.

What does it mean if my answer is negative?

A negative rate means that quantity is decreasing. For example, if \frac{dh}{dt} = -2 m/s, the height is falling at 2 m/s. Always include the sign—it tells you the direction of change.

Do I have to use the chain rule formula \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} explicitly?

No, you can just differentiate the constraint equation directly (implicitly) with respect to time using Implicit Differentiation. The chain rule formula is what you get when you rearrange that result, but both methods give the same answer. Use whichever feels clearer to you.

What if the variables don't have units like m and m/s—can I still use Related Rates?

Yes. Related Rates works with any quantities that change with time—areas, temperatures, angles, populations, concentrations—as long as they're related by an equation. The method doesn't depend on what you're measuring or the units you use.

Related Rates

Related Rates connects the rate of change of one quantity to the rate of change of another quantity through implicit differentiation.

\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}

Solve a problem with Related Rates

Type the problem. The solver will use Related Rates where Related Rates is the right tool, and tell you when it is not.

What each symbol means

What Related Rates takes

Symbol	Meaning
$V$	The quantity being tracked (volume, area, temperature, etc.), which changes with time through another variable; confusing it with a rate like $\frac{dV}{dt}$ is a dimensional error.
$r$	The mediating quantity (radius, height, distance) that changes with time; treating it as constant breaks related rates because you need the chain rule to connect changes in $r$ to changes in $V$ .
$t$	Time, the independent variable in every related rates problem; forgetting that everything depends on $t$ means missing the chain rule step $\frac{dr}{dt}$ .

When to use it

Use Related Rates when multiple quantities that change with time are related by an equation, and you need to find one rate from another.

Level

Usually taught in: Calculus I · Appears on: AP Calculus

Worked examples

1. Ladder sliding down a wall

Problem

A ladder 5 meters long leans against a vertical wall. The base of the ladder slides away from the wall at 2 m/s. How fast is the top of the ladder sliding down the wall when the base is 3 meters from the wall?

$x^2 + y^2 = 25$
Set up the relationship using the Pythagorean theorem, where $x$ is the distance from the wall and $y$ is the height.
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
Differentiate both sides of the constraint with respect to time $t$ using the chain rule.
$2(3)(2) + 2(4)\frac{dy}{dt} = 0$
Substitute the known values: $x = 3$ m, $y = 4$ m (from $3^2 + y^2 = 25$ ), and $\frac{dx}{dt} = 2$ m/s.
$\frac{dy}{dt} = -\frac{3}{2}$
Simplify to solve for $\frac{dy}{dt}$ : divide both sides by $8$ and simplify.

Answer: $\frac{dy}{dt} = -\frac{3}{2} \text{ m/s}$

This is the classic related rates example because the ladder's fixed length creates a constraint that connects the two rates. The negative sign indicates the top is falling as the base moves outward. It demonstrates how a physical constraint (fixed length) translates into an equation relating two rates.

2. Water filling a conical tank

Problem

A conical tank with height 10 m and base radius 4 m fills with water at 3 m³/s. How fast is the water level rising when the water depth is 2 m?

$V = \frac{1}{3}\pi r^2 h$
Write the volume formula for a cone with radius $r$ and height $h$ .
$r = \frac{2h}{5}$
Use similar triangles to relate radius to height: the ratio $\frac{r}{h}$ equals $\frac{4}{10}$ throughout the cone.
$V = \frac{1}{3}\pi \left(\frac{2h}{5}\right)^2 h = \frac{4\pi h^3}{75}$
Substitute the relationship to express $V$ as a function of $h$ alone: $\frac{1}{3}\pi \cdot \frac{4h^2}{25} \cdot h = \frac{4\pi h^3}{75}$ .
$\frac{dV}{dt} = \frac{4\pi \cdot 3h^2}{75}\frac{dh}{dt} = \frac{4\pi h^2}{25}\frac{dh}{dt}$
Differentiate both sides with respect to time using the chain rule: $\frac{d}{dt}\left(\frac{4\pi h^3}{75}\right) = \frac{4\pi}{75} \cdot 3h^2 \cdot \frac{dh}{dt}$ .
$3 = \frac{16\pi}{25}\frac{dh}{dt}$
Substitute $h = 2$ and $\frac{dV}{dt} = 3$ : $\frac{4\pi (2)^2}{25}\frac{dh}{dt} = \frac{16\pi}{25}\frac{dh}{dt}$ .
$\frac{dh}{dt} = \frac{75}{16\pi}$
Solve for the rate by multiplying both sides by $\frac{25}{16\pi}$ .

Answer: $\frac{dh}{dt} = \frac{75}{16\pi} \text{ m/s}$

This problem is harder because you must first eliminate one variable using a geometric constraint (similar triangles), then differentiate. It shows why constraint equations are essential when you have multiple changing quantities. The water forms a cone similar to the tank, so the radius and height are always proportional.

3. Thermal expansion of a spherical vessel

Problem

A spherical steel vessel expands as it heats. Its radius is currently 10 cm and increasing at 0.2 cm/s. The thermal energy stored in the vessel is

Q = 5V

, where

V

is the volume in cm³. How fast is the stored energy increasing at this moment?

$V = \frac{4}{3}\pi r^3$
Write the volume of a sphere in terms of its radius $r$ .
$Q = 5V = \frac{20\pi r^3}{3}$
Substitute the volume formula into the energy equation to express $Q$ as a function of $r$ .
$\frac{dQ}{dt} = \frac{dQ}{dr} \cdot \frac{dr}{dt}$
Write the related rates formula: the rate of change of $Q$ with respect to time equals the rate of change of $Q$ with respect to $r$ , multiplied by the rate of change of $r$ with respect to time.
$\frac{dQ}{dr} = 20\pi r^2$
Differentiate $Q = \frac{20\pi r^3}{3}$ with respect to $r$ : $\frac{d}{dr}\left(\frac{20\pi r^3}{3}\right) = \frac{20\pi}{3} \cdot 3r^2 = 20\pi r^2$ .
$\frac{dQ}{dt} = 20\pi(10)^2 \cdot 0.2 = 400\pi$
Substitute $r = 10$ cm and $\frac{dr}{dt} = 0.2$ cm/s: $20\pi \cdot 100 \cdot 0.2 = 400\pi$ .

Answer: $\frac{dQ}{dt} = 400\pi \text{ J/s}$

This example directly applies the related rates formula shown at the top of the page and demonstrates how it works in a real physical context. Unlike the ladder and cone problems, we explicitly compute $\frac{dQ}{dr}$ and $\frac{dr}{dt}$ separately, then multiply them together—this is the chain rule in its most direct form for related rates.

Common mistakes

Where Related Rates usually goes wrong

The mistake	Why it is wrong	The fix
Students differentiate $V = \frac{4}{3}\pi r^3$ to get $\frac{dV}{dr} = 4\pi r^2$ , then forget to multiply by $\frac{dr}{dt}$ , writing $\frac{dV}{dt} = 4\pi r^2$ instead.	The chain rule requires multiplying by $\frac{dr}{dt}$ because $r$ depends on $t$ . Without this step, you're finding how $V$ changes with respect to $r$ , not with respect to time.	Always include $\frac{dr}{dt}$ : write $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$ . This is the definition of related rates.
Students substitute numerical values (like $r = 10$ and $\frac{dr}{dt} = 0.2$ ) into the original equation before differentiating, then try to differentiate a number instead of a variable.	Differentiation requires variables; if you substitute numbers too early, you eliminate the variables you need to differentiate with respect to time.	Differentiate the equation in its symbolic form first. Only substitute numbers after you have written the chain rule form $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$ and are ready to calculate the final answer.
In the cone problem, students write $V = \frac{1}{3}\pi r^2 h$ and then differentiate it directly, treating both $r$ and $h$ as independent variables: $\frac{dV}{dt} = \frac{2\pi rh}{3}\frac{dr}{dt} + \frac{\pi r^2}{3}\frac{dh}{dt}$ , without using similar triangles to relate $r$ and $h$ .	The problem states that the shape of the cone constrains $r$ and $h$ : as the water fills, the ratio $\frac{r}{h}$ stays constant. Ignoring this constraint means you have too many unknowns to solve the problem.	Before differentiating, use geometric constraints like similar triangles to write $r$ in terms of $h$ (or vice versa), so $V$ depends on only one variable. Then differentiate that single-variable equation.

Tips and when to use something else

Always write the constraint equation first (like $x^2 + y^2 = 25$ or $r = \frac{2h}{5}$ ), then differentiate it. This is where the chain rule kicks in.
If you have multiple variables, use Implicit Differentiation or substitute to reduce to one variable—don't try to differentiate an equation with too many unknowns.
Related Rates and Implicit Differentiation are inseparable: every related rates problem is really implicit differentiation applied to a physical situation where time is the underlying variable.
Check your signs: negative rates (things shrinking or moving in a direction) are real and matter for the final answer.

Related formulas

Chain RuleThe Chain Rule tells you how to differentiate a composite function—when one function is plugged into another—by multiplying the outer and inner derivatives.Implicit DifferentiationUse implicit differentiation to find dy/dx when an equation defines y implicitly instead of as an explicit function of x.Power RuleThe Power Rule states that the derivative of x^n is nx^{n-1}, making it the essential tool for differentiating any polynomial expression.Definition of the DerivativeThe Definition of the Derivative finds the instantaneous rate of change at a point using limits; the foundation for all derivative calculations.

Frequently asked questions

Can I use Related Rates if I only know some of the rates?: Yes—that's the whole point. Related Rates connects several rates through one equation, so if you know all but one of them, you can solve for the missing one. Make sure you have enough information: all variables' values at the moment in question, all but one rate, and the geometric or physical relationship that connects the variables.
What does it mean if my answer is negative?: A negative rate means that quantity is decreasing. For example, if $\frac{dh}{dt} = -2$ m/s, the height is falling at 2 m/s. Always include the sign—it tells you the direction of change.
Do I have to use the chain rule formula $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$ explicitly?: No, you can just differentiate the constraint equation directly (implicitly) with respect to time using Implicit Differentiation. The chain rule formula is what you get when you rearrange that result, but both methods give the same answer. Use whichever feels clearer to you.
What if the variables don't have units like m and m/s—can I still use Related Rates?: Yes. Related Rates works with any quantities that change with time—areas, temperatures, angles, populations, concentrations—as long as they're related by an equation. The method doesn't depend on what you're measuring or the units you use.

Need a different method?

The full solver is not scoped to one formula — type any problem and it will pick the method.

Open the math solver

Reviewed 2026-09-18