Related Rates Solver

Math Calculus • Applications of Derivatives

Written by STEM Calculators Team Published December 23, 2025 Updated February 24, 2026

2. Related Rates Solver

Differentiate an equation with respect to time and solve for a target rate (e.g. \(dy/dt\)). Chain rule: \(\dfrac{d}{dt}F(v_1(t),v_2(t),\dots)=\sum \dfrac{\partial F}{\partial v}\dfrac{dv}{dt}\).

Inputs

Relation (equation)

Use = for an equation. If you omit “=”, we assume the expression equals \(0\). Constants: pi, e. Functions: sin, cos, tan, ln, log, sqrt, abs, exp. Implicit multiplication allowed: 2x, (x+1)(y-1), 3sin(x). Trig powers: cos^2(2x).

Solve for

Chooses which \(\dfrac{d(\cdot)}{dt}\) to isolate.

Time unit

Used for displaying rate units.

Missing value helper Prefer positive solution (length-like)

Numeric evaluation: if exactly one variable value is blank, we attempt to solve it from the relation (within your range). If more than one is blank, we still show the symbolic formula.

Solve range min

Solve range max

Output format

Presets

Click a preset to fill inputs and compute.

Quick actions

Variables & rates

Values are the instant values \(v\). Known rates are \(dv/dt\). Leave blank if unknown. The solver always shows the symbolic isolated rate; numeric evaluation needs enough values/rates.

variable	value	unit	known rate \(d(\cdot)/dt\)	rate unit
Enter a relation to populate variables.

If the diagram is plotting \(X\) on the horizontal axis and \(Y\) on the vertical axis, and you know both \(dX/dt\) and \(dY/dt\), then the tangent slope is \(\dfrac{dY}{dX}=\dfrac{(dY/dt)}{(dX/dt)}\) (when \(dX/dt\neq 0\)).

Diagram

Drag to pan • wheel/pinch to zoom • plots the implicit curve \(F(X,Y)=0\) in your chosen plane.

Plot horizontal \(X\)

Plot vertical \(Y\)

If the relation has more variables, the rest are fixed to your entered values (or 0 if blank).

Result

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2. Related Rates Solver — Theory

Related rates problems connect how multiple quantities change over time using differentiation and the chain rule.

1) The idea

If two (or more) variables are linked by a relation, changing one forces the others to change. We differentiate the relation with respect to time \(t\) and solve for the desired rate.

\[ \frac{dy}{dt} = \frac{dy}{dx}\cdot \frac{dx}{dt} \]

This is a special case of the multivariable chain rule.

2) Chain rule for implicit relations

Suppose a relation can be written as:

\[ F(x(t),y(t),z(t),\dots)=0. \]

Differentiating with respect to time gives:

\[ \frac{dF}{dt} = \frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt} + \frac{\partial F}{\partial z}\frac{dz}{dt} + \cdots = 0. \]

This equation is typically linear in the rates \(\dfrac{dx}{dt},\dfrac{dy}{dt},\dots\), so we can isolate the one we want.

3) Standard workflow

Draw a diagram and label variables (especially for geometry).
Write a relation between variables (often from Pythagorean theorem, area/volume formulas, etc.).
Differentiate with respect to \(t\), using chain rule for each variable.
Substitute numerical values at the instant (including computed missing values like \(y\)).
Solve for the desired rate and attach units.

4) Classic example: ladder sliding

A ladder of length \(5\) m leans against a wall. Let \(x\) be the distance of the base from the wall and \(y\) be the height on the wall. Then:

\[ x^2 + y^2 = 25. \]

Differentiate w.r.t. time:

\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0. \]

Solve for \(\dfrac{dy}{dt}\):

\[ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}. \]

If \(x=3\) m, then \(y=\sqrt{25-9}=4\) m. If \(\dfrac{dx}{dt}=0.2\) m/s:

\[ \frac{dy}{dt} = -\frac{3}{4}\cdot 0.2 = -0.15\ \text{m/s}. \]

The negative sign means the top of the ladder moves down as the base slides out.

5) Units

Rates carry “per time” units. For example, if \(x\) is in meters and \(t\) is in seconds, then \(\dfrac{dx}{dt}\) is in m/s. Keeping units consistent helps catch mistakes.

6) Advanced notes

Multiple chains: if \(y\) depends on \(r\) and \(r\) depends on \(t\), you may see nested use of chain rule.
More variables: some problems involve three or more changing quantities (e.g. cone volume, expanding sphere, related angles).
Geometry often introduces a hidden step: compute a missing value at the instant (like \(y\) in the ladder problem).

Frequently Asked Questions

What does a related rates solver compute?

It computes an unknown time rate of change such as dy/dt by differentiating a relation that links multiple variables and then isolating the requested rate. The result can be symbolic or numeric depending on how many values and known rates you provide.

How does implicit differentiation work for related rates?

If a relation is written as F(x(t), y(t), ...) = 0, differentiating with respect to time gives dF/dt = (∂F/∂x)(dx/dt) + (∂F/∂y)(dy/dt) + ... = 0. This equation is then solved for the target rate.

Why is my related rate negative?

A negative sign means the variable is decreasing at that instant with respect to time, given the direction of change of the other quantities. For example, in the sliding ladder relation, the height decreases as the base moves away from the wall.

How does the missing value helper work in this calculator?

If exactly one variable value is left blank, the calculator attempts to solve that value from the relation within your specified solve range before evaluating the target rate. If more than one value is missing, it still provides the symbolic isolated rate.

What does the diagram show in the related rates solver?

The diagram plots the implicit curve F(X,Y)=0 for your chosen horizontal variable X and vertical variable Y. Any additional variables in the relation are fixed to the values you entered (or to 0 if left blank) when drawing the curve.