Related Rates

What are related rates?

Related rates are the rates of change, with respect to time, of several quantities that are tied together by an equation, so that knowing how fast some of them change tells you how fast another one does. The process of finding one of them from the others is a related rates problem.

Suppose air is pumped into a spherical balloon. The volume V and the radius r both grow with time t, and always V=\frac43π r^3. Because r is a function of t, the chain rule differentiates the equation: (dV)/(dt)=(dV)/(dr)·(dr)/(dt)=4π r^2(dr)/(dt). This is a related rates equation. It is useful because the rate of the volume is easy to measure, by metering the air, while the rate of the radius is not. If air goes in at 100 cm^3/s, then when r=5 cm (dr)/(dt)=(100)/(4π·5^2)=\frac1π≈0.318 cm/s.

Why it is the chain rule. A formula such as A=π r^2 is true at every moment, so it really says A(t)=π r(t)^2. The derivative of r(t)^2 is 2 r(t) r'(t): an outside function, the square, and an inside function, r(t). Every related rates equation is made this way. Differentiate each term of the equation, and each variable q that depends on time brings the factor (dq)/(dt), which is its rate of change.

The related rates equations of common formulas.

The cylinder row uses the product rule, because r and h both change. The last row divides by 2 after differentiating, and it is the equation of the cruiser in Example 3 below.

How do you solve a related rates problem?

Solve a related rates problem in four steps: name the variables and draw a picture, write an equation that relates them, differentiate it with respect to t, and only then substitute the numbers of the moment and solve for the unknown rate.

1. Name and draw. Let t be time and give a letter to every quantity that changes with it, assuming each is a differentiable function of t. Quantities that stay fixed, such as a distance of 150 m, need no name. A picture showing the geometry is usually the key.
2. Relate. Write an equation between the variables, using geometry (Pythagoras, similar triangles, trigonometry) or a formula for area or volume. Combine equations until only the variables you know something about, and the one you want, are left.
3. Differentiate. Differentiate both sides with respect to t, using the chain rule on every variable. This is the related rates equation.
4. Substitute and solve. Put in the values and rates of the moment. You may need the equation from step 2 to find a missing value. Then solve for the unknown rate, with its sign and its units.

Example 1: water filling a cone. Water runs into a conical tank at 0.25 m^3/min. The tank stands point down, is 3 m high and has a base radius of 1.5 m. How fast is the water level rising when the water is 1.8 m deep?

Let V be the volume of the water, x the radius of its surface and y its depth. The water is a cone, V=\frac13π x^2y, but nothing is known about x or (dx)/(dt), so eliminate x. Similar triangles give \frac x y=(1.5)/(3), that is x=\frac y2, and so V=\frac13π\left(\frac y2\right)^2y=(π y^3)/(12). Differentiate with respect to t: (dV)/(dt)=(π y^2)/(4)(dy)/(dt). Now put in y=1.8 and (dV)/(dt)=0.25: 0.25=(π(1.8)^2)/(4)(dy)/(dt) (dy)/(dt)=(1)/(3.24π)≈0.098 m/min. The level is rising at about 0.098 m per minute. Move the depth below: the same inflow raises the level more and more slowly, because the surface keeps widening.

What do worked related rates examples look like?

Five more problems show the four steps with a tangent, a distance, a moving tangent line, a fast jet and a rope over a pulley, and each answer comes from differentiating one equation.

Example 2: a balloon and a range finder. A balloon rises straight up from a level field and is tracked by a range finder 150 m from the liftoff point. When the elevation angle is \fracπ4, the angle is increasing at 0.14 rad/min. How fast is the balloon rising? Let θ be the angle and y the height. Then y=150\tanθ, and with the derivative of the tangent (dy)/(dt)=150\sec^2θ (dθ)/(dt) (dy)/(dt)=150·2·0.14=42 m/min, because \sec^2\fracπ4=2.

Example 3: a cruiser chasing a car. A police cruiser approaches a right-angled intersection from the north, chasing a car that has turned the corner and is moving east. When the cruiser is 0.6 km north of the intersection and the car is 0.8 km east of it, the distance between them is increasing at 30 km/h. The cruiser is moving at 100 km/h. How fast is the car going? Let x be the car's position, y the cruiser's and s the distance between them, so s^2=x^2+y^2 and, at that moment, s=√(0.8^2+0.6^2)=1 km. Differentiating, s(ds)/(dt)=x(dx)/(dt)+y(dy)/(dt) 1·30=0.8(dx)/(dt)+0.6·(-100). The cruiser is coming closer, so (dy)/(dt)=-100, and (dx)/(dt)=(30+60)/(0.8)=112.5 km/h.

Example 4: a sliding tangent line. A particle moves clockwise at a constant rate around a circle of radius 10 m centered at the origin, from (0,10) to (10,0) in 30 s. The tangent at the particle meets the x-axis at Q, a distance x from the origin. How fast is Q moving when x=20 m? Let θ be the angle from the x-axis to the radius through the particle. Then x\cosθ=10, and a quarter turn in half a minute makes (dθ)/(dt)=-π rad/min, negative because θ is decreasing. The product rule gives (dx)/(dt)\cosθ-x\sinθ (dθ)/(dt)=0. At x=20 we have \cosθ=\frac12 and \sinθ=(√3)/(2), so \frac12(dx)/(dt)=20·(√3)/(2)·(-π) and (dx)/(dt)=-20√3 π≈-108.8 m/min. The minus sign says that Q moves toward the origin.

Example 5: a jet and a radar station. A jet flies at a constant altitude of 10 km toward a radar station on an island. When first detected, the line of sight makes an angle of 30^\circ with sea level and is turning upward at \frac13 degree per second. How fast is the jet approaching? With x in kilometers, x=10\cotθ, so (dx)/(dt)=-10\csc^2θ (dθ)/(dt). Here the units need care. Angles in derivatives are in radians, and the answer is wanted in km/h, so (dθ)/(dt)=\frac13·(π)/(180)·3600=(20π)/(3) rad/h. With \csc^2\fracπ6=4, (dx)/(dt)=-10·4·(20π)/(3) (dx)/(dt)=-(800π)/(3)≈-837.8 km/h.

Example 6: a weight on a rope. A rope runs over a pulley P that is 6 m above a point O and lifts a weight W; the worker holds the other end at the height of O and walks away from the line under the pulley at 1.2 m/s. The rope is 13.5 m long. How fast is the weight rising when the worker is x=6.3 m from the line? Let h be the height of W above O. The rope has 6-h m from W to P and z m from P to the hand, so z=7.5+h, and Pythagoras gives z^2=6^2+x^2: x^2+36=(7.5+h)^2 2x(dx)/(dt)=2(7.5+h)(dh)/(dt). When x=6.3 the equation gives (7.5+h)^2=36+39.69=75.69, so 7.5+h=8.7, and (dh)/(dt)=(x)/(7.5+h)(dx)/(dt) (dh)/(dt)=(6.3)/(8.7)·1.2≈0.869 m/s.

The tool below does the same steps for these problems, or for an equation of your own. Leave one value empty and it is worked out from the equation, as the angle was in Example 4.

Why does a sliding ladder speed up as it falls?

A ladder whose foot is pulled away from a wall at a steady speed has a top that falls at \frac xy times the speed of the foot, a ratio that grows as the top comes down. Let the ladder have the length L, its foot x from the wall and its top y above the floor. The ladder, the wall and the floor make a right triangle, so for all t x^2+y^2=L^2, and L is constant. Differentiating with respect to t, 2x(dx)/(dt)+2y(dy)/(dt)=0 ⇒ (dy)/(dt)=-\frac xy (dx)/(dt). For a 5 m ladder with the foot 3 m from the wall (so the top is y=√(25-9)=4 m high) pulled at 2 m/s, the top falls at \frac34·2=1.5 m/s.

Because \frac xy is small near the wall and large near the floor, the top is slow at first and fast at the end. The two speeds are equal when x=y, that is at x=(L)/(√2), where the ladder is at 45^\circ. Beyond that the top moves faster than the foot, and as y→0 the ratio \frac xy grows without bound. A real ladder cannot do this: it leaves the wall before the top reaches the floor, and the equation describes only a ladder that stays in contact with it. Try the sliders below; the dashed line marks x=(L)/(√2).

When do you put the numbers in?

Differentiate first and substitute afterwards: until the very end every changing quantity is a function of time, and a number put in too early is a constant, whose rate of change is 0. Take the balloon again. If you set r=5 in V=\frac43π r^3 at the start, you get V=(500π)/(3), a constant, and differentiating gives (dV)/(dt)=0, which says that the balloon does not grow. The numbers you are given belong to one moment, while the equation holds at every moment. Differentiate the equation that holds at every moment, then read it at the one moment you care about.

What you may put in before differentiating is only what never changes or always holds: - the constants of the problem: the 150 m to the range finder, the 10 km altitude of the jet, the length of the ladder, the 6 m height of the pulley; - relations between the variables, which you use to remove a variable you have no information about, as the similar triangles removed x in Example 1.

Three checks on the answer. - Sign. A quantity that is decreasing has a negative rate. The cruiser's y falls, so (dy)/(dt)=-100 and not 100; a negative answer, like -108.8 m/min for Q, says that the quantity is decreasing, and its speed is the absolute value. - Units. The unit of the answer is the unit of the quantity per unit of time: m/min, km/h, cm^3/s. If the data mix minutes with hours, or degrees with radians, convert first, as in Example 5. - Radians. The derivative formulas for \sin, \cos, \tan and the others hold only for angles in radians. A rate in degrees per second must be converted before it is used in \sec^2θ (dθ)/(dt).

How do you solve related rates problems in code?

With SymPy, declare each quantity as a function of t, differentiate the equation with sp.diff, put in the numbers of the moment and solve for the unknown rate. The function below does the four steps of this article and reproduces Examples 1, 2, 3 and 5 with exact answers: (25)/(81π)≈0.0982, 42, (225)/(2) and -(800π)/(3). The last lines check the ladder without differentiating anything: they move the foot at 2 m/s, time the top over a tiny step, and get the -1.5 m/s of the formula.

import math
import sympy as sp

t = sp.symbols('t')


def solve_rate(relation, quantities, values, rates, unknown):
    """Differentiate relation (left - right = 0) in time, put in one moment's numbers, solve for d(unknown)/dt."""
    d = {q: sp.Symbol('d' + q.func.__name__) for q in quantities}
    dF = sp.diff(relation, t)                       # step 3: the related rates equation
    for q in quantities:
        dF = dF.subs(sp.Derivative(q, t), d[q])     # dq/dt becomes a plain symbol
    dF = dF.subs(values).subs({d[q]: r for q, r in rates.items()})   # step 4: put in the moment
    return sp.simplify(sp.solve(dF, d[unknown])[0])


y, V, theta, x, s = (sp.Function(n)(t) for n in ('y', 'V', 'theta', 'x', 's'))

# Example 1: V = pi y^3 / 12, dV/dt = 1/4, y = 9/5
print(solve_rate(V - sp.pi * y**3 / 12, [V, y], {y: sp.Rational(9, 5)}, {V: sp.Rational(1, 4)}, y))
# 25/(81*pi)   (0.0982)

# Example 2: y = 150 tan(theta), theta = pi/4, dtheta/dt = 0.14
print(solve_rate(y - 150 * sp.tan(theta), [y, theta], {theta: sp.pi / 4}, {theta: sp.Rational(14, 100)}, y))
# 42

# Example 3: s^2 = x^2 + y^2 with x = 0.8, y = 0.6, s = 1, dy/dt = -100, ds/dt = 30
print(solve_rate(s**2 - x**2 - y**2, [s, x, y], {s: 1, x: sp.Rational(4, 5), y: sp.Rational(3, 5)}, {y: -100, s: 30}, x))
# 225/2        (112.5)

# Example 5: x = 10 cot(theta), theta = pi/6, dtheta/dt = 20 pi / 3 rad per hour
print(solve_rate(x - 10 * sp.cot(theta), [x, theta], {theta: sp.pi / 6}, {theta: 20 * sp.pi / 3}, x))
# -800*pi/3    (-837.76)

# the ladder, timed without any derivative: L = 5, foot at 3 m moving at 2 m/s
L = 5.0
foot = lambda time: 3 + 2 * time
top = lambda time: math.sqrt(L**2 - foot(time) ** 2)
dt = 1e-6
print(round((top(dt) - top(0)) / dt, 4), -(3 / 4) * 2)       # -1.5 -1.5

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const double PI = acos(-1.0);

// The related rates answers by really moving the quantities in time and timing the unknown.
// Example 1: water fills a cone, V = pi y^3 / 12, dV/dt = 0.25, at y = 1.8
double yCone(double t) {
  double v = PI * 1.8 * 1.8 * 1.8 / 12 + 0.25 * t;     // the volume at time t
  return pow(12 * v / PI, 1.0 / 3);                    // the depth that holds it
}
// Example 2: y = 150 tan(theta), theta = pi/4 + 0.14 t
double yBalloon(double t) { return 150 * tan(PI / 4 + 0.14 * t); }
// Example 5: x = 10 cot(theta), theta = pi/6 + (20 pi / 3) t   (t in hours)
double xJet(double t) { return 10 / tan(PI / 6 + 20 * PI / 3 * t); }
// the ladder: L = 5, the foot at 3 m moving at 2 m/s
double top(double t) { return sqrt(25 - (3 + 2 * t) * (3 + 2 * t)); }

int main() {
  double h = 1e-6;
  printf("%.4f\n", (yCone(h) - yCone(-h)) / (2 * h));          // 0.0982  (25 / (81 pi))
  printf("%.4f\n", (yBalloon(h) - yBalloon(-h)) / (2 * h));    // 42
  printf("%.2f\n", (xJet(h / 1000) - xJet(-h / 1000)) / (2 * h / 1000));   // -837.76  (-800 pi / 3)
  printf("%.4f %.4f\n", (top(h) - top(-h)) / (2 * h), -(3.0 / 4) * 2);     // -1.5 -1.5
  return 0;
}

/* output:
0.0982
42.0000
-837.76
-1.5000 -1.5000
*/

Where do related rates come from?

Related rates are the original problem of Newton's calculus: given a relation between quantities that flow with time, find the relation between their rates of flow.

• Fluents and fluxions. In his Method of Fluxions, written in 1671 and published in 1736, Isaac Newton called a quantity that changes with time a fluent and its rate of change a fluxion, written with a dot, \dot x. His first problem is the one of this article: the relation of the fluents to each other is given, and the relation of their fluxions is wanted.
• Two notations. Leibniz published his differential notation (dx)/(dt) in 1684, and it prevailed in textbooks, because it shows which quantity is differentiated with respect to which and makes the chain rule look like a cancelation of fractions. Newton's dot survives in physics and engineering, where \dot x is the derivative of a position with respect to time.
• Time as the variable. Newton thought of time as the quantity that flows evenly and of everything else as depending on it. This is the reason why in a related rates problem there is no x-axis with a function on it: all the quantities are functions of t, and the equation between them is a constraint that holds at every moment.

What mistakes do people make in related rates problems?

The most common mistakes in related rates problems are the nine below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: related rates at a glance

• Idea: quantities tied by an equation have rates that are tied too; differentiate the equation with respect to t and each variable q brings (dq)/(dt).
• Strategy: name the variables and draw, relate them by an equation, differentiate in t, then substitute the values of the moment and solve.
• Order: differentiate first, substitute second; only constants and relations that always hold may go in earlier.
• Removing a variable: use similar triangles, Pythagoras or trigonometry, as x=\frac y2 for the cone.
• Sign, units, radians: a decreasing quantity has a negative rate; convert units; angles in radians.
• Ladder: x^2+y^2=L^2 gives (dy)/(dt)=-\frac xy(dx)/(dt), faster than the foot once x>y.
• Code: sp.diff(relation, t) and sp.solve for the unknown rate.

Glossary: key terms

Related rates problem

A problem in which several quantities change with time and are tied by an equation, and the rate of one of them is found from the rates of the others at a given moment.

Related rates equation

The equation obtained by differentiating an equation between changing quantities with respect to time, so that it contains their rates of change.

Rate of change with respect to time

The derivative of a quantity with respect to time t, written dq over dt, which says how fast the quantity changes at one moment.

Similar triangles

Triangles with equal angles, whose sides are in the same ratio; used in related rates to remove a variable, as the radius of the water surface in a cone.

Fluent and fluxion

Newton's names for a quantity that changes with time and for its rate of change; a fluxion is what we now call a time derivative and write with a dot.

Constant of the problem

A quantity that does not change with time, such as the length of a ladder, whose rate is zero and which may be put into the equation before differentiating.

Frequently asked questions

References

1. Joel Hass, Christopher Heil, Maurice D. Weir et al. (2023). Thomas' Calculus: Early Transcendentals in SI Units (15th ed.), Section 3.10: Related Rates. Pearson.
2. James Stewart (2015). Calculus (8th ed.), section 3.9: Related Rates. Cengage Learning.
3. Isaac Newton (1736). The Method of Fluxions and Infinite Series (translated by John Colson). London.
4. Gottfried Wilhelm Leibniz (1684). Nova methodus pro maximis et minimis, itemque tangentibus. Acta Eruditorum.

Related articles

• The Chain Rule: The chain rule explained: how to find the inside and outside of a composite function, the power chain rule, nested functions, second derivatives, and code.
• Implicit Differentiation: Implicit differentiation explained: find dy/dx without solving for y, tangent and normal lines, second derivatives, rational powers, pitfalls, and code.
• The Derivative as a Rate of Change: The derivative as a rate of change: velocity, speed, acceleration and jerk, free fall, marginal cost and revenue, sensitivity to change, and rates from data in code.
• Linearization and Differentials: Linearization and differentials: the tangent-line approximation L(x) = f(a) + f′(a)(x − a), df = f′(x) dx, estimates, errors, percentage change and code.