Should I draw a diagram?

Almost always. Labelling the changing lengths on a sketch is how you find the relationship between them, and it makes clear which quantities are fixed and which are moving.

Why can I not just substitute the numbers at the start?

Because the values given describe one instant, not the whole motion. Treating a moving quantity as a constant removes its derivative, and with it the rate you were asked to find.

What units should the answer have?

The units of the quantity divided by time: centimetres per second for a level that is falling, metres per second for a distance, radians per second for an angle. Carrying units through the final line is a cheap check that you differentiated the right thing.

MATH 100 related rates practice

In a related rates problem two quantities change together, you are told how fast one of them is changing, and you are asked about the other. The calculus is a single line of implicit differentiation with respect to time. Everything difficult happens before and after it.

The reliable procedure is the same every time: name the variables, write a relationship between them that holds at every instant, differentiate both sides with respect to time, and only then substitute the values at the particular instant you are asked about.

That ordering matters more than anything else in the topic. Substituting first freezes a variable that is supposed to be moving, and the equation collapses to nothing useful.

Start practising Browse all 26 questions

In this topic

Taught: Before the midterm
Questions: 26
Total marks: 77
Suggested time: 121 min
Difficulty: Introductory to Capstone
Show the 7 hardest

Key results

Rates through the chain rule: $\frac{d}{dt}\,f\bigl(x(t)\bigr) = f'\bigl(x(t)\bigr)\,\frac{dx}{dt}, \qquad \text{e.g. } \frac{d}{dt}\,x^{2} = 2x\,\frac{dx}{dt}$
When $x$ is a differentiable function of $t$ and $f$ is differentiable at $x(t)$ .
Pythagoras in time: $x^{2} + y^{2} = z^{2} \implies x\,\frac{dx}{dt} + y\,\frac{dy}{dt} = z\,\frac{dz}{dt}$
When $x^{2} + y^{2} = z^{2}$ holds at every instant and $x$ , $y$ and $z$ are differentiable functions of $t$ .
Circle and sphere: $A = \pi r^{2}, \qquad V = \tfrac{4}{3}\pi r^{3}, \qquad S = 4\pi r^{2}$
For a circle of radius $r$ (area $A$ ) and a sphere of radius $r$ (volume $V$ , surface area $S$ ).
Cone: $V = \tfrac{1}{3}\pi r^{2} h$
For a cone of base radius $r$ and height $h$ ; in a cone of fixed shape, $r/h$ stays constant, by similar triangles.
Percentage rates of change: $R = k\,P^{a}Q^{b} \implies \frac{R'}{R} = a\,\frac{P'}{P} + b\,\frac{Q'}{Q}$
For constants $k \neq 0$ , $a$ and $b$ , with $P$ and $Q$ positive and differentiable.
An angle: $\tan\theta = \frac{y}{x},\ \ x \text{ fixed} \implies \sec^{2}\theta\,\frac{d\theta}{dt} = \frac{1}{x}\,\frac{dy}{dt}$
With $\theta$ in radians, so its rate is in radians per unit of time; when $x$ changes too, differentiate $\tan\theta = y/x$ by the quotient rule.

Read the theoryUBC Math 100 §5 Related ratesCLP-1 §3.2 Related rates

Worked example

Standard3 marks · 5 min

A 5 m ladder leans against a vertical wall. The foot of the ladder slides away from the wall at 0.6 m/s. At what rate is the height of the top of the ladder changing when the foot is 3 m from the wall?

Give a signed rate in metres per second, taking upward as positive.

Strategy

Write a relationship between the two changing lengths that holds at every instant, differentiate it with respect to time, and only then substitute the values at the instant of interest.

Working

$x^{2} + y^{2} = 5^{2} = 25$
Pythagoras. The ladder length is constant, which is what makes this equation usable.
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
Differentiate with respect to time. The constant on the right gives zero.
$x = 3 \implies y = \sqrt{25 - 9} = 4$
Find the missing length at this instant, using the original relationship.
$\frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} = -\frac{3}{4}(0.6)$
Solve for the unknown rate, then substitute.
$\frac{dy}{dt} = -0.45 \ \text{m/s}$
The negative sign says the top is descending, as it must be while the foot slides away from the wall.

Answer

\frac{dy}{dt} = -0.45 \ \text{m/s}

Check

After 0.01 s the foot is at 3.006 m and the top at about 3.99549 m, 0.0045 m lower, so it falls at about 0.45 m/s.

Where this goes wrong

Substituting x = 3 into the Pythagorean relationship before differentiating. That makes x a constant, so dx/dt drops out and 9 + y² = 25 differentiates to 2y·dy/dt = 0, which wrongly says the top is not moving. Substitute only after differentiating.

On the paper

Draw and label the picture, write an equation that holds at every instant, and differentiate it with respect to $t$ before putting in the values at the instant asked. Give the rate with its sign and its units.

Leans on

Implicit differentiation, from Implicit differentiation and inverse trigonometric functions

Your turn

Standard3 marks · 5 min

At noon, ship A is 30 km due north of a port and sailing north at 20 km/h, while ship B is 40 km due east of the same port and sailing east at 15 km/h. How fast is the distance between the ships increasing at noon?

Give the answer in kilometres per hour.

Answer this question

What you should be able to do

Name variables and rates explicitly
Write down what each letter means and which rate you are given and want. Related rates questions are mostly translation, and an unlabelled diagram is where translations go wrong.
Find a relationship that holds at every instant
Pythagoras for a sliding ladder, a volume formula for a filling tank, similar triangles for a shadow. The relationship must stay true as things move, not just at the moment of interest.
Eliminate variables using the fixed geometry
A cone of fixed shape gives a constant ratio between radius and depth, which lets you write the volume in one variable before differentiating.
Differentiate with respect to time
Every variable produces its own rate through the chain rule. A squared term becomes $2x\,\dfrac{dx}{dt}$ , and a product of two varying quantities needs the product rule.
Substitute the instant last, and interpret the sign
Put the numbers in only once the differentiation is complete. A negative rate means the quantity is decreasing, and saying so is part of the answer.

Where marks are usually lost

Substituting the given instant before differentiating
Putting $x = 3$ into $x^{2} + y^{2} = 25$ first makes $x$ a constant, its derivative zero, and the whole equation useless. Differentiate, then substitute.
Treating a changing dimension as fixed
In a draining cone the radius of the water surface shrinks with the depth. Using the container radius throughout gives the wrong rate.
Answering the wrong rate
A shadow question may ask how fast the tip moves, not how fast the shadow lengthens. They are different quantities, and both appear among the plausible-looking numbers.
Dropping the sign of the given rate
Water draining out is a negative $\dfrac{dV}{dt}$ . Entering it as positive produces a level that rises, which the question has already told you it does not.
Forgetting the chain rule on a squared term
Differentiating $r^{2}$ with respect to time gives $2r\,\dfrac{dr}{dt}$ , not $2r$ . Without the rate factor there is nothing to solve for.

Practice questions

5 of the 26, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Browse all 26 questions

Common questions

Should I draw a diagram?: Almost always. Labelling the changing lengths on a sketch is how you find the relationship between them, and it makes clear which quantities are fixed and which are moving.
Why can I not just substitute the numbers at the start?: Because the values given describe one instant, not the whole motion. Treating a moving quantity as a constant removes its derivative, and with it the rate you were asked to find.
What units should the answer have?: The units of the quantity divided by time: centimetres per second for a level that is falling, metres per second for a distance, radians per second for an angle. Carrying units through the final line is a cheap check that you differentiated the right thing.

Needs
Implicit differentiation and inverse trigonometric functions
Differentiating relations, tangent lines to curves, and the derivatives of inverse and inverse trigonometric functions.
30 questions
Differentiation rules
Power, sum, product and quotient rules, tangent lines and higher-order derivatives.
23 questions
Optimization
Building an objective function and justifying the extreme value.
18 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Leans on

Your turn

What you should be able to do

Name variables and rates explicitly

Find a relationship that holds at every instant

Eliminate variables using the fixed geometry

Differentiate with respect to time

Substitute the instant last, and interpret the sign

Where marks are usually lost

Substituting the given instant before differentiating

Treating a changing dimension as fixed

Answering the wrong rate

Dropping the sign of the given rate

Forgetting the chain rule on a squared term

Practice questions

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Common questions

Related topics

Implicit differentiation and inverse trigonometric functions

Differentiation rules

Optimization