MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: related rates (MATH 140)

This sheet is not a summary of section 3.9 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on related rates in MATH 140 at McGill University, and which precise gesture avoids each loss.

Related rates problems look like word problems, and they are lost like word problems: not in the derivative, which is always a chain rule, but in the figure, the relation and the order of the steps. Every value below is exact and done by hand, as on the exam.

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The thread of the chapter

Letters for everything that changes, numbers only for what stays fixed: the relation must hold at EVERY instant, it is differentiated with respect to time, and only THEN are the values of the instant substituted. A number put in too early is a rate set to zero, and every rate keeps its sign and its radians.

This chapter is part of MATH 140, Calculus 1 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Related rates
Chain rule
Differentiation with respect to time
Pythagorean theorem
Similar triangles
Volume of a cone
Volume of a sphere
Angular rate of change
Radian measure
Signed rate of change

The essentials

The five gestures, always in this order

• Draw the figure. Label with a LETTER every quantity that changes ( $x$ , $y$ , $h$ , $\theta$ ) and with a NUMBER only what stays fixed (the ladder, the height of the lamp, the size of the tank).
• Write the relation that holds at EVERY instant: Pythagoras, similar triangles, a trigonometric ratio or a volume formula. If it has one variable too many, eliminate it now, by another relation valid at every instant.
• Differentiate with respect to $t$ . Each variable is a function of $t$ , so each term gets the chain rule: $\frac{d}{dt}\left(y^2\right) = 2y\frac{dy}{dt}$ , $\frac{d}{dt}\cos\theta = -\sin\theta\,\frac{d\theta}{dt}$ .
• Only NOW substitute the values of the instant, including the ones you compute from the relation ( $y = 6$ from $x = 2.5$ and the ladder).
• Conclude with a sentence: the rate, its sign read as increasing or decreasing, and its unit.

Same ladder, same foot speed

0.3

m/s: the top falls at

\frac{1}{8}

m/s when

x = 2.5

but at

\frac{18}{25}

m/s when

x = 6

. The rate depends on the instant.

The relation and the differentiated relation are worth marks even if the arithmetic fails. Writing them before any number is what a marker looks for first.

Where the relation comes from

• Two lengths at right angles and a third joining them: Pythagoras, $x^2 + y^2 = z^2$ , then $x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$ .
• A fixed length and an angle: a trigonometric ratio with the fixed side, $\tan\theta = \frac{h}{4}$ , then $\sec^2\theta\,\frac{d\theta}{dt} = \frac{1}{4}\frac{dh}{dt}$ , the angle in radians.
• Two triangles sharing an angle (a light ray, the section of a cone): similar triangles, $\frac{r}{h} = \frac{R}{H}$ , used to ELIMINATE a variable before differentiating.
• A volume or an area: the formula of the solid, reduced to ONE variable first. Then $\frac{dV}{dt} = A\frac{dh}{dt}$ , where $A$ is the area of the free surface.
• $\sec^2\theta = 1 + \tan^2\theta$ gives $\sec^2\theta$ from the figure without ever computing $\theta$ , and without a calculator.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The standard relations, differentiated with respect to time

Read a line as: in this situation, write this relation, and its derivative with respect to $t$ is in the last column. The red lines are derivatives that students write and that do not exist.

Situation	Relation	Differentiated in t
ladder $6.5$	$x^2 + y^2 = 6.5^2$	$xx' + yy' = 0$
Example: $x = 2.5$ , $y = 6$ , $x' = 0.3$ : $y' = -\frac{2.5}{6}(0.3) = -\frac{1}{8}$ m/s.
sphere	$V = \frac{4}{3}\pi r^3$	$V' = 4\pi r^2 r'$
Example: $V' = 72\pi$ , $r = 6$ : $r' = \frac{72\pi}{144\pi} = \frac{1}{2}$ cm/s.
cone, $r = \frac{h}{3}$	$V = \frac{\pi h^3}{27}$	$V' = \frac{\pi h^2}{9}h'$
Example: $V' = \frac{\pi}{2}$ , $h = 3$ : $h' = \frac{1}{2}$ m/min.
shadow	$s = \frac{2}{3}x$	$s' = \frac{2}{3}x'$
Example: $x' = 1.2$ : $s' = 0.8$ m/s at every position.
angle	$\tan\theta = \frac{h}{4}$	$\sec^2\theta\,\theta' = \frac{h'}{4}$
Example: $h = 4$ , $h' = 0.8$ : $\sec^2\theta = 2$ , $\theta' = \frac{1}{10}$ rad/s.
area $\frac{1}{2}xy$	$A = \frac{1}{2}xy$	$A' = \frac{1}{2}x'y'$ not a rule
Example: Ladder at $x = 2.5$ : $\frac{1}{2}(0.3)\left(-\frac{1}{8}\right) = -\frac{3}{160}$ , but the true rate is $+\frac{119}{160}$ m²/s. What to do: Product rule: $A' = \frac{1}{2}(x'y + xy') = \frac{1}{2}\left(1.8 - 0.3125\right) = \frac{119}{160}$ .
cone, $r$ frozen at $1$	$V = \frac{\pi}{3}h$	$V' = \frac{\pi}{3}h'$ not a rule
Example: Gives $h' = \frac{3}{2}$ m/min, three times the true $\frac{1}{2}$ . What to do: Eliminate $r$ by $r = \frac{h}{3}$ , valid at every instant, then differentiate $V = \frac{\pi h^3}{27}$ .

Every correct line has the same structure: each variable is differentiated, and each derivative carries its factor $\frac{d(\cdot)}{dt}$ by the chain rule. A line where a variable has no rate next to it has frozen that variable.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Substituting the values of the instant before differentiating

the whole question

What not to write

“At $x = 2.5$ : $2.5^2 + y^2 = 42.25$ , so $2y\frac{dy}{dt} = 0$ and the top does not move.”

What to write

“ $x^2 + y^2 = 42.25$ at every instant, so $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ . At $x = 2.5$ , $y = 6$ : $\frac{dy}{dt} = -\frac{2.5}{6}(0.3) = -\frac{1}{8}$ m/s.”

Differentiating the relation follows the curve

x^2 + y^2 = 42.25

and its tangent of slope

-\frac{5}{12}

(2.5, 6)

; substituting first differentiates the flat line

y = 6

, of slope

0

Why: A value substituted early turns a variable into a constant, and the derivative of a constant is $0$ : the rate you were given disappears. The figure shows it: the relation is a curve with a slope, the frozen number is a flat line.

2. Dropping the sign of a decreasing quantity

1 mark, and every later line that uses the rate

What not to write

“The top slides down at $\frac{1}{8}$ m/s, so $\frac{dy}{dt} = \frac{1}{8}$ .”

What to write

“ $\frac{dy}{dt} = -\frac{1}{8}$ m/s: $y$ decreases, the top slides down at $\frac{1}{8}$ m/s.”

Why: A rate is the derivative of a named quantity, not a speed. In the two cars problem, taking $+60$ for a car that approaches the crossroads turns a stationary distance into one growing at $72$ km/h.

3. Freezing the radius of a cone instead of eliminating it

the whole question

What not to write

“ $V = \frac{1}{3}\pi r^2 h$ with $r = 1$ , so $\frac{dV}{dt} = \frac{\pi}{3}\frac{dh}{dt}$ and $\frac{dh}{dt} = \frac{3}{2}$ m/min.”

What to write

“By similar triangles $r = \frac{h}{3}$ , so $V = \frac{\pi h^3}{27}$ and $\frac{dV}{dt} = \frac{\pi h^2}{9}\frac{dh}{dt}$ ; at $h = 3$ , $\frac{dh}{dt} = \frac{1}{2}$ m/min.”

Why: $r$ grows with $h$ ; its value at the instant is not a relation. Only a relation valid at every instant, here the similar triangles, may be substituted before differentiating.

4. Believing that a constant inflow gives a constant rise

2 marks

What not to write

“The pump gives $\frac{\pi}{2}$ m³/min, so the level rises at the same speed all the time.”

What to write

“ $\frac{dV}{dt} = A\frac{dh}{dt}$ , so $\frac{dh}{dt} = \frac{1}{A}\frac{dV}{dt}$ : in the cone, $\frac{1}{2}$ m/min at $h = 3$ and $\frac{1}{8}$ m/min at $h = 6$ .”

Same inflow

\frac{\pi}{2}

m³/min: in the cylinder the level is a straight line; in the cone it climbs fast, then slows. At

t = 2

, where the cone's surface also has area

\pi

, both rise at

\frac{1}{2}

m/min.

Why: The level rises at the inflow divided by the area of the free surface. It is constant only when that area is, in a cylinder or a box.

5. Putting the diameter where the radius goes

2 marks

What not to write

“The diameter is $24$ cm, so $72\pi = 4\pi(24)^2\frac{dr}{dt}$ and $\frac{dr}{dt} = \frac{1}{32}$ .”

What to write

“ $D = 24$ means $r = 12$ : $72\pi = 4\pi(144)\frac{dr}{dt}$ , $\frac{dr}{dt} = \frac{1}{8}$ , and $\frac{dD}{dt} = 2\frac{dr}{dt} = \frac{1}{4}$ cm/s.”

Why: The instant must be translated into the variable of your relation before it is substituted, and the rate asked must be the rate of the quantity named in the question.

6. Differentiating an angle in degrees

the whole numerical answer

What not to write

“The beam turns at $3$ rev/min, that is $1080$ degrees per minute, so $\frac{dx}{dt} = 2\sec^2\theta \times 1080$ .”

What to write

“ $3$ rev/min $= 6\pi$ rad/min, so $\frac{dx}{dt} = 2\sec^2\theta\,(6\pi) = 24\pi$ km/min at $\theta = \frac{\pi}{4}$ .”

Why: $\frac{d}{d\theta}\tan\theta = \sec^2\theta$ is proved with angles in radians. In degrees an extra factor $\frac{\pi}{180}$ appears; converting first avoids it.

7. Answering the lengthening of the shadow when the tip is asked

2 marks

What not to write

“The shadow grows at $0.8$ m/s, so its tip moves at $0.8$ m/s.”

What to write

“The tip is at $x + s = \frac{5}{3}x$ from the post, so it moves at $\frac{5}{3}(1.2) = 2$ m/s.”

Why: The length of the shadow and the position of its tip are two different quantities; their rates differ by the speed of the walker. Name the quantity asked before differentiating anything.

8. Adding the speeds of two cars on perpendicular roads

the whole question

What not to write

“A drives at $60$ km/h and B at $45$ km/h, so the distance between them changes at $105$ km/h, or $15$ .”

What to write

“ $z\frac{dz}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt} = 8(45) + 6(-60) = 0$ : at this instant the distance is stationary.”

Why: Speeds add only along the line that joins the cars. On perpendicular roads, only Pythagoras differentiated with signed rates gives the rate of the distance.

Which method to choose

Which relation, by what the figure shows

Look at the figure before writing anything: its shape picks the relation

Three figures, three relations: two legs and a hypotenuse give Pythagoras; a fixed side

d

and an angle give

\tan\theta = \frac{h}{d}

; a ray over a post and a walker gives two similar triangles.

If two moving lengths at right angles, and the segment between their ends → Pythagoras, then differentiate each square by the chain rule

Example: ladder $x^2 + y^2 = 6.5^2$ ; cars $x^2 + y^2 = z^2$ ; kite $x^2 + 45^2 = s^2$
If a fixed length, a moving length, and an angle asked or given → the trigonometric ratio that contains the fixed side, angle in radians

Example: $\tan\theta = \frac{h}{4}$ ; lighthouse $x = 2\tan\theta$ , $\frac{dx}{dt} = 2\sec^2\theta\,\frac{d\theta}{dt}$
If a light ray over an object, or the section of a cone → similar triangles, to eliminate a variable before differentiating

Example: $\frac{s}{1.8} = \frac{x + s}{4.5}$ gives $s = \frac{2}{3}x$ ; cone $r = \frac{h}{3}$
If a volume or an area with two variables → reduce to one variable by a relation valid at every instant, then differentiate

Example: $V = \frac{1}{3}\pi r^2 h$ with $r = \frac{h}{3}$ becomes $\frac{\pi h^3}{27}$
If a solid whose shape changes with the level (sloped bottom) → write the volume piece by piece, and choose the piece of the instant

Example: pool: $V = 20h + 10h^2$ for $h \le 2$ , $V = 80 + 60(h - 2)$ above
If a product of two moving quantities (an area) → product rule, never the product of the rates

Example: $A = \frac{1}{2}xy$ , $A' = \frac{1}{2}(x'y + xy')$

Two branches can apply to one problem: the kite uses Pythagoras for the string and a trigonometric ratio for the angle. Write the relation you use, and why the figure gives it, on the first line.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up a related rates problem

When to use it: Any question that gives one rate and asks for another at a precise instant

1 Draw and label the figure: letters for what changes, numbers for what is fixed. State the given rate with its sign, $\frac{dx}{dt} = 0.3$ m/s, and the rate asked, $\frac{dy}{dt}$ when $x = 2.5$ .
2 Write the relation valid at every instant, naming its source: “by Pythagoras”, “by similar triangles”. Eliminate any extra variable now.
3 Differentiate with respect to $t$ , naming the chain rule, with every variable followed by its rate.
4 Compute the missing values of the instant from the relation, then substitute everything and solve for the rate asked.
5 Conclude in a sentence: the value, its sign read in words, the unit.

Concluding sentence

“By Pythagoras, $x^2 + y^2 = 6.5^2$ at every instant. Differentiating with respect to $t$ , $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ . When $x = 2.5$ , $y = 6$ , so $\frac{dy}{dt} = -\frac{2.5}{6}(0.3) = -\frac{1}{8}$ m/s: the top slides down at $\frac{1}{8}$ m/s.”

The trap: Substituting $x = 2.5$ into the relation before differentiating: the rate vanishes and the whole solution collapses.

Marking: Typically 2 marks for the figure and variables, 3 for the relation, 2 for the differentiation, 2 for the substitution and the value, 1 for the sentence with sign and unit.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The sign matches the motion

Look at the figure: is the quantity asked growing or shrinking? The sign of your answer must say the same.

A ladder whose foot slides out cannot have a top that climbs: $\frac{dy}{dt} = +\frac{1}{8}$ is wrong on sight.
The unit is the unit of the quantity over time

A rate of volume is in cubic units per time, a rate of angle in radians per time. A unit that does not fit reveals a missing factor.

$\frac{dh}{dt}$ in m³/min means the area of the surface was never divided out.
The level rate is the inflow over the surface

For any tank, the level rate times the area of the free surface must give back the inflow.

Cone at $h = 3$ : $\pi \times \frac{1}{2} = \frac{\pi}{2}$ ✓; the frozen radius answer $\frac{3}{2}$ gives $\frac{3\pi}{2}$ , three times the pump.
Two relations, one answer

When the figure offers two relations (tan and sin, trigonometry and Pythagoras), redo the key rate with the second one.

Lighthouse: $2\sec\theta\tan\theta\,(6\pi) = 12\sqrt 2\,\pi$ and $\frac{x}{z}\frac{dx}{dt} = \frac{2}{2\sqrt 2}(24\pi) = 12\sqrt 2\,\pi$ .

The typical problem, taken apart

The coffee filter and the pot: two containers, one flow

Coffee drains from a conical filter, vertex down, $12$ cm tall with a top radius of $6$ cm, into a cylindrical pot of radius $4$ cm, at the rate of $2\pi$ cm³/s. At the instant the coffee in the filter is $8$ cm deep, how fast is its level falling in the filter, and how fast is the level rising in the pot?

No calculator. Every step must be justified as on a MATH 140 final.

The filter's coffee has depth

h

and surface radius

r = \frac{h}{2}

; the pot's level

y

rises in a cylinder of fixed radius

4

cm.

Step 1

Variables: $h$ , the depth in the filter, $y$ , the level in the pot, both functions of $t$ in seconds. Fixed: $12$ , $6$ , $4$ . Given: the filter loses $2\pi$ cm³/s and the pot gains $2\pi$ cm³/s.

Why

The same flow has two signs: $-2\pi$ for the filter, $+2\pi$ for the pot. Writing both now avoids the sign error at the end.

Step 2

Similar triangles in the filter: $\frac{r}{h} = \frac{6}{12}$ , so $r = \frac{h}{2}$ at every instant, and $V_f = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$ .

Why

The cone formula has two variables and one rate is given: the extra variable is eliminated by a relation valid at every instant, never by its value at $h = 8$ .

Step 3

Differentiate: $\frac{dV_f}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$ . At $h = 8$ : $-2\pi = 16\pi\frac{dh}{dt}$ , so $\frac{dh}{dt} = -\frac{1}{8}$ cm/s.

Why

Differentiate first, substitute after. The negative sign says the level falls, which is what the question asked.

Step 4

In the pot, $V_p = \pi(4)^2 y = 16\pi y$ , so $\frac{dV_p}{dt} = 16\pi\frac{dy}{dt}$ , and $2\pi = 16\pi\frac{dy}{dt}$ gives $\frac{dy}{dt} = \frac{1}{8}$ cm/s.

Why

A cylinder has a constant cross-section, so its level rate does not depend on the instant: no value of $y$ is needed.

Step 5

Check: at $h = 8$ the filter's surface radius is $r = 4$ , equal to the pot's radius; the two surfaces have the same area $16\pi$ , so the two levels must move at the same speed, in opposite directions. $\frac{1}{8}$ and $-\frac{1}{8}$ ✓.

Why

$\frac{dV}{dt} = A\frac{dh}{dt}$ in both containers gives a check in one line, without redoing the computation.

The conclusion, written out

“At the instant the filter holds $8$ cm of coffee, its level falls at $\frac{1}{8}$ cm/s, $\frac{dh}{dt} = -\frac{1}{8}$ , and the level in the pot rises at $\frac{1}{8}$ cm/s.”

The classic mistake on this problem: Replacing $r$ by $4$ in $V_f = \frac{1}{3}\pi r^2 h$ before differentiating, which gives $\frac{dh}{dt} = -\frac{3}{8}$ , three times too fast; or forgetting that the filter LOSES the coffee and giving $+\frac{1}{8}$ for both levels.

Learn by heart

• Letters for what changes, numbers for what is fixed.
• Relation valid at EVERY instant, then $\frac{d}{dt}$ , THEN the values of the instant.
• Every variable differentiated gets its rate: $\frac{d}{dt}\left(y^2\right) = 2y\frac{dy}{dt}$ .
• Extra variable: eliminate it by a relation (similar triangles), never by its value.
• Decreasing quantity: negative derivative. Angles in radians: $1$ rev $= 2\pi$ rad.
• $\frac{dV}{dt} = A\frac{dh}{dt}$ : the level rate is the flow over the area of the surface.
• $\sec^2\theta = 1 + \tan^2\theta$ : read it on the figure, no angle needed.

Frequently asked questions

When do I plug in the numbers in a related rates problem?

Only after differentiating. Write the relation that is true at every instant, differentiate it with respect to time, and only then replace the variables by their values at the instant asked. A value put in before differentiating turns a variable into a constant, and its rate becomes zero, which is why the ladder would seem not to move.

How do I handle a cone in a related rates problem?

The cone volume has two variables, the radius and the height of the liquid. Use similar triangles between the liquid and the whole cone to write the radius as a fixed multiple of the height, substitute that relation into the volume, and only then differentiate. Never replace the radius by its value at the instant.

Why is my related rate negative?

Because the quantity it measures is decreasing. A negative derivative is not an error: the top of a sliding ladder moves down, a draining tank loses volume, a car approaching a crossroads gets closer. Read the sign in words in your conclusion, and check it against the figure.

Do I have to use radians in related rates with angles?

Yes. The derivatives of sine, cosine and tangent hold only for angles measured in radians, so convert every angle rate first: one revolution is two pi radians. If the question wants degrees per second, convert your final answer back by multiplying by 180 over pi.

Practise it

Corrected exercises: Related rates, MATH 140 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Logarithmic differentiation Next sheet Linear approximation and differentials

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-related-rates. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: related rates (MATH 140)

The essentials

The five gestures, always in this order

Where the relation comes from

The rules in table form

The standard relations, differentiated with respect to time

The mistakes that cost marks

1. Substituting the values of the instant before differentiating

2. Dropping the sign of a decreasing quantity

3. Freezing the radius of a cone instead of eliminating it

4. Believing that a constant inflow gives a constant rise

5. Putting the diameter where the radius goes

6. Differentiating an angle in degrees

7. Answering the lengthening of the shadow when the tip is asked

8. Adding the speeds of two cars on perpendicular roads

Which method to choose

Which relation, by what the figure shows

How the answer is expected to be written

Writing up a related rates problem

Check before you hand in

The typical problem, taken apart

The coffee filter and the pot: two containers, one flow

Learn by heart

Frequently asked questions

When do I plug in the numbers in a related rates problem?

How do I handle a cone in a related rates problem?

Why is my related rate negative?

Do I have to use radians in related rates with angles?

Corrected exercises: Related rates, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?