MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: related rates (MATH 203)

This sheet is not a summary of section 3.10 of Thomas: you already have the course notes. It answers one question only, what makes students lose marks on related rates in MATH 203 at Concordia University, and which precise gesture avoids each loss.

In a related rates problem the differentiation is almost never the difficulty: it is always a chain rule in time. The marks go in the algebra that comes before and after it, the algebra of the instant. A scientific calculator is allowed on the exam; it computes a square root, it does not tell you which one to take.

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The thread of the chapter

The calculus of a related rates problem is one line, the chain rule in time; the marks are lost in the algebra of the instant: rewriting the relation in the variable of the question before differentiating, computing the companion values from the relation, and isolating a rate that always appears to the first power, with its sign.

This chapter is part of MATH 203, Calculus I (Concordia)

The essentials

The method, in this order

• Draw the figure. A LETTER for every quantity that changes ( $x$ , $y$ , $h$ , $D$ , $\theta$ ), a NUMBER only for what stays fixed (the ladder, the depth of the tank, the altitude of the plane).
• Write the relation that holds at EVERY instant: Pythagoras, similar triangles, a trigonometric ratio, a formula of geometry or of physics.
• Rewrite it in the variable the question names, and eliminate any extra variable by another relation valid at every instant.
• Differentiate with respect to $t$ : every variable gets its rate by the chain rule, $\frac{d}{dt}\left(y^2\right) = 2y\frac{dy}{dt}$ .
• Only NOW substitute the values of the instant, including the companion values computed from the relation; isolate the rate; conclude with its sign and its unit.

Ladder

2.6

m, top falling at

0.25

m/s: the foot moves at

0.6

m/s when the top is

2.4

m high, at

0.104

m/s when it is

1

m high. The instant decides, and it must be read on the right variable.

The relation and its derivative are worth marks even when the arithmetic fails: write them on the first two lines, before any number of the instant.

The algebra of the instant: three places where the marks go

• BEFORE differentiating: rewrite. $S = 4\pi\left(\frac{D}{2}\right)^2 = \pi D^2$ ; $r = \frac{3h}{4}$ gives $r^2 = \frac{9h^2}{16}$ ; $\frac{1.8u}{u - 2} = 1.8 + \frac{3.6}{u - 2}$ ; $\frac{1}{R} = R^{-1}$ .
• AT the instant: compute the missing values from the relation. $x^2 + 2.4^2 = 2.6^2$ gives $x = \sqrt{6.76 - 5.76} = 1$ ; $D^2 = 9^2 + 8^2 + 12^2$ gives $D = 17$ . Translate the instant into your variables: $6$ m from the actor is $u = 8$ from the wall.
• AFTER differentiating: the unknown rate appears to the first power. Collect it on one side and divide: $\frac{dx}{dt} = -\frac{y}{x}\frac{dy}{dt}$ , $\frac{dR}{dt} = R^2\left(\frac{R_1'}{R_1^2} + \frac{R_2'}{R_2^2}\right)$ .
• Round only at the end: keep $\sqrt{3400}$ or $\frac{2}{15\pi}$ in the calculator, then give the decimals the question asks for.

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.

Differentiating each kind of term with respect to time

Read a line as: this term in the relation, rewritten this way, has this derivative in $t$ . The red lines are derivatives that students write and that do not exist.

Term	Rewrite first	Derivative in t
$x^2$	$x^2$	$2x\,x'$
Example: Ladder: $2(1)x' + 2(2.4)(-0.25) = 0$ gives $x' = 0.6$ m/s.
$\frac{1}{R}$	$R^{-1}$	$-\frac{R'}{R^2}$
Example: $R = 24$ , $R' = -0.024$ : $-\frac{-0.024}{576} = \frac{1}{24000}$ , which is $\frac{0.2}{1600} - \frac{0.3}{3600}$ ✓.
$PV$	$PV$	$P'V + PV'$
Example: $20(600) + 150(-80) = 12000 - 12000 = 0$ ✓.
$V^{1.4}$	$V^{1.4}$	$1.4V^{0.4}V'$
Example: In $PV^{1.4} = C$ : $V' = -\frac{V}{1.4P}P' = -\frac{600(20)}{210} \approx -57.14$ cm³/min.
$\frac{1.8u}{u - 2}$	$1.8 + \frac{3.6}{u - 2}$	$-\frac{3.6\,u'}{(u - 2)^2}$
Example: $u = 8$ , $u' = -0.5$ : $-\frac{3.6(-0.5)}{36} = 0.05$ m/s.
$PV$	$PV$	$P'V'$ not a rule
Example: $20 \times (-80) = -1600 \ne 0$ , although $PV$ is constant. What to do: Product rule: $P'V + PV' = 0$ , then isolate $V' = -\frac{V}{P}P'$ .
$\frac{1}{R}$	$R^{-1}$	$\frac{1}{R'}$ not a rule
Example: $\frac{1}{0.3} + \frac{1}{-0.2} = -\frac{5}{3}$ gives $R' = -0.6$ , twenty-five times the true $-0.024$ . What to do: Power rule on $R^{-1}$ : $-\frac{R'}{R^2}$ , then multiply by $-R^2$ to isolate $R'$ .

On every correct line, each variable is followed by its own rate. A line where a rate multiplies another rate, or sits in a denominator, has invented a rule.

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 $y = 2.4$ : $x^2 + 5.76 = 6.76$ , so $2x\frac{dx}{dt} = 0$ and the foot does not move.”

What to write

“ $x^2 + y^2 = 6.76$ at every instant, so $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ . At $y = 2.4$ , $x = 1$ : $\frac{dx}{dt} = -\frac{2.4}{1}(-0.25) = 0.6$ m/s.”

Why: A value put in early turns a variable into a constant, whose derivative is $0$ : the rate you were given disappears from the equation.

2. Splitting the square root of a difference

2 marks, and every value computed after it

What not to write

“The top is $2.4$ m high, so the foot is $2.6 - 2.4 = 0.2$ m from the wall, and $\frac{dx}{dt} = 3$ m/s.”

What to write

“ $x = \sqrt{2.6^2 - 2.4^2} = \sqrt{6.76 - 5.76} = 1$ m, so $\frac{dx}{dt} = 0.6$ m/s.”

Left, legs

1

and

2.4

close the

2.6

m ladder. Right, the companion value

2.6 - 2.4 = 0.2

gives a hypotenuse of

2.41

: it is not the same ladder.

Why: $\sqrt{a^2 - b^2} \ne a - b$ . The figure settles it: legs $0.2$ and $2.4$ close a triangle whose hypotenuse is $2.41$ , not the $2.6$ m ladder.

3. Using the diameter where the similar triangles need the radius

the whole question

What not to write

“The top is $6$ m across and $4$ m high, so $\frac{r}{h} = \frac{6}{4}$ and $r = 1.5h$ .”

What to write

“The top radius is $3$ m, so $\frac{r}{h} = \frac{3}{4}$ , $r = \frac{3h}{4}$ and $V = \frac{3\pi h^3}{16}$ .”

The water's radius

r

and the tank's radius

R = 3

are both measured from the axis: the proportion is

\frac{r}{h} = \frac{3}{4}

, never

\frac{6}{4}

Why: $r$ is the half-width of the water, so it is compared with the half-width of the tank. With the diameter the volume is four times too large and the level rate four times too small.

4. Squaring only the numerator of a fraction

the whole question

What not to write

“ $r = \frac{3h}{4}$ , so $r^2 = \frac{3h^2}{16}$ and $V = \frac{\pi h^3}{16}$ .”

What to write

“ $r^2 = \left(\frac{3h}{4}\right)^2 = \frac{9h^2}{16}$ , so $V = \frac{1}{3}\pi \cdot \frac{9h^2}{16} \cdot h = \frac{3\pi h^3}{16}$ .”

Why: $\left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}$ : every factor is squared. The volume comes out three times too small, and the level rate three times too large, $-0.1273$ instead of $-0.0424$ m/min at $h = 2$ .

5. Writing a negative rate as positive

1 mark, and every later line that uses the rate

What not to write

“The top slides down at $0.25$ m/s, so $\frac{dy}{dt} = 0.25$ .”

What to write

“ $y$ decreases, so $\frac{dy}{dt} = -0.25$ m/s; the top slides down at $0.25$ m/s.”

Why: A rate is the derivative of a named quantity, not a speed. With the wrong sign the foot of the ladder comes out moving toward the wall, which the figure forbids.

6. Taking the product of the rates for the derivative of a product

the whole question

What not to write

“ $PV = C$ , so $\frac{dP}{dt}\frac{dV}{dt} = 0$ : the volume must be constant.”

What to write

“ $\frac{dP}{dt}V + P\frac{dV}{dt} = 0$ , so $\frac{dV}{dt} = -\frac{V}{P}\frac{dP}{dt} = -\frac{600}{150}(20) = -80$ cm³/min.”

Why: The derivative of a product is not the product of the derivatives. The same slip appears on an area $A = \frac{1}{2}xy$ or on the law of cosines.

7. Taking the reciprocal of a derivative for the derivative of a reciprocal

the whole question

What not to write

“ $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ , so $\frac{1}{R'} = \frac{1}{0.3} + \frac{1}{-0.2}$ and $R' = -0.6$ ohm/s.”

What to write

“ $-\frac{R'}{R^2} = -\frac{R_1'}{R_1^2} - \frac{R_2'}{R_2^2}$ , so $R' = 576\left(\frac{0.3}{3600} - \frac{0.2}{1600}\right) = -0.024$ ohm/s.”

Why: Rewrite $\frac{1}{R}$ as $R^{-1}$ and use the power rule. If one resistance were constant, the false line would divide by zero.

8. Leaving a rate in the time unit of the data

1 to 2 marks

What not to write

“ $\frac{d\theta}{dt} = 72$ , so the antenna turns at $72$ rad/min.”

What to write

“ $t$ is in hours, so $\frac{d\theta}{dt} = 72$ rad/h $= 1.2$ rad/min $= 0.02$ rad/s, about $1.15$ degrees per second.”

Why: A derivative has the unit of the quantity over the unit of $t$ . Speeds in km/h give rates per hour; convert at the end, and convert km/h to m/s by dividing by $3.6$ before mixing with metres.

Which method to choose

What to do with the relation before differentiating, by its form

Look at the relation you have just written: its form tells you the algebra to do first

If one variable too many (a cone with $r$ and $h$ ) → eliminate it by similar triangles, a relation valid at every instant

Example: $\frac{r}{h} = \frac{3}{4}$ , $V = \frac{1}{3}\pi\left(\frac{3h}{4}\right)^2h = \frac{3\pi h^3}{16}$
If a formula in a variable other than the one the question names → rewrite it in that variable

Example: question on the diameter: $S = 4\pi\left(\frac{D}{2}\right)^2 = \pi D^2$
If the variable on top and at the bottom of a fraction → divide first, or use the quotient rule in the right order

Example: $\frac{1.8u}{u - 2} = 1.8 + \frac{3.6}{u - 2}$
If reciprocals or a square root → negative or fractional exponent, then power rule with the chain rule

Example: $\frac{1}{R} = R^{-1}$ ; $\sqrt{x^2 + 64} = (x^2 + 64)^{1/2}$
If a product of two varying quantities → product rule, never the product of the rates

Example: $PV = C$ : $P'V + PV' = 0$
If a constant added (a height, a fixed side) → it differentiates to $0$ but stays in the relation for the companion values

Example: $D^2 = x^2 + y^2 + 144$ : $D = 17$ , not $\sqrt{145}$

If no branch applies, the relation is already ready: differentiate it as it is. And whatever the branch, no value of the instant enters before the differentiation.

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 Label the figure: letters for what changes, numbers for what is fixed. State the given rate WITH its sign, $\frac{dy}{dt} = -0.25$ m/s, and the rate asked.
2 Write the relation valid at every instant and name its source: “by Pythagoras”, “by similar triangles”. Rewrite it in the variable of the question.
3 Differentiate with respect to $t$ , naming the chain rule, the product rule or the quotient rule used.
4 Compute the companion values of the instant from the relation, substitute, and isolate the rate.
5 Conclude in a sentence: the value, its sign read in words, the unit, and the rounding asked.

Concluding sentence

“By Pythagoras, $x^2 + y^2 = 2.6^2$ at every instant. Differentiating with respect to $t$ , $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ . When $y = 2.4$ , $x = \sqrt{6.76 - 5.76} = 1$ , so $\frac{dx}{dt} = -\frac{2.4}{1}(-0.25) = 0.6$ m/s: the foot moves away from the wall at $0.6$ m/s.”

The trap: Computing the companion value as $2.6 - 2.4$ , or substituting $y = 2.4$ before differentiating.

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 foot that slides away cannot have $\frac{dx}{dt} < 0$ ; a draining tank cannot have $\frac{dh}{dt} > 0$ .
The companion values satisfy the relation

Put the values of the instant back into the relation before using them. Ten seconds with the calculator.

$1^2 + 2.4^2 = 6.76 = 2.6^2$ ✓; with $x = 0.2$ , $0.04 + 5.76 = 5.8 \ne 6.76$ : the value is wrong.
Nudge the instant with the calculator

Move the given variable by its rate times a small time step, recompute the other one from the relation, and divide the change by the step.

$y = 2.4 - 0.25(0.01) = 2.3975$ gives $x = \sqrt{6.76 - 2.3975^2} \approx 1.00598$ : the foot moved $0.00598$ m in $0.01$ s, about $0.6$ m/s ✓.
The unit is the unit of the quantity over the unit of time

A rate of volume is in cubic units per time, an angle rate in radians per time, in the time unit of the data.

$\frac{dh}{dt}$ in m³/min means the area was never divided out; $72$ computed from km/h is per HOUR.

The typical problem, taken apart

The runner and home plate

A softball diamond is a square $18$ m on a side. A runner goes from second base to third base at $7.5$ m/s. How fast is her distance to home plate changing at the instant she is $13.5$ m from third base?

Scientific calculator allowed. Every step must be justified as on a MATH 203 final.

The runner is

x

from third base on the side from second to third, perpendicular to the side from third to home: her distance

D

to home is a hypotenuse.

Step 1

Variables: $x$ , the distance from the runner to third base, and $D$ , her distance to home plate, in metres, functions of $t$ in seconds. Fixed: the side $18$ m. She runs TOWARD third base, so $x$ decreases: $\frac{dx}{dt} = -7.5$ m/s.

Why

The sign is decided by the figure before any computation; a positive $7.5$ would send her back toward second base.

Step 2

The side from second to third is perpendicular to the side from third to home, so by Pythagoras $D^2 = x^2 + 18^2 = x^2 + 324$ at every instant.

Why

Naming the source of the relation, “by Pythagoras”, is worth a mark on its own, and it holds at every instant, which is what allows the differentiation.

Step 3

Differentiate with respect to $t$ , chain rule on both squares: $2D\frac{dD}{dt} = 2x\frac{dx}{dt}$ , so $\frac{dD}{dt} = \frac{x}{D}\frac{dx}{dt}$ .

Why

The constant $324$ differentiates to $0$ but stays in the relation: it will be needed for the companion value.

Step 4

Companion value: $D = \sqrt{13.5^2 + 324} = \sqrt{182.25 + 324} = \sqrt{506.25} = 22.5$ m. Then $\frac{dD}{dt} = \frac{13.5}{22.5}(-7.5) = 0.6(-7.5) = -4.5$ m/s.

Why

The square root of a sum is computed as one number, never split into $13.5 + 18$ ; the rate is isolated before the calculator is used.

Step 5

Check by nudging: after $0.01$ s, $x = 13.425$ and $D = \sqrt{13.425^2 + 324} \approx 22.4551$ , a change of about $-0.0449$ m in $0.01$ s, that is $-4.49$ m/s ✓.

Why

Ten seconds with the calculator confirm both the sign and the size, without redoing the calculus.

The conclusion, written out

“At the instant she is $13.5$ m from third base, the runner's distance to home plate decreases at $4.5$ m/s: $\frac{dD}{dt} = -4.5$ m/s.”

The classic mistake on this problem: Taking $\frac{dx}{dt} = +7.5$ , which gives a distance that increases at $4.5$ m/s, or computing $D = 13.5 + 18 = 31.5$ , which gives $-3.21$ m/s.

Learn by heart

• Letters for what changes, numbers for what is fixed.
• Relation at EVERY instant, rewritten in the variable asked, then $\frac{d}{dt}$ , THEN the values of the instant.
• Companion values come from the relation: $\sqrt{a^2 - b^2} \ne a - b$ .
• Similar triangles compare radius with radius: half the diameter.
• $\left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}$ ; $(PV)' = P'V + PV'$ ; $\left(\frac{1}{R}\right)' = -\frac{R'}{R^2}$ .
• Decreasing quantity: negative rate. Unit of a rate: quantity over the time unit of the data.

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 becomes a constant, its rate becomes zero, and the rate you were given disappears from the equation.

How do I find the missing value at the instant in a related rates problem?

From the relation itself, not from the question. If the ladder's top is 2.4 m high and the ladder is 2.6 m long, the foot is at the square root of 6.76 minus 5.76, that is 1 m, and not at 2.6 minus 2.4. Put the value back into the relation to check it before you use it.

Why do I get the wrong answer for a conical tank?

Usually because of the algebra before the derivative. Use the radius of the top, half the diameter, in the similar triangles, and square the whole fraction when you substitute the radius into the cone volume: three h over four, squared, is nine h squared over sixteen. Then differentiate the volume, which now depends on the depth alone.

Can I use my calculator for related rates in MATH 203?

Yes, the approved scientific calculator is allowed, and decimals are expected when the context asks for them. Use it only at the end: keep exact square roots and fractions in memory until the last line, then round as the question says. It is also a good check: nudge the instant by a small time step and see whether the change matches your rate.

Practise it

Corrected exercises: Related rates, MATH 203 at Concordia

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

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-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 203)

The essentials

The method, in this order

The algebra of the instant: three places where the marks go

The rules in table form

Differentiating each kind of term with respect to time

The mistakes that cost marks

1. Substituting the values of the instant before differentiating

2. Splitting the square root of a difference

3. Using the diameter where the similar triangles need the radius

4. Squaring only the numerator of a fraction

5. Writing a negative rate as positive

6. Taking the product of the rates for the derivative of a product

7. Taking the reciprocal of a derivative for the derivative of a reciprocal

8. Leaving a rate in the time unit of the data

Which method to choose

What to do with the relation before differentiating, by its form

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 runner and home plate

Learn by heart

Frequently asked questions

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

How do I find the missing value at the instant in a related rates problem?

Why do I get the wrong answer for a conical tank?

Can I use my calculator for related rates in MATH 203?

Corrected exercises: Related rates, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?