Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: related rates (201-SN2-RE)

Q: How do you solve a related rates problem?

Draw a diagram, name the quantities that depend on time, write the known rates and the rate you are looking for, then write a relation between these quantities that holds at every instant. Differentiate this relation with respect to time and substitute the numerical values only at the end.

Q: Why should you not plug in the values before differentiating?

Because a numerical value describes a single instant. By substituting it, you turn a function of time into a constant, and the derivative of a constant is zero. The solution then proves that the quantity does not change, which the problem explicitly contradicts.

Q: What do you do when two unknowns remain in the equation?

It is the sign that a geometric relation has not been used. In a cone, the similar triangles link the radius to the height; under a street lamp, it is also similar triangles. Write this relation, use it to eliminate a variable, then differentiate.

Q: Why does the level rise more slowly in a cone that is filling up?

Because the free surface widens as the level rises. The same volume per minute has to spread over a larger cross-section, so it raises the level by less. The rising speed is inversely proportional to the area of the cross-section, hence to the square of the height in a cone.

Q: Do you have to convert angles to radians?

Yes, always. The differentiation formulas for trigonometric functions hold only in radians, because they rest on a limit that equals one only in that unit. One revolution is two pi radians, and one degree is pi over one hundred eighty radians. Forgetting the conversion throws the result off by a factor of about fifty-seven.

This sheet does not explain the chain rule again: you know it from the previous chapter. It deals with what decides the mark on related rates in 201-SN2-RE (formerly 201-NYA), where students lose their marks before the first derivative: in translating the problem into a relation, in the similar triangles they did not see, and in the quantity whose rate they computed when it was not the one asked for.

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

A related rate is a chain rule whose intermediate variable is TIME: you differentiate a relation that holds at every instant, never a numerical value. Substituting before differentiating turns a function into a constant, and a constant has a zero derivative.

This chapter is part of 201-SN2-RE, Differential Calculus

The essentials

What varies, what does not

• A VARIABLE quantity gets the name of a function of time, $x(t)$ , and will be differentiated.
• A CONSTANT quantity gets a number, and will come out of the derivative or disappear.
• The relation to differentiate must hold at EVERY instant. “ $x=3$ ” does not: it is a value, it cannot be differentiated.
• Differentiate the LITERAL relation first, substitute the numbers only afterwards.

The ladder in two successive positions: its length,

5

, is the same at both instants, while

x

and

y

have both changed. Only the relation

x^{2}+y^{2}=25

, true at both instants, can be differentiated.

Setting out this list in two columns before computing earns the method mark and avoids the central error of the chapter.

The method, in six steps

• 1. Draw a DIAGRAM and name the quantities.
• 2. Separate what varies from what stays fixed.
• 3. Write the known rates and the rate you are looking for in the notation $\frac{d\,\cdot}{dt}$ .
• 4. Write a relation between the quantities that holds at every instant, and keep only one variable in it if possible.
• 5. Differentiate this relation with respect to $t$ .
• 6. Only then substitute the values of the instant in question, and conclude with the unit and the direction.

If TWO variables remain and only one rate is known, a geometric relation has not been used: similar triangles, Pythagoras, a proportion.

Time derivatives to know by heart

• $\frac{d}{dt}\left(x^{2}\right)=2x\frac{dx}{dt}$ and $\frac{d}{dt}\left(x^{3}\right)=3x^{2}\frac{dx}{dt}$ .
• $\frac{d}{dt}(xy)=x\frac{dy}{dt}+y\frac{dx}{dt}$ : two variable quantities, hence the product rule.
• $\frac{d}{dt}\left(\tan\theta\right)=\sec^{2}\theta\cdot\frac{d\theta}{dt}$ , in RADIANS.
• Sphere: $V=\frac{4}{3}\pi r^{3}$ gives $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$ ; the surface area $S=4\pi r^{2}$ gives $\frac{dS}{dt}=8\pi r\frac{dr}{dt}$ .
• Cone with a fixed proportion $r=kh$ : $V=\frac{\pi k^{2}h^{3}}{3}$ gives $\frac{dV}{dt}=\pi k^{2}h^{2}\frac{dh}{dt}$ .

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 numerical value before differentiating

the whole question

What not to write

“ $r=5$ , so $V=\frac{4}{3}\pi(5)^{3}=\frac{500\pi}{3}$ , so $\frac{dV}{dt}=0$ .”

What to write

“ $V=\frac{4}{3}\pi r^{3}$ , so $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$ ; substituting $r=5$ and $\frac{dV}{dt}=100$ , we get $\frac{dr}{dt}=\frac{1}{\pi}$ .”

Why: Substituting first replaces a function of time by a constant, whose derivative is zero: the solution then proves that the volume does not change, which the problem contradicts. A value describes one instant, a relation describes all instants, and only a relation can be differentiated.

2. Keeping two variables for lack of seeing the similar triangles

the whole question: one equation in two unknowns cannot be solved

What not to write

“ $V=\frac{1}{3}\pi r^{2}h$ , so $\frac{dV}{dt}=\frac{\pi}{3}\left(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}\right)$ , and I am stuck.”

What to write

“The triangles are similar, so $\frac{r}{h}=\frac{3}{6}$ and $r=\frac{h}{2}$ ; then $V=\frac{\pi h^{3}}{12}$ and $\frac{dV}{dt}=\frac{\pi h^{2}}{4}\frac{dh}{dt}$ .”

At two different heights, the ratio of radius to height is the SAME: this proportion

r=\frac{h}{2}

is what eliminates a variable, and missing it is what gets the solution stuck.

Why: The equality written first is correct, but it contains two unknown rates for a single equation. The similarity is the second equation, and using it BEFORE differentiating halves the work. The general rule: two unknowns always signal an unused geometric relation.

3. Computing the rate of a quantity other than the one asked for

the whole question, even though the computation is right

What not to write

“The speed of the tip of the shadow is $\frac{ds}{dt}=\frac{3}{7}(1.5)\approx 0.64$ m/s.”

What to write

“ $\frac{ds}{dt}\approx 0.64$ m/s is the speed at which the shadow LENGTHENS; the tip of the shadow is at the distance $x+s$ , so it moves at $1.5+0.64=\frac{15}{7}\approx 2.14$ m/s.”

Two distances:

s

, the length of the shadow, grows at

0.64

m/s;

x+s

, the position of its tip, grows at

1.5+0.64\approx 2.14

m/s. Reread the question before concluding.

Why: Two very similar wordings refer to two different quantities. Underlining in the problem the exact quantity whose rate is asked for, before any computation, removes this loss for sure. The same trap exists between direct distance and horizontal distance, between volume and height, between length and position.

4. Differentiating an angle measured in degrees

1 mark, and every numerical value that follows

What not to write

“The angle turns by $30$ degrees per second, so $\frac{d\theta}{dt}=30$ .”

What to write

“ $\frac{d\theta}{dt}=30\times\frac{\pi}{180}=\frac{\pi}{6}$ rad/s.”

Why: The formulas $\left(\tan\theta\right)'=\sec^{2}\theta$ and $(\sin\theta)'=\cos\theta$ rest on $\lim_{u\to 0}\frac{\sin u}{u}=1$ , which equals $\frac{\pi}{180}$ in degrees. The error is a factor of about $57.3$ , which makes the answer absurd and easy to spot from its order of magnitude.

5. Taking a negative rate for a computational error

1 mark if you change the sign, more if you alter the computation

What not to write

“I get $\frac{dy}{dt}=-0.45$ , there must be a sign mistake.”

What to write

“ $\frac{dy}{dt}=-0.45$ m/s: the top of the ladder is going DOWN at $0.45$ m/s.”

The foot moves away (

\frac{dx}{dt}=0.6>0

), the top slides down (

\frac{dy}{dt}=-0.45<0

). The minus sign is the answer: it says that

y

decreases.

Why: The sign is information, not an anomaly: it says that the quantity decreases. The opposite error costs just as much, writing $\frac{dV}{dt}=20$ when the problem says “loses $20$ cm $^{3}$ per minute”. The verbs “lose”, “melt”, “descend”, “drain” impose a negative rate.

6. Treating a varying quantity as a constant

1 to 2 marks, and sometimes a result of the opposite sign

What not to write

“The volume of the cylinder is $\pi r^{2}h$ , so $\frac{dV}{dt}=2\pi rh\frac{dr}{dt}$ .”

What to write

“If $h$ varies too, $\frac{dV}{dt}=\pi\left(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}\right)$ .”

Why: A missing term can reverse the conclusion: with $r=6$ , $h=10$ , $\frac{dr}{dt}=0.3$ and $\frac{dh}{dt}=-0.5$ , the incomplete formula announces $36\pi$ whereas the true value is $18\pi$ . Step 2 of the method, separating variable from constant, exists precisely for this.

7. Believing that a constant flow rate gives a constant rise

the whole interpretation question

What not to write

“Water is poured in at $2$ m $^{3}$ /min, so the level always rises at the same speed.”

What to write

“In a cone, $\frac{dh}{dt}=\frac{4}{\pi h^{2}}\cdot\frac{dV}{dt}$ : the rise is four times slower at $4$ m than at $2$ m.”

Constant flow rate of

2

^{3}

/min into the cone: the height

h(t)

rises more and more slowly, because

\frac{dh}{dt}=\frac{8}{\pi h^{2}}

; at

h=4

, four times more slowly than at

h=2

Why: The rising speed is inversely proportional to the AREA of the free surface. It is constant only in a cylinder or a right prism. This is the most frequent interpretation question at the end of a problem, and it is answered with a formula, not with an impression.

Which method to choose

Which relation to write, depending on the configuration

The geometry of the problem points to the relation, and at CEGEP there are only five families of them.

If two perpendicular lengths and a distance between their ends → Pythagoras $x^{2}+y^{2}=z^{2}$ , then $x\frac{dx}{dt}+y\frac{dy}{dt}=z\frac{dz}{dt}$

Example: the ladder against the wall, two vehicles on perpendicular roads
If two nested triangles sharing an angle → similar triangles: set two ratios equal, which gives a LINEAR relation between the quantities

Example: the filling cone, the shadow under a street lamp
If an angle appears in the problem or in the question → right-triangle trigonometry, almost always $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

Example: the lighthouse sweeping a coast, the angle of elevation of a plane
If a solid fills, empties, inflates or melts → volume formula, after reducing to ONE variable with similar triangles

Example: spherical balloon, V-shaped trough, conical tank
If two paths forming an angle that is not a right angle → law of cosines $z^{2}=x^{2}+y^{2}-2xy\cos\alpha$ , differentiated with the product rule on $xy$

Example: two vehicles on roads at $60$ degrees

When the two objects start from the same point at constant speed, the speed at which they move apart is simply the magnitude of the relative velocity: $\sqrt{60^{2}+80^{2}}=100$ km/h, with no differentiation at all.

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.

A related rates problem, written up for the marker

When to use it: Any problem where one quantity varies in time and the rate of change of another is asked for.

1 Draw the diagram and name: “Let $x(t)$ be the distance from the foot to the wall and $y(t)$ the height reached on the wall.”
2 Write the data and the unknown in Leibniz notation: “Given $\frac{dx}{dt}=0.6$ m/s; find $\frac{dy}{dt}$ when $x=3$ .”
3 Write the relation and say why it holds at every instant: “The ladder, the wall and the ground form a right triangle, so $x^{2}+y^{2}=25$ at every instant.”
4 Differentiate the relation with respect to $t$ , without any numerical value.
5 Substitute, then conclude with a sentence that contains the unit AND the direction of change.

Concluding sentence

When the foot is $3$ m from the wall, $\frac{dy}{dt}=-0.45$ m/s: the top of the ladder is sliding down at $0.45$ metre per second.

The trap: Writing the numerical values into the relation as early as the third step. The marker looks for a literal relation that is differentiated, and the substitution last: it is the order that is graded, not only the final number.

Marking: Usually 1 mark for the diagram and notation, 1 mark for the relation, 1 mark for the differentiation, 1 mark for the substitution and the conclusion.

Check before you hand in

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

Check the sign against the situation described

Ask yourself whether the quantity you are looking for physically increases or decreases, then check that the sign you obtained says the same thing.

The top of the ladder slides down, so $\frac{dy}{dt}$ must be negative. A positive result signals an error in isolating, almost always a sign lost when moving a term to the other side.
Check that the units are consistent

A speed fed into a volume relation must produce a volume per unit of time, not a length. Write the units next to the numbers when you substitute.

$\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$ : cm $^{2}$ multiplied by cm/s does give cm $^{3}$ /s ✓.
Test a nearby instant

Recompute the quantity you are looking for at a slightly different instant and check that the change you observe has the right order of magnitude.

At $x=3$ we find $y=4$ ; at $x=3.1$ we find $y\approx 3.923$ , a drop of $0.077$ for a displacement of $0.1$ made in $\frac{0.1}{0.6}$ s, so about $-0.46$ m/s ✓.
Check that only one unknown remains

Before substituting, count the unknown rates in the differentiated equation. If there are two, a geometric relation is missing.

$\frac{dV}{dt}=\frac{\pi}{3}\left(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}\right)$ contains two unknowns: the similarity $r=\frac{h}{2}$ has not been used.

The typical problem, taken apart

The street lamp and the shadow, as on the exam

A street lamp is $6$ m tall. A $1.8$ m person walks away from it in a straight line at $1.5$ m/s.

How fast is the length of their shadow increasing, and how fast is the tip of the shadow moving away from the lamp?

Two different quantities are involved:

s

, the length of the shadow, and

x+s

, the distance from the lamp to the tip of the shadow. The question does not ask the same thing depending on the wording.

Step 1

Let $x(t)$ be the distance from the person to the foot of the lamp and $s(t)$ the length of their shadow. Given: $\frac{dx}{dt}=1.5$ m/s. The heights $6$ and $1.8$ are CONSTANT.

Why

Separate variable from constant on the very first line: the two heights will never be differentiated, and spotting them avoids introducing ghost rates.

Step 2

The large triangle (lamp, ground, tip of the shadow) and the small one (person, ground, tip of the shadow) are similar. Their bases are not the same: the large one has base $x+s$ , the small one $s$ . Hence $\frac{6}{x+s}=\frac{1.8}{s}$ .

Why

This is where the question is decided. Writing $\frac{6}{x}=\frac{1.8}{s}$ amounts to placing the lamp where the person stands, and the rest of the problem becomes wrong despite flawless computations.

Step 3

Cross-multiplying: $6s=1.8(x+s)$ , so $4.2s=1.8x$ and $s=\frac{3}{7}x$ .

Why

Reduce to ONE variable before differentiating, exactly as the similar triangles of the cone eliminate the radius. The relation obtained is linear, which already announces a constant rate.

Step 4

Differentiate with respect to $t$ : $\frac{ds}{dt}=\frac{3}{7}\cdot\frac{dx}{dt}=\frac{3}{7}(1.5)=\frac{4.5}{7}\approx 0.64$ m/s.

Why

Since the relation is linear, no position needs to be substituted: the rate does not depend on $x$ . That is what sets this problem apart from the cone, where the relation was cubic.

Step 5

The tip of the shadow is at the distance $x+s$ from the lamp, so its speed is $\frac{d}{dt}(x+s)=1.5+0.64=\frac{15}{7}\approx 2.14$ m/s.

Why

Second quantity, second rate. Reading the question word for word when you conclude is the only protection against the most frequent error of the chapter.

Step 6

Check: the two speeds are in the ratio $\frac{15/7}{4.5/7}=\frac{10}{3}$ , and the tip of the shadow necessarily moves faster than the person, who is in turn faster than the lengthening of the shadow. The order $0.64<1.5<2.14$ is consistent ✓.

Why

A qualitative check catches inversions: a shadow lengthening faster than its tip moves would be geometrically impossible.

The conclusion, written out

The shadow lengthens at $\frac{4.5}{7}\approx 0.64$ m/s, and its tip moves away from the lamp at $\frac{15}{7}\approx 2.14$ m/s. Both speeds are constant and do not depend on the person's position.

The classic mistake on this problem: Answering $0.64$ m/s to the question “how fast is the tip of the shadow moving”. The computation is right, the quantity is not, and the whole question is lost.

Learn by heart

• You differentiate a RELATION that holds at every instant, never a value. Substitution comes last.
• $\frac{d}{dt}\left(x^{2}\right)=2x\frac{dx}{dt}$ : the factor $\frac{dx}{dt}$ comes from the chain rule and must not be forgotten.
• Two unknowns in the differentiated equation: a similarity or a Pythagoras has not been used.
• Angles are differentiated in RADIANS. One revolution is $2\pi$ , and $1°=\frac{\pi}{180}$ .
• A negative rate means that the quantity decreases: it is information, not a mistake.
• Length of the shadow and position of its tip, direct distance and horizontal distance: read the quantity asked for.
• A constant flow rate does not mean a constant rise, except with a constant cross-section.

Frequently asked questions

How do you solve a related rates problem?

Draw a diagram, name the quantities that depend on time, write the known rates and the rate you are looking for, then write a relation between these quantities that holds at every instant. Differentiate this relation with respect to time and substitute the numerical values only at the end.

Why should you not plug in the values before differentiating?

Because a numerical value describes a single instant. By substituting it, you turn a function of time into a constant, and the derivative of a constant is zero. The solution then proves that the quantity does not change, which the problem explicitly contradicts.

What do you do when two unknowns remain in the equation?

It is the sign that a geometric relation has not been used. In a cone, the similar triangles link the radius to the height; under a street lamp, it is also similar triangles. Write this relation, use it to eliminate a variable, then differentiate.

Why does the level rise more slowly in a cone that is filling up?

Because the free surface widens as the level rises. The same volume per minute has to spread over a larger cross-section, so it raises the level by less. The rising speed is inversely proportional to the area of the cross-section, hence to the square of the height in a cone.

Do you have to convert angles to radians?

Yes, always. The differentiation formulas for trigonometric functions hold only in radians, because they rest on a limit that equals one only in that unit. One revolution is two pi radians, and one degree is pi over one hundred eighty radians. Forgetting the conversion throws the result off by a factor of about fifty-seven.

Practise it

Corrected exercises: Related rates, 201-SN2-RE

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.

16 corrected exercises
160 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc1-related-rates. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: related rates (201-SN2-RE)

The essentials

What varies, what does not

The method, in six steps

Time derivatives to know by heart

The mistakes that cost marks

1. Substituting the numerical value before differentiating

2. Keeping two variables for lack of seeing the similar triangles

3. Computing the rate of a quantity other than the one asked for

4. Differentiating an angle measured in degrees

5. Taking a negative rate for a computational error

6. Treating a varying quantity as a constant

7. Believing that a constant flow rate gives a constant rise

Which method to choose

Which relation to write, depending on the configuration

How the answer is expected to be written

A related rates problem, written up for the marker

Check before you hand in

The typical problem, taken apart

The street lamp and the shadow, as on the exam

Learn by heart

Frequently asked questions

How do you solve a related rates problem?

Why should you not plug in the values before differentiating?

What do you do when two unknowns remain in the equation?

Why does the level rise more slowly in a cone that is filling up?

Do you have to convert angles to radians?

Corrected exercises: Related rates, 201-SN2-RE

See also

Looking for a 201-SN2-RE tutor in Montreal?