Differential Calculus for Social Science 201-103 • CEGEP, Montreal

Revision sheet: related rates (201-103)

Q: Why do you have to differentiate before substituting in a related-rates problem?

Because a substituted value becomes a constant, and the derivative of a constant is zero. If you write that the price is twenty dollars before differentiating, the price stops varying in the calculation, and the whole term that describes its decline disappears. So you write the relation with letters, differentiate, and substitute only at the end.

Q: What is the chain factor in a differentiation with respect to time?

It is the speed of change of every quantity that is not time itself. Differentiating the square of a quantity with respect to time gives twice that quantity multiplied by its own speed of change. Forgetting this factor gives a wrong result, and the error is larger the faster the quantity varies.

Q: How do you know whether a per capita indicator is increasing?

You compare growth rates in percent, never absolute changes. The indicator increases if and only if the numerator grows faster, in percent, than the denominator. Comparing billions of dollars with thousands of people makes no sense, because they are different units.

Q: Why can a birth rate fall without there being fewer births?

Because it is a ratio between births and the total population. If the population grows, for example through immigration, the denominator gets bigger and the ratio falls, even when the number of births stays the same. Commenting on an indicator therefore always means looking at both of its terms.

Q: What is the difference between a stock and a flow?

A stock is a quantity measured at an instant, such as a population or an amount saved. A flow is a speed, measured per unit of time, such as births per year or dollars per month. The units always tell them apart: a flow contains a per year or a per month, a stock never does.

Q: When should you use relative rates rather than direct differentiation?

As soon as the answer is expected in percent, or the statement gives only percentages with no absolute values. The method is then shorter, since rates add for a product and subtract for a quotient. When both methods are possible, doing both serves as a check.

In Social Science, related rates are almost never about sliding ladders or inflating balloons: they are about revenue whose price falls while its volume rises, a population density whose numerator and denominator move together, a crude rate that its denominator alone is enough to make vary.

The method is nonetheless the same everywhere, and it comes down to five steps whose order is not negotiable. This sheet gives those five steps, the faster route of relative rates, the nine mistakes that come back every exam period, and the units check that catches almost all of them.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Differentiate the RELATION, then substitute: a value substituted too early becomes a constant, its derivative is zero, and the result is wrong without a single line of the calculation being wrong.

This chapter is part of 201-103, Differential Calculus for Social Science

The essentials

The five steps, in this order

• Name the variables that depend on time, and write down what you know: values and rates, with their units.
• Write the RELATION that links these variables to each other, without putting a single numerical value in it.
• Differentiate the relation with respect to TIME, each quantity bringing its chain factor.
• Substitute then, and only then, the numerical values of the instant considered.
• Conclude with a sentence that includes the UNITS, and check that reading them makes sense.

The two curves pass through the same point at the marked instant, and only one has a slope: replacing

n

with

40

before differentiating amounts to following the horizontal line.

The only order that causes trouble is that of steps 3 and 4. A value substituted before differentiating becomes a constant, and a constant has a zero derivative.

What differentiating with respect to time brings out

• Every quantity that is not $t$ brings a chain factor: $\dfrac{d}{dt}\left(n^{2}\right)=2n\,\dfrac{dn}{dt}$ .
• Product: $\dfrac{dz}{dt}=\dfrac{dx}{dt}\,y+x\,\dfrac{dy}{dt}$ .
• Quotient: $\dfrac{dz}{dt}=\dfrac{x'y-xy'}{y^{2}}$ , with the order imposed in the numerator.
• Chain of several links: $\dfrac{dW}{dt}=\dfrac{dW}{d\rho}\times\dfrac{d\rho}{dc}\times\dfrac{dc}{dt}$ , and the intermediate units cancel in pairs.
• $\dfrac{dA}{dt}$ and $\dfrac{dA}{dr}$ are not measured in the same unit: it is the variable of differentiation that fixes the denominator.

Write $\dfrac{dn}{dt}$ in full rather than $n'$ throughout the chapter: the notation reminds you by itself what you are differentiating with respect to.

The short route: relative rates

• Product: $\dfrac{(xy)'}{xy}=\dfrac{x'}{x}+\dfrac{y'}{y}$ . Relative rates ADD.
• Quotient: $\dfrac{(x/y)'}{x/y}=\dfrac{x'}{x}-\dfrac{y'}{y}$ . They SUBTRACT.
• A quotient increases if and only if $\dfrac{x'}{x}>\dfrac{y'}{y}$ : it is the order of the PERCENTAGES that decides, never that of the absolute values.
• A stock is differentiated into its flows: $\dfrac{dP}{dt}=B-D+M$ , births minus deaths plus net migration.
• A crude rate $\dfrac{1000B}{P}$ falls as soon as $P$ grows, even with $B$ constant: the denominator alone is enough to move the indicator.

The population is a STOCK, measured in people; births, deaths and migration are FLOWS, measured in people per year. The derivative of the stock is the sum of the flows coming in minus those going out.

On a composite indicator, this route is almost always faster than direct differentiation, and it gives, as a bonus, the result in percent, which is the form asked 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 a relation with respect to time

The reference values for all the examples: $p=20$ dollars, $q=500$ units, $\dfrac{dp}{dt}=-0.50$ dollar per week and $\dfrac{dq}{dt}=30$ units per week. The last column gives the rate obtained.

Relation	What you apply	Rate obtained
$z=xy$	the product rule	$x'y+xy'$
Example: $(-0.50)(500)+(20)(30)=-250+600=350$ dollars per week.
$z=\dfrac{x}{y}$	the quotient rule	$\dfrac{x'y-xy'}{y^{2}}$
Example: $\dfrac{(-0.50)(500)-(20)(30)}{500^{2}}=\dfrac{-850}{250{,}000}=-0.0034$ dollar per unit per week.
$z=n^{2}$	the chain rule	$2n\,\dfrac{dn}{dt}$
Example: With $n=40$ and $\dfrac{dn}{dt}=3$ : $2\times 40\times 3=240$ per unit of time.
$P$ stock, $B$ , $D$ , $M$ flows	the algebraic sum	$B-D+M$
Example: $900-620+150=430$ people per year.
relative rate of $xy$	the sum of the rates	$\dfrac{x'}{x}+\dfrac{y'}{y}$
Example: $-2.5$ percent $+\,6$ percent $=3.5$ percent per week, that is $350$ dollars out of $10{,}000$ .
$\dfrac{d}{dt}\left(n^{2}\right)=2n$	the power alone	does not exist nonexistent rule
Example: With $n=40$ and $\dfrac{dn}{dt}=3$ , the true value is $240$ , whereas this expression would give $80$ . What to do: Multiply by $\dfrac{dn}{dt}$ : every quantity other than $t$ brings its chain factor.

Lines 1 and 5 give the same result by two routes, $350$ dollars per week and $3.5$ percent of $10{,}000$ dollars. Doing both is the surest check of the chapter.

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 before differentiating

the whole question, and the error always gives zero

What not to write

“At this instant $n=40$ , so $A=n^{2}=1600$ , and $\dfrac{dA}{dt}=0$ .”

What to write

“ $A=n^{2}$ , so $\dfrac{dA}{dt}=2n\,\dfrac{dn}{dt}$ . Substituting afterwards, $2\times 40\times 3=240$ .”

Why: A substituted value becomes a constant, and a constant does not vary. That is why step 4 comes after step 3, and never before.

2. Forgetting the chain factor

2 marks, and a result that is wrong by a factor equal to the forgotten rate

What not to write

“ $\dfrac{d}{dt}\left(n^{2}\right)=2n=80$ .”

What to write

“ $\dfrac{d}{dt}\left(n^{2}\right)=2n\,\dfrac{dn}{dt}=2\times 40\times 3=240$ .”

Why: You differentiate with respect to $t$ , not with respect to $n$ . Each time a quantity other than time appears, its own speed of variation comes out as a factor.

3. Comparing absolute values to decide whether a quotient rises

2 marks, and sometimes a reversed conclusion

What not to write

“Gross domestic product rises by $3$ billion and the population by $40{,}000$ people, so the product per capita rises.”

What to write

“The quotient increases if and only if $\dfrac{x'}{x}>\dfrac{y'}{y}$ : you must compare the PERCENTAGES, not billions with people.”

Why: Two changes expressed in different units cannot be compared. Relative rates, on the other hand, have no unit and compare directly.

4. Believing that a crude rate moves only through its numerator

2 marks of interpretation, and it is the typical reasoning question

What not to write

“The number of births is stable, so the birth rate is too.”

What to write

“The crude rate is $\dfrac{1000B}{P}$ : with $B$ constant, it FALLS as soon as $P$ increases, without a single birth having been lost.”

The population grows in the same way for both curves, and yet the indicator rises in one case and falls in the other: only the numerator tells them apart.

Why: An indicator is a ratio, and a ratio has two terms. Half of the faulty commentary in Social Science comes from looking at only one of them.

5. Adding the relative rates of a quotient

2 marks, and a reversed sign in the conclusion

What not to write

“Sales grow by $3$ percent and the number of customers by $5$ percent, so the average basket grows by $8$ percent.”

What to write

“The average basket is a QUOTIENT: its relative rate is $3-5=-2$ percent. It FALLS.”

The average basket is the QUOTIENT of sales by the number of customers: its rate is

3-5=-2

percent (green). Adding,

3+5=8

(red dashed), treats the quotient as a product.

Why: $\ln\dfrac{x}{y}=\ln x-\ln y$ , so on differentiating the rates subtract. Adding amounts to treating a quotient as a product.

6. Mixing up the units of $\dfrac{dA}{dt}$ and $\dfrac{dA}{dr}$

1 mark per answer, and the error comes back in every question

What not to write

“The density changes by $12$ , so the city is becoming denser by $12$ people per square kilometre.”

What to write

“ $\dfrac{dD}{dt}$ is measured in people per square kilometre AND PER YEAR. $\dfrac{dD}{dr}$ would be measured in people per square kilometre per kilometre.”

Why: The unit of a derivative is that of the quantity divided by that of the VARIABLE OF DIFFERENTIATION. Changing the variable changes the unit, hence the meaning of the sentence.

7. Reading an unemployment rate without its participation rate

2 marks of interpretation, and it is the commentary question that is expected

What not to write

“The unemployment rate has fallen, so employment is improving.”

What to write

“The unemployment rate has fallen, but you must look at the participation rate: if people have stopped looking, the denominator has shrunk, and the indicator falls without a single job having been created.”

Why: This is the textbook case of related rates in Social Science: two opposite ways of lowering the same ratio, one good, the other bad.

8. Believing that the same effort produces the same gain everywhere

2 marks, and a useless public-policy recommendation

What not to write

“Lowering the utilization rate by two points always saves the same waiting time.”

What to write

“The wait varies as $\dfrac{1}{1-\rho}$ , so its sensitivity is $\dfrac{1}{(1-\rho)^{2}}$ : it equals $4$ at $50$ percent utilization and about $59$ at $87$ percent.”

The two marked points are on the same curve: on the left it rises gently, on the right it steepens, and the slope there is nearly fifteen times greater.

Why: A derivative is local. On a sharply curved graph, the same change in input gives changes in output of very different sizes depending on where you stand.

9. Confusing a stock and a flow

1 mark, and a concluding sentence that means nothing

What not to write

“The population of the municipality is $430$ people per year.”

What to write

“The population is a STOCK, measured in people; $430$ people per year is a FLOW, that is, $\dfrac{dP}{dt}$ .”

Why: The units say it by themselves: a stock has no “per year” in its unit, a flow always has one. It is the quickest check on an answer.

Which method to choose

Which relation to write, depending on what the statement gives

Step 2 is the only one where you have to think. Once the relation is written, everything else is mechanical.

If a revenue, a total budget, an overall expense → a PRODUCT, price times quantity

Example: $R=pq$ , so $R'=p'q+pq'$
If an average, a “per capita” or “per student” indicator, a rate → a QUOTIENT, and the relative-rate route is the shortest

Example: $\dfrac{\text{GDP}}{\text{population}}$ , relative rate $3-1.2=1.8$ percent
If a population, a headcount, a reserve that fills and empties → a STOCK, differentiated into the algebraic sum of its flows

Example: $\dfrac{dP}{dt}=B-D+M=900-620+150=430$
If an area, a volume, an urbanized surface → a geometric formula, with the chain factor on each quantity

Example: $A=\pi r^{2}$ gives $\dfrac{dA}{dt}=2\pi r\,\dfrac{dr}{dt}$
If a cascade of effects, each quantity driving another → a CHAIN of several links, to be written out in full before calculating

Example: $\dfrac{dW}{dt}=\dfrac{dW}{d\rho}\times\dfrac{d\rho}{dc}\times\dfrac{dc}{dt}$

the intermediate units cancel in pairs, which gives a free check

If the statement gives no relation, it is almost always in the vocabulary: “per capita” means quotient, “total” means product or sum.

Direct differentiation or relative rates: which to choose

The two routes give the same result. The choice depends on what the statement asks for and what it provides.

If the statement asks for an answer in units, dollars per week for example → direct differentiation, product or quotient

Example: $R'=350$ dollars per week
If the statement asks for an answer in PERCENT per unit of time → relative rates, added or subtracted

Example: $-2.5+6=3.5$ percent per week
If the statement gives only percentages, with no absolute value → relative rates, NECESSARILY: direct differentiation is impossible

Example: growth of $3$ percent and of $1.2$ percent
If the statement only asks whether the indicator rises or falls → compare the two relative rates, without computing the value

Example: $3>1.2$ , so the product per capita increases
If the statement gives values AND rates, and asks for a check → do BOTH routes: they must agree

Example: $3.5$ percent of $10{,}000$ gives back $350$

When both routes are possible, the second is faster and directly gives the form asked for in Social Science. The first remains indispensable as soon as a unit is expected.

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 in five steps

When to use it: As soon as the statement gives a speed of change and asks for another, at a precise instant.

1 Name the variables and list what is known, values AND rates, with the units of each.
2 Write the relation that links them, WITHOUT any numerical value in it.
3 Differentiate the relation with respect to $t$ , letting each $\dfrac{d\cdot}{dt}$ appear.
4 Now substitute the values of the instant considered.
5 Conclude with a sentence that has the units and interpret the sign.

Concluding sentence

“Let $R=pq$ . Differentiating with respect to time, $\dfrac{dR}{dt}=\dfrac{dp}{dt}\,q+p\,\dfrac{dq}{dt}$ . At this instant, $\dfrac{dR}{dt}=(-0.50)(500)+(20)(30)=350$ dollars per week: revenue is increasing, despite the drop in price.”

The trap: Writing the relation with the numbers already in it, for example $R=20q$ . The price then becomes a constant, its rate disappears, and a whole term is missing from the result.

Marking: 1 mark for the variables and their units, 1 mark for the literal relation, 2 marks for the differentiation, 1 mark for the interpreted conclusion.

Writing up a chain of several links

When to use it: As soon as one quantity drives a second that drives a third, and the statement asks for the effect of the first on the last.

1 Write the complete chain in Leibniz notation BEFORE calculating anything.
2 Give the unit of each link and check that the intermediate units cancel.
3 Compute each link separately, saying which one you are handling.
4 Multiply, then give the result with the unit that remains.
5 Interpret: say what investing one unit actually brings.

Concluding sentence

“ $\dfrac{dW}{dt}=\dfrac{dW}{d\rho}\times\dfrac{d\rho}{dc}\times\dfrac{dc}{dt}$ . The days per unit of $\rho$ , the units of $\rho$ per dollar and the dollars per year cancel to leave days per year.”

The trap: Adding the links instead of multiplying them. The units check detects it at once, since a sum of quantities with different units makes no sense.

Marking: 1 mark for the chain written out, 1 mark per link computed, 1 mark for the final unit and the interpretation.

Check before you hand in

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

The units, read aloud

Write the unit of the result, unit of the quantity divided by unit of the variable of differentiation, and read the whole sentence. If it cannot be said, the calculation is wrong.

“ $350$ dollars per week” can be read; “ $350$ dollars times weeks” cannot, and signals a multiplication in place of a division.
The two routes must agree

Compute by direct differentiation, then by relative rates, and compare. The two results must be the same number expressed in two ways.

$350$ dollars per week, and $3.5$ percent of $10{,}000$ dollars. A gap signals a term forgotten in the product rule.
The expected sign

Before any calculation, ask yourself whether the quantity should rise or fall according to the statement. The sign obtained must match, otherwise reread the order of the numerator of the quotient.

If the price falls faster in percent than the volume rises, revenue must fall. A positive result would then be inconsistent.
The chain factor, counted

Count how many quantities other than $t$ appear in the relation. There must be as many factors $\dfrac{d\cdot}{dt}$ in the derivative.

$A=\pi r^{2}$ contains a single varying quantity, so the derivative contains exactly one $\dfrac{dr}{dt}$ . Zero factors signals that the chain rule was forgotten.
The order of magnitude in percent

Convert the result to a percentage of the current level and ask whether it is plausible for a social phenomenon over the period considered.

A population that would grow by $40$ percent per year almost always signals a time-unit conversion error.

The typical problem, taken apart

The price falls, the volume rises: what does revenue do

A product currently sells for $20$ dollars a unit, at a rate of $500$ units per week.

The price falls by $0.50$ dollar per week and the quantity sold increases by $30$ units per week.

Does weekly revenue increase or decrease, and by how much? From what rate of price decline would it decrease?

Step 1

Variables: $p$ the price in dollars, $q$ the quantity in units per week, $R$ the revenue in dollars per week. Known: $p=20$ , $q=500$ , $\dfrac{dp}{dt}=-0.50$ dollar per week, $\dfrac{dq}{dt}=30$ units per week.

Why

Listing the units at the first step lets you know, at the end, what the result must be measured in without having to think.

Step 2

Relation: $R=pq$ , written without any number.

Why

This is where the question is decided. Writing $R=20q$ would make the price a constant and remove a whole term from the derivative.

Step 3

Differentiation with respect to time, product rule: $\dfrac{dR}{dt}=\dfrac{dp}{dt}\,q+p\,\dfrac{dq}{dt}$ .

Why

Both quantities depend on time, so each brings its term. A single term would signal that one of the two has been frozen.

Step 4

Substitution: $\dfrac{dR}{dt}=(-0.50)(500)+(20)(30)=-250+600=350$ .

Why

The first term is the price effect, negative, the second the volume effect, positive. Naming them makes the conclusion much easier to interpret.

Step 5

Units: dollars per unit times units per week, plus dollars times units per week, gives dollars per week.

Why

The check confirms that the product rule has been applied properly: two terms with different units could not be added.

Step 6

Check by relative rates: $\dfrac{-0.50}{20}+\dfrac{30}{500}=-0.025+0.060=0.035$ , that is $3.5$ percent per week. Now $3.5$ percent of $R=10{,}000$ dollars is indeed $350$ dollars.

Why

The two routes agree, which validates the calculation. It is also the form of answer that a Social Science statement asks for most often.

Step 7

Threshold: revenue would decrease if $\dfrac{dp}{dt}\,q+p\,\dfrac{dq}{dt}<0$ , that is $500\,\dfrac{dp}{dt}<-600$ , so $\dfrac{dp}{dt}<-1.20$ dollars per week.

Why

The inequality is solved on the differentiated relation, not on a new one. The threshold question is almost always the last, and it is worth the most marks.

Step 8

Checking the threshold: with $\dfrac{dp}{dt}=-1.20$ , we get $(-1.20)(500)+(20)(30)=-600+600=0$ .

Why

Plugging the value found back into the differentiated relation is the quickest check on a threshold question.

The conclusion, written out

“ $\dfrac{dR}{dt}=(-0.50)(500)+(20)(30)=350$ dollars per week: revenue INCREASES by $350$ dollars per week, that is $3.5$ percent, the volume effect outweighing the price effect. It would decrease if the price fell by more than $1.20$ dollars per week.”

The classic mistake on this problem: Concluding that revenue falls because the price falls. The sign of the result depends on BOTH effects, and here the second is more than twice as large as the first.

Learn by heart

• Differentiate the RELATION, then substitute. Never the reverse.
• Every quantity that is not $t$ brings its factor $\dfrac{d\cdot}{dt}$ : $\dfrac{d}{dt}\left(n^{2}\right)=2n\,\dfrac{dn}{dt}$ .
• Product: $x'y+xy'$ . Quotient: $\dfrac{x'y-xy'}{y^{2}}$ , the top is differentiated first.
• Relative rates ADD for a product and SUBTRACT for a quotient.
• A quotient rises if and only if the relative rate of the top exceeds that of the bottom, in PERCENT.
• A stock is measured in units, a flow in units PER unit of time. The unit always tells them apart.
• A crude rate moves through its denominator as much as through its numerator.
• The unit of a derivative is that of the quantity divided by that of the VARIABLE OF DIFFERENTIATION.

Frequently asked questions

Why do you have to differentiate before substituting in a related-rates problem?

Because a substituted value becomes a constant, and the derivative of a constant is zero. If you write that the price is twenty dollars before differentiating, the price stops varying in the calculation, and the whole term that describes its decline disappears. So you write the relation with letters, differentiate, and substitute only at the end.

What is the chain factor in a differentiation with respect to time?

It is the speed of change of every quantity that is not time itself. Differentiating the square of a quantity with respect to time gives twice that quantity multiplied by its own speed of change. Forgetting this factor gives a wrong result, and the error is larger the faster the quantity varies.

How do you know whether a per capita indicator is increasing?

You compare growth rates in percent, never absolute changes. The indicator increases if and only if the numerator grows faster, in percent, than the denominator. Comparing billions of dollars with thousands of people makes no sense, because they are different units.

Why can a birth rate fall without there being fewer births?

Because it is a ratio between births and the total population. If the population grows, for example through immigration, the denominator gets bigger and the ratio falls, even when the number of births stays the same. Commenting on an indicator therefore always means looking at both of its terms.

What is the difference between a stock and a flow?

A stock is a quantity measured at an instant, such as a population or an amount saved. A flow is a speed, measured per unit of time, such as births per year or dollars per month. The units always tell them apart: a flow contains a per year or a per month, a stock never does.

When should you use relative rates rather than direct differentiation?

As soon as the answer is expected in percent, or the statement gives only percentages with no absolute values. The method is then shorter, since rates add for a product and subtract for a quotient. When both methods are possible, doing both serves as a check.

Practise it

Corrected exercises: Related rates, 201-103

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.

11 corrected exercises
110 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Implicit differentiation Next sheet Differentials and linear approximation

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-103-related-rates. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Looking for a 201-103 tutor in Montreal?

Get in touch for a first session. We go back over the moves that cost marks on the calculus exam, then we test them on exercises at the level of real CEGEP tests.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: related rates (201-103)

The essentials

The five steps, in this order

What differentiating with respect to time brings out

The short route: relative rates

The rules in table form

Differentiating a relation with respect to time

The mistakes that cost marks

1. Substituting the values before differentiating

2. Forgetting the chain factor

3. Comparing absolute values to decide whether a quotient rises

4. Believing that a crude rate moves only through its numerator

5. Adding the relative rates of a quotient

6. Mixing up the units of dAdt\dfrac{dA}{dt}dtdA​ and dAdr\dfrac{dA}{dr}drdA​

7. Reading an unemployment rate without its participation rate

8. Believing that the same effort produces the same gain everywhere

9. Confusing a stock and a flow

Which method to choose

Which relation to write, depending on what the statement gives

Direct differentiation or relative rates: which to choose

How the answer is expected to be written

Writing up a related-rates problem in five steps

Writing up a chain of several links

Check before you hand in

The typical problem, taken apart

The price falls, the volume rises: what does revenue do

Learn by heart

Frequently asked questions

Why do you have to differentiate before substituting in a related-rates problem?

What is the chain factor in a differentiation with respect to time?

How do you know whether a per capita indicator is increasing?

Why can a birth rate fall without there being fewer births?

What is the difference between a stock and a flow?

When should you use relative rates rather than direct differentiation?

Corrected exercises: Related rates, 201-103

See also

Looking for a 201-103 tutor in Montreal?

6. Mixing up the units of $\dfrac{dA}{dt}$ and $\dfrac{dA}{dr}$