Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: the differential and linear approximation (201-SN2-RE)

This sheet does not prove $dy=f'(x)\,dx$ again: that formula fits on one line and nobody loses marks on it. It deals with what really costs marks on the third test of 201-SN2-RE (formerly 201-NYA): telling $dy$ apart from $\Delta y$ , choosing the right base point, and turning a measurement error into a relative error without being off by a factor.

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

Linear approximation replaces the curve by its TANGENT LINE: it is excellent near the base point and less and less accurate as you move away from it. The whole chapter is about knowing which side of “near” you are on, and not confusing the actual change with the change the tangent line predicts.

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

The essentials

Three quantities you must not confuse

• $\Delta x=dx$ : the actual step in the variable. Both notations mean the same thing.
• $\Delta y=f(x+\Delta x)-f(x)$ : the ACTUAL change in the function, measured on the curve.
• $dy=f'(x)\,dx$ : the change PREDICTED by the tangent line. It is an approximation of $\Delta y$ .
• Linear approximation: $f(x+\Delta x)\approx f(x)+f'(x)\,\Delta x$ , in other words $L(x)=f(a)+f'(a)(x-a)$ .

Between

x=4

and

x=9

, the curve rises from

2

3

, so

\Delta y=1

, while the tangent line rises to

3.25

, so

dy=1.25

. The gap of

0.25

is the error of the linear approximation.

$dy$ and $\Delta y$ coincide exactly for a LINEAR function, and only for it: the tangent line is then the curve itself.

The two errors, and the one that matters

• ABSOLUTE error on $y$ : $\left|dy\right|=\left|f'(x)\right|\left|dx\right|$ , in the unit of $y$ .
• RELATIVE error on $y$ : $\left|\frac{dy}{y}\right|$ , with no unit, often expressed as a percentage.
• Power: if $y=x^{n}$ , then $\frac{dy}{y}=n\frac{dx}{x}$ . The relative error is MULTIPLIED by the exponent.
• Volume of a sphere: $\frac{dV}{V}=3\frac{dr}{r}$ ; area: $\frac{dS}{S}=2\frac{dr}{r}$ .

In science, the relative error is the one that means something: an error of $1$ mm on $1$ m is excellent, on $1$ cm it is unacceptable.

The linearizations to know near zero

• $\sqrt{1+x}\approx 1+\frac{x}{2}$ , $\frac{1}{1+x}\approx 1-x$ , $(1+x)^{n}\approx 1+nx$ .
• $e^{x}\approx 1+x$ , $\ln(1+x)\approx x$ , $\sin x\approx x$ , $\tan x\approx x$ , $\cos x\approx 1$ .
• All valid for $x$ CLOSE TO ZERO, and all obtained from the same formula $L(x)=f(0)+f'(0)x$ .

Near

0

, each curve (solid) and its linearization (dashed) coincide:

e^{x}\approx 1+x

\sqrt{1+x}\approx 1+\frac{x}{2}

. They separate as soon as you move away from

0

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. Confusing the differential with the actual change

1 to 2 marks, depending on whether the question asked for an estimate or a value

What not to write

“ $dy=1.25$ , so $f(9)=2+1.25=3.25$ .”

What to write

“ $L(9)=2+1.25=3.25$ is only an ESTIMATE; the exact value is $f(9)=\sqrt{9}=3$ , and the approximation overestimates by $0.25$ .”

Why: $dy$ lives on the tangent line, $\Delta y$ on the curve. The two coincide only for a linear function. When the question says “estimate”, give $dy$ ; when it says “compute”, give $\Delta y$ , and the question often asks for both so you can compare.

2. Choosing a base point far away

the whole question, since the estimate becomes absurd

What not to write

“To estimate $\sqrt{101}$ , I linearize at $a=100$ ... no, at $a=0$ , it is simpler.”

What to write

“I linearize at the nearest KNOWN point: $a=100$ , because $\sqrt{100}=10$ is exact and $101$ is very close to it.”

Why: The base point must meet two conditions: $f(a)$ and $f'(a)$ can be computed exactly, and $a$ is close to the target value. At $a=0$ , the linearization of $\sqrt{x}$ does not even exist, since the derivative is infinite there. The right reflex is to look for the nearest perfect square, perfect cube or special angle.

3. Using the approximation far from the base point

the whole question, and an answer outside the interval

\left[-1, 1\right]

What not to write

“ $\sin x\approx x$ , so $\sin 3\approx 3$ .”

What to write

“ $\sin x\approx x$ only holds for $x$ close to $0$ ; $\sin 3\approx 0.141$ , and the approximation is off by more than $2$ .”

The line

y=x

and the curve

y=\sin x

are indistinguishable between

-0.3

and

0.3

, then drift apart for good: at

x=3

, the line predicts

3

while the sine equals

0.14

Why: A linear approximation has no range of validity printed on it: it is up to the student to set it. The check that saves you: if the answer falls outside the possible values of the function, like a sine greater than $1$ , it is certainly wrong.

4. Forgetting to multiply the relative error by the exponent

2 marks, and a scientifically wrong conclusion

What not to write

“The radius is known to within $0.5\%$ , so the volume is too.”

What to write

“ $V=\frac{4}{3}\pi r^{3}$ gives $\frac{dV}{V}=3\frac{dr}{r}$ , so the volume is known to within $1.5\%$ .”

Why: The relative error is multiplied by the exponent, because the logarithmic derivative of $r^{n}$ is $\frac{n}{r}$ . That is what makes volume measurements much less precise than length measurements, and it is the most useful result of the chapter in the lab.

5. Confusing absolute and relative error

1 interpretation mark, often the last one of the question

What not to write

“The error on the volume is $62.8$ , so it is huge.”

What to write

“The absolute error is $62.8$ cm $^{3}$ for a volume of $4189$ cm $^{3}$ , a relative error of $1.5\%$ , which is excellent.”

Why: An absolute number cannot be judged on its own: it must be compared with the quantity measured. That is exactly why questions almost always ask for both, and why the conclusion must quote the percentage.

6. Not saying in which direction the approximation is wrong

1 mark, the interpretation one

What not to write

“The approximation gives $3.25$ instead of $3$ , the error is $0.25$ .”

What to write

“Since the curve is CONCAVE DOWN, the tangent line lies above it: the approximation OVERESTIMATES, here by $0.25$ .”

Concave down curve (

\sqrt{x}

, blue): the tangent line lies above, the approximation OVERESTIMATES. Concave up curve (

x^{2}

, orange): the tangent line lies below, it UNDERESTIMATES. The sign of

f''

decides.

Why: Concavity decides the direction of the error, and saying so costs nothing: $f''<0$ gives a tangent line above, hence an overestimate, $f''>0$ a tangent line below, hence an underestimate. It is the only qualitative information you can give without computing the exact value, and therefore precisely the one the question is after.

7. Writing $dx$ with the wrong sign

2 marks, since the estimate starts off on the wrong side

What not to write

“To estimate $\sqrt{98}$ from $100$ , I take $dx=2$ .”

What to write

“ $dx=98-100=-2$ , so $dy=\frac{-2}{20}=-0.1$ and $\sqrt{98}\approx 9.9$ .”

\sqrt{98}

lies to the LEFT of

100

dx=-2

and the tangent line gives

9.9

(green). With

dx=+2

, you read

10.1

, the estimated value of

\sqrt{102}

(red).

Why: $dx$ is the SIGNED step between the target value and the base point, in that order. The check is immediate: $98<100$ so $\sqrt{98}<10$ , and an estimate greater than $10$ is wrong before it is even checked.

Which method to choose

Which base point to choose

Two conditions, always the same: $f(a)$ and $f'(a)$ must be computable exactly, and $a$ must be as close as possible to the target value.

If you are estimating a square root → take the nearest PERFECT SQUARE

Example: $\sqrt{101}$ is linearized at $a=100$ , $\sqrt{24}$ at $a=25$
If you are estimating a cube root → take the nearest PERFECT CUBE

Example: $\sqrt[3]{28}$ is linearized at $a=27$
If you are estimating a trigonometric value → take the nearest SPECIAL angle, in radians

Example: $\sin 31°$ is linearized at $a=\frac{\pi}{6}$ , with $dx=\frac{\pi}{180}$
If you are estimating an exponential or a logarithm → take $a=0$ for $e^{x}$ and $\ln(1+x)$ , or $a=1$ for $\ln x$

Example: $\ln(1.02)\approx 0.02$
If it is a propagation of measurement error → the base point is the MEASUREMENT itself, and $dx$ is the uncertainty

Example: $r=10$ cm measured to $\pm 0.05$ cm gives $a=10$ and $dx=0.05$

If no convenient point is close, linear approximation is not the expected tool: the question then calls for an exact calculation or Newton's method.

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.

An error propagation, written up for the marker

When to use it: The question gives a measurement with an uncertainty and asks for the error on a quantity that depends on it.

1 Write the relation between the quantities: “The volume of the sphere is $V=\frac{4}{3}\pi r^{3}$ .”
2 Differentiate: “ $dV=4\pi r^{2}\,dr$ .”
3 Substitute the measurement and the uncertainty: “With $r=10$ cm and $dr=0.05$ cm, $dV=4\pi(100)(0.05)=20\pi$ cm $^{3}$ .”
4 Give the ABSOLUTE error with its unit, then the RELATIVE error as a percentage.
5 Conclude with a sentence that interprets the percentage.

Concluding sentence

The volume is about $4189$ cm $^{3}$ to within $63$ cm $^{3}$ , a relative error of $1.5\%$ , three times that of the radius.

The trap: Stopping at the absolute error. The number $62.8$ says nothing on its own; it is the ratio to $4189$ that lets you judge the measurement, and that is what the marking scheme expects on the last line.

Marking: Usually 1 mark for the differential, 1 mark for the substitution, 1 mark for the relative error expressed as a percentage.

Check before you hand in

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

Compare the estimate with the calculator

Compute the exact value on the calculator and check that the estimate is close to it, on the right side, and with a plausible gap.

$\sqrt[3]{28}\approx 3.0370$ by linearization, $3.03659$ on the calculator: a gap of $0.0004$ , consistent with a step of $1$ from $27$ ✓.
Check the direction of the error with concavity

Compute the sign of $f''$ at the base point. Concave down, the tangent line is above and the estimate is too high; concave up, it is too low.

$f(x)=\sqrt{x}$ has $f''(x)=-\frac{1}{4}x^{-3/2}<0$ : every tangent line estimate is too high, as $3.25>3$ confirms.
Check which side of the answer you should land on

Ask yourself whether the target value is larger or smaller than $f(a)$ , before any calculation, and check that the estimate goes the same way.

$98<100$ so $\sqrt{98}<10$ : an estimate of $10.1$ signals a $dx$ with the wrong sign.
Check the unit of each error

The absolute error carries the unit of the computed quantity, the relative error has none. A relative error with a unit, or a percentage greater than $100$ , signals a mix-up between the two.

$dV=62.8$ cm $^{3}$ and $\frac{dV}{V}=0.015$ , with no unit. Writing $0.015$ cm $^{3}$ gives away a badly formed ratio.

The typical problem, taken apart

Estimating a cube root without a calculator

Estimate $\sqrt[3]{28}$ by linear approximation, then say whether the estimate is too large or too small, and by how much.

Around

x=27

, the curve

y=\sqrt[3]{x}

and its tangent line practically coincide: that is what makes the estimate reliable at

28

and much less so at

36

Step 1

Let $f(x)=x^{1/3}$ and choose $a=27$ , the perfect cube closest to $28$ , with $f(27)=3$ exactly.

Why

The choice of base point is the only decision in the problem. Taking $a=8$ would give $f(8)=2$ , just as exact, but the step $dx=20$ would ruin the precision.

Step 2

$f'(x)=\frac{1}{3}x^{-2/3}=\frac{1}{3\sqrt[3]{x^{2}}}$ , so $f'(27)=\frac{1}{3\times 9}=\frac{1}{27}$ .

Why

The derivative can be computed exactly because $27^{2/3}=9$ is an integer. That is the second condition for a good base point, and it matters as much as closeness.

Step 3

$dx=28-27=1$ , so $dy=f'(27)\,dx=\frac{1}{27}\approx 0.0370$ .

Why

$dx$ is the SIGNED step, target value minus base point. Here it is positive, so the estimate will be greater than $3$ , which makes sense since $28>27$ .

Step 4

$\sqrt[3]{28}\approx f(27)+dy=3+\frac{1}{27}=\frac{82}{27}\approx 3.0370$ .

Why

Add the exact value at the base point and the predicted change. Writing the exact fraction $\frac{82}{27}$ before rounding is what the marking scheme expects: the decimal alone proves nothing.

Step 5

Direction of the error: $f''(x)=-\frac{2}{9}x^{-5/3}<0$ for $x>0$ , so the curve is concave down and the tangent line lies ABOVE it. The estimate is too high.

Why

Concavity answers without computing the exact value, which is exactly what the question asks. It is a free mark for anyone who thinks of differentiating twice.

Step 6

Check: the exact value is $3.03659$ , the estimate $3.03704$ , the gap is $0.00045$ , positive ✓ as concavity predicted. The relative error is $0.015\%$ .

Why

Two independent checks: the sign of the gap confirms the concavity, and its smallness confirms that the base point was well chosen.

The conclusion, written out

$\sqrt[3]{28}\approx\frac{82}{27}\approx 3.0370$ . Since the function is concave down, the tangent line lies above the curve: the estimate OVERESTIMATES the exact value, here by about $0.00045$ .

The classic mistake on this problem: Taking $a=30$ “because it is a round number”. Neither $30^{1/3}$ nor $30^{-2/3}$ can be computed by hand: the base point must be a perfect CUBE, not a round number.

Learn by heart

• $dy=f'(x)\,dx$ is the change PREDICTED by the tangent line; $\Delta y$ is the actual change on the curve.
• $L(x)=f(a)+f'(a)(x-a)$ : the tangent line, written as a function.
• $dx$ is the SIGNED step, target value minus base point. It can be negative.
• Relative error: $\frac{dy}{y}$ . For $y=x^{n}$ , it equals $n\frac{dx}{x}$ , so it is multiplied by the exponent.
• $f''<0$ : tangent line above, the estimate is too high. $f''>0$ : it is too low.
• Base point: the CLOSEST perfect square, perfect cube or special angle, never the roundest number.
• $\sqrt{1+x}\approx 1+\frac{x}{2}$ , $(1+x)^{n}\approx 1+nx$ , $e^{x}\approx 1+x$ , $\ln(1+x)\approx x$ , $\sin x\approx x$ , for $x$ near $0$ .

Frequently asked questions

What is the difference between dy and delta y?

Delta y is the actual change in the function, measured on the curve between two points. Dy is the change that the tangent line predicts over the same step. The two are close when the step is small, and coincide exactly only for a linear function, whose tangent line is the curve itself.

How do you choose the base point of a linear approximation?

It must meet two conditions. The value of the function and of its derivative must be computable there exactly, and it must be as close as possible to the target value. For a square root, take the nearest perfect square, for a cube root the nearest perfect cube, for a sine the nearest special angle.

How do you know whether a linear approximation overestimates or underestimates?

The sign of the second derivative tells you. If it is negative, the curve is concave down and the tangent line lies above it, so the estimate is too large. If it is positive, the curve is concave up and the tangent line lies below it, so the estimate is too small. No calculation of the exact value is needed.

Why is the relative error on a volume three times that on the radius?

Because the volume depends on the cube of the radius. Differentiating, the ratio of the error to the volume is three times the ratio of the error to the radius. It is the general result for a power: the relative error is multiplied by the exponent, which makes volume measurements clearly less precise than length measurements.

How far does the approximation sin x equals x remain valid?

It stays within one percent as long as the angle in radians is below about zero point two five, roughly fourteen degrees. Beyond that it gets worse quickly, and for an angle of three radians it predicts three while the sine is zero point one four. A sine value greater than one always signals that the approximation has been pushed too far.

Practise it

Corrected exercises: Differentials and linear approximation, 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-differentials-linear-approximation. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: the differential and linear approximation (201-SN2-RE)

The essentials

Three quantities you must not confuse

The two errors, and the one that matters

The linearizations to know near zero

The mistakes that cost marks

1. Confusing the differential with the actual change

2. Choosing a base point far away

3. Using the approximation far from the base point

4. Forgetting to multiply the relative error by the exponent

5. Confusing absolute and relative error

6. Not saying in which direction the approximation is wrong

7. Writing dxdxdx with the wrong sign

Which method to choose

Which base point to choose

How the answer is expected to be written

An error propagation, written up for the marker

Check before you hand in

The typical problem, taken apart

Estimating a cube root without a calculator

Learn by heart

Frequently asked questions

What is the difference between dy and delta y?

How do you choose the base point of a linear approximation?

How do you know whether a linear approximation overestimates or underestimates?

Why is the relative error on a volume three times that on the radius?

How far does the approximation sin x equals x remain valid?

Corrected exercises: Differentials and linear approximation, 201-SN2-RE

See also

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7. Writing $dx$ with the wrong sign