MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: linear approximation and differentials (MATH 140)

This sheet is not a summary of section 3.10 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on linear approximation and differentials in MATH 140 at McGill University, and which precise gesture avoids each loss.

The chapter looks like arithmetic, and that is the danger: the numbers are easy, and the marks go to what surrounds them, the centre, the formula, the side of the error and the exponent of a propagated error. Every value below is exact and done by hand, as on the exam.

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The thread of the chapter

The tangent line is exact AT its point and wrong everywhere else: an estimate by $L(x) = f(a) + f'(a)(x - a)$ earns its marks only with the right CENTRE (near the target, $f$ and $f'$ exact there, angles in radians), the factor $(x - a)$ , and the SIDE of the error read on the sign of $f''$ ; a propagated error is $dy$ , and the relative error is multiplied by the exponent.

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

Linear approximation
Linearization
Tangent line
Differentials
Concavity
Overestimate and underestimate
Propagated error
Relative error
Percentage error
Small-angle approximation
Binomial approximation

The essentials

The tangent line as a function, and the two rises it separates

• Linearization of $f$ at $a$ : $L(x) = f(a) + f'(a)(x - a)$ . It is the tangent line at $(a, f(a))$ , and $f(x) \approx L(x)$ only for $x$ NEAR $a$ .
• Test of any linearization in two seconds: $L(a) = f(a)$ . The line $2 + \frac{1}{4}x$ fails it for $\sqrt x$ at $4$ , since it gives $3$ at $x = 4$ .
• The centre $a$ is a point close to the target where $f(a)$ and $f'(a)$ are EXACT: $4$ for $\sqrt{4.1}$ , $8$ for $\sqrt[3]{7.9}$ , $1$ for $\arctan 1.02$ , $0$ for $e^{-0.03}$ .
• Differentials: $dx$ is an increment and $dy = f'(x)\,dx$ is the rise along the TANGENT; $\Delta y = f(x + dx) - f(x)$ is the rise along the CURVE; $\Delta y \approx dy$ .
• For $y = x^3$ at $2$ : $dx = 0.1$ gives $dy = 1.2$ and $\Delta y = 1.261$ ; $dx = 0.01$ gives $dy = 0.12$ and $\Delta y = 0.120601$ . The gap shrinks a hundred times when $dx$ shrinks ten times.

With

dx = 1

from

(1, 1)

, the tangent rises by

dy = \frac{1}{2}

L(2) = 1.5

, the curve only by

\Delta y = \sqrt 2 - 1

: the gap between the two is the error of the approximation.

A marker looks for three things in an approximation question: the centre named, $L$ written with $(x - a)$ , and the side of the error justified by $f''$ . The number itself is usually the smallest part of the mark.

The side of the error, and the propagated error

• $f'' < 0$ on the interval between $a$ and $x$ : concave down, the curve is below its tangent, $L$ OVERESTIMATES. $f'' > 0$ : concave up, $L$ UNDERESTIMATES.
• The side does not depend on whether $x$ is left or right of $a$ , as long as $f''$ keeps its sign between them. At an inflection point ( $\sin x$ at $0$ ) the tangent crosses the curve and the side changes.
• Near $0$ , radians: $e^x \approx 1 + x$ (under), $\ln(1 + x) \approx x$ (over), $\sin x \approx x$ (over for $x > 0$ ), $(1 + x)^k \approx 1 + kx$ .
• Propagated error: a measurement $x \pm dx$ gives $y = f(x)$ with maximum error $|dy| = |f'(x)|\,dx$ , relative error $\left|\frac{dy}{y}\right|$ .
• For $y = Cx^k$ : $\frac{dy}{y} = k\frac{dx}{x}$ . The cube TRIPLES the relative error, the square doubles it, the square root halves it.

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 small-x linearizations, with their side

Read a line as: near $x = 0$ , the function of the first column is replaced by its tangent line, and the curve lies on the side given in the last column. The red line is a linearization that does not exist.

Function	Tangent at 0	Curve vs tangent
$e^x$	$1 + x$	above: under
Example: $e^{-0.03} \approx 0.97$ , too small since $(e^x)'' = e^x > 0$ .
$\ln(1 + x)$	$x$	below: over
Example: $\ln 1.02 \approx 0.02$ , too large; the error is less than $0.0002$ .
$\sin x$	$x$	below for x > 0
Example: $\sin 0.01 \approx 0.01$ , too large; $\sin(-0.01) \approx -0.01$ , too small.
$(1 + x)^{1/2}$	$1 + \frac{x}{2}$	below: over
Example: $\sqrt{1.06} \approx 1.03$ , and $1.03^2 = 1.0609 > 1.06$ .
$(1 + x)^{-3}$	$1 - 3x$	above: under
Example: $\frac{1}{1.01^3} \approx 0.97$ , and $0.97 \times 1.030301 = 0.99939197 < 1$ .
$\sqrt x$	none at 0	vertical tangent no linearization
Example: $f'(x) = \frac{1}{2\sqrt x}$ has no value at $0$ ; centred at $1$ instead, $\sqrt{4.1} \approx 2.55$ , off by more than $0.5$ . What to do: Centre at the nearest perfect square: $\sqrt{4.1} \approx 2 + \frac{0.1}{4} = 2.025$ .

Every line holds for $x$ in radians and $x$ SMALL: $\ln(1 + x) \approx x$ at $x = 1$ gives $\ln 2 \approx 1$ instead of about $0.69$ .

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. Dropping the factor (x - a) from the linearization

the whole question: every value that follows is wrong

What not to write

“ $L(x) = 2 + \frac{1}{4}x$ , so $\sqrt{4.1} \approx 3.025$ .”

What to write

“ $L(x) = 2 + \frac{1}{4}(x - 4)$ , so $\sqrt{4.1} \approx 2 + \frac{0.1}{4} = 2.025$ .”

Why: Without $(x - a)$ the line has the right slope and passes through the wrong point. Test $L(a) = f(a)$ before using $L$ : here $L(4)$ must be $2$ .

2. Announcing the side of the error without the second derivative

1 mark, the justification of the side

What not to write

“The tangent is below the curve, so $\ln 1.02 \approx 0.02$ is an underestimate.”

What to write

“ $(\ln(1 + x))'' = -\frac{1}{(1 + x)^2} < 0$ : the curve is concave down and lies below its tangent, so $0.02$ is an OVERestimate.”

Same tangent

y = x

at the origin, two curves:

e^x - 1

(

f'' > 0

) lies above it, so

L

underestimates;

\ln(1 + x)

(

f'' < 0

) lies below, so

L

overestimates.

Why: The side is a consequence of the sign of $f''$ between $a$ and $x$ , not a general rule. Two functions with the same tangent $y = x$ at $0$ can lie on opposite sides of it.

3. Using the tangent far from its point of contact

the whole value, and the credibility of the answer

What not to write

“With $L(x) = 2 + \frac{1}{4}(x - 4)$ , $\sqrt{25} \approx 7.25$ .”

What to write

“ $25$ is far from $4$ : centre at the nearest perfect square instead, or here simply $\sqrt{25} = 5$ .”

x = 25

the tangent at

(4, 2)

gives

7.25

while the curve is at

5

: the line has drifted far above the concave curve.

Why: The error grows much faster than the distance: about $0.0002$ at distance $0.1$ , $0.014$ at distance $1$ , $2.25$ at distance $21$ . A linearization is a local statement.

4. Feeding degrees to sin x close to x

the whole question, and an impossible value since sine never exceeds 1

What not to write

“ $\sin x \approx x$ , so $\sin 2^\circ \approx 2$ .”

What to write

“ $2^\circ = \frac{\pi}{90}$ rad, so $\sin 2^\circ \approx \frac{\pi}{90}$ , about $0.035$ .”

Why: $(\sin x)' = \cos x$ only in radians; every trigonometric linearization is built on it. Convert the angle first.

5. Carrying the relative error unchanged through a power

2 marks, the whole relative error

What not to write

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

What to write

“ $\frac{dV}{V} = \frac{4\pi r^2\,dr}{\frac{4}{3}\pi r^3} = 3\frac{dr}{r} = 1.5\%$ .”

Why: For $y = Cx^k$ , $\frac{dy}{y} = k\frac{dx}{x}$ : the exponent multiplies the relative error. With $r = 10$ and $dr = 0.05$ , $dV = 20\pi$ against $V = \frac{4000\pi}{3}$ .

6. Confusing dy with Delta y

1 mark, and the definition question of the midterm

What not to write

“For $y = x^2$ at $x = 3$ with $dx = 0.5$ : $dy = \Delta y = 3.5^2 - 9 = 3.25$ .”

What to write

“ $dy = 2x\,dx = 3$ is the rise along the tangent; $\Delta y = 3.25$ is the rise along the curve; $\Delta y \approx dy$ .”

Why: $dy$ is computed with the derivative at the starting point and is linear in $dx$ ; $\Delta y$ is the exact change. They differ by $(dx)^2 = 0.25$ here.

7. Losing the sign of a negative increment

1 to 2 marks

What not to write

“ $\sqrt{3.96} \approx 2 + \frac{1}{4}(0.04) = 2.01$ .”

What to write

“ $x - a = 3.96 - 4 = -0.04$ , so $\sqrt{3.96} \approx 2 - 0.01 = 1.99$ .”

Why: A number below the centre has a root below $f(a)$ when $f$ increases: $2.01 > 2$ for $3.96 < 4$ is impossible, and a two second check catches it.

8. Counting a coat of thickness t once in the side of a cube

half the answer

What not to write

“The paint is $0.05$ cm thick, so $ds = 0.05$ and $dV = 3(20)^2(0.05) = 60$ cm $^3$ .”

What to write

“The coat adds $t$ on both opposite faces: $ds = 2t = 0.1$ , so $dV = 3(400)(0.1) = 120$ cm $^3$ , the area $6s^2$ times $t$ .”

Why: Read the increment on a picture before differentiating. The check $dV = (\text{total area}) \times t = 6(400)(0.05) = 120$ confirms it.

Which method to choose

Which centre and which formula, by the FORM of the question

Look at the number or the quantity in the statement: its form picks the function and the centre

If a root of a number close to a perfect power, $\sqrt{4.1}$ , $\sqrt[3]{7.9}$ → $f(x) = x^{1/n}$ centred at the perfect power

Example: $\sqrt[3]{7.9} \approx 2 - \frac{0.1}{12} = \frac{239}{120}$
If $1 +$ something small, raised to a power, including $\frac{1}{1 + h}$ → $(1 + x)^k \approx 1 + kx$ , with $k$ and its sign

Example: $\frac{1}{1.01^3} \approx 1 - 3(0.01) = 0.97$
If $\ln$ of a number near $1$ , $e$ to a small power, $\sin$ or $\tan$ of a small angle → the small-x rules at $0$ , angles converted to radians

Example: $\ln 1.02 \approx 0.02$ ; $e^{-0.03} \approx 0.97$ ; $\sin 2^\circ \approx \frac{\pi}{90}$
If a function at a point near a value where it is known exactly, $\arctan 1.02$ → centre at that value, NOT at $0$

Example: $\arctan 1.02 \approx \frac{\pi}{4} + \frac{1}{2}(0.02) = \frac{\pi}{4} + 0.01$
If only $f(a)$ , $f'(a)$ and the sign of $f''$ are given → $L$ from the data, side from the sign of $f''$ on the interval

Example: $f(1) = 2$ , $f'(1) = 3$ , $f'' < 0$ : $f(1.2) \approx 2.6$ , too large
If a measured quantity with an error, and a computed quantity → $dy = f'(x)\,dx$ , then $\frac{dy}{y}$ ; for a power, $k\frac{dx}{x}$ directly

Example: $r = 10 \pm 0.05$ : $dV = 20\pi$ , $1.5\%$
If an interval that must CONTAIN the value → tangent on one side, chord of the same concave curve on the other

Example: $\frac{56}{11} < \sqrt{26} < \frac{51}{10}$

MATH 140 stops at the tangent line: no quadratic term, no error bound of Taylor. Those are MATH 141. When no branch applies, the answer is not a linear approximation.

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.

Approximating a number with a linearization

When to use it: Any question that says “use a linear approximation (or differentials) to estimate” a number, with or without “is it an over or underestimate”

1 Name the function and the centre: $f(x) = \sqrt[3]{x}$ , $a = 27$ , chosen close to the target with $f(a)$ exact.
2 Compute $f(a)$ and $f'(a)$ exactly, naming the rule used for $f'$ (power rule, chain rule with the inner function named).
3 Write $L(x) = f(a) + f'(a)(x - a)$ with the numbers, and check $L(a) = f(a)$ .
4 Evaluate $L$ at the target, keeping the sign of $x - a$ , and give the value as a fraction or a short decimal.
5 Compute $f''$ , state its sign on the interval between $a$ and the target, and conclude over or underestimate.

Concluding sentence

“Since $f''(x) = -\frac{2}{9}x^{-5/3} < 0$ for $x > 0$ , $f$ is concave down on $[26.7, 27]$ and its graph lies below its tangent line there, so $L(26.7) = \frac{269}{90}$ is an overestimate of $\sqrt[3]{26.7}$ .”

The trap: Stopping at the number: without the centre named and without $f''$ , the question loses about half its marks even with the right value.

Marking: Typically 1 mark for the choice of f and a, 2 for f(a) and f'(a), 2 for L, 2 for the value, 3 for the side with its justification.

Estimating a propagated error

When to use it: A measurement given “to within” or “with a maximum error of”, and a quantity computed from it

1 Write the formula $y = f(x)$ of the computed quantity in terms of the MEASURED one (the radius, not the diameter, if the radius is measured).
2 Differentiate: $dy = f'(x)\,dx$ , with $x$ the measured value and $dx$ the maximum error.
3 Give $dy$ exactly, with units, as the maximum error of $y$ .
4 Divide by $y$ for the relative error, and multiply by $100$ for the percentage.

Concluding sentence

“With $r = 10$ cm and $dr = 0.05$ cm, $dV = 4\pi r^2\,dr = 20\pi$ cm $^3$ ; the relative error is $\frac{dV}{V} = 3\frac{dr}{r} = 0.015$ , that is $1.5\%$ .”

The trap: Answering with $\Delta V$ computed exactly: it is not wrong, but the question asks for the differential, and on a harder formula it cannot be done by hand.

Marking: Typically 2 marks for the formula, 3 for dy, 2 for its value with units, 3 for the relative and percentage errors.

Check before you hand in

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

L(a) = f(a)

Plug the centre into your linearization: it must return f(a) exactly. This catches a dropped factor (x - a) in two seconds.

$L(x) = 2 + \frac{1}{4}x$ gives $L(4) = 3 \ne 2$ : the formula is wrong before any number is computed.
Square or cube the estimate

For a root, raise the estimate to the power and compare with the number: this confirms the value AND the side, in exact arithmetic.

$2.025^2 = 4.100625 > 4.1$ : the estimate of $\sqrt{4.1}$ is too large, as $f'' < 0$ predicted. A result below $4.1$ would mean an error in the side or in the value.
The estimate is on the right side of f(a)

If the target is below the centre and f increases, the estimate must be below f(a); if f decreases, above.

$\sqrt{3.96}$ must be below $2$ : $2.01$ is impossible, $1.99$ is plausible.
The relative error carries the exponent

For a power formula, the relative error of the result is the exponent times that of the measurement. Compare with your answer.

Sphere, $0.5\%$ on $r$ : the volume must come out at $1.5\%$ , the area at $1\%$ . A volume at $0.5\%$ signals a missing factor $3$ .

The typical problem, taken apart

The cube root of 26.7, with its side

Use a linear approximation to estimate $\sqrt[3]{26.7}$ . Is your estimate too large or too small? Justify.

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

The curve

y = \sqrt[3]{x}

bends below its tangent at

(27, 3)

on both sides, so at

x = 26.7

the tangent gives a value slightly too large.

Step 1

$f(x) = x^{1/3}$ and $a = 27$ , the perfect cube next to $26.7$ .

Why

The centre must be close to the target AND give exact values: $\sqrt[3]{27} = 3$ and $\sqrt[3]{27^2} = 9$ are integers.

Step 2

By the power rule, $f'(x) = \frac{1}{3}x^{-2/3}$ , so $f(27) = 3$ and $f'(27) = \frac{1}{3 \cdot 9} = \frac{1}{27}$ .

Why

Naming the rule earns the method mark; the exact value $\frac{1}{27}$ is why $27$ was chosen.

Step 3

$L(x) = 3 + \frac{1}{27}(x - 27)$ , and $L(27) = 3$ as it must.

Why

The factor $(x - 27)$ is the whole tangent line; the check $L(a) = f(a)$ guards it.

Step 4

$x - a = 26.7 - 27 = -0.3$ , so $\sqrt[3]{26.7} \approx 3 - \frac{0.3}{27} = 3 - \frac{1}{90} = \frac{269}{90}$ .

Why

The increment is negative and the function increases: the estimate must be below $3$ , which it is.

Step 5

$f''(x) = -\frac{2}{9}x^{-5/3} < 0$ for $x > 0$ : concave down on $[26.7, 27]$ , so the estimate is too large. Check: $269^3 = 19\,465\,109$ and $26.7 \times 90^3 = 19\,464\,300$ , so $\left(\frac{269}{90}\right)^3 > 26.7$ .

Why

The side is a separate question, answered by $f''$ and confirmed by cubing in exact arithmetic. The two must agree.

The conclusion, written out

“With $f(x) = \sqrt[3]{x}$ and $a = 27$ , $L(x) = 3 + \frac{1}{27}(x - 27)$ , so $\sqrt[3]{26.7} \approx \frac{269}{90}$ . Since $f'' < 0$ for $x > 0$ , the graph lies below its tangent and the estimate is too large.”

The classic mistake on this problem: Writing $L(x) = 3 + \frac{1}{27}x$ , or $3 + \frac{0.3}{27}$ with the sign of the increment lost; both give a value above $3$ for a number below $27$ .

Learn by heart

• $L(x) = f(a) + f'(a)(x - a)$ , exact at $a$ , valid only NEAR $a$ ; check $L(a) = f(a)$ .
• Centre: close to the target, with $f(a)$ and $f'(a)$ exact; never where $f'$ does not exist.
• $f'' < 0$ between $a$ and $x$ : overestimate. $f'' > 0$ : underestimate. Inflection: the side changes.
• Near $0$ , in RADIANS: $e^x \approx 1 + x$ , $\ln(1 + x) \approx x$ , $\sin x \approx x$ , $(1 + x)^k \approx 1 + kx$ .
• $dy = f'(x)\,dx$ along the tangent, $\Delta y$ along the curve, $\Delta y \approx dy$ .
• Propagated error $dy$ , relative $\frac{dy}{y}$ ; for $Cx^k$ the relative error is multiplied by $|k|$ .
• Concave down: below the tangents, above the chords. Tangent and chord trap a value between them.

Frequently asked questions

How do I know if a linear approximation is an overestimate or an underestimate?

Look at the second derivative on the interval between the centre and the point you estimate. If it is negative there, the curve is concave down and lies below its tangent line, so the approximation is too large. If it is positive, the curve lies above the tangent and the approximation is too small. At an inflection point the answer changes from one side to the other.

How do I choose the point a for a linear approximation in MATH 140?

Pick a point close to the number you want, where the function and its derivative take exact, simple values: the nearest perfect square for a square root, the nearest perfect cube for a cube root, zero for a small exponent or logarithm of a number near one, one for the arctangent of a number near one. Never choose a point where the derivative does not exist.

What is the difference between dy and delta y?

Delta y is the exact change of the function when x moves by dx, measured along the curve. The differential dy is f prime of x times dx, the change measured along the tangent line. For a small dx the two are close, and the gap between them shrinks faster than dx itself, which is exactly why the tangent line is a good local approximation.

How do you find the relative error of a volume from the error on a radius?

Differentiate the volume formula to get dV equal to the derivative times the error on the radius, then divide by the volume. For a sphere this gives three times the relative error of the radius, because the radius is cubed. In general the relative error of a power of x is the exponent times the relative error of x: a square doubles it, a square root halves it.

Practise it

Corrected exercises: Linear approximation and differentials, 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

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

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Revision sheet: linear approximation and differentials (MATH 140)

The essentials

The tangent line as a function, and the two rises it separates

The side of the error, and the propagated error

The rules in table form

The small-x linearizations, with their side

The mistakes that cost marks

1. Dropping the factor (x - a) from the linearization

2. Announcing the side of the error without the second derivative

3. Using the tangent far from its point of contact

4. Feeding degrees to sin x close to x

5. Carrying the relative error unchanged through a power

6. Confusing dy with Delta y

7. Losing the sign of a negative increment

8. Counting a coat of thickness t once in the side of a cube

Which method to choose

Which centre and which formula, by the FORM of the question

How the answer is expected to be written

Approximating a number with a linearization

Estimating a propagated error

Check before you hand in

The typical problem, taken apart

The cube root of 26.7, with its side

Learn by heart

Frequently asked questions

How do I know if a linear approximation is an overestimate or an underestimate?

How do I choose the point a for a linear approximation in MATH 140?

What is the difference between dy and delta y?

How do you find the relative error of a volume from the error on a radius?

Corrected exercises: Linear approximation and differentials, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?