MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: linearization and differentials (MATH 203)

Q: How do I know if a linearization overestimates or underestimates in MATH 203?

Look at the sign of the second derivative between the centre and the point you estimate. If it is negative, the curve is concave down and lies below its tangent line, so the linearization is too large. If it is positive, the curve lies above the tangent and the linearization is too small. Then check with your calculator: the actual error must have that sign.

Q: How do I choose the point a for a linearization?

Choose a point close to the number you want where the function and its derivative are known exactly. It does not have to be a simple number: for e to the 0.7, use ln 2, where the exponential and its derivative both equal 2. For a root, use the nearest perfect power; for a trigonometric function, a standard angle in radians.

Q: When can I use (1 + x) to the k is about 1 + kx?

Only when the expression starts with 1 and the product k times x is small. For a number like the square root of 9.18, factor the 9 out first to get 3 times the square root of 1.02. And check the product k times x, not x alone: for 1.08 to the fifth, x is small but 5 times 0.08 is 0.4, and the error is almost five percent.

Q: What is the difference between relative error and percentage error?

The relative error is the maximum error divided by the value, a pure fraction such as 0.02. The percentage error is the same fraction multiplied by 100, here 2 percent. Writing 0.02 percent is a common slip that claims a precision a hundred times better than the measurement allows.

This sheet is not a summary of section 3.11 of Thomas' Calculus: you have the textbook and the course notes. It answers one question only, what makes students lose marks on linearization and differentials in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The chapter is short and its calculus is easy, which is exactly why it is marked on the algebra: an exponent at the centre, a sign, a factored constant, a simplified quotient. A scientific calculator is allowed, so a complete answer also compares the estimate with the calculator value and states the side of the error.

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 and tells you which chapter to tackle next. Create your space, an email is enough.

The thread of the chapter

The calculus of $L(x) = f(a) + f'(a)(x - a)$ is one line and the marks are lost in the ALGEBRA around it: a negative or fractional exponent evaluated at the centre ( $16^{-3/4} = \frac{1}{8}$ ), the sign of $x - a$ , the constant factored out before $(1 + x)^k \approx 1 + kx$ , and $\frac{dy}{y}$ simplified to $k\frac{dx}{x}$ before the numbers; then the side of the error, read on $f''$ and confirmed by the calculator.

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

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

Linearization
Standard linear approximation
Differentials
Absolute change
Relative change
Percentage error
Propagated error
Concavity
Overestimate and underestimate
Fractional exponents
Implicit differentiation

The essentials

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

• Linearization of $f$ at $a$ : $L(x) = f(a) + f'(a)(x - a)$ . Thomas calls $f(x) \approx L(x)$ the standard linear approximation of $f$ at $a$ . It holds only for $x$ NEAR $a$ .
• Two-second test of any linearization: $L(a) = f(a)$ . The formula $(2 + x)^{-2} \approx 1 - 2x$ fails it, since it gives $1$ at $x = 0$ instead of $\frac{1}{4}$ .
• The centre is a point close to the target where $f(a)$ AND $f'(a)$ are exact: $16$ for $\sqrt[4]{17}$ , $\ln 2$ for $e^{0.7}$ , $\frac{\pi}{4}$ for $\tan 44^\circ$ , $\frac{1}{2}$ for $\arcsin 0.52$ . It does not have to look simple.
• Differentials: $dy = f'(x)\,dx$ is the change along the TANGENT, $\Delta y = f(x + dx) - f(x)$ the change along the CURVE, and $\Delta y = dy + \varepsilon\,dx$ with $\varepsilon \to 0$ .
• For $y = x^{3/2}$ at $4$ : $dx = 0.1$ gives $dy = 0.3$ and $\Delta y = 0.301867$ ; $dx = 0.01$ gives $dy = 0.03$ and $\Delta y = 0.030019$ .

From

(1, 1)

with

dx = 1

, the tangent rises by

dy = 2

L(2) = 3

and the curve by

\Delta y = 3

f(2) = 4

: the gap

TQ

is the error of the approximation.

A marker looks for the centre named, $L$ written with $(x - a)$ , the value, the side justified by $f''$ , and, since the calculator is allowed, the actual error. The number itself is rarely more than a third of the mark.

The side of the error, and the propagated error

• $f'' > 0$ between $a$ and $x$ : concave up, the curve is ABOVE its tangent, $L$ UNDERestimates. $f'' < 0$ : concave down, $L$ OVERestimates.
• The side does not change with the direction of the step, as long as $f''$ keeps its sign between $a$ and $x$ : $\sqrt[4]{17}$ and $\sqrt[4]{15.5}$ are both overestimated.
• Propagated error: a measurement $x$ with maximum error $dx$ gives $y = f(x)$ with maximum error $|dy| = |f'(x)|\,dx$ , relative error $\left|\frac{dy}{y}\right|$ , percentage error $100\left|\frac{dy}{y}\right|$ .
• For $y = cx^k$ : $\frac{dy}{y} = k\frac{dx}{x}$ . A square doubles the relative error, a cube triples it, a square root (the period of a pendulum) halves it.
• Two measured factors: $d(uv) = u\,dv + v\,du$ , so $\frac{d(uv)}{uv} = \frac{du}{u} + \frac{dv}{v}$ , and in the worst case the relative errors add.

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.

Negative and fractional exponents at the centre

Read a line as: the power in the first column, rewritten root first, gives the value in the last column. These values are what $f'(a)$ needs at a perfect-power centre. The red line is a rule that does not exist, and it is the most expensive line of the chapter.

Power	Root first	Value
$16^{-3/4}$	$\frac{1}{(\sqrt[4]{16})^3} = \frac{1}{2^3}$	$\frac{1}{8}$
Example: $f(x) = \sqrt[4]{x}$ : $f'(16) = \frac{1}{4} \cdot \frac{1}{8} = \frac{1}{32}$ , and $\sqrt[4]{17} \approx 2.03125$ .
$4^{-5/2}$	$\frac{1}{(\sqrt 4)^5} = \frac{1}{2^5}$	$\frac{1}{32}$
Example: $g(x) = x^{-3/2}$ : $g'(4) = -\frac{3}{2} \cdot \frac{1}{32} = -\frac{3}{64}$ , and $4.2^{-3/2} \approx 0.115625$ .
$81^{-3/4}$	$\frac{1}{(\sqrt[4]{81})^3} = \frac{1}{3^3}$	$\frac{1}{27}$
Example: $f'(81) = \frac{1}{4} \cdot \frac{1}{27} = \frac{1}{108}$ , and $\sqrt[4]{80} \approx 3 - \frac{1}{108} = 2.990741$ .
$8^{2/3}$	$(\sqrt[3]{8})^2 = 2^2$	$4$
Example: $h(x) = x^{2/3}$ : $h(8) = 4$ and $h'(8) = \frac{2}{3} \cdot 8^{-1/3} = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$ .
$8^{-1/3}$	$-\sqrt[3]{8}$	a negative number rule that does not exist
Example: $8^{-1/3} = -2$ would give $h'(8) = -\frac{4}{3}$ , a falling tangent for a rising curve; the calculator gives $8^{-1/3} = 0.5$ . What to do: A negative exponent is a RECIPROCAL: $8^{-1/3} = \frac{1}{8^{1/3}} = \frac{1}{2}$ .

Take the root first: $16^{3/4} = 2^3 = 8$ is mental arithmetic, $\sqrt[4]{16^3} = \sqrt[4]{4096}$ is not.

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. Reading a negative exponent as a negative number

the whole question: every value after it is wrong

What not to write

“ $f'(16) = \frac{1}{4} \cdot 16^{-3/4} = \frac{1}{4}(-8) = -2$ .”

What to write

“ $16^{-3/4} = \frac{1}{(\sqrt[4]{16})^3} = \frac{1}{8}$ , so $f'(16) = \frac{1}{32}$ and $L(x) = 2 + \frac{1}{32}(x - 16)$ .”

Why: The fourth root increases, so its slope cannot be negative. A negative exponent means a reciprocal; a sign never appears from an exponent.

2. Applying (1 + x) to the k to a number that does not start with 1

the whole value

What not to write

“ $\sqrt{9.18} = (1 + 8.18)^{1/2} \approx 1 + \frac{8.18}{2} = 5.09$ .”

What to write

“ $\sqrt{9.18} = \sqrt 9\,\sqrt{1.02} = 3(1 + 0.02)^{1/2} \approx 3(1.01) = 3.03$ .”

x = 9.18

the tangent at

1

gives

5.09

while the tangent at

9

gives

3.03

, next to the curve: factoring out the

9

is choosing the right tangent.

Why: Without factoring, the rule is the tangent of $\sqrt x$ at $1$ used eight units away. Factor the constant so that what remains is $1 +$ something small: $(c + x)^k = c^k\left(1 + \frac{x}{c}\right)^k$ .

3. Losing the sign of the increment x - a

1 to 2 marks

What not to write

“ $\sqrt[4]{15.5} \approx 2 + \frac{0.5}{32} = 2.015625$ .”

What to write

“ $x - a = 15.5 - 16 = -0.5$ , so $\sqrt[4]{15.5} \approx 2 - \frac{0.5}{32} = 1.984375$ .”

Why: A number below the centre has a fourth root below $2$ . Writing $x - a$ with its value, sign included, before multiplying prevents the slip.

4. Feeding degrees to the derivative of tan

the whole question, and an absurd value for an acute angle

What not to write

“ $\tan 44^\circ \approx 1 + 2(44 - 45) = -1$ .”

What to write

“ $h = -1^\circ = -\frac{\pi}{180}$ rad, so $\tan 44^\circ \approx 1 - \frac{2\pi}{180} = 1 - \frac{\pi}{90} = 0.965093$ .”

Why: $(\tan x)' = \sec^2 x$ holds only in radians, so the increment must be in radians. The calculator, in degree mode, gives $\tan 44^\circ = 0.965689$ .

5. Announcing the side of the error backwards

1 mark, the justification of the side

What not to write

“ $\tan x$ is concave up near $\frac{\pi}{4}$ , so the tangent is above and $0.965093$ overestimates $\tan 44^\circ$ .”

What to write

“ $(\tan x)'' = 2\sec^2 x\tan x > 0$ on $\left(0, \frac{\pi}{2}\right)$ : concave up, the tangent lies BELOW the curve, so $0.965093$ is an underestimate.”

On both sides of

\left(\frac{\pi}{4}, 1\right)

the curve

\tan x

bends up away from its tangent: concave up means the tangent is below and every estimate is too small.

Why: Concave up, tangent under, estimate under. Concave down, tangent over, estimate over. The calculator confirms: $0.965093 < 0.965689$ .

6. Writing a relative error as the wrong percentage

1 mark, and a precision claimed a hundred times too good

What not to write

“ $\frac{dV}{V} = 0.02$ , so the volume is known to within $0.02\%$ .”

What to write

“ $\frac{dV}{V} = 0.02 = \frac{2}{100}$ , so the volume is known to within $2\%$ .”

Why: A relative error is a fraction; the percentage is that fraction times $100$ . Say which one you give: “relative error $0.02$ ” or “percentage error $2\%$ ”.

7. Putting the error of a diameter or a circumference on the radius

half the question: the error comes out doubled

What not to write

“The diameter is $2.4 \pm 0.01$ m, so $r = 1.2$ with $dr = 0.01$ and $dV = 2\pi rh\,dr = 0.084\pi$ .”

What to write

“ $dr = \frac{dD}{2} = 0.005$ m, so $dV = 2\pi(1.2)(3.5)(0.005) = 0.042\pi$ m $^3$ ; better, write $V = \frac{\pi}{4}D^2h$ with the measured $D$ .”

Why: The error belongs to the MEASURED quantity. For a circumference $C = 2\pi r$ , $dr = \frac{dC}{2\pi}$ : a tape error of $0.5$ cm is only $0.0796$ cm on the radius.

8. Using the exponent the wrong way in the reverse question

the whole answer: the correction is four times too small

What not to write

“The clock gains $30$ s a day and $T = 2\pi\sqrt{\frac{L}{g}}$ , so the rod must be lengthened by $\frac{1}{2} \cdot \frac{30}{86\,400} = \frac{1}{5760}$ .”

What to write

“ $\frac{dT}{T} = \frac{1}{2}\frac{dL}{L}$ and we need $\frac{dT}{T} = \frac{1}{2880}$ , so $\frac{dL}{L} = \frac{2}{2880} = \frac{1}{1440}$ , about $0.069\%$ .”

Why: The exponent multiplies the relative error going FROM the measurement TO the result; going back, divide by it. Dividing by $\frac{1}{2}$ means doubling: on $0.994$ m, $dL = 0.69$ mm.

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, the centre and the formula

If a root or a power of a number close to a perfect power, $\sqrt[4]{17}$ , $4.2^{-3/2}$ → $f(x) = x^{k}$ centred at the perfect power, exponent evaluated root first

Example: $\sqrt[4]{17} \approx 2 + \frac{1}{32} = 2.03125$
If a constant plus something small, raised to a power: $\sqrt{9.18}$ , $(2.06)^{-2}$ → factor the constant, then $(1 + u)^k \approx 1 + ku$ , checking that $ku$ is small

Example: $\frac{1}{2.06^2} = \frac{1}{4}(1.03)^{-2} \approx \frac{1}{4}(0.94) = 0.235$
If $e$ , $\tan$ , $\arcsin$ or $\ln$ of a number near a value where they are exact → centre at that value, even if it is $\ln 2$ or $e$ ; angles in radians

Example: $e^{0.7} \approx 2 + 2(0.7 - \ln 2) = 2.013706$
If a point of a curve given by an equation → implicit differentiation for the slope, then $L$ ; if the tangent is vertical, linearize $x$ as a function of $y$

Example: $x^3 + y^3 = 9$ at $(1, 2)$ : $y(1.1) \approx 2 - \frac{0.1}{4} = 1.975$
If one measured quantity with an error, and a computed quantity → write $y$ in terms of the MEASURED quantity, then $dy = f'(x)\,dx$ and $\frac{dy}{y}$

Example: $V = \frac{C^3}{6\pi^2}$ , $C = 75 \pm 0.5$ : $\frac{dV}{V} = 3\frac{dC}{C} = 2\%$
If two measured quantities multiplied → $d(uv) = u\,dv + v\,du$ ; in the worst case the relative errors add, each times its exponent

Example: $V = \frac{\pi}{4}D^2h$ : $\frac{1}{120} + \frac{1}{175}$ , about $1.40\%$

MATH 203 stops at the tangent line: no quadratic term, no error bound. The calculator measures the actual error; the sign of $f''$ predicts its side.

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 linearization (or differentials) to approximate” a number, with or without “is it an over or underestimate” and “compare with your calculator”

1 Name the function and the centre: $f(x) = \sqrt[4]{x}$ , $a = 81$ , close to the target with $f(a)$ and $f'(a)$ exact.
2 Compute $f(a)$ and $f'(a)$ exactly, rewriting every negative or fractional exponent root first.
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, writing $x - a$ with its sign.
5 Compute $f''$ , state its sign between $a$ and the target, conclude over or underestimate.
6 Compare with the calculator: actual error $|f(x) - L(x)|$ , on the side predicted.

Concluding sentence

“Since $f''(x) = -\frac{3}{16}x^{-7/4} < 0$ for $x > 0$ , $f$ is concave down between $80$ and $81$ and lies below its tangent, so $L(80) = \frac{323}{108}$ overestimates $\sqrt[4]{80}$ ; the calculator gives $2.990698$ , an error of $0.000043$ .”

The trap: Stopping at the number: without the centre named, the side justified and the comparison, the question loses about half its marks.

Marking: Typically 1 mark for f and a, 2 for f(a) and f'(a), 2 for L, 1 for the value, 2 for the side with its reason, 2 for the comparison with the calculator.

Estimating a propagated error

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

1 Write the computed quantity in terms of the MEASURED one: $V = \frac{C^3}{6\pi^2}$ if the circumference is measured.
2 Differentiate: $dy = f'(x)\,dx$ , with $x$ the measured value and $dx$ the maximum error, both in the same unit.
3 Give $dy$ with its unit as the maximum error of $y$ .
4 Divide by $y$ , simplifying the quotient BEFORE the numbers, for the relative error; multiply by $100$ for the percentage.

Concluding sentence

“With $C = 75$ cm and $dC = 0.5$ cm, $dV = \frac{C^2}{2\pi^2}\,dC = 142.48$ cm $^3$ , and $\frac{dV}{V} = 3\frac{dC}{C} = 0.02$ , that is $2\%$ .”

The trap: Differentiating the textbook formula $V = \frac{4}{3}\pi r^3$ and plugging in the error of the circumference as if it were $dr$ .

Marking: Typically 2 marks for the formula in the measured variable, 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 the linearization: it must return f(a). This catches a forgotten factored constant or a dropped (x - a) in two seconds.

$(2 + x)^{-2} \approx 1 - 2x$ gives $1$ at $x = 0$ instead of $\frac{1}{4}$ : wrong before any number is computed.
The slope has the sign of the curve

An increasing function has a positive f'(a). A negative slope for a root or a positive power means an exponent read as a sign.

$f'(16) = -2$ for $\sqrt[4]{x}$ is impossible; $\frac{1}{32}$ is plausible.
The calculator agrees, on the predicted side

The actual error should be small and have the sign announced by f''. A large error or the wrong sign means a wrong centre, slope or side.

$\sqrt[4]{17} \approx 2.03125$ and the calculator gives $2.030543$ : smaller, as $f'' < 0$ predicted, error $0.000707$ .
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 your answer with that multiple.

Tape error $\frac{0.5}{75}$ , about $0.67\%$ : the volume must come out at $2\%$ , the area at $1.33\%$ .

The typical problem, taken apart

The fourth root of 80, with its side and the calculator

Use a linearization to approximate $\sqrt[4]{80}$ . Is your estimate too large or too small? Justify, then compare with your calculator.

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

The curve

y = \sqrt[4]{x}

bends below its tangent at

(81, 3)

on both sides, so at

x = 80

the tangent gives a value slightly too large.

Step 1

$f(x) = x^{1/4}$ and $a = 81$ , the perfect fourth power next to $80$ .

Why

The centre must be close to the target AND give exact values: $\sqrt[4]{81} = 3$ is an integer.

Step 2

By the power rule $f'(x) = \frac{1}{4}x^{-3/4}$ , and $81^{-3/4} = \frac{1}{(\sqrt[4]{81})^3} = \frac{1}{27}$ , so $f'(81) = \frac{1}{108}$ .

Why

The algebra step of the question: root first, then the cube, then the reciprocal. A negative exponent never makes the slope negative.

Step 3

$L(x) = 3 + \frac{1}{108}(x - 81)$ , and $L(81) = 3$ as it must.

Why

The factor $(x - 81)$ is the whole tangent line; the check $L(a) = f(a)$ guards it.

Step 4

$x - a = 80 - 81 = -1$ , so $\sqrt[4]{80} \approx 3 - \frac{1}{108} = \frac{323}{108} = 2.990741$ .

Why

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

Step 5

$f''(x) = -\frac{3}{16}x^{-7/4} < 0$ for $x > 0$ : concave down, the estimate is too large. Calculator: $\sqrt[4]{80} = 2.990698$ , error $0.000043$ .

Why

The side is predicted by $f''$ and checked by the calculator: the two must agree, and here they do.

The conclusion, written out

“With $f(x) = \sqrt[4]{x}$ and $a = 81$ , $L(x) = 3 + \frac{1}{108}(x - 81)$ , so $\sqrt[4]{80} \approx \frac{323}{108} = 2.990741$ . Since $f'' < 0$ for $x > 0$ , the estimate is too large; the calculator gives $2.990698$ , an actual error of $0.000043$ .”

The classic mistake on this problem: Writing $81^{-3/4} = -27$ and a slope of $-\frac{27}{4}$ , or losing the sign of $80 - 81$ ; both give a value above $3$ for a number below $81$ .

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, $f(a)$ and $f'(a)$ exact; it may be $\ln 2$ , $e$ or $\frac{\pi}{4}$ . Radians.
• $a^{-m/n} = \frac{1}{(\sqrt[n]{a})^m}$ : root, power, reciprocal. Never a negative number.
• $(1 + x)^k \approx 1 + kx$ when $kx$ is small; factor first: $(c + x)^k = c^k\left(1 + \frac{x}{c}\right)^k$ .
• $f'' > 0$ : underestimate. $f'' < 0$ : overestimate.
• $dy = f'(x)\,dx$ along the tangent, $\Delta y$ along the curve, $\Delta y = dy + \varepsilon\,dx$ .
• Relative error $\frac{dy}{y}$ , percentage $100\frac{dy}{y}$ ; for $cx^k$ , multiply by $k$ ; going back, divide by $k$ .

Frequently asked questions

How do I know if a linearization overestimates or underestimates in MATH 203?

Look at the sign of the second derivative between the centre and the point you estimate. If it is negative, the curve is concave down and lies below its tangent line, so the linearization is too large. If it is positive, the curve lies above the tangent and the linearization is too small. Then check with your calculator: the actual error must have that sign.

How do I choose the point a for a linearization?

Choose a point close to the number you want where the function and its derivative are known exactly. It does not have to be a simple number: for e to the 0.7, use ln 2, where the exponential and its derivative both equal 2. For a root, use the nearest perfect power; for a trigonometric function, a standard angle in radians.

When can I use (1 + x) to the k is about 1 + kx?

Only when the expression starts with 1 and the product k times x is small. For a number like the square root of 9.18, factor the 9 out first to get 3 times the square root of 1.02. And check the product k times x, not x alone: for 1.08 to the fifth, x is small but 5 times 0.08 is 0.4, and the error is almost five percent.

What is the difference between relative error and percentage error?

The relative error is the maximum error divided by the value, a pure fraction such as 0.02. The percentage error is the same fraction multiplied by 100, here 2 percent. Writing 0.02 percent is a common slip that claims a precision a hundred times better than the measurement allows.

Practise it

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

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Get in touch for a first session. Linearization comes back on the final as a full question, and most of its marks go to the algebra: exponents, signs, factoring, percentages.

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Revision sheet: linearization and differentials (MATH 203)

The essentials

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

The side of the error, and the propagated error

The rules in table form

Negative and fractional exponents at the centre

The mistakes that cost marks

1. Reading a negative exponent as a negative number

2. Applying (1 + x) to the k to a number that does not start with 1

3. Losing the sign of the increment x - a

4. Feeding degrees to the derivative of tan

5. Announcing the side of the error backwards

6. Writing a relative error as the wrong percentage

7. Putting the error of a diameter or a circumference on the radius

8. Using the exponent the wrong way in the reverse question

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 fourth root of 80, with its side and the calculator

Learn by heart

Frequently asked questions

How do I know if a linearization overestimates or underestimates in MATH 203?

How do I choose the point a for a linearization?

When can I use (1 + x) to the k is about 1 + kx?

What is the difference between relative error and percentage error?

Corrected exercises: Linearization and differentials, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?