This deck covers local linearity and the linearization of a function at a point, then differentials and propagated measurement error, using concavity to decide whether an estimate comes out too high or too low, and Newton's Method together with its failure modes. It targets the classic errors: centering at a point whose value you do not know, confusing the differential with the true change, getting the over-or-under direction backwards, and iterating Newton's Method with the previous function value instead of the previous x-value.
Subject: Calculus I · 130 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 12
One idea - the tangent line is a good stand-in for the curve - worn three different ways.
Objectives
By the end of this deck you can:
Warm-up
Discussion prompt
Before we open Linear Approximation, Differentials, and Newton's Method: without looking back, what was the main idea of Related Rates, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck handles two or more quantities that change in time and are linked by an equation. It gives the master procedure and then works the seven classic problems: the ladder, the ripple, the balloon, the cone tank, the separating vehicles, the streetlight shadow, and the angle of elevation. It targets the four errors that cost the most points - substituting the instant's numbers before differentiating, dropping the chain-rule rate factor, using a relation that holds only at one instant, and getting the sign or the units of a rate wrong.
Section
Part 1
Concept
Take any curve that has a derivative at a point, and zoom in on that point. The curve flattens. Zoom far enough and it becomes visually indistinguishable from a straight line.
That straight line is the tangent line. This is the whole deck in one sentence.
Counterexample
Discussion prompt
Take any curve that has a derivative at a point, and zoom in on that point. The curve flattens. Zoom far enough and it becomes visually indistinguishable from a straight line.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Picture it
Figure (svg): Three panels showing the same curve and its tangent line at increasing zoom levels; by the third panel the curve and the line overlap.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
This is why a map of your neighborhood can be flat even though the Earth is round. Locally, curvature is invisible.
Intuition
Figure (svg): Three panels showing the same curve and its tangent line at increasing zoom levels; by the third panel the curve and the line overlap.
This is why a map of your neighborhood can be flat even though the Earth is round. Locally, curvature is invisible.
local linearity — The property that a differentiable function looks like a straight line when you look at a small enough piece of it. Having a derivative at a point IS this property.
Concept
To pin down a line you need a point and a slope. At the center you already have both.
The height comes from the function, the slope comes from the derivative.
\[ \text{point: } (a,\, f(a)) \qquad \text{slope: } f'(a) \]
So the tangent line agrees with the curve in both value and steepness right at that spot. That is why it is the best straight-line copy.
Analogy
Discussion prompt
Explain The tangent line already knows two things by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
To pin down a line you need a point and a slope. At the center you already have both.
Picture it
Animation
Shows: The tangent line already knows two things — a rendered Manim animation.
Rendered with Manim.
Takeaway: Nothing else is needed, which is why the formula is so short.
Picture it
Animation
Shows: A tangent line hugging a square-root curve near the point of contact.
Close in, the two are hard to tell apart.
Takeaway: Near the point of contact the line and the curve are nearly indistinguishable, which is what makes the tangent usable as a calculator.
Concept
Point-slope form, with the point and slope from the previous slide:
\[ y - f(a) = f'(a)\,(x - a) \]
Solve for the height and you have a formula you can plug numbers into.
\[ y = f(a) + f'(a)\,(x - a) \]
Nothing new has happened yet. The new part is deciding to use this line in place of the function.
Explain it
Discussion prompt
Explain Writing the tangent line down to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Solve for the height and you have a formula you can plug numbers into.
Concept
When we use that tangent line as a stand-in for the function, we give it a name and a letter.
\[ L(x) = f(a) + f'(a)\,(x - a) \]
linearization of f at a — The tangent-line function L, used as an approximation to f for inputs x near the center a. It is the best first-order copy of f at that point.
The claim we are making is an approximation, not an equality. The wavy equals sign is doing real work: it says close to, and only for inputs near the center.
\[ f(x) \approx L(x) \quad \text{for } x \text{ near } a \]
Definition probe
Sort into buckets
Every line below is part of the definition of local linearity or of linearization of f at a — one or the other, never both. Put each where it belongs.
Picture it
Animation
Shows: The linearization formula written as a tangent line.
Nothing new is being introduced here.
Takeaway: The linearization is the tangent line, renamed for the job it is doing. Recognising that saves memorising a separate formula.
Intuition
You can evaluate a line in your head. Addition and one multiplication. That is it.
Square roots, fifth powers, sines of odd angles, logarithms - none of those are head-arithmetic. But near a point you know, the line does the job.
\[ \sqrt{9.2}, \quad (1.02)^5, \quad \sin 31^{\circ} \]
The trade is honest: you give up exactness and you get an answer you can compute. Calculus tells you how much you gave up.
Ranking
Put in order
Put the moves of Worked example: estimate the square root of 9.2 into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The function is the square root.
Worked example
Estimate this without a calculator.
\[ \sqrt{9.2} \]
Name the function and pick the center a = 9
Why: The function is the square root. Nine is the nearest input whose square root we know exactly, and the target is only 0.2 away from it.
Compute the two ingredients: the height and the slope at the center
Why: The linearization needs f(a) and f prime of a, nothing else.
\[ f(x)=\sqrt{x} \;\Rightarrow\; f(9)=3, \qquad f'(x)=\frac{1}{2\sqrt{x}} \;\Rightarrow\; f'(9)=\frac{1}{6} \]
Assemble the linearization
Why: Drop the two ingredients into the tangent-line formula centered at 9.
\[ L(x) = 3 + \frac{1}{6}(x-9) \]
Evaluate at the target input
Why: The change from the center is 0.2, so the tangent line rises by one sixth of 0.2.
\[ L(9.2) = 3 + \frac{0.2}{6} = 3 + \frac{1}{30} = 3.03\overline{3} \]
Verify by squaring the estimate
Why: Squaring should land close to 9.2. It lands slightly above, which flags the estimate as a slight overestimate - exactly what a concave-down curve should do.
\[ \left(\tfrac{91}{30}\right)^2 = \tfrac{8281}{900} = 9.2011\overline{1} \;>\; 9.2 \]
| quantity | value |
|---|---|
| linear estimate | 3.033333 |
| estimate squared | 9.201111 |
| target | 9.200000 |
| true square root of 9.2 | 3.033150 |
| error | 0.000183 too high |
Picture it
Animation
Shows: Each line of the worked example "estimate the square root of 9.2", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Squaring should land close to 9.2. It lands slightly above, which flags the estimate as a slight overestimate - exactly what a concave-down curve should do.
Pattern
Steps 4 and 5 collapse into one formula worth memorizing:
\[ f(a + \Delta x) \approx f(a) + f'(a)\,\Delta x \]
Concept
A good center has to satisfy two conditions, and students usually only check one.
A center that is close but not friendly forces you to use a calculator for the very thing you were trying to avoid. A center that is friendly but far gives a bad estimate.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Estimating the square root of 9.2 and choosing the center at 10 because 10 feels like a round number.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The formula opens with the square root of 10, which is exactly the kind of number you cannot produce by hand.
Choose the nearest input whose value you genuinely know: a perfect square.
Why: The formula opens with the square root of 10, which is exactly the kind of number you cannot produce by hand. You have traded one unknown root for another.
Trap
Estimating the square root of 9.2 and choosing the center at 10 because 10 feels like a round number.
\[ L(x) = \sqrt{10} + \frac{1}{2\sqrt{10}}(x-10) \]
Now you are stuck
Why: The formula opens with the square root of 10, which is exactly the kind of number you cannot produce by hand. You have traded one unknown root for another.
Push through with a calculator and it is worse anyway
Why: Even granting yourself the value 3.162278, the center is 0.8 away instead of 0.2, so the estimate drifts much further from the truth.
\[ L(9.2) \approx 3.162278 - \frac{0.8}{6.324555} = 3.035787 \]
Choose the nearest input whose value you genuinely know: a perfect square.
\[ L(x) = 3 + \frac{1}{6}(x-9) \]
Both ingredients are exact and hand-sized
Why: The square root of 9 is 3 and the slope is one sixth. Nothing needs a calculator.
And the estimate is fourteen times more accurate
Why: Centering at 9 misses by 0.000183; centering at 10 misses by 0.002637. Closer center, smaller error.
\[ L(9.2) = 3.033333, \qquad \sqrt{9.2} = 3.033150 \]
Notation
Annotate
From Trap: centering where you do not know the value — read this one piece at a time. What is each part doing?
On: \( L(x) = \sqrt{10} + \frac{1}{2\sqrt{10}}(x-10) \)
Step zero
Discussion prompt
Worked example: estimate 1.02 raised to the fifth power — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Take the fifth-power function and center at a = 1
Answer:
Worked example
A quantity grows by two percent per year for five years. Estimate the total growth factor.
\[ (1.02)^5 \]
Take the fifth-power function and center at a = 1
Why: One raised to any power is 1, and 1.02 is very close to 1. Friendly and close.
Get the height and the slope at the center
Why: The power rule gives the derivative; evaluating at 1 keeps every number an integer.
\[ f(x)=x^5 \Rightarrow f(1)=1, \qquad f'(x)=5x^4 \Rightarrow f'(1)=5 \]
Assemble and evaluate
Why: The change from the center is 0.02, and the slope multiplies it by 5.
\[ L(x) = 1 + 5(x-1) \;\Rightarrow\; L(1.02) = 1 + 5(0.02) = 1.10 \]
Read that in words: five years of two percent growth is roughly ten percent total. That is the shortcut everyone in finance uses.
Verify against the exact expansion
Why: Squaring twice and multiplying once gives the exact value by hand, and the estimate sits just below it - the signature of a concave-up curve.
\[ 1.02^2 = 1.0404,\quad 1.02^4 = 1.08243216,\quad 1.02^5 = 1.1040808032 \]
| estimate | exact | error | direction |
|---|---|---|---|
| 1.100000 | 1.104081 | 0.004081 | underestimate |
Picture it
Animation
Shows: Each line of the worked example "estimate 1.02 raised to the fifth power", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Squaring twice and multiplying once gives the exact value by hand, and the estimate sits just below it - the signature of a concave-up curve.
Check
You want to estimate this by hand with a linearization.
\[ \sqrt{50} \]
Check your understanding
Which center a should you use?
Answer: A
Why: 49 is the nearest perfect square, so f(49) = 7 and the slope 1/(2 times 7) = 1/14 are both exact and easy. The target is only 1 away, so the estimate 7 + 1/14 = 7.0714 is very close to the true value 7.0711.
Estimation
Predict first
Estimate the sine of an angle you do not have memorized, using one you do.
Commit before you compute: what does Worked example: estimate the sine of 31 degrees come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against the true value
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The true sine of 31 degrees is 0.5150381, so the estimate is high by about 0.00008 - eight hundredths of one tenth of a percent.
Worked example
Estimate the sine of an angle you do not have memorized, using one you do.
\[ \sin 31^{\circ} \]
Convert to radians and center at the known angle
Why: The derivative formulas for trig functions are only true in radians. Thirty degrees is one sixth of pi and its sine and cosine are both memorized.
\[ 31^{\circ} = \frac{\pi}{6} + \frac{\pi}{180}, \qquad a = \frac{\pi}{6} \]
Get the height and the slope at the center
Why: The derivative of sine is cosine, and both values at one sixth of pi are exact.
\[ f\!\left(\tfrac{\pi}{6}\right)=\tfrac{1}{2}, \qquad f'\!\left(\tfrac{\pi}{6}\right)=\cos\tfrac{\pi}{6}=\tfrac{\sqrt{3}}{2} \]
Assemble the linearization
Why: Same template as always: height plus slope times the change from the center.
\[ L(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{6}\right) \]
The change from the center is one degree, in radians
Why: One degree is pi over 180, about 0.0174533 radians. That is the small nudge the tangent line rides along.
\[ L\!\left(\tfrac{31\pi}{180}\right) = 0.5 + (0.8660254)(0.0174533) = 0.5151157 \]
Verify against the true value
Why: The true sine of 31 degrees is 0.5150381, so the estimate is high by about 0.00008 - eight hundredths of one tenth of a percent. Sine is concave down on this interval, so an overestimate is exactly what we predicted.
| estimate | true value | error |
|---|---|---|
| 0.5151157 | 0.5150381 | 0.0000776 too high |
Picture it
Animation
Shows: Each line of the worked example "estimate the sine of 31 degrees", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The true sine of 31 degrees is 0.5150381, so the estimate is high by about 0.00008 - eight hundredths of one tenth of a percent. Sine is concave down on this interval, so an overestimate is exactly what we predicted.
Concept
The tangent line matches the curve perfectly at the center. Step away and the gap opens up.
For the square-root linearization centered at 9, watch the error as the target moves away:
| target x | linear estimate | true value | error |
|---|---|---|---|
| 9.2 | 3.033333 | 3.033150 | 0.000183 |
| 10 | 3.166667 | 3.162278 | 0.004389 |
| 16 | 4.166667 | 4.000000 | 0.166667 |
| 25 | 5.666667 | 5.000000 | 0.666667 |
The error grows roughly like the square of the distance from the center. Double the distance, quadruple the error.
Pattern
Step through it
Step through Accuracy falls off fast with distance one row at a time. What is driving the change, and what would the row after the last one be?
Anomaly
Predict first
A student writes this, and it looks reasonable:
Using the square-root linearization built at 9 to estimate the square root of 25, because the formula still accepts the input.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The true square root of 25 is 5.
Rebuild the linearization at a center near the new target. Twenty-five is itself a perfect square.
Why: The true square root of 25 is 5. Nothing in the algebra complains - a linear formula happily extrapolates forever. Only you can enforce the word 'near'.
Trap
Using the square-root linearization built at 9 to estimate the square root of 25, because the formula still accepts the input.
\[ L(25) = 3 + \frac{25-9}{6} = 3 + \frac{16}{6} = 5.6\overline{6} \]
The formula answered, and the answer is wrong by 13 percent
Why: The true square root of 25 is 5. Nothing in the algebra complains - a linear formula happily extrapolates forever. Only you can enforce the word 'near'.
Rebuild the linearization at a center near the new target. Twenty-five is itself a perfect square.
\[ L(x) = 5 + \frac{1}{10}(x-25) \;\Rightarrow\; L(25) = 5 \]
A linearization is a local tool, so re-center for each neighborhood
Why: If the target moved, the center should move with it. Recentering costs one derivative evaluation and buys back all the accuracy.
Break the constraint
Discussion prompt
The rule this trap just fixed:
If the target moved, the center should move with it. Recentering costs one derivative evaluation and buys back all the accuracy.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
The true square root of 25 is 5. Nothing in the algebra complains - a linear formula happily extrapolates forever. Only you can enforce the word 'near'.
Elimination
Eliminate the wrong options
Using a linearization centered at 8, what is the estimate?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The cube root of 8 is 2 and the derivative one third times x to the negative two thirds evaluates to 1/(3 times 4) = 1/12 at x = 8. So L(8.6) = 2 + 0.6/12 = 2.05, which is close to the true value 2.04880.
Check
Build the linearization of the cube-root function at a friendly center and estimate this value.
\[ \sqrt[3]{8.6} \]
Check your understanding
Using a linearization centered at 8, what is the estimate?
Answer: A
Why: The cube root of 8 is 2 and the derivative one third times x to the negative two thirds evaluates to 1/(3 times 4) = 1/12 at x = 8. So L(8.6) = 2 + 0.6/12 = 2.05, which is close to the true value 2.04880.
Section
Part 2
Concept
Start at an input and nudge it. Two things move: the curve's height, and the tangent line's height. They are not the same number.
The true change in the function has a name and so does the tangent-line change.
\[ \Delta y = f(x + \Delta x) - f(x) \qquad\text{versus}\qquad dy = f'(x)\,dx \]
Everything in this section is about keeping those two straight.
Picture it
Figure (svg): A curve with its tangent line at a point; from the shifted input, one vertical segment reaches the tangent line labelled dy and a taller one reaches the curve labelled true change.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
The horizontal nudge is the same for both. The tangent line answers with a straight-line rise; the curve answers with its own, slightly different rise.
Intuition
Figure (svg): A curve with its tangent line at a point; from the shifted input, one vertical segment reaches the tangent line labelled dy and a taller one reaches the curve labelled true change.
The horizontal nudge is the same for both. The tangent line answers with a straight-line rise; the curve answers with its own, slightly different rise.
The differential is the tangent line's answer. It is the prediction; the true change is the outcome.
Concept
We simply declare the input nudge to be a variable in its own right, and define the output nudge from it.
\[ dx = \text{any real number you choose}, \qquad dy = f'(x)\,dx \]
differential — The tangent-line change dy produced by an input change dx. It is the linear approximation to the true change, written as a product of the derivative and the input nudge.
Divide both sides and the Leibniz notation you have been writing since day one stops being a symbol and becomes an actual quotient.
\[ \frac{dy}{dx} = f'(x) \]
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of local linearity, linearization of f at a, differential as Linear Approximation, Differentials, and Newton's Method uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Picture it
Animation
Shows: The differential relation and its approximation to the true change.
The approximation decays as fast as the curve bends.
Takeaway: The differential estimates the true change, and the estimate degrades exactly as fast as the curve bends away from its tangent.
Step zero
Discussion prompt
Worked example: the differential versus the true change — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Compute the differential
Answer:
Worked example
Take the squaring function at the input 3, and nudge the input by one tenth.
\[ f(x)=x^2, \qquad x = 3, \qquad dx = \Delta x = 0.1 \]
Compute the differential
Why: The derivative at 3 is 6, and the differential is just that slope times the nudge.
\[ dy = f'(3)\,dx = 6(0.1) = 0.6 \]
Compute the true change
Why: The true change asks the function itself, at both endpoints, and subtracts.
\[ \Delta y = f(3.1) - f(3) = 9.61 - 9 = 0.61 \]
Name the gap exactly
Why: Expanding the true change shows the differential is the first piece and the leftover is the square of the nudge - which is why small nudges make the approximation good.
\[ (3+h)^2 - 9 = 6h + h^2 \;\Rightarrow\; \Delta y - dy = h^2 = 0.01 \]
Verify the pattern by shrinking the nudge
Why: Cutting the nudge by a factor of ten cuts the gap by a factor of one hundred, confirming the leftover really is the square of the nudge.
| dx | dy | true change | gap |
|---|---|---|---|
| 0.1 | 0.6 | 0.61 | 0.01 |
| 0.01 | 0.06 | 0.0601 | 0.0001 |
| 0.001 | 0.006 | 0.006001 | 0.000001 |
Picture it
Animation
Shows: Each line of the worked example "the differential versus the true change", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Cutting the nudge by a factor of ten cuts the gap by a factor of one hundred, confirming the leftover really is the square of the nudge.
Trap
Writing the differential and the true change as if they were the same quantity.
\[ \Delta y \;=\; dy \;=\; f'(3)(0.1) \;=\; 0.6 \quad \text{\small(claimed exact)} \]
The equals sign is a lie by 0.01
Why: The function really moved 0.61. Calling 0.6 exact means every error estimate built on it silently loses the leftover term, and in an error-analysis problem the leftover is the entire point.
Keep them as two named quantities and connect them with an approximation, not an equality.
\[ \Delta y \approx dy, \qquad \Delta y = 0.61, \qquad dy = 0.6 \]
Say which one the question wants
Why: 'Use differentials to estimate' means report dy. 'Find the exact change' means report the difference of two function values. They are different questions with different answers.
Translation
\( \Delta y \;=\; dy \;=\; f'(3)(0.1) \;=\; 0.6 \quad \text{\small(claimed exact)} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
Every physical measurement comes with a tolerance. You measured an edge as 10 centimeters, give or take half a millimeter.
You then compute something from it - a volume, an area, a resistance. How much uncertainty did you inherit?
That inherited uncertainty is called propagated error, and the differential is exactly the tool that estimates it: the measurement tolerance is your dx.
Estimation
Predict first
A metal cube's edge is measured as 10 centimeters, with a possible measurement error of at most 0.05 centimeters. Estimate the resulting error in the computed volume.
Commit before you compute: what does Worked example: propagated error in a cube's volume come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against the exact computation
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Cubing the largest possible edge and subtracting the nominal volume gives 15.075 cubic centimeters, so the differential estimate of 15 is right to within half a percent of itself.
Worked example
A metal cube's edge is measured as 10 centimeters, with a possible measurement error of at most 0.05 centimeters. Estimate the resulting error in the computed volume.
Write the quantity you compute as a function of the quantity you measure
Why: The volume depends only on the edge, so the edge is the input carrying the uncertainty.
\[ V(x) = x^3, \qquad x = 10, \qquad dx = 0.05 \]
Take the differential
Why: The differential converts an input tolerance into an output tolerance by multiplying by the derivative.
\[ dV = 3x^2\,dx \]
Substitute the measured value and the tolerance
Why: Three times one hundred is three hundred, times five hundredths is fifteen. Units are cubic centimeters because the volume is a volume.
\[ dV = 3(10)^2(0.05) = 15 \ \text{cm}^3 \]
So a half-millimeter uncertainty on the edge becomes a fifteen cubic centimeter uncertainty on the volume. That is a lot of metal for a very small ruler error.
Verify against the exact computation
Why: Cubing the largest possible edge and subtracting the nominal volume gives 15.075 cubic centimeters, so the differential estimate of 15 is right to within half a percent of itself.
\[ (10.05)^3 - (10)^3 = 1015.075125 - 1000 = 15.075125 \]
Concept
Fifteen cubic centimeters sounds huge or tiny depending on how big the cube is. The honest way to report error is as a fraction of the quantity itself.
\[ \text{relative error} = \frac{dV}{V}, \qquad \text{percent error} = \frac{dV}{V}\times 100\% \]
For the cube, the answer is small and reassuring.
\[ \frac{dV}{V} = \frac{15}{1000} = 0.015 = 1.5\% \]
The denominator is the computed quantity, not the measurement. That is the single most common slip here.
Picture it
Animation
Shows: The error grows as you walk away — a rendered Manim animation.
Rendered with Manim.
Takeaway: Near the contact point the line is excellent. Far away it is useless.
Anomaly
Predict first
A student writes this, and it looks reasonable:
You found a volume error of 15 cubic centimeters for a cube of edge 10. Reporting the relative error by dividing by the edge.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The number is absurd on its face, and the units give it away: cubic centimeters divided by centimeters is not a pure ratio, so it cannot be a percentage of anything.
Relative error compares a quantity to itself. The volume error belongs over the volume.
Why: The number is absurd on its face, and the units give it away: cubic centimeters divided by centimeters is not a pure ratio, so it cannot be a percentage of anything.
Trap
You found a volume error of 15 cubic centimeters for a cube of edge 10. Reporting the relative error by dividing by the edge.
\[ \frac{dV}{x} = \frac{15}{10} = 1.5 = 150\% \]
A 150 percent error on a measurement you trust to half a millimeter
Why: The number is absurd on its face, and the units give it away: cubic centimeters divided by centimeters is not a pure ratio, so it cannot be a percentage of anything.
Relative error compares a quantity to itself. The volume error belongs over the volume.
\[ \frac{dV}{V} = \frac{15}{1000} = 0.015 = 1.5\% \]
Check the units cancel
Why: Cubic centimeters over cubic centimeters is dimensionless, which is what a percentage has to be. If the units do not cancel, you divided by the wrong thing.
\[ \frac{\text{cm}^3}{\text{cm}^3} = 1 \quad\text{(dimensionless)} \]
Notation
Annotate
From Trap: dividing by the measurement instead of the quantity — read this one piece at a time. What is each part doing?
On: \( \frac{\text{cm}^3}{\text{cm}^3} = 1 \quad\text{(dimensionless)} \)
Ranking
Put in order
Put the moves of Worked example: percent error in a circle's area into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Area depends on the radius alone, so the radius carries all of the uncertainty.
Worked example
A circular plot is measured to have a radius of 10 meters, accurate to within 2 centimeters. Estimate the absolute and percent error in the computed area.
Write the computed quantity in terms of the measured one
Why: Area depends on the radius alone, so the radius carries all of the uncertainty. Convert the tolerance into meters first so the units match.
\[ A(r) = \pi r^2, \qquad r = 10\ \text{m}, \qquad dr = 0.02\ \text{m} \]
Take the differential and substitute
Why: The derivative of the area with respect to the radius is the circumference, which is a nice sanity check on its own: growing a disc by a thin ring adds circumference times thickness.
\[ dA = 2\pi r\,dr = 2\pi(10)(0.02) = 0.4\pi \approx 1.2566\ \text{m}^2 \]
Convert to relative error
Why: Dividing by the area itself lets pi cancel, which is a strong hint that relative error is the more natural quantity.
\[ \frac{dA}{A} = \frac{2\pi r\,dr}{\pi r^2} = \frac{2\,dr}{r} = \frac{2(0.02)}{10} = 0.004 = 0.4\% \]
Notice the shape of that answer: a two tenths of a percent error in the radius became a four tenths of a percent error in the area. The exponent 2 doubled it.
Verify against the exact area difference
Why: Computing both areas exactly gives 1.2578 square meters, within one tenth of a percent of the differential estimate 1.2566. The differential is doing its job.
\[ \pi(10.02)^2 - \pi(10)^2 = \pi(0.4004) \approx 1.2578\ \text{m}^2 \]
Picture it
Animation
Shows: Each line of the worked example "percent error in a circle's area", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Computing both areas exactly gives 1.2578 square meters, within one tenth of a percent of the differential estimate 1.2566. The differential is doing its job.
Concept
Whenever the computed quantity is a constant times a power of the measured one, the relative errors are linked by that power.
\[ y = kx^n \;\Rightarrow\; \frac{dy}{y} = \frac{nkx^{n-1}\,dx}{kx^n} = n\,\frac{dx}{x} \]
Length errors double for areas and triple for volumes. It is worth memorizing.
| computed from a length | power | percent error multiplier |
|---|---|---|
| perimeter or circumference | 1 | 1 times |
| area or surface area | 2 | 2 times |
| volume | 3 | 3 times |
Comparison
Comparison matrix
From The exponent multiplies the percent error: refill the percent error multiplier column from what you know. The rest of the table is as it appeared.
| computed from a length | power | percent error multiplier |
|---|---|---|
| perimeter or circumference | 1 | 1 times |
| area or surface area | 2 | 2 times |
| volume | 3 | 3 times |
Pattern
Prediction
Predict first
Approximately what percent error should you report in the computed volume?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: About 3 percent
Why: Volume is a constant times the radius cubed, so the relative error in the volume is 3 times the relative error in the radius: 3 times 1 percent equals about 3 percent. The differential gives dV/V = 3 dr/r.
Check
A ball bearing's radius is measured with a possible error of 1 percent. You use that radius to compute the bearing's volume.
Check your understanding
Approximately what percent error should you report in the computed volume?
Answer: A
Why: Volume is a constant times the radius cubed, so the relative error in the volume is 3 times the relative error in the radius: 3 times 1 percent equals about 3 percent. The differential gives dV/V = 3 dr/r.
Commit first
Predict first
Which statement correctly describes those two numbers?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: 0.12 is the rise along the tangent line and 0.120601 is the rise along the curve; the gap of 0.000601 is the approximation error.
Why: The differential dV = 0.12 is what the tangent line predicts; the exact difference of cubes 0.120601 is what the function actually did. Their difference, 0.000601, is precisely the leftover higher-order terms 3x(dx) squared plus (dx) cubed.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
A cube's edge grows from 2 centimeters to 2.01 centimeters. Two numbers come out of the calculation.
\[ dV = 3(2)^2(0.01) = 0.12, \qquad \Delta V = (2.01)^3 - 2^3 = 0.120601 \]
Check your understanding
Which statement correctly describes those two numbers?
Answer: A
Why: The differential dV = 0.12 is what the tangent line predicts; the exact difference of cubes 0.120601 is what the function actually did. Their difference, 0.000601, is precisely the leftover higher-order terms 3x(dx) squared plus (dx) cubed.
Section
Part 3
Concept
Near the center, the curve does not weave back and forth across its tangent line. It stays on one side.
Which side depends on how the curve bends - and bending is exactly what the second derivative measures.
So you can tell whether your estimate is too big or too small without knowing the true answer. That is a genuinely useful superpower on an exam.
Picture it
Animation
Shows: Concavity decides which way the estimate errs — a rendered Manim animation.
Rendered with Manim.
Takeaway: A concave-up curve sits above its tangent, so the linearization under-estimates.
Picture it
Figure (svg): Left panel: a concave-up parabola with a horizontal tangent line lying below it. Right panel: a concave-down arc with a horizontal tangent line lying above it.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
A cup that holds water sits above every one of its tangent lines. Turn it over and it spills - and now it sits below every tangent line.
Intuition
Figure (svg): Left panel: a concave-up parabola with a horizontal tangent line lying below it. Right panel: a concave-down arc with a horizontal tangent line lying above it.
A cup that holds water sits above every one of its tangent lines. Turn it over and it spills - and now it sits below every tangent line.
Since the linearization IS the tangent line, its output is below the true value in the first picture and above it in the second.
Concept
Compute the second derivative at the center and read off the answer.
| second derivative near a | shape | tangent line is | linear estimate is |
|---|---|---|---|
| positive | concave up | below the curve | an underestimate |
| negative | concave down | above the curve | an overestimate |
| zero and changing sign | inflection | crosses the curve | no guarantee |
Read the last two columns together and it stops being memorization: the tangent line is below the curve, so the number it reports is below the true number.
Trade off
Comparison matrix
From The rule, in one table: every row here is a choice with a cost. Fill the tangent line is column, then say which row you would actually pick and what you give up for it.
| second derivative near a | shape | tangent line is | linear estimate is |
|---|---|---|---|
| positive | concave up | below the curve | an underestimate |
| negative | concave down | above the curve | an overestimate |
| zero and changing sign | inflection | crosses the curve | no guarantee |
Hypothesis
Predict first
Worked example: over or under, decided in advance is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: First, the square root centered at 9
Why: Differentiate twice. The second derivative is negative for every positive input, so the square-root curve is concave down everywhere on its domain.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Two estimates from earlier in this deck. Decide the direction of each error before looking up the truth.
First, the square root centered at 9
Why: Differentiate twice. The second derivative is negative for every positive input, so the square-root curve is concave down everywhere on its domain.
\[ f(x)=\sqrt{x},\quad f''(x) = -\frac{1}{4}x^{-3/2}, \quad f''(9) = -\frac{1}{108} < 0 \]
Conclude: overestimate
Why: Concave down means the tangent line rides above the curve, so the linearization reports a value larger than the truth.
Second, the exponential centered at 0
Why: The natural exponential is its own derivative twice over, and it is always positive, so the curve is concave up everywhere.
\[ g(x)=e^{x},\quad g''(x)=e^{x} > 0, \qquad L(x) = 1 + x \]
Conclude: underestimate
Why: Concave up means the tangent is trapped below the curve, so the linearization reports a value smaller than the truth.
Verify both predictions against the true values
Why: Both predictions land: the square-root estimate really is high and the exponential estimate really is low, and neither check required knowing the answer in advance.
| estimate | value | true value | predicted | actual |
|---|---|---|---|---|
| square root of 9.2 | 3.033333 | 3.033150 | over | over |
| e to the 0.1 | 1.100000 | 1.105171 | under | under |
Picture it
Animation
Shows: Each line of the worked example "over or under, decided in advance", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both predictions land: the square-root estimate really is high and the exponential estimate really is low, and neither check required knowing the answer in advance.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Reasoning from whether the function is increasing instead of from how it bends.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Increasing versus decreasing tells you nothing about which side of the tangent the curve is on.
Ask only one question: what is the sign of the second derivative near the center?
Why: Increasing versus decreasing tells you nothing about which side of the tangent the curve is on. That is a first-derivative fact answering a second-derivative question.
Trap
Reasoning from whether the function is increasing instead of from how it bends.
'The square root is increasing, so the tangent line must fall behind it - underestimate.'
Why: Increasing versus decreasing tells you nothing about which side of the tangent the curve is on. That is a first-derivative fact answering a second-derivative question.
The claim fails immediately
Why: The estimate 3.033333 is larger than the true 3.033150, so it was an overestimate. Both of these functions are increasing, yet they miss in opposite directions.
| function | increasing? | concavity | estimate misses |
|---|---|---|---|
| square root at 9 | yes | down | high |
| exponential at 0 | yes | up | low |
Ask only one question: what is the sign of the second derivative near the center?
Compute the second derivative and read its sign
Why: Negative means concave down means the tangent is above means overestimate. Positive means the mirror image. Increasing or decreasing never enters the argument.
Check it against the picture every time
Why: Sketch a quick bowl or dome with a tangent line. The sketch settles the direction in two seconds and cannot be misremembered.
\[ f''>0 \Rightarrow L(x) \le f(x), \qquad f''<0 \Rightarrow L(x) \ge f(x) \]
Comparison
Comparison matrix
From Trap: getting the direction backwards: refill the estimate misses column from what you know. The rest of the table is as it appeared.
| function | increasing? | concavity | estimate misses |
|---|---|---|---|
| square root at 9 | yes | down | high |
| exponential at 0 | yes | up | low |
Pattern
If the second derivative changes sign between the center and the target, say so and stop. An inflection point in the way means there is no guarantee either direction.
Edge cases
Discussion prompt
Pattern: deciding over versus under works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
If the second derivative changes sign between the center and the target, say so and stop. An inflection point in the way means there is no guarantee either direction.
Prediction
Predict first
Is the estimate 0.1 an overestimate or an underestimate, and why?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: Overestimate: the second derivative is negative, so the curve bends below its tangent line.
Why: The second derivative of the natural logarithm is negative one over x squared, which is negative at every point of the domain, so the curve is concave down and its tangent lies above it. Indeed the true value is 0.0953102, less than the estimate 0.1.
Check
The natural logarithm is linearized at the center 1, which gives a very clean formula and a very clean estimate.
\[ L(x) = x - 1 \;\Rightarrow\; \ln(1.1) \approx 0.1 \]
Check your understanding
Is the estimate 0.1 an overestimate or an underestimate, and why?
Answer: A
Why: The second derivative of the natural logarithm is negative one over x squared, which is negative at every point of the domain, so the curve is concave down and its tangent lies above it. Indeed the true value is 0.0953102, less than the estimate 0.1.
Section
Part 4
Concept
You have formulas for linear and quadratic equations. Beyond that, almost nothing solves in closed form.
There is no algebra move that isolates x in either of these:
\[ x^3 - x - 1 = 0, \qquad \cos x = x \]
Roots still exist. We just need a way to hunt them down numerically, to as many decimal places as we want.
Picture it
Figure (svg): A curve crossing the horizontal axis, with a starting point marked above the axis, its tangent line drawn down to the axis, and the crossing point of the tangent marked closer to the true root than the start was.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
You cannot solve the curve equals zero. But you can always solve line equals zero - that is one step of algebra.
Intuition
Figure (svg): A curve crossing the horizontal axis, with a starting point marked above the axis, its tangent line drawn down to the axis, and the crossing point of the tangent marked closer to the true root than the start was.
You cannot solve the curve equals zero. But you can always solve line equals zero - that is one step of algebra.
So: replace the curve by its tangent line at your current guess, solve the easy equation, and take that answer as your new guess. Then do it again.
Concept
Start with the linearization at the current guess, and set it equal to zero.
\[ L(x) = f(x_n) + f'(x_n)\,(x - x_n) = 0 \]
Isolate the change from the current guess. One subtraction, one division.
\[ x - x_n = -\frac{f(x_n)}{f'(x_n)} \]
Call that solution the next guess and you have the whole method.
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
Nothing new was invented. Newton's Method is the linearization, applied over and over.
Picture it
Animation
Shows: Newton's formula falls out of the linearization — a rendered Manim animation.
Rendered with Manim.
Takeaway: Solve the tangent line for its own root, then repeat.
Concept
Newton's Method — An iteration that improves a root estimate by replacing the function with its tangent line and jumping to where that line crosses the horizontal axis. Each new guess is the old guess minus the function value divided by the derivative value, both evaluated at the old guess.
Say it as a correction: new guess equals old guess plus a correction, and the correction is how far off you are divided by how fast the function is moving.
That reading explains both failure modes in advance. If the function is barely moving, the division blows the correction up. If the function value is already zero, the correction is zero and you stop.
Step zero
Discussion prompt
Worked example: the square root of 2 in three iterations — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write the iteration for this particular function
Answer:
Worked example
Find the positive root of this equation, starting from a first guess of 1.5.
\[ f(x) = x^2 - 2, \qquad x_0 = 1.5 \]
Write the iteration for this particular function
Why: The derivative is 2x, so the formula specializes to something you can run in your head with fractions.
\[ x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n} \]
First iteration from 1.5
Why: The function value at 1.5 is 0.25 and the slope is 3, so the tangent hits the axis one twelfth of a unit to the left.
\[ x_1 = 1.5 - \frac{0.25}{3} = \frac{17}{12} = 1.416667 \]
Second iteration
Why: The function value has dropped from 0.25 to under 0.007, so the correction is much smaller than the first one.
\[ x_2 = \frac{17}{12} - \frac{1/144}{17/6} = \frac{577}{408} = 1.414216 \]
Third iteration
Why: The function value is now six millionths, so the correction is about two millionths and we have run out of decimal places on a calculator.
\[ x_3 = \frac{577}{408} - \frac{1/166464}{577/204} = 1.414213562 \]
Verify by squaring the final iterate
Why: Squaring the answer must give back 2. It agrees to nine decimal places, and the table shows the error shrinking from eight hundredths to two millionths in three steps.
| n | x sub n | f at x sub n | f prime at x sub n | next guess |
|---|---|---|---|---|
| 0 | 1.500000000 | 0.250000000 | 3.000000000 | 1.416666667 |
| 1 | 1.416666667 | 0.006944444 | 2.833333333 | 1.414215686 |
| 2 | 1.414215686 | 0.000006007 | 2.828431373 | 1.414213562 |
\[ (1.414213562)^2 = 1.999999999\ldots \approx 2 \quad\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the square root of 2 in three iterations", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Squaring the answer must give back 2. It agrees to nine decimal places, and the table shows the error shrinking from eight hundredths to two millionths in three steps.
Concept
Look at the errors from that run. Each one is roughly the square of the previous one.
| iterate | value | error |
|---|---|---|
| start | 1.500000000 | 0.0857864 |
| after 1 step | 1.416666667 | 0.0024531 |
| after 2 steps | 1.414215686 | 0.0000021 |
| after 3 steps | 1.414213562 | under a billionth |
quadratic convergence — Near a simple root, each Newton iteration roughly squares the error, which doubles the number of correct decimal places every step. It is why three or four iterations usually suffice.
This is why calculators use Newton's Method internally for square roots and reciprocals. A handful of multiplications gets you full machine precision.
Trade off
Comparison matrix
From The digits double every step: every row here is a choice with a cost. Fill the error column, then say which row you would actually pick and what you give up for it.
| iterate | value | error |
|---|---|---|
| start | 1.500000000 | 0.0857864 |
| after 1 step | 1.416666667 | 0.0024531 |
| after 2 steps | 1.414215686 | 0.0000021 |
| after 3 steps | 1.414213562 | under a billionth |
Estimation
Predict first
This cubic has no factorable root. Find its real root to six decimal places, starting from 1.5.
Commit before you compute: what does Worked example: a root of x cubed minus x minus 1 come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by substituting the final iterate back into the function
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The function value at the answer is about two ten-millionths, which is as close to zero as six decimal places can report.
Worked example
This cubic has no factorable root. Find its real root to six decimal places, starting from 1.5.
\[ f(x) = x^3 - x - 1, \qquad f'(x) = 3x^2 - 1, \qquad x_0 = 1.5 \]
Confirm a root is nearby before iterating
Why: The function is negative at 1 and positive at 2, so a continuous function must cross zero between them. Starting inside that bracket is a good habit.
\[ f(1) = -1 < 0, \qquad f(2) = 5 > 0 \]
First iteration
Why: The function value at 1.5 is 0.875 and the slope is 5.75, so the tangent lands about 0.152 to the left.
\[ x_1 = 1.5 - \frac{0.875}{5.75} = 1.347826 \]
Second iteration
Why: The function value has fallen by a factor of nine, so the next correction is nearly seven times smaller than the first.
\[ x_2 = 1.347826 - \frac{0.100682}{4.449905} = 1.325200 \]
Third iteration
Why: The function value is now two thousandths and the correction is under five ten-thousandths - the iterates have stopped changing in the sixth decimal place.
\[ x_3 = 1.325200 - \frac{0.002058}{4.268468} = 1.324718 \]
Verify by substituting the final iterate back into the function
Why: The function value at the answer is about two ten-millionths, which is as close to zero as six decimal places can report. The root is 1.324718.
| n | x sub n | f at x sub n | f prime at x sub n | next guess |
|---|---|---|---|---|
| 0 | 1.500000 | 0.875000 | 5.750000 | 1.347826 |
| 1 | 1.347826 | 0.100682 | 4.449905 | 1.325200 |
| 2 | 1.325200 | 0.002058 | 4.268468 | 1.324718 |
\[ f(1.324718) \approx 0.0000002 \approx 0 \quad\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a root of x cubed minus x minus 1", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The function value at the answer is about two ten-millionths, which is as close to zero as six decimal places can report. The root is 1.324718.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Feeding the previous function value into the front of the formula, because it was the last number you wrote down.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The function value at 1.5 is 0.25, and putting that in front gives a first guess of about 0.167 - nowhere near the root at 1.414.
The leading term is always the previous x-value. The function value only ever appears inside the fraction.
Why: The function value at 1.5 is 0.25, and putting that in front gives a first guess of about 0.167 - nowhere near the root at 1.414.
Trap
Feeding the previous function value into the front of the formula, because it was the last number you wrote down.
\[ x_{n+1} = f(x_n) - \frac{f(x_n)}{f'(x_n)} \quad \text{\small(wrong)} \]
Run it on the square-root-of-2 problem from 1.5
Why: The function value at 1.5 is 0.25, and putting that in front gives a first guess of about 0.167 - nowhere near the root at 1.414.
\[ x_1 = 0.25 - \frac{0.25}{3} = 0.166667 \]
The second step throws it clear off the map
Why: From 0.167 the function value is about minus 1.97 and the slope is only one third, so the correction is enormous. The iterates wander instead of converging.
| step | guess produced | distance from the root |
|---|---|---|
| 1 | 0.166667 | 1.247547 |
| 2 | 3.944444 | 2.530231 |
The leading term is always the previous x-value. The function value only ever appears inside the fraction.
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
Run it correctly from 1.5
Why: Starting from 1.5 and subtracting the correction 0.25 over 3 gives 1.4167 - still an x-value, still on the number line where the root lives.
\[ x_1 = 1.5 - \frac{0.25}{3} = 1.416667 \]
And it closes in immediately
Why: Each guess is an input, each correction is a small nudge to that input. Keeping the two roles separate is the whole discipline here.
| step | guess produced | distance from the root |
|---|---|---|
| 1 | 1.416667 | 0.002453 |
| 2 | 1.414216 | 0.000002 |
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
If the tangent line at your guess is horizontal, it never meets the axis. There is no next guess.
Algebraically the formula divides by zero; geometrically the tangent runs parallel to the axis forever.
\[ f(x) = x^2 - 1, \quad x_0 = 0 \;\Rightarrow\; f'(0) = 0 \]
A near-miss is almost as bad: a starting point where the slope is very small produces a gigantic correction that flings the guess far away.
Explain it
Discussion prompt
Explain Failure one: the derivative is zero to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
If the tangent line at your guess is horizontal, it never meets the axis. There is no next guess.
Concept
Newton's Method is only guaranteed to work when you start close enough to a root. From far away, anything can happen.
None of these announce themselves. This is why you always sketch, or bracket the root with a sign change, before you start iterating.
Analogy
Discussion prompt
Explain Failure two: a bad starting guess by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Newton's Method is only guaranteed to work when you start close enough to a root. From far away, anything can happen.
Step zero
Discussion prompt
Worked example: an iteration that cycles forever — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: First iteration from 0
Answer:
Worked example
This cubic has a perfectly good real root, but one particular starting guess traps the method.
\[ f(x) = x^3 - 2x + 2, \qquad f'(x) = 3x^2 - 2, \qquad x_0 = 0 \]
First iteration from 0
Why: The function value is 2 and the slope is negative 2, so the correction is a full unit to the right.
\[ x_1 = 0 - \frac{2}{-2} = 1 \]
Second iteration from 1
Why: The function value is 1 and the slope is 1, so the correction is a full unit back to the left - landing exactly where we started.
\[ x_2 = 1 - \frac{1}{1} = 0 \]
Name what happened: a two-cycle
Why: The method now alternates between 0 and 1 forever. Nothing diverges and nothing errors out, so a program with no iteration cap would simply hang.
| n | x sub n |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 0 |
| 3 | 1 |
| 4 | 0 |
Fix it by restarting near an actual sign change
Why: The function is negative at minus 2 and positive at minus 1, so the root lives between them. Starting at minus 2 converges in three steps.
\[ f(-2) = -2 < 0, \qquad f(-1) = 3 > 0 \]
| n | x sub n |
|---|---|
| 0 | -2.000000 |
| 1 | -1.800000 |
| 2 | -1.769948 |
| 3 | -1.769293 |
Verify the recovered root
Why: Substituting the final iterate gives a function value of about five hundred-thousandths, essentially zero, so the real root is about negative 1.769292. The method was never broken - the starting point was.
\[ f(-1.769293) \approx 0.00000 \quad\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "an iteration that cycles forever", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Substituting the final iterate gives a function value of about five hundred-thousandths, essentially zero, so the real root is about negative 1.769292. The method was never broken - the starting point was.
Concept
Two habits turn Newton's Method from a gamble into a reliable tool.
Stop when consecutive guesses agree to the precision you need, and confirm by checking that the function value is genuinely near zero.
\[ |x_{n+1} - x_n| < \text{tolerance} \quad\text{and}\quad |f(x_{n+1})| \text{ small} \]
Counterexample
Discussion prompt
Two habits turn Newton's Method from a gamble into a reliable tool.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Stop when consecutive guesses agree to the precision you need, and confirm by checking that the function value is genuinely near zero.
Pattern
The engine of every row is the same single line:
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
Check
Run a single iteration by hand. No calculator needed.
\[ f(x) = x^2 - 5, \qquad x_0 = 2 \]
Check your understanding
What is the first Newton iterate?
Answer: A
Why: The function value at 2 is negative 1 and the derivative is 4, so the iterate is 2 minus negative one quarter, which is 2.25. That is already within 0.014 of the true square root of 5, namely 2.236068.
Check
A student runs Newton's Method and reports the following list of iterates, which continues in the same alternating way.
\[ f(x) = x^3 - 2x + 2, \quad x_0 = 0: \qquad 0,\; 1,\; 0,\; 1,\; 0,\; \ldots \]
Check your understanding
What went wrong?
Answer: A
Why: From 0 the iteration produces 1, and from 1 it produces 0 again, so the two guesses map to each other forever. The method is working exactly as defined - the starting point is the problem, and restarting near a sign change at negative 2 converges to the root at about negative 1.769292.
Section
Part 5
Concept
Every tool in this deck is the tangent line answering a different question.
| question you are asking | tool | what the tangent line gives you |
|---|---|---|
| What is the function's value near here? | linearization | the height of the line at the new input |
| How much does the output move? | differential | the rise of the line over a small run |
| Where does the function hit zero? | Newton's Method | where the line crosses the axis, then repeat |
Learn the tangent line once and you have learned all three. That is why they live in the same chapter.
Comparison
Comparison matrix
From Linearization, differentials, Newton: the same tangent line: refill the tool column from what you know. The rest of the table is as it appeared.
| question you are asking | tool | what the tangent line gives you |
|---|---|---|
| What is the function's value near here? | linearization | the height of the line at the new input |
| How much does the output move? | differential | the rise of the line over a small run |
| Where does the function hit zero? | Newton's Method | where the line crosses the axis, then repeat |
Picture it
Animation
Shows: Linearization, differentials, Newton — a rendered Manim animation.
Rendered with Manim.
Takeaway: Recognising them as one object removes two formulas from the list.
Ranking
Put in order
Put the moves of Worked example: estimate, then improve into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Sixteen is the nearest perfect square, so both the value 4 and the slope one eighth are exact.
Worked example
Estimate this by hand, say whether the estimate is high or low, and then improve it with one Newton step.
\[ \sqrt{17} \]
Linearize the square root at the center 16
Why: Sixteen is the nearest perfect square, so both the value 4 and the slope one eighth are exact.
\[ L(x) = 4 + \frac{1}{8}(x-16) \;\Rightarrow\; L(17) = 4.125 \]
Call the direction from concavity
Why: The second derivative of the square root is negative everywhere on its domain, so the curve is concave down and the tangent rides above it. The estimate 4.125 must be too big.
\[ f''(x) = -\tfrac{1}{4}x^{-3/2} < 0 \;\Rightarrow\; \text{overestimate} \]
Hand the estimate to Newton's Method as a starting guess
Why: Finding the square root of 17 is the same as solving x squared minus 17 equals zero, and 4.125 is already a very good start.
\[ g(x) = x^2 - 17, \qquad x_0 = 4.125 \]
Take one iteration
Why: The function value at 4.125 is 0.015625 and the slope is 8.25, so the correction is about 0.0019 to the left - which agrees with the concavity call that we were too high.
\[ x_1 = 4.125 - \frac{0.015625}{8.25} = 4.1231061 \]
Verify by squaring both answers
Why: The linear estimate squares to 17.015625, off by fifteen thousandths. One Newton step squares to 17.0000036, off by less than four millionths. Same tangent line, one extra turn of the crank, four hundred times the accuracy.
| estimate | value | its square | distance from 17 |
|---|---|---|---|
| linearization | 4.1250000 | 17.015625 | 0.015625 |
| one Newton step | 4.1231061 | 17.0000036 | 0.0000036 |
| true value | 4.1231056 | 17.000000 | 0 |
Picture it
Animation
Shows: Each line of the worked example "estimate, then improve", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The linear estimate squares to 17.015625, off by fifteen thousandths. One Newton step squares to 17.0000036, off by less than four millionths. Same tangent line, one extra turn of the crank, four hundred times the accuracy.
Pattern
In all four cases the first move is identical: find the function, the center, and the derivative at the center.
Real world
Discussion prompt
Outside this lesson: where does Linear Approximation, Differentials, and Newton's Method actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: which tool does this problem want? is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck covers local linearity and the linearization of a function at a point, then differentials and propagated measurement error, using concavity to decide whether an estimate comes out too high or too low, and Newton's Method together with its failure modes. It targets the classic errors: centering at a point whose value you do not know, confusing the differential with the true change, getting the over-or-under direction backwards, and iterating Newton's Method with the previous function value instead of the previous x-value.
Elimination
Eliminate the wrong options
What is the first Newton iterate, rounded to six decimal places?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The function value at 3 is 27 minus 20 equals 7, and the derivative 3x squared is 27, so the iterate is 3 minus 7 over 27, which is 2.740741. The true cube root of 20 is 2.714418, so one step already closes most of the gap.
Check
One iteration, one starting guess, one careful substitution.
\[ f(x) = x^3 - 20, \qquad x_0 = 3 \]
Check your understanding
What is the first Newton iterate, rounded to six decimal places?
Answer: A
Why: The function value at 3 is 27 minus 20 equals 7, and the derivative 3x squared is 27, so the iterate is 3 minus 7 over 27, which is 2.740741. The true cube root of 20 is 2.714418, so one step already closes most of the gap.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Local Linearity and the Linearization · Differentials: the calculus of small nudges · Too Big or Too Small? Concavity Decides · Newton's Method: the same idea, on repeat · One Idea, Three Costumes. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
Everything here came from one fact: a differentiable curve looks like its tangent line up close.
\[ L(x) = f(a) + f'(a)(x-a), \qquad dy = f'(x)\,dx, \qquad x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
| if you remember one thing | it is this |
|---|---|
| the center | must be friendly and close, or the estimate is worthless |
| the differential | is the tangent's prediction, not the function's true change |
| the direction | comes from concavity, never from increasing or decreasing |
| the Newton step | starts from the previous x-value, never from a function value |
Next up: indeterminate forms and a rule for the limits that this chapter's approximations cannot settle.
Want this taught 1-on-1? Alexander tutors Calculus I — $55/session, free consultation.