The linearization of a function at a point and its use for approximating values, the differential as the change along the tangent, the direction of the error decided by concavity and its second-order growth with the step, and relative and percentage error in propagated measurements.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 4 — Applications of Derivatives
Linear Approximations and Differentials
Objectives
Five outcomes. The first is Section 3.1's tangent line read as a tool rather than a description.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 308-316 — the section these objectives are drawn from
Warm-up
Section 3.4 estimated the change in a quantity by multiplying its rate by a small step, and noticed the estimate was slightly low for a curve bending upward.
Discussion prompt
Section 3.1 found the tangent to the square root at 9 to be y equals x over 6 plus 3 halves. Use it to estimate the square root of 10, and say what the estimate really is.
Hint: Substitute 10 into the tangent line rather than into the curve.
Answer:
\[ y = \tfrac{x}{6} + \tfrac{3}{2} \;\Longrightarrow\; \text{at } x = 10: \; \tfrac{10}{6} + \tfrac{3}{2} = 3.1\overline{6} \]
The true value is about 3.1623, so the estimate is high by 0.0044 — a relative error of about a tenth of a percent.
What has been computed is the height of the tangent line at 10, standing in for the height of the curve. That substitution is the whole method of this section, and the rest is about how good it is and when.
Concept
Near the point of contact, a differentiable function is almost indistinguishable from its tangent. Replacing the function by that line gives an approximation which is easy to compute and whose error can be described.
linearization — The linear function whose graph is the tangent to f at a. It equals f of a plus f prime of a times the step from a, and it is the best linear approximation to f near that point.
\[ L(x) = f(a) + f'(a)(x-a) \approx f(x) \]
This is not a new object. It is the tangent line from Section 3.1, written in point-slope form and read as a recipe: start from a value you know, and travel along the slope.
Figure (svg): The linearization formula, with each piece labelled by where it comes from
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 308-310
Section
Section 1
Concept
Choose a nearby point where both the value and the derivative are easy. The linearization then estimates the function at any nearby input by stepping along the tangent from that anchor.
the anchor point — The input a at which the value and derivative are known exactly. It should be as close as possible to the target while still being easy to evaluate — usually a perfect square, a whole number or a landmark angle.
\[ f(x) \approx f(a) + f'(a)(x-a) \]
The choice of anchor matters more than anything else. Estimating the root of 10 from 9 works because 9 is a perfect square one unit away; anchoring at 4 would give a far worse estimate for the same effort.
Figure (svg): A curve with its tangent at a point, and the approximation error opening up away from contact
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 308-311 — linear approximation of a function at a point
Picture it
The square root near nine.
Figure (svg): A curve with its tangent at a point, and the approximation error opening up away from contact
One unit from contact the two graphs are barely distinguishable, and the estimate is right to three decimal places. Ten units away they have separated visibly.
Worked example
Example 4.7. Anchor at the nearest easy point.
\[ \text{Estimate } \sqrt{10} \text{ using a linear approximation.} \]
Choose the anchor
Why: The nearest perfect square.
\[ a = 9, f(9) = 3 \]
Compute the derivative there
Why: One over twice the root.
\[ f'(9) = \frac{1}{6} \]
Write the linearization
Why: Value plus slope times step.
\[ L(x) = 3 + (\frac{1}{6}) (x - 9) \]
Evaluate at the target
Why: The step is 1.
\[ L(10) = 3 + \frac{1}{6} \]
State
Why: As a decimal.
\[ \text{about } 3.1667 \]
Figure (svg): The solution to Worked example approximating a root shown as a ladder of expressions, one row per legal move
\[ \sqrt{10} \approx 3 + \tfrac16 \approx 3.1667 \]
Verify: compute the error and its relative size
Why: The true value is 3.16228, so the estimate is high by 0.0044 — a relative error of 0.14 percent, from a calculation needing no more than a division by 6. The estimate is high because the square root bends downward, so its tangent lies above it, which the next-but-one idea makes into a rule. Anchoring at 4 instead would have given 2 plus 6 over 4, which is 3.5 — far worse, because the step is six times longer.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 309-310
Matching
Nearest point where the value is easy.
Match the pairs
Why: In every case the anchor is the nearest input at which both the value and the derivative are trivial. Since the error grows with the square of the step, closeness is worth far more than any convenience gained by a rounder anchor.
Worked example
Checkpoint 4.7. A landmark angle as the anchor.
\[ \text{Estimate } \sin(0.1) \text{ using a linear approximation, in radians.} \]
Choose the anchor
Why: The nearest angle with known values.
\[ a = 0, \sin 0 = 0 \]
Compute the derivative there
Why: Cosine of 0.
\[ f'(0) = 1 \]
Write the linearization
Why: Value plus slope times step.
\[ L(x) = 0 + 1(x - 0) = x \]
Evaluate
Why: The step is 0.1.
\[ L(0.1) = 0.1 \]
Compare
Why: The true value.
\[ \sin(0.1) = 0.099833 \]
Figure (svg): The solution to Worked example a trigonometric estimate shown as a ladder of expressions, one row per legal move
\[ \sin x \approx x \text{ for small } x \]
Verify: recognise the result and check the direction of the error
Why: The linearization of sine at the origin is simply x, which is the small-angle approximation used throughout physics — and it is nothing more than this section's method applied at a convenient anchor. The estimate is high by 0.000167, and sine bends downward for small positive x, so the tangent lies above it exactly as expected. Note also that this only works in radians: in degrees the slope at 0 would be pi over 180, not 1.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 310-311
Trap
\[ \text{estimate } \sqrt{10} \text{ from } a = 4 \]
Use a familiar anchor rather than a near one
Why: The student picks 4 because its root is 2.
\[ L(10) = 2 + \tfrac14(6) = 3.5 \quad \text{(error } 0.34\text{)} \]
The step is six units long, and the error is nearly eighty times larger than anchoring at 9 would have given.
\[ a = 9: \quad L(10) = 3 + \tfrac16 \approx 3.1667 \quad \text{(error } 0.0044\text{)} \]
Choose the NEAREST point at which the value is easy
Why: The error grows with the square of the step, so distance is expensive.
Both anchors are equally easy to compute with, so nothing was gained by the poorer choice. Since the error grows quadratically, a step six times longer produces roughly thirty-six times the error — and here it produced eighty, because the curve is bending more sharply over the wider interval.
Fill the middle
The square root anchored at nine.
Fill in the blanks
L(x) = 3 + \tfrac16(x - 9) \;\Longrightarrow\; L(10) = 3 + 1/6
Why: The step is 1, so the correction is one sixth. The estimate 3.1667 is high by 0.0044, since the square root's tangent lies above the curve.
Ranking
Approximating a value linearly.
Put in order
Why: Step b is where most of the accuracy is won or lost, since the error is quadratic in the step. Step e costs one second and turns an estimate into an estimate with a known direction of error, which is often what an application needs.
Prediction
Commit before reasoning.
Predict first
What object is the linearization of f at a?
Correct: The tangent line, written in point-slope form.
\[ y - f(a) = f'(a)(x-a) \;\Longleftrightarrow\; L(x) = f(a) + f'(a)(x-a) \]
Why: Nothing new is being defined. Section 3.1's tangent equation was y minus f of a equals f prime of a times x minus a, and rearranging gives exactly the linearization. The only change is one of attitude: instead of describing the curve's direction, the line is being used to stand in for the curve's values. A quadratic fit would be a better approximation and is a genuine idea, but it belongs to Taylor polynomials rather than here.
Section
Section 2
Concept
The differential dy is the change in height along the tangent when the input changes by dx. It approximates the true change in the function, and the two agree in the limit as the step shrinks.
the differential — For a differentiable function, dy equals f prime of x times dx. It is the rise along the tangent line over a step dx, and it approximates the actual change delta y in the function.
\[ dy = f'(x)\,dx \approx \Delta y = f(x+dx) - f(x) \]
The notation finally makes dy by dx look like a genuine quotient, and here it is one: dy divided by dx really is f prime of x. That is why Leibniz shaped the symbol as he did.
Figure (svg): The differential dy against the true change, with the gap between them marked
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 311-313 — differentials
Picture it
The differential and the true change, for the squaring function.
Figure (svg): The differential dy against the true change, with the gap between them marked
The green segment is dy, travelling along the tangent, and the true change reaches one unit higher. The red gap is the error, and it is exactly the square of the step for this function.
Worked example
Example 4.9. The differential against the true change.
\[ \text{For } y = x^{2} \text{ at } x = 2 \text{ with } dx = 1, \text{ find } dy \text{ and } \Delta y. \]
Compute the derivative
Why: Power rule.
\[ \,dy / \,dx = 2 x,\text{ so } 4\text{ at } x = 2 \]
Compute the differential
Why: Slope times step.
\[ \,dy = 4(1) = 4 \]
Compute the actual change
Why: The function's values.
\[ f(3) - f(2) = 9 - 4 = 5 \]
Compare
Why: The differential is low.
\[ \text{error } 1 \]
Note the error's form
Why: For this function it is exactly the square of the step.
\[ (\,dx) ^{2} = 1 \]
Figure (svg): The solution to Worked example computing a differential shown as a ladder of expressions, one row per legal move
\[ dy = 4, \qquad \Delta y = 5 \]
Verify: repeat with a smaller step and watch the error
Why: With dx equal to 0.1: dy is 0.4 and the true change is 4.41 minus 4, which is 0.41 — an error of 0.01, exactly the square of the step. With dx equal to 0.01 the error is 0.0001. The error is second order in the step, which is why the approximation is excellent for small steps and poor for large ones. For the squaring function the error is exactly dx squared; in general it is approximately half the second derivative times dx squared.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 312-313
Fill the middle
The squaring function at x equal to 2, stepping one unit.
Fill in the blanks
dy = f'(2)\,dx = 4(1) = 4
Why: The differential is 4, while the true change is 5. The gap of 1 is exactly the square of the step for this function, which is why the approximation improves so sharply as the step shrinks.
Worked example
Checkpoint 4.9. The same method, applied.
\[ \text{Use a differential to estimate } (1.02)^{5}. \]
Identify the function and anchor
Why: The fifth power, near 1.
\[ f(x) = x ^{5}, a = 1 \]
Compute the value and derivative there
Why: Both easy.
\[ f(1) = 1, f'(1) = 5 \]
Compute the differential
Why: Slope times step.
\[ \,dy = 5(0.02) = 0.1 \]
Add to the anchor's value
Why: Estimate.
\[ 1 + 0.1 = 1.1 \]
Compare with the true value
Why: By direct computation.
\[ 1.10408 \]
Figure (svg): The solution to Worked example using a differential to estimate shown as a ladder of expressions, one row per legal move
\[ (1.02)^{5} \approx 1.1 \]
Verify: check the direction and the size of the error
Why: The estimate is low by 0.004, and the fifth power bends upward, so its tangent lies below the curve — the direction is as the concavity predicts. The relative error is about 0.4 percent from a mental calculation. Note that the exponent 5 became the magnification factor: a 2 percent step in x produced a 10 percent change in the output, which is the propagation effect the last idea makes precise.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 313-313
Error analysis
A student computes a change in area.
Annotate
On: \( A = x^{2}, \; x: 2 \to 3 \;\Longrightarrow\; \Delta A = dA = 2(2)(1) = 4 \)
The differential is an approximation to the change, exact only in the limit. For a step of 1 on this function the gap is a fifth of the answer, which is far too large to ignore — though at a step of 0.01 it would be a hundredth of a percent.
Sorting
Distinguish the two vertical distances.
Sort into buckets
Sort each quantity.
The two agree in the limit as the step shrinks to zero, which is exactly the definition of the derivative read backwards. For any finite step they differ, and the gap is what the second derivative controls.
Two truths and a lie
All three are about differentials.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The differential travels along the tangent and the true change along the curve, so they differ for any finite step. For the squaring function at 2 with a step of 1 they are 4 and 5, a 20 percent discrepancy. Only in the limit do they coincide.
Prediction
Commit before reasoning.
Predict first
Section 3.2 warned that dy/dx is not really a fraction. What changes here?
Correct: dy and dx are now separately defined, and their quotient really is f prime of x.
\[ dy = f'(x)\,dx \;\Longrightarrow\; \frac{dy}{dx} = f'(x) \quad \text{genuinely} \]
Why: Section 3.2's warning was about the derivative symbol taken as a whole, where no separate du existed to cancel. Here dx is chosen freely and dy is DEFINED as f prime of x times dx, so dividing gives the derivative exactly. This retrospectively justifies Leibniz's notation and is why substitution in Section 5.5 will manipulate dx and dy as though they were quantities — because with this definition, they are.
Section
Section 3
Concept
A curve bending upward lies above every one of its tangent lines, so a linear estimate undershoots. A curve bending downward lies below them, so the estimate overshoots. The sign of the second derivative decides which.
concavity and estimation error — When the second derivative is positive the graph bends upward and lies above its tangents, so linear estimates are too small. When it is negative the graph bends downward and estimates are too large.
\[ f'' > 0 \Rightarrow \text{undershoot}; \qquad f'' < 0 \Rightarrow \text{overshoot} \]
The direction is the same on both sides of the anchor, which is worth noticing. It depends on how the curve bends, not on which way you step from the point.
Figure (svg): The error's direction decided by concavity, shown for an upward and a downward bending curve
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 310-314 — accuracy of the approximation
Picture it
Two curves with tangents at the same relative position.
Figure (svg): The error's direction decided by concavity, shown for an upward and a downward bending curve
On the left the tangent runs beneath the curve on both sides; on the right it runs above. One sign of one derivative settles which case you are in.
Worked example
Example 4.10. Check the second derivative before computing.
\[ \text{Will the linear estimate of } \sqrt{10} \text{ from } a=9 \text{ be too high or too low?} \]
Compute the first derivative
Why: The root's derivative.
\[ f'(x) = (\frac{1}{2}) x ^{-\frac{1}{2}} \]
Differentiate again
Why: Reduce the exponent.
\[ f''(x) = -(\frac{1}{4}) x ^{-\frac{3}{2}} \]
Determine the sign
Why: Negative for every positive x.
\[ f'' < 0 \]
Conclude
Why: The curve bends downward, so the tangent is above it.
Check against the numbers
Why: Estimate 3.1667 against true 3.1623.
Figure (svg): The solution to Worked example predicting the direction shown as a ladder of expressions, one row per legal move
\[ f'' < 0 \;\Longrightarrow\; \text{the estimate overshoots} \]
Verify: confirm the prediction was made before the true value was known
Why: The sign of the second derivative was determined from the formula alone, with no reference to the actual value of the root of 10 — and it correctly predicted the direction. That is the practical value: in a real application the true value is unknown, so knowing that an estimate is an overestimate rather than an underestimate is genuine information. For a safety margin it is often the more important half of the answer.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 313-314
Sorting
Determine the sign of the second derivative.
Sort into buckets
Sort each linear estimate.
The three overshooting cases are all functions that grow ever more slowly — roots, sines and logarithms — while the two undershooting ones accelerate. That correspondence between growth behaviour and bending is what Section 4.5 will formalise.
Worked example
Checkpoint 4.10. A curve bending upward.
\[ \text{Will the estimate of } (1.02)^{5} \text{ from } a = 1 \text{ be too high or too low?} \]
Differentiate twice
Why: Power rule twice.
\[ f' = 5 x ^{4}, f'' = 20 x ^{3} \]
Determine the sign near the anchor
Why: Positive for positive x.
\[ f'' > 0 \]
Conclude
Why: Bending upward, so the tangent is below.
Check
Why: Estimate 1.1 against true 1.10408.
Figure (svg): The solution to Worked example the other direction shown as a ladder of expressions, one row per legal move
\[ f'' > 0 \;\Longrightarrow\; \text{the estimate undershoots} \]
Verify: test the claim that the direction is the same on both sides
Why: Estimating at x equal to 0.98 gives 1 minus 0.1, or 0.9, while the true value is 0.90392 — again an underestimate, on the other side of the anchor. The direction depends on the concavity rather than on which way the step goes, which is what makes it a reliable rule. Chapter 4's Section 4.5 will develop concavity properly; here only its sign is needed.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 314-314
Trap
\[ \text{stepping right undershoots, so stepping left must overshoot} \]
Assume the error flips with the direction of the step
Why: The student expects symmetry about the anchor.
For the fifth power, estimates at 1.02 and at 0.98 are BOTH too low. The direction did not flip.
\[ f'' > 0 \;\Longrightarrow\; \text{the curve is above its tangent on BOTH sides} \]
Read the direction from the concavity alone
Why: A curve bending upward lies above its tangent everywhere near the point, left and right alike.
The picture makes it obvious once seen: a tangent to a bowl-shaped curve touches at one point and lies beneath it in both directions. Only at an inflection point, where the concavity changes, does the tangent cross the curve — and Section 4.5 is where those are hunted.
Fill the middle
The square root, differentiated twice.
Fill in the blanks
f'(x) = \tfrac12 x^-3/2 \;\Longrightarrow\; f''(x) = -\tfrac14 x^___}
Why: Negative one half minus one is negative three halves, and the coefficient picks up a minus. The result is negative for every positive x, so the square root bends downward and its tangent lies above it.
Two truths and a lie
All three are about the error's direction.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Concavity is a property of the curve, not of the direction of travel, so a bowl-shaped curve lies above its tangent on both sides and every linear estimate from that point undershoots. Testing the fifth power at 0.98 and 1.02 confirms it: both estimates are low.
Prediction
Commit before reasoning.
Predict first
An engineer estimates a load-bearing capacity linearly and knows the function bends downward. What follows?
Correct: The estimate is too high, so the true capacity is lower than computed.
\[ f'' < 0 \;\Longrightarrow\; L(x) > f(x) \text{ near } a \]
Why: A downward-bending function lies below its tangent, so the linear estimate overstates it — and in a safety calculation that is exactly the dangerous direction. Knowing the sign of the error without knowing its size is often decisive: it tells you whether the approximation is conservative or optimistic. That is why the second derivative's sign is worth the ten seconds it costs, even when no error bound is computed.
Section
Section 4
Concept
The error in a linear approximation is roughly proportional to the square of the step. Halving the step quarters the error, which is why these estimates are excellent nearby and worthless far away.
second-order error — The gap between a function and its linearization is approximately half the second derivative times the square of the step. Because the step is squared, small steps give disproportionately small errors.
\[ f(x) - L(x) \approx \tfrac{1}{2}f''(a)(x-a)^{2} \]
The quadratic growth explains both the method's power and its limits. A step of a hundredth gives an error of order a ten-thousandth; a step of ten gives an error of order a hundred.
Figure (svg): The error growing with the square of the step, tabulated
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 313-315 — the accuracy of linear approximation
Picture it
The error tabulated against the step.
Figure (svg): The error growing with the square of the step, tabulated
Each row's step is a fifth or a half of the previous one and the error falls by the square of that factor. For the squaring function the relationship is exact rather than approximate.
Worked example
Example 4.11. Tabulate and read the pattern.
\[ \text{For } f(x)=x^{2} \text{ at } a=2, \text{ compute the error for } dx = 1, 0.5, 0.1. \]
Compute the differential in general
Why: Slope times step.
\[ \,dy = 4 \,dx \]
Compute the true change in general
Why: Expand.
\[ (2 + \,dx) ^{2} - 4 = 4 \,dx + (\,dx) ^{2} \]
Subtract
Why: The difference.
\[ \text{error } = (\,dx) ^{2} \]
Tabulate
Why: Three steps.
\[ 1, 0.25, 0.01 \]
Read the pattern
Why: The square of the step, exactly.
Figure (svg): The solution to Worked example the error's growth measured shown as a ladder of expressions, one row per legal move
\[ \Delta y - dy = (dx)^{2} \]
Verify: compare with the general second-order formula
Why: The general estimate says the error is about half the second derivative times the step squared. Here the second derivative is 2, so half of it is 1, and the predicted error is exactly the step squared — matching the exact computation. For the squaring function the formula is exact because the third and higher derivatives vanish; for other functions it is an approximation that improves as the step shrinks.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 314-315
Fill the middle
For the squaring function, where the error is exactly the square of the step.
Fill in the blanks
\text0.0001 0.1 \Rightarrow \text___ 0.01; \quad \text___ 0.01 \Rightarrow \text___ ___
Why: The error is the square of the step, so a tenfold smaller step gives a hundredfold smaller error. That quadratic improvement is what makes linear approximation so effective at short range.
Worked example
Checkpoint 4.11. Bound the error before relying on it.
\[ \text{Estimate } \sqrt{10} \text{ and bound the error using the second derivative.} \]
Write the second derivative
Why: From the earlier computation.
\[ f''(x) = -(\frac{1}{4}) x ^{-\frac{3}{2}} \]
Bound its size on the interval
Why: Largest in magnitude at the left end.
\[ | f'' | \le \frac{1}{108}\text{ on } [9, 10] \]
Apply the second-order bound
Why: Half the bound times the step squared.
\[ \text{error } \le(\frac{1}{2}) (\frac{1}{108}) (1) \]
Evaluate
Why: About 0.0046.
\[ \text{at most } 0.005 \]
Compare with the actual error
Why: 0.0044.
Figure (svg): The solution to Worked example judging whether an estimate is trustworthy shown as a ladder of expressions, one row per legal move
\[ |\text{error}| \le \tfrac12\cdot\tfrac{1}{108}\cdot 1 \approx 0.0046 \]
Verify: check the bound is honest rather than lucky
Why: The actual error, 0.0044, is just under the bound of 0.0046, so the bound is tight but valid. It was obtained without knowing the true value — only from the second derivative's size on the interval — which is what makes it useful in practice. Note that the bound needed the maximum of the second derivative's magnitude over the whole interval, not just at the anchor, since the curve bends throughout the step.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 315-315
Error analysis
A student reasons about how a linear estimate degrades.
Annotate
On: \( \text{step } 0.1 \text{ gives error } 0.01, \text{ so step } 1 \text{ gives error } 0.1 \)
The quadratic growth is what makes these approximations both powerful and dangerous. Small steps are far better than linear reasoning suggests, and large steps are far worse.
Sorting
Judge by the step size and the curvature.
Sort into buckets
Sort each situation.
The linear case is worth dwelling on: with a zero second derivative the error term vanishes identically, so the approximation is exact for any step whatever. Everything else is a matter of how much bending happens over the interval.
Two truths and a lie
All three are about the error.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it badly misjudges the method in both directions. The error is proportional to the SQUARE of the step, so short steps are far better than linear reasoning predicts and long steps far worse. For the squaring function the error is exactly the step squared, which makes the pattern easy to verify.
Prediction
Commit before reasoning.
Predict first
Why is the error second order rather than first order in the step?
Correct: Because the tangent already matches both the value and the slope, so the leading leftover involves the second derivative.
\[ f(x) = f(a) + f'(a)(x-a) + \tfrac12 f''(c)(x-a)^{2} \]
Why: The linearization is built to agree with f in value and in first derivative at the anchor, so those two contributions cancel exactly. The first surviving discrepancy comes from the second derivative, and it enters multiplied by the square of the step. This is the first term of a Taylor expansion, and it explains both why the method works so well nearby and why a quadratic approximation would do better still.
Section
Section 5
Concept
A measured quantity carries an error, and computing with it propagates that error. The differential converts a small measurement error into the resulting error in the computed quantity, and relative errors magnify by the exponent involved.
relative and percentage error — The relative error is the absolute error divided by the quantity, and the percentage error is that times a hundred. Relative errors are what compare across scales, since an absolute error means nothing without a size to compare it with.
\[ \frac{dV}{V} = 3\frac{dr}{r} \quad \text{for a sphere} \]
The magnification by the exponent is the practical headline. A one percent error in a measured radius becomes a three percent error in the computed volume, because the volume depends on the cube.
Figure (svg): Absolute against relative error, showing why the second is usually what matters
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 315-316 — calculating the amount of error
Picture it
One centimetre, on a rod and on a road.
Figure (svg): Absolute against relative error, showing why the second is usually what matters
The absolute error is identical and the two situations are not comparable at all. Only the relative error distinguishes a ruinous measurement from a negligible one.
Worked example
Example 4.12. The differential does the propagation.
\[ \text{A sphere's radius is measured as } 10 \text{ cm with a possible error of } 0.1 \text{ cm. Bound the error in the computed volume.} \]
Write the volume relation
Why: The sphere.
\[ V = (\frac{4}{3}) \pi r ^{3} \]
Take the differential
Why: The chain rule factor is the surface area.
Substitute the measured values
Why: Radius 10, error 0.1.
Evaluate
Why: About 125.7.
Express as a relative error
Why: Divide by the volume.
Figure (svg): How a measurement error in a radius propagates into the volume, magnified by the power
\[ dV = 40\pi \approx 126 \text{ cm}^{3}, \quad \frac{dV}{V} = 3\% \]
Verify: check the relative error against the exponent rule
Why: The radius carries a 1 percent error and the volume a 3 percent one — exactly three times, because the volume depends on the cube of the radius. That relationship comes straight from differentiating: the exponent 3 comes down and becomes the magnification factor. The absolute figure of 126 cubic centimetres sounds large but is only 3 percent of a volume of about 4189, which is why the relative form is the more informative one.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 315-316
Fill the middle
A sphere, whose volume depends on the cube of the radius.
Fill in the blanks
\frac3___ = ___\,\frac______
Why: The exponent 3 becomes the magnification factor, so a 1 percent error in the radius gives a 3 percent error in the volume. The same happens for any power: the exponent is the multiplier.
Worked example
Checkpoint 4.12. How precisely must you measure?
\[ \text{How accurately must a sphere's radius be measured for the volume to be within } 1\%? \]
Write the relative error relation
Why: From the propagation rule.
Substitute the required tolerance
Why: One percent on the volume.
\[ 0.01 = 3(\,dr / r) \]
Solve for the radius's relative error
Why: Divide by 3.
\[ \,dr / r = \frac{1}{300} \]
Express as a percentage
Why: About a third of a percent.
\[ 0.33 \% \]
Interpret for a 10 cm sphere
Why: Multiply by the radius.
\[ \text{within } 0.033 \text{cm} \]
Figure (svg): The solution to Worked example working backwards from a tolerance shown as a ladder of expressions, one row per legal move
\[ \frac{dr}{r} = \frac{1}{300} \approx 0.33\% \]
Verify: notice the direction of the demand
Why: The tolerance on the radius is three times TIGHTER than the tolerance on the volume, because errors are magnified going forwards and so must be shrunk going backwards. This is the practically important direction: a specification on a computed quantity translates into a much stricter demand on the measured one. For a cube, where volume also goes as the third power, the same factor of three applies.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 316-316
Trap
\[ dV = 126 \text{ cm}^{3} \]
Report the absolute error alone
Why: The student gives the figure without context.
Is that a serious error? The number alone cannot say — it depends entirely on how large the volume is.
\[ dV = 126 \text{ cm}^{3}, \quad \frac{dV}{V} = 3\% \]
Report the relative error alongside the absolute one
Why: Only the relative figure is comparable across scales.
A 126 cubic centimetre error is 3 percent of this sphere and would be a millionth of a percent of a swimming pool. Scientific work reports relative or percentage error for exactly this reason, and a specification is almost always stated that way too.
Matching
The exponent is the factor.
Match the pairs
Why: The last row runs the propagation backwards, and the factor inverts: computing a side from a measured volume divides the relative error by three. That is why a volume measurement can give a surprisingly precise length even when the volume itself is roughly known.
Sorting
Which figure answers the question asked?
Sort into buckets
Sort each statement.
Specifications are almost always relative and raw measurements almost always absolute, so converting between them is routine work. The conversion is a single division, and forgetting to do it is what makes an error figure uninterpretable.
Prediction
Commit before reasoning.
Predict first
You need a computed volume accurate to 2 percent. How accurately must the radius be measured?
Correct: About 0.67 percent — the tolerance tightens by a factor of three.
\[ \frac{dV}{V} = 3\frac{dr}{r} = 0.02 \;\Longrightarrow\; \frac{dr}{r} = 0.0067 \]
Why: Since the volume's relative error is three times the radius's, the radius's must be a third of the target, which is 2 over 3 percent. The demand on the input is always stricter than the specification on the output when a power is involved, and by exactly the exponent's factor. Six percent goes the wrong way entirely and would give an 18 percent volume error.
Comparison
Fill the blanks. Every row is a different reading of the same tangent line.
Comparison matrix
| Object | What it is | What it is used for |
|---|---|---|
| L(x) | the tangent line as a function | approximating f(x) near a |
| dy | the rise along the tangent over a step dx | estimating the change in f |
| delta y | the actual change in f | the truth that dy approximates |
| f'' | the bending | the direction and size of the error |
The last row is what turns an estimate into a usable one. Without it you have a number; with it you have a number, a direction of error and a bound.
Pattern
Given a value to approximate or an error to propagate.
Step one dominates the accuracy, since the error grows with the square of the step. Step four costs one differentiation and turns a bare estimate into one whose error direction is known.
Stewart, Calculus: Early Transcendentals 8e, §3.10 Linear Approximations and Differentials §3.10, pp. 251-258
Check
The linearization. Anchor nearby.
Check your understanding
Estimate sqrt(10) using a linear approximation at a = 9.
Answer: A
Why: L(x) = 3 + (1/6)(x - 9), so L(10) = 3 + 1/6.
Check
Differential against actual change.
Check your understanding
For y = x^2 at x = 2 with dx = 1, what are dy and delta y?
Answer: A
Why: The differential is the slope 4 times the step 1; the true change is 9 - 4 = 5.
Check
Propagation. The exponent magnifies.
Check your understanding
A sphere's radius carries a 1 percent error. What is the relative error in the volume?
Answer: A
Why: dV/V = 3 dr/r, because the volume depends on the cube of the radius.
Real world
A surveyor measures the angle of elevation to the top of a tower as 32 degrees, standing 100 metres away. The angle is accurate to plus or minus half a degree, and the horizontal distance is exact.
Discussion prompt
Compute the tower's height, propagate the angular error into a height error, and say whether measuring from further away would help.
Hint: Height is 100 times the tangent of the angle, and the error propagates through the derivative of tangent.
Answer:
\[ h = 100\tan\theta \;\Longrightarrow\; h(32^\circ) = 100\tan(0.5585) \approx 62.5 \text{ m} \]
The differential propagates the angular error, and the angle must be in radians for the derivative to be right:
\[ dh = 100\sec^{2}\theta\,d\theta, \qquad d\theta = 0.5^\circ = 0.00873 \text{ rad} \]
\[ dh = 100(1.390)(0.00873) \approx 1.21 \text{ m} \]
So the height is 62.5 plus or minus 1.2 metres, a relative error of about 1.9 percent — from an angular error of only half a degree. The secant squared factor is what magnifies it.
Measuring from further away makes it worse, not better. At 200 metres the angle would be about 17.4 degrees, where secant squared is smaller — but the distance factor of 200 more than compensates, and the height error rises to about 1.9 metres. The angle shrinks faster than the sensitivity falls.
The general lesson is the one from the sphere: an error in a measured quantity is magnified by the derivative of whatever is computed from it, and the magnification can be large. Here it is the trigonometry rather than an exponent, but the mechanism is identical.
Commit first
Answer, then rate your confidence honestly.
Predict first
A linear estimate is made of a function that bends upward. What can you say about it?
Correct: It is too low — the curve lies above its tangent.
\[ f'' > 0 \;\Longrightarrow\; f(x) > L(x) \text{ for } x \ne a \text{ nearby} \]
Why: A positive second derivative means the graph is bowl-shaped, and a tangent to a bowl touches at one point and runs beneath it on both sides. So every linear estimate from that point undershoots, whichever direction you step. Knowing this requires only the sign of the second derivative and not the true value, which is precisely what makes it useful in an application where the true value is unknown.
Explain it
They think a linear approximation is just a rough guess and cannot see why anyone would trust it.
Discussion prompt
In four sentences or fewer, show them why the estimate is better than they expect.
Hint: Have them halve the step and watch the error.
Answer:
Ask them to estimate the square of 2.1 from the tangent at 2: they get 4.4, and the true value is 4.41, so the error is 0.01. Now try 2.01: the estimate is 4.04 and the truth is 4.0401, an error of 0.0001.
The step shrank by a factor of ten and the error by a factor of a hundred, because the error is proportional to the SQUARE of the step. That is why these estimates are trustworthy close in and worthless far out — and why the whole method is about choosing an anchor near the target.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For anchors, take the nearest input where the value is easy — closeness beats convenience because the error is quadratic. For dy and delta y, remember one travels along the tangent and the other along the curve. For direction, compute the second derivative's sign and read it as bowl or dome. For propagation, remember the exponent is the magnification factor. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw the square root curve near x equal to 9 with its tangent, mark the point of contact, the true value at 10 and the estimate at 10, and label the gap between them. Write the linearization formula beside it with each of its three pieces labelled. Below, draw the parabola at x equal to 2 with a step of one, marking dx, dy and the true change, and write the numerical values of all three. Beneath that, tabulate the error for steps of 1, 0.5 and 0.1 and write one sentence saying how the error scales. In the lower half, draw two small pictures side by side of a bowl-shaped and a dome-shaped curve, each with a tangent, and write which way the estimate errs in each case. At the bottom, take a sphere of radius 10 with a measurement error of 0.1, compute the volume error both absolutely and as a percentage, and write the relation between the two relative errors with the exponent circled. In a margin, write the second-order error formula.
If your error table does not fall by a factor of a hundred between the steps of 1 and 0.1, recheck it — that hundredfold drop from a tenfold shorter step is the whole reason the method is worth having.
Recap
Five things, and together they turn a tangent line into a working tool.
| If you see | Then |
|---|---|
| A value to approximate | Anchor at the nearest easy point |
| f'' > 0 | The estimate undershoots |
| f'' < 0 | The estimate overshoots |
| A step ten times shorter | An error a hundred times smaller |
| A linear function | The approximation is exact |
| An absolute error alone | Convert it to relative before judging it |
| A quantity raised to a power | The relative error is multiplied by the exponent |
Section 4.3 changes the question. Instead of asking what a function does near one point, it asks where a function attains its largest and smallest values — and the answer turns out to depend on exactly the points where the derivative vanishes or fails to exist.
OpenStax Calculus Volume 1, §4.2 Linear Approximations and Differentials §4.2, pp. 308-316 — everything on these slides traces back here
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