Change and approximation

6 visual steps · 3 worked examples · 5 exercises with solutions

First, let’s see the idea.

What is the error when replacing a curve by its tangent?

Move forward one step at a time. Change the values and observe what happens.

How far does the tangent work?xf(x)-0,60,671,953,224,5-441220a = 1,5
Drawing description

The contact point (1,5; 2,25) is marked on the parabola x².

Step 1 of 6

Choose what you already know

Change a and follow the green point along the curve.

For f(x)=x², we know height a² at input a. The green point marks this starting information.

Starting point
(1,5; 2,25)

What if you change this?

0,53
-11

Before moving on

What largest change magnitude guarantees error no greater than 0,01?

Compare with the explanation

|h|≤0,1 because h²≤0,01.

You decide when to mark this lesson as studied.

How far does the tangent work?

Slide the drawing to see the details.

Explore the full explanationThe foundations, principles, worked examples and all the exercises, at your own pace.

What you will understand

  • Deduce the linear approximation from the derivative.
  • Distinguish exact change, differential and error.
  • Estimate sensitivity and recognise local limitations.

Before you start

You do not need to know everything by heart. The explanation revisits the necessary ideas and shows how they connect. If a foundation is still unclear, you can return to these lessons:

Review the meaning of rate

If a function f is differentiable at a, the quotient [f(a+h)−f(a)]/h approaches f′(a) as h→0. For small h, we can therefore write f(a+h)−f(a) approximately equal to f′(a)h. The product has the correct unit: the derivative's unit is output per input and h has the input unit. The final estimate returns to the output unit.

The derivative does not say that the rate is exactly constant in a neighbourhood. It says that a constant rate provides the best first-order description sufficiently near the point. If a length changes by 0,02 cm, the area of a square changes in a way depending on the initial length. The same increment can have very different effects in different systems or different regions of the same function.

Construct the linearisation

The tangent line at a is L(x)=f(a)+f′(a)(x−a). It is called the linearisation of f at that point. To estimate f(a+h), choose a base a where f(a) and f′(a) are easy to calculate. Closeness must be assessed on the function's scale, not just by the number of decimal places. For √x, approximating 36,4 with a line at 36 is reasonable; using the same line for an input near zero is not.

There is a precise way to express the approximation: f(a+h)=f(a)+f′(a)h+r(h), where r(h)/h→0. The remainder r(h) becomes small compared with h. This does not imply that the remainder is zero or by itself provide a numerical bound for a chosen h. Quantitative control of this error requires additional information about the function on the interval considered.

Exact change, differential and error

The exact change is Δy=f(a+h)−f(a). The differential dy=f′(a)h is the change predicted by the line. The signed estimation error is L(a+h)−f(a+h); the absolute error is its magnitude. A relative error divides the absolute error by the magnitude of the exact value, when that value is nonzero. Multiplying by 100 gives the percentage error. The denominator used should be stated, especially when comparing changes rather than total values.

For f(x)=x², we have Δy=2ah+h² and dy=2ah. The difference is h², directly revealing the neglected term. At a=10 and h=0,1, the actual change is 2,01, whereas the differential is 2. The absolute error in the change is 0,01; relative to the change it is about 0,498%, but relative to the final value 102,01 it is about 0,0098%. The two percentages answer different questions.

Curvature and error control

When f has a continuous second derivative on the interval between a and a+h, and |f″(x)|≤M there, the linearisation error is at most M h²/2. The idea is that the rate may move away from f′(a), but curvature controls that change; accumulating the deviation produces a quadratic term. Here we use this result as an extension, not as an automatic consequence of the existence of just one derivative.

The sign of curvature also guides the prediction. If f″>0 throughout the small interval, the curve lies above the tangent: L underestimates f. If f″<0, the tangent overestimates. For √x, the second derivative is −1/(4x3/2) for x>0, so the tangent approximation overestimates. Near zero, the magnitude of curvature grows greatly, warning that the same horizontal distance can generate a larger error.

Sensitivity and measurement uncertainty

If a square's side ℓ has small uncertainty Δℓ, the area A=ℓ² has approximate change 2ℓΔℓ. Dividing by A gives ΔA/A approximately 2Δℓ/ℓ. For V=ℓ³, the relative uncertainty is approximately 3Δℓ/ℓ. These relationships assume small relative changes and a suitable geometric model; they do not replace exact intervals when a guarantee is required.

A symmetric input tolerance may produce an asymmetric output tolerance. If ℓ=10±0,1 cm, the area lies exactly between 9,9²=98,01 and 10,1²=102,01 cm². The differential estimate would be 100±2 cm². It explains the scale of the effect well, but the exact endpoints show the small quadratic term. If the function stops being monotonic within the uncertainty interval, interior points must also be studied, rather than merely substituting the endpoints.

Critical application and precision

Linearisation is useful for mental calculation, instrument calibration and analysing small disturbances. It also explains why we should not assign physical meaning to every decimal place a calculator produces. An input measured with uncertainty carries that uncertainty into the output; more digits in the operation do not remove the initial limitation.

Before using an estimate, define the acceptable tolerance and check the domain along the interval. A model with a denominator near zero may greatly amplify a small change. A linearisation may even predict negative values for an always-positive quantity when extrapolated too far. The solution is to return to the original function, change the reference point or use a richer approximation. The method's usefulness depends on recognising this limit, not applying it at any distance.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Example 1

Estimate a root

Estimate √36,4 by linearisation at 36 and verify the error's direction.

  1. For f(x)=√x, f(36)=6 and f′(36)=1/12. The increment is h=0,4.
  2. The prediction is 6+0,4/12=6,033333… .
  3. Since f″<0 on the interval, the tangent overestimates. The directly calculated value is approximately 6,033241… .
Result

√36,4≈6,03333, with absolute error close to 0,0000921.

Check the result

The positive difference between estimate and direct value agrees with downward concavity.

Example 2

A cubic tolerance

A cube has side 5 cm with tolerance ±0,02 cm. Estimate the maximum change in volume.

  1. The nominal volume is 5³=125 cm³. The derivative with respect to the side is 3ℓ²=75 cm².
  2. The differential has magnitude 75·0,02=1,5 cm³. The relative estimate is 1,5/125=1,2%.
  3. Calculate the exact endpoints: 4,98³=123,505992 and 5,02³=126,506008 cm³.
Result

The estimate is about ±1,5 cm³; the exact interval is [123,505992;126,506008] cm³.

Check the result

The exact tolerances are slightly asymmetric. A rigorous guarantee can use the exact interval.

Example 3

Bound an approximation's error

Use the tangent to f(x)=x² at a=3 to estimate f(3,2) and give an error bound.

  1. f(3)=9 and f′(3)=6, so L(3,2)=9+6·0,2=10,2.
  2. The second derivative is always 2; we can take M=2. The bound is 2·0,2²/2=0,04.
  3. The exact value is 10,24. The convex curve lies above the tangent.
Result

Estimate 10,2; absolute error 0,04, equal to the bound in this case.

Check the result

The identity (3+0,2)²=9+1,2+0,04 makes the neglected term visible.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Exercise 1

An estimate has absolute error 0,2 on an exact value of 40. What is the percentage error?
A hint
Divide by the stated reference value.
See the worked solution
The relative error is 0,2/40=0,005; multiplying by 100 gives 0,5%. Absolute and percentage errors are distinct numbers with different interpretations.

Exercise 2

Linearise 1/x at a=4.
A hint
First calculate the point and slope.
See the worked solution
f(4)=1/4 and f′(4)=−1/16. Therefore L(x)=1/4−(x−4)/16, a local estimate valid near 4; it must not cross the singularity x=0.

Exercise 3

Estimate 1/4,08 using the preceding linearisation.
A hint
The increment is 0,08.
See the worked solution
L(4,08)=0,25−0,08/16=0,245. The exact value is about 0,245098. The estimate is below it because 1/x has positive second derivative for x>0.

Exercise 4

A circular area uses r=20 cm, with Δr=0,1 cm. Estimate ΔA.
A hint
Differentiate A=πr² with respect to r.
See the worked solution
dA=2πrΔr=2π·20·0,1=4π≈12,566 cm². The exact change for an increase is 4,01π, including the term π(0,1)².

Exercise 5

If |f″|≤8 and |h|=0,05, what is the linear error bound?
A hint
Use M h²/2 and keep the square.
See the worked solution
The absolute error is at most 8·0,05²/2=0,01, in the output unit, provided the second derivative bound holds throughout the interval between a and a+h.

Mistakes worth avoiding

The tangent coincides with the function near the point.
In general it only approximates it; there is a remainder that needs assessment.
An increment of 0,1 is always small.
Scale, proximity to singularities and curvature determine the error.
Relative and absolute uncertainty have the same units.
Absolute uncertainty has the quantity's unit; relative uncertainty is a dimensionless ratio.

The essentials to take with you

  • L(x)=f(a)+f′(a)(x−a) is a local tool.
  • Distinguish exact Δy from approximate dy.
  • Curvature can indicate the sign and bound of the error.
  • An input tolerance can become asymmetric in the output.

If you still cannot explain one of these ideas, return to the corresponding section and work through an example again. Recognising the difficulty is part of studying.

References for further study

Text and examples prepared for this learning path. These sources help check the principles and continue studying:

The grade organization is a suggested study sequence. Grade 13 provides general extension and preparation for higher education. Pedagogical review by teachers remains a stage of the project.