Linear approximation

Close to the point of tangency, a differentiable function is almost its tangent line, so the tangent line estimates a value you cannot compute by hand. The skill is in choosing the point: one near the value you want, where the function and its derivative are easy to evaluate. The sign of the second derivative between that point and the value you want then says whether the estimate is too high or too low.

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In this topic

Taught: Before the final
Questions: 14
Total marks: 33
Suggested time: 61 min
Difficulty: Introductory to Capstone
Show the 4 hardest

Key results

Linear approximation: $f(x) \approx L(x) = f(a) + f'(a)(x - a)$
When $f$ is differentiable at $a$ and $x$ is close to $a$ .
Estimating a change: $f(a + \Delta x) - f(a) \approx f'(a)\,\Delta x$
When $f$ is differentiable at $a$ and $\Delta x$ is small; with $\Delta x = 1$ this is the marginal estimate.
Too high or too low: $\begin{aligned} f'' > 0 &\implies L(x) < f(x) \\ f'' < 0 &\implies L(x) > f(x) \end{aligned}$
For $x \neq a$ , when $f''$ keeps that sign on the whole interval between $a$ and $x$ , not only at $a$ : concave up, the line underestimates; concave down, it overestimates.

Read the theoryUBC Math 100 §3.3.3 Tangent lines and linear approximationsUBC Math 100 §9.2 Linear approximationCLP-1 §3.4 Taylor polynomials

Worked example

Standard2 marks · 4 min

At a distance of $d$ metres from a small lamp, the illuminance is $I(d) = \dfrac{900}{d^{2}}$ lux. A reader sitting 3 m from the lamp moves back to 3.1 m. Use a linear approximation to estimate the change in illuminance.

Give the change in lux, with its sign.

Strategy

Near d = 3 the graph is close to its tangent line, so a small step in distance changes the illuminance by about the slope at 3 times the step. Moving away from the lamp makes it darker, so expect a negative answer.

Working

$I'(d) = -1800\,d^{-3} = -\frac{1800}{d^{3}}$
Power rule on 900d⁻².
$I'(3) = -\frac{1800}{27} = -\frac{200}{3}$
The rate of change at 3 m, in lux per metre.
$I(3.1) - I(3) \approx -\frac{200}{3}\cdot 0.1 = -\frac{20}{3} \approx -6.67$
Slope times the step of 0.1 m. The exact change is about −6.35 lux, so the estimate is close.

Answer

I(3.1) - I(3) \approx -\frac{20}{3} \approx -6.67 \text{ lux}

Check

I(3) = 100 lux and I(3.1) = 900/9.61 ≈ 93.65 lux, a real drop of about 6.35 lux.

Where this goes wrong

Giving the change as positive. The light dims with distance, so the derivative is negative and the change is a drop of about 6.67 lux.

On the paper

Write $L(x) = f(a) + f'(a)(x - a)$ with your $a$ , and say from the sign of $f''$ whether the estimate is too high or too low.

Leans on

The power rule for negative powers, from Differentiation rules

Your turn

Standard2 marks · 4 min

Kleiber’s law models a mammal’s resting metabolic rate as $R(M) = 70M^{3/4}$ kilocalories per day, where $M$ is its mass in kilograms. A sheep of 81 kg loses 2 kg over the winter. Use a linear approximation at $M = 81$ to estimate its resting metabolic rate at 79 kg.

Give the answer in kilocalories per day.

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Practice questions

5 of the 14, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory1 mark · 2 min
Use a linear approximation of $f(x) = \sqrt{x}$ at $x = 25$ to estimate $\sqrt{26}$ .
Strategy
Near a point where the function is easy to evaluate, replace it by its tangent line. 25 is a perfect square next door to 26.
Working
$f(25) = 5, \qquad f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{10}$
Value and slope at the base point.
$\sqrt{26} \approx L(26) = 5 + \frac{1}{10}(26 - 25) = 5.1$
The true value is 5.0990…, so the estimate is slightly high, as expected for a concave function.
Answer
$\sqrt{26} \approx 5.1$
The answer is 5.1.
Check
5.1² = 26.01, just 0.01 above 26, so 5.1 is a slight overestimate of √26.
Where this goes wrong
Using the derivative at 26 rather than at 25, which defeats the purpose: the whole point is that everything is evaluated at the easy point 25.
Leans on
The derivative of √x, from Differentiation rules
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Use a linear approximation to estimate $(3.002)^{4}$ .
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Introductory1 mark · 2 min
The linear approximation of a function $f$ about $x = 2$ is $L(x) = 4x - 7$ . What are $f(2)$ and $f'(2)$ ?
Strategy
The tangent line shares the function’s value and slope at the base point, and nothing else. Read both off the line.
Working
$f'(2) = \text{slope of } L = 4$
Coefficient of x.
$f(2) = L(2) = 4(2) - 7 = 1$
The line agrees with f at the base point.
Answer
$f(2) = 1,\quad f'(2) = 4$
The answer is f of 2 is 1 and f prime of 2 is 4.
Check
The line through (2, 1) with slope 4 is y = 1 + 4(x − 2) = 4x − 7, the L given.
Where this goes wrong
Reading f(2) as the intercept −7. The intercept is L(0), the value at x = 0, which tells you nothing about f; only the value at the base point 2 is shared.
Leans on
Slope-intercept form of a line, from Functions, graphs and inverses
Try a similar question
Use a linear approximation of $f(x) = \sqrt{x}$ at $x = 25$ to estimate $\sqrt{26}$ .
Answer this question
Standard2 marks · 3 min
Use a linear approximation to estimate $(3.002)^{4}$ .
Strategy
Choose the nearest point where the power is easy, linearise there, and step 0.002 along the tangent line.
Working
$f(x) = x^{4}, \quad f(3) = 81, \quad f'(x) = 4x^{3}, \quad f'(3) = 108$
Value and slope at the base point.
$(3.002)^{4} \approx 81 + 108(0.002) = 81 + 0.216 = 81.216$
The exact value is 81.2162…; the linear estimate is accurate to three decimal places.
Answer
$(3.002)^{4} \approx 81.216$
The answer is 81.216.
Check
x⁴ is concave up, so the tangent estimate should be a little low, and indeed 3.002⁴ ≈ 81.21622.
Where this goes wrong
Applying the 0.002 to the base value, 81 + 0.002, forgetting that the change in output is the slope times the change in input.
Leans on
The power rule, from Differentiation rules
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Use a linear approximation of $f(x) = \sqrt{x}$ at $x = 25$ to estimate $\sqrt{26}$ .
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Challenging4 marks · 7 min
The equation $y^{3} + xy = 10$ defines $y$ implicitly as a function of $x$ near the point $(1, 2)$ . Use the tangent-line approximation at that point to estimate $y$ when $x = 1.1$ .
Strategy
A linear approximation only needs the value and the slope at the base point. The value is given; the slope comes from implicit differentiation, since there is no formula for y.
Working
$3y^{2}\,y' + y + x\,y' = 0 \quad\Longrightarrow\quad y' = -\frac{y}{3y^{2} + x}$
Product rule on xy; solve for y′.
$y'\big|_{(1, 2)} = -\frac{2}{12 + 1} = -\frac{2}{13}$
Check first that (1, 2) is on the curve: 8 + 2 = 10.
$y(1.1) \approx 2 - \frac{2}{13}(0.1) = 2 - \frac{1}{65} = \frac{129}{65} \approx 1.98462$
Tangent-line estimate. The true value is 1.98462 to five places as well, so give the estimate exactly.
Answer
$y(1.1) \approx 2 - \frac{1}{65} = \frac{129}{65}$
The answer is 2 minus 1 over 65, which is 129 over 65, about 1.9846.
Check
Putting y = 129/65 ≈ 1.9846 back in, y³ + 1.1y ≈ 9.9999, within 0.0002 of 10.
Where this goes wrong
Trying to solve the cubic for y at x = 1.1. That is exactly the calculation linear approximation is meant to avoid, and it is not what was asked.
Leans on
Implicit differentiation, from Implicit differentiation and inverse trigonometric functions
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The point $(1, 2)$ lies on the curve below. Find $\dfrac{dy}{dx}$ at that point. $x^{2} + xy + y^{2} = 7$
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Capstone5 marks · 10 min
Let $f(x) = \log\!\left(1 + x^{2}\right)$ . Use the linear approximation of $f$ at $x = 1$ to estimate $f(1.1)$ and $f(0.9)$ . Then decide, for each estimate, whether it is an overestimate or an underestimate, and justify your answer.
Show your working and your reasoning, then compare against the model solution.
Strategy
Build the tangent line at 1, where f and f′ are exact, and step 0.1 to each side. Then check the concavity on each step: 1 is an inflection point of this curve, so the two estimates err in opposite directions.
Working
$f(1) = \log 2, \qquad f'(x) = \frac{2x}{1 + x^{2}}, \qquad f'(1) = 1$
Chain rule: the derivative of log u is u′/u.
$L(x) = \log 2 + (x - 1): \qquad f(1.1) \approx \log 2 + \tfrac{1}{10}, \qquad f(0.9) \approx \log 2 - \tfrac{1}{10}$
Steps of +0.1 and −0.1 from the base point.
$f''(x) = \frac{2(1 + x^{2}) - 2x\cdot 2x}{(1 + x^{2})^{2}} = \frac{2(1 - x^{2})}{(1 + x^{2})^{2}}$
Quotient rule on 2x/(1 + x²); the denominator is positive, so f″ has the sign of 1 − x².
$1 < x \leq 1.1:\ f'' < 0 \quad\Longrightarrow\quad \log(2.21) < \log 2 + \tfrac{1}{10}$
Concave down, so the graph lies below its tangent line and this estimate is an overestimate.
$0.9 \leq x < 1:\ f'' > 0 \quad\Longrightarrow\quad \log(1.81) > \log 2 - \tfrac{1}{10}$
Concave up, so the graph lies above its tangent line and this estimate is an underestimate.
Answer
$f(1.1) \approx \log 2 + \frac{1}{10}\ (\text{over}), \qquad f(0.9) \approx \log 2 - \frac{1}{10}\ (\text{under})$
The answer is f(1.1) is about log 2 plus one tenth, an overestimate; f(0.9) is about log 2 minus one tenth, an underestimate..
What earns the marks
Finds f(1) = log 2 and f′(1) = 1, so L(x) = log 2 + (x − 1).
Estimates f(1.1) ≈ log 2 + 1/10 and f(0.9) ≈ log 2 − 1/10.
Finds f″(x) = 2(1 − x²)/(1 + x²)² and notes that f″(1) = 0, so the second derivative at the base point decides nothing.
Uses the sign of f″ on each side of 1: negative for x > 1, so log 2 + 1/10 is an overestimate; positive for −1 < x < 1, so log 2 − 1/10 is an underestimate.
Check
log 2.21 ≈ 0.79299 sits just below log 2 + 0.1 ≈ 0.79315, and log 1.81 ≈ 0.59333 just above log 2 − 0.1 ≈ 0.59315.
Where this goes wrong
Deciding both from f″(1). It is 0 because 1 is an inflection point, so on its own it says nothing: the tangent line crosses the graph there, and the sign of f″ on each side of 1 decides.
Leans on
Concavity and inflection points, from Curve sketching
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A drug’s effect at concentration $c$ (in mg/L), as a fraction of its greatest possible effect, is modelled by $E(c) = \dfrac{c^{2}}{1 + c^{2}}$ . The tangent line at $c = 1$ gives the estimate $E(1.2) \approx 0.6$ . Which statement is correct?
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Browse all 14 questions

Needs
The definition of the derivative
Secants and tangents, average and instantaneous velocity, the difference quotient and differentiability.
27 questions
Needs
The chain rule and logarithms
Compositions, nested rules, the natural logarithm and logarithmic differentiation.
30 questions
Leads to
Taylor polynomials
Quadratic and higher approximations, the Maclaurin polynomials of the standard functions and, for flavour A, bounds on the error.
16 questions
Leads to
Newton’s method
Following the tangent line towards a root: an iteration or two by hand, and the starting points where it fails.
13 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Leans on

Your turn

Practice questions

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

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Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

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Strategy

Working

Answer

What earns the marks

Check

Where this goes wrong

Leans on

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Related topics

The definition of the derivative

The chain rule and logarithms

Taylor polynomials

Newton’s method