Exercise 0.1: Compute a differential and estimate with the tangent line: ten direct calculations

For rows a) to c), compute the differential $dy=f'(a)\,dx$ at the given point $a$ and for the given increment $dx$ . For rows d) to j), estimate the requested number with the tangent line at the given base point $a$ , that is, with $L(a+dx)=f(a)+f'(a)\,dx$ . No comparison with the exact value is required.

One method for all ten rows: name the function $f$ , the base point $a$ and the increment $dx$ WITH ITS SIGN, compute $f'(a)$ , then multiply by $dx$ . For an estimate, add $f(a)$ , which the very choice of base point makes known without a calculator. The calculator, in radian mode, is only for the final value of the last two rows.

a) $f(x)=2x^{3}-5x+1$ : $dy$ at $x=2$ with $dx=0.1$ .
b) $f(x)=\frac{4}{x}$ : $dy$ at $x=5$ with $dx=-0.25$ .
c) $f(x)=x^{\frac{2}{3}}$ : $dy$ at $x=27$ with $dx=-0.9$ .
d) $\sqrt{63.2}$ , with the tangent line to $y=\sqrt{x}$ at $a=64$ .
e) $\sqrt[3]{27.54}$ , with the tangent line to $y=\sqrt[3]{x}$ at $a=27$ .
f) $(2.98)^{4}$ , with the tangent line to $y=x^{4}$ at $a=3$ .
g) $\frac{1}{9.9}$ , with the tangent line to $y=\frac{1}{x}$ at $a=10$ .
h) $\ln(0.96)$ , with the tangent line to $y=\ln x$ at $a=1$ .
i) $e^{0.97}$ , with the tangent line to $y=e^{x}$ at $a=1$ .
j) $\sin(0.5)$ , with the tangent line to $y=\sin x$ at $a=\frac{\pi}{6}$ .

Type your answers, the page tells you right or wrong 0/10

a)

dy

b)

dy

c)

dy

d)

\sqrt{63.2}\approx

e)

\sqrt[3]{27.54}\approx

f)

(2.98)^{4}\approx

g)

\frac{1}{9.9}\approx

h)

\ln(0.96)\approx

i)

e^{0.97}\approx

j)

\sin(0.5)\approx

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Answers

a) $dy=(6x^{2}-5)\,dx$ ; $dy=19\times 0.1=1.9$
b) $dy=-\frac{4}{x^{2}}\,dx$ ; $dy=0.04$
c) $dy=\frac{2}{3\sqrt[3]{x}}\,dx$ ; $dy=-0.2$
d) $L=8+\frac{1}{16}(-0.8)$ ; $\sqrt{63.2}\approx 7.95$
e) $L=3+\frac{1}{27}(0.54)$ ; $\sqrt[3]{27.54}\approx 3.02$
f) $L=81+108(-0.02)$ ; $(2.98)^{4}\approx 78.84$
g) $L=\frac{1}{10}-\frac{1}{100}(-0.1)$ ; $\frac{1}{9.9}\approx 0.101$
h) $L=0+1\times(-0.04)$ ; $\ln(0.96)\approx-0.04$
i) $L=e+e(-0.03)=0.97\,e$ ; $e^{0.97}\approx 2.637$
j) $L=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(0.5-\frac{\pi}{6}\right)$ ; $\sin(0.5)\approx 0.4796$

The move that decides everything is to write $dx$ WITH ITS SIGN before computing anything: $dx=x_{\text{target}}-a$ . For $\sqrt{63.2}$ at $a=64$ , $dx=-0.8$ , and that sign is what puts the estimate BELOW $8$ . The rest is a derivative evaluated at a convenient point, then a multiplication.

a)

$f'(x)=6x^{2}-5$ , so $f'(2)=19$ and $dy=19\times 0.1=1.9$ . Pitfall: computing the actual change $\Delta y=f(2.1)-f(2)\approx 2.022$ , which is not what is asked.

b)

$f'(x)=-\frac{4}{x^{2}}$ , so $f'(5)=-\frac{4}{25}$ and $dy=-\frac{4}{25}\times(-0.25)=\frac{1}{25}=0.04$ . Two minus signs: the function is decreasing and we move backward, so $y$ increases. Pitfall: losing one of the two signs and answering $-0.04$ .

c)

$f'(x)=\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3\sqrt[3]{x}}$ , so $f'(27)=\frac{2}{3\times 3}=\frac{2}{9}$ and $dy=\frac{2}{9}\times(-0.9)=-0.2$ . Pitfall: the exponent $-\frac{1}{3}$ puts a cube root IN THE DENOMINATOR, not a product with $\sqrt[3]{27}$ .

d)

$f(x)=\sqrt{x}$ , $f'(x)=\frac{1}{2\sqrt{x}}$ , $a=64$ , $dx=63.2-64=-0.8$ . So $L=\sqrt{64}+\frac{1}{16}\times(-0.8)=8-0.05=7.95$ . Pitfall: taking $dx=+0.8$ and answering $8.05$ , which is above $8$ even though $63.2<64$ .

e)

$f(x)=\sqrt[3]{x}=x^{\frac{1}{3}}$ , $f'(x)=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{x^{2}}}$ , so $f'(27)=\frac{1}{3\times 9}=\frac{1}{27}$ . With $dx=0.54$ : $L=3+\frac{0.54}{27}=3+0.02=3.02$ . Pitfall: $\sqrt[3]{27^{2}}=9$ , not $3$ .

f)

$f(x)=x^{4}$ , $f'(x)=4x^{3}$ , $f'(3)=108$ , $dx=-0.02$ . So $L=81+108\times(-0.02)=81-2.16=78.84$ . Pitfall: basing the approximation at $a=2.98$ itself, which brings back the very calculation you wanted to avoid.

g)

$f(x)=\frac{1}{x}$ , $f'(x)=-\frac{1}{x^{2}}$ , $f'(10)=-\frac{1}{100}$ , $dx=-0.1$ . So $L=\frac{1}{10}+\left(-\frac{1}{100}\right)(-0.1)=0.1+0.001=0.101$ . The denominator decreases, so the fraction increases: the sign of the correction can be checked before computing anything.

h)

$f(x)=\ln x$ , $f'(x)=\frac{1}{x}$ , $f(1)=0$ and $f'(1)=1$ , $dx=-0.04$ . So $L=0+1\times(-0.04)=-0.04$ . The logarithm of a number less than $1$ is negative: the estimate must be negative too.

i)

$f(x)=e^{x}$ , $f'(x)=e^{x}$ , so $f(1)=f'(1)=e$ and $dx=-0.03$ . So $L=e+e\times(-0.03)=0.97\,e\approx 2.637$ . Pitfall: basing the approximation at $0$ with $1+x$ , which would give $1.97$ , much too far from the base point.

j)

$f(x)=\sin x$ , $f'(x)=\cos x$ , $f\left(\frac{\pi}{6}\right)=\frac{1}{2}$ , $f'\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$ , and $dx=0.5-\frac{\pi}{6}\approx-0.02360$ . So $L=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(0.5-\frac{\pi}{6}\right)\approx 0.5-0.02044=0.4796$ . Pitfall: taking $dx=0.5$ , as if you started from $0$ , or leaving the calculator in degrees.

Check: the calculator gives $\sqrt{63.2}\approx 7.9498$ , $\sqrt[3]{27.54}\approx 3.0199$ , $(2.98)^{4}\approx 78.862$ , $\frac{1}{9.9}\approx 0.10101$ , $\ln(0.96)\approx-0.0408$ , $e^{0.97}\approx 2.638$ and $\sin(0.5)\approx 0.4794$ : every estimate lands within a few thousandths, above the true value when the curve is concave down (roots, logarithm, sine) and below it when the curve is concave up ( $x^{4}$ , $\frac{1}{x}$ , $e^{x}$ ). An estimate on the wrong side almost always signals a $dx$ with the wrong sign.

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Exercise 1: The differential and the actual change

The figure shows $y=x^{2}$ and its tangent line at the point $(3,9)$ . Over a step $dx$ , the curve rises by $\Delta y$ and the tangent line by $dy$ : the whole chapter lies in the gap between these two rises.

a) Define the differential $dy$ for $y=f(x)$ , then compute $dy$ for $y=x^{2}$ at $x=3$ with $dx=0.1$ .
b) Compute the actual change $\Delta y$ under the same conditions. What is the gap with $dy$ , and exactly where does it come from?
c) Redo the calculation with $dx=0.01$ then $dx=0.001$ . How does the gap $\Delta y-dy$ behave compared with $dx$ ?
d) Using the figure, explain what $dy$ and $\Delta y$ represent geometrically, and why the approximation improves as $dx$ decreases.

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Answers

a) $dy=f'(x)\,dx=0.6$
b) $\Delta y=0.61$ ; gap $(dx)^{2}=0.01$
c) Gaps $10^{-4}$ and $10^{-6}$ : proportional to $(dx)^{2}$
d) $dy$ along the tangent line, $\Delta y$ along the curve

a)

Definition: $dy=f'(x)\,dx$ , where $dx$ is an arbitrary (not necessarily infinitesimal) increment of the variable. Here $f(x)=x^{2}$ , so $f'(x)=2x$ and $f'(3)=6$ . With $dx=0.1$ : $dy=6\times 0.1=0.6$ . Note that $dy$ depends on TWO quantities: the point $x$ where you stand and the increment $dx$ you allow yourself.

b)

$\Delta y=f(3.1)-f(3)=9.61-9=0.61$ . The gap is $\Delta y-dy=0.61-0.6=0.01$ . It comes from an exact algebraic calculation: $\Delta y=(3+dx)^{2}-3^{2}=6\,dx+(dx)^{2}$ , while $dy=6\,dx$ . So the gap is exactly $(dx)^{2}=(0.1)^{2}=0.01$ ✓. In other words, the differential captures the LINEAR part of the change and drops the quadratic term. That is the precise meaning of “the derivative linearizes the function”.

c)

With $dx=0.01$ : $dy=0.06$ , $\Delta y=(3.01)^{2}-9=0.0601$ , gap $=0.0001=(dx)^{2}$ . With $dx=0.001$ : $dy=0.006$ , $\Delta y=0.006001$ , gap $=10^{-6}=(dx)^{2}$ . The gap is the SQUARE of $dx$ : when $dx$ is divided by 10, the gap is divided by 100. In proportion to $dx$ itself, the relative error $\frac{\Delta y-dy}{dx}=dx$ goes to zero. That is the property behind every practical use of the differential: the error made is not only small, it is negligible COMPARED WITH $dx$ .

d)

On the figure, the parabola and its tangent line at $(3,9)$ separate very slowly. Starting from $x=3$ and moving forward by $dx$ , the rise along the TANGENT LINE is $dy$ , and the rise along the CURVE is $\Delta y$ . Since the tangent is a straight line, its rise is proportional to $dx$ , which makes $dy$ easy to compute; the curve bends slightly upward above it, hence $\Delta y>dy$ here. The approximation improves as $dx$ decreases because, closer and closer to the point of tangency, the curve and its tangent line become visually indistinguishable: that is exactly what differentiability says, namely that a differentiable function is LOCALLY a straight line. On the figure, the gap is only visible at the ends of the plotted interval.

Exercise 2: Linear approximation: computing without a calculator

Replacing a function by its tangent line lets you estimate in your head values that would otherwise need a calculator. All the work lies in choosing the right base point.

a) Write the linearization $L(x)$ of $f(x)=\sqrt{x}$ at $a=4$ , then estimate $\sqrt{4.1}$ . Compare with the exact value and give the error.
b) Estimate $\sin(0.1)$ by linearization at $a=0$ . Which famous approximation from physics do you recover?
c) Establish the general approximation $(1+x)^{n}\approx 1+nx$ for small $x$ , then estimate $\sqrt[3]{1.03}$ and $\frac{1}{1.02}$ .
d) To estimate $\sqrt{4.1}$ , why choose $a=4$ rather than $a=5$ or $a=4.1$ ? State the two criteria for the choice.

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Answers

a) $L(x)=2+\frac{x-4}{4}$ ; $2.025$ , error $1.54\times 10^{-4}$
b) $\sin(0.1)\approx 0.1$ : small angles
c) $(1+x)^{n}\approx 1+nx$ ; $1.01$ and $0.98$
d) A close and convenient point: a perfect square

a)

$f(x)=\sqrt{x}$ , $f'(x)=\frac{1}{2\sqrt{x}}$ , so $f(4)=2$ and $f'(4)=\frac{1}{4}$ . The linearization is $L(x)=2+\frac{1}{4}(x-4)$ . Estimate: $\sqrt{4.1}\approx L(4.1)=2+\frac{0.1}{4}=2.025$ . Exact value: $\sqrt{4.1}\approx 2.0248457$ . The absolute error is about $0.000154$ , a relative error of $0.0076$ percent. For mental arithmetic, that is remarkable: three correct decimals with no tool at all.

b)

$f(x)=\sin x$ , $f'(x)=\cos x$ , with $f(0)=0$ and $f'(0)=1$ : $L(x)=0+1\cdot(x-0)=x$ . So $\sin(0.1)\approx 0.1$ , compared with the exact value $0.0998334$ , an error of $1.67\times 10^{-4}$ . This is the SMALL-ANGLE approximation, $\sin\theta\approx\theta$ (in radians), which is used to linearize the equation of the simple pendulum in physics and without which the pendulum would not have a constant period. It is the most rewarding example of the chapter, because it shows how a mathematical approximation can change an entire physical model.

c)

With $f(x)=(1+x)^{n}$ at $a=0$ : $f(0)=1$ and $f'(x)=n(1+x)^{n-1}$ gives $f'(0)=n$ . So $L(x)=1+nx$ , that is, $(1+x)^{n}\approx 1+nx$ for small $x$ . Applications. For $\sqrt[3]{1.03}=(1+0.03)^{1/3}\approx 1+\frac{0.03}{3}=1.01$ (exact value $1.009901$ , error $10^{-4}$ ). For $\frac{1}{1.02}=(1+0.02)^{-1}\approx 1-0.02=0.98$ (exact value $0.980392$ , error $4\times 10^{-4}$ ). This single formula covers roots, reciprocals and powers: it is the most useful one in the chapter and deserves to be known by heart.

d)

Two criteria. First, $a$ must be CLOSE to the target value, because the error grows with the distance $|x-a|$ : $a=5$ would be ten times farther from $4.1$ than $a=4$ , and the estimate would suffer ( $L(4.1)$ built at $a=5$ would give $2.236+\frac{-0.9}{2(2.236)}\approx 2.035$ , clearly worse). Second, $a$ must be CONVENIENT, meaning that $f(a)$ and $f'(a)$ can be computed exactly and in your head: that is what rules out $a=4.1$ , for which $f(a)=\sqrt{4.1}$ is precisely the unknown quantity you are looking for. Choosing $a=4.1$ would make the reasoning circular. The right base point is therefore the nearest perfect square, perfect cube or special angle: that is always how the question is set on an exam.

Exercise 3: Propagation of measurement errors

No instrument measures exactly. The differential answers the question every lab report asks: what error does my measurement uncertainty produce on the computed quantity?

a) The radius of a sphere is measured as $10$ cm with an uncertainty of $\pm 0.05$ cm. Estimate the absolute error on the volume using the differential.
b) Compute the relative error on the volume, as a percentage. Compare it with the relative error on the radius.
c) The edge of a cube is measured as $5.00$ cm with an uncertainty of $\pm 0.02$ cm. Give the absolute and relative errors on the volume, then on the total surface area.
d) Why use $dV$ rather than $\Delta V$ for this kind of calculation, even though $\Delta V$ would be exact? Check on the sphere of part a) that the gap is negligible.

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Answers

a) $dV=20\pi\approx 62.83$ cm³
b) $1.5$ % against $0.5$ %: factor $3$
c) $1.5$ cm³ ( $1.2$ %); $1.2$ cm² ( $0.8$ %)
d) $\Delta V\approx 63.14$ cm³: negligible gap

a)

$V=\frac{4}{3}\pi r^{3}$ , so $\frac{dV}{dr}=4\pi r^{2}$ and $dV=4\pi r^{2}\,dr$ . With $r=10$ and $dr=0.05$ : $dV=4\pi(100)(0.05)=20\pi\approx 62.83$ cm³. The volume itself is $V=\frac{4}{3}\pi(1000)\approx 4188.79$ cm³. So we write $V=4188.8\pm 62.8$ cm³. Notice that $4\pi r^{2}$ is the SURFACE AREA of the sphere: the error on the volume is the area times the thickness $dr$ , which is exactly the thin shell added, and the geometric interpretation fits. The figure shows the mechanism on a generic curve: an error $dx$ on the measurement becomes an error $dy\approx f'(a)\,dx$ on the result, read along the tangent line.

b)

Relative error on the volume: $\frac{dV}{V}=\frac{4\pi r^{2}dr}{\frac{4}{3}\pi r^{3}}=\frac{3\,dr}{r}$ . Numerically: $\frac{3\times 0.05}{10}=0.015$ , that is $1.5$ percent. The relative error on the radius is $\frac{dr}{r}=\frac{0.05}{10}=0.005$ , that is $0.5$ percent. The ratio is exactly 3: the relative error is TRIPLED going from the radius to the volume. A direct check: $\frac{62.83}{4188.79}=0.0150$ ✓.

c)

Volume of the cube: $V=a^{3}$ , so $dV=3a^{2}da=3(25)(0.02)=1.5$ cm³, on a volume $V=125$ cm³, a relative error of $\frac{1.5}{125}=1.2$ percent (or directly $3\times\frac{0.02}{5}=3\times 0.4$ percent). Total surface area: $A=6a^{2}$ , so $dA=12a\,da=12(5)(0.02)=1.2$ cm², on $A=150$ cm², that is $0.8$ percent (or $2\times 0.4$ percent). The factor 3 shows up for the volume and the factor 2 for the area, in line with the exponents.

d)

We use $dV$ because it is LINEAR in $dr$ , so immediate to compute and to interpret, whereas $\Delta V$ requires expanding a cube and yields a cumbersome expression with no gain in meaning. Check on the sphere: $\Delta V=\frac{4}{3}\pi\left(10.05^{3}-10^{3}\right)=\frac{4}{3}\pi(1015.075-1000)=\frac{4}{3}\pi(15.0751)\approx 63.14$ cm³, against $dV\approx 62.83$ cm³. The gap is $0.31$ cm³, half a percent of the error itself, and $0.007$ percent of the volume. It is negligible twice over: first because it is tiny, then and above all because the measurement uncertainty $\pm 0.05$ is itself only known approximately. Refining the error calculation beyond the precision of the uncertainty would make no physical sense, and that is the decisive argument in the lab.

Exercise 4: The table of standard approximations and their range of validity

Five linearizations come up everywhere in science. Knowing them by heart saves time; knowing HOW FAR they remain acceptable keeps you from using them where they lie.

a) Establish the five standard approximations near 0: $\sin x$ , $\tan x$ , $e^{x}$ , $\ln(1+x)$ and $\sqrt{1+x}$ .
b) Compute the relative error of $\sin x\approx x$ for $x=0.1$ then $x=0.5$ radian. Does the ratio between the two errors surprise you?
c) Up to what value of $x$ does the approximation $\sin x\approx x$ stay accurate to within 1 percent? Convert to degrees.
d) Rank the five approximations from most to least accurate at $x=0.1$ , and explain the link with the second derivative.

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Answers

a) $x$ , $x$ , $1+x$ , $x$ , $1+\frac{x}{2}$
b) $0.167$ % and $4.29$ %: factor $5^{2}$
c) $x\le 0.245$ rad, that is $14.0°$
d) $\sqrt{1+x}$ , $\sin x$ , $\tan x$ , $e^{x}$ , $\ln(1+x)$

a)

Each one is the linearization $f(x)\approx f(0)+f'(0)x$ at $a=0$ . For $\sin$ : $f(0)=0$ , $f'(0)=\cos 0=1$ , so $\sin x\approx x$ . For $\tan$ : $f(0)=0$ , $f'(0)=\sec^{2}0=1$ , so $\tan x\approx x$ . For the exponential: $f(0)=1$ , $f'(0)=1$ , so $e^{x}\approx 1+x$ . For the logarithm: $f(0)=\ln 1=0$ , $f'(x)=\frac{1}{1+x}$ so $f'(0)=1$ , hence $\ln(1+x)\approx x$ . For the root: $f(0)=1$ , $f'(x)=\frac{1}{2\sqrt{1+x}}$ so $f'(0)=\frac{1}{2}$ , hence $\sqrt{1+x}\approx 1+\frac{x}{2}$ . This last one is the case $n=\frac{1}{2}$ of the general formula $(1+x)^{n}\approx 1+nx$ , which sums them all up for powers.

b)

For $x=0.1$ : $\sin(0.1)=0.0998334$ , the approximation gives $0.1$ , the absolute error is $1.666\times 10^{-4}$ and the RELATIVE error $\frac{1.666\times 10^{-4}}{0.0998334}\approx 0.167$ percent. For $x=0.5$ : $\sin(0.5)=0.4794255$ , the absolute error is $0.0205745$ and the relative error $\approx 4.29$ percent. The argument was multiplied by 5, but the relative error by about 26, that is $5^{2}$ . This should not be a surprise: the linearization error behaves like $(x-a)^{2}$ , so multiplying the distance to the base point by five multiplies the error by 25. The quadratic behaviour of the error is the central fact of the chapter.

c)

The relative error of $\sin x\approx x$ is approximately $\frac{x^{2}}{6}$ , since $\sin x\approx x-\frac{x^{3}}{6}$ to degree 3. Require $\frac{x^{2}}{6}\leq 0.01$ , that is $x^{2}\leq 0.06$ and $x\leq 0.245$ radian. Direct check: $\sin(0.245)=0.242556$ , the relative error is $\frac{0.245-0.242556}{0.242556}\approx 1.008$ percent, right at the threshold ✓. Conversion: $0.245$ rad $=0.245\times\frac{180}{\pi}\approx 14.0^{\circ}$ . So remember that the small-angle approximation is reliable to within 1 percent up to about FOURTEEN DEGREES, which is far more generous than intuition suggests, and which explains why it is used without a second thought for a pendulum of small amplitude.

d)

Relative errors at $x=0.1$ : $\sqrt{1+x}$ gives $0.114$ percent; $\sin x$ , $0.167$ percent; $\tan x$ , $0.334$ percent; $e^{x}$ , $0.468$ percent; $\ln(1+x)$ , $4.92$ percent. Ranking from most to least accurate: $\sqrt{1+x}$ , $\sin x$ , $\tan x$ , $e^{x}$ , $\ln(1+x)$ . Two factors explain it. The ABSOLUTE error is about $\frac{|f''(0)|}{2}x^{2}$ : $\frac{x^{2}}{8}$ for the root, $\frac{x^{2}}{2}$ for the exponential and for the logarithm; for $\sin$ and $\tan$ , $f''(0)=0$ and the error only starts at degree 3, as $\frac{x^{3}}{6}$ and $\frac{x^{3}}{3}$ , hence how well they hold up. The RELATIVE error then divides by the value of the function: about $1$ for the root and the exponential, but only $0.1$ for $\ln(1+x)$ , $\sin x$ and $\tan x$ . The logarithm suffers from both handicaps, an absolute error of order $x^{2}$ divided by a small value: its relative error is about $\frac{x}{2}$ , that is $5$ percent, ten times that of the exponential for an almost identical absolute error.

Exercise 5: Estimating a change in a formula

The differential is not only for measurement errors: it also answers the question “if I change this parameter by so much, by how much does the result change?”, which is the engineer's question.

a) A pendulum of length $L=1.00$ m has period $T=2\pi\sqrt{\frac{L}{g}}$ with $g=9.81$ m/s². The rod is lengthened by 3 percent. Estimate the change in period with the differential, then compare with the exact change.
b) The kinetic energy of a 1200 kg car is $E=\frac{1}{2}mv^{2}$ . Its speed goes from 20 m/s to 20.4 m/s. Estimate $\Delta E$ with the differential and compare with the exact value.
c) For this same energy, a 2 percent increase in speed produces what increase in energy, as a percentage? Recover the result without computing any energy.
d) Why does the differential give a value slightly DIFFERENT from the exact change here, and in which direction is it wrong for these two functions?
e) A copper pipe $3$ m long has an inner radius of $2$ cm and a wall $0.15$ cm thick. Estimate the volume of copper with the differential of $V=\pi r^{2}h$ with respect to $r$ , the length staying fixed. Then compute the exact volume and the relative error of the estimate.
f) Redo the estimate of part e) computing the lateral area at the MEAN radius of the wall, $2.075$ cm. What do you notice? Explain it with a factorization.

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Answers

a) $dT\approx 0.0301$ s against $\Delta T\approx 0.0299$ s
b) $dE=9600$ J against $\Delta E=9696$ J
c) $4$ %
d) Energy concave up: underestimate; period concave down: overestimate
e) $dV=180\pi\approx 565.5$ cm³ against $186.75\pi\approx 586.7$ cm³: $3.6$ %
f) $186.75\pi\approx 586.7$ cm³: exact, because $R^{2}-r^{2}=(R-r)(R+r)$

a)

$T=2\pi\sqrt{\frac{L}{g}}$ gives $T(1.00)=2\pi\sqrt{\frac{1}{9.81}}\approx 2.00607$ s. Since the length appears with exponent $\frac{1}{2}$ , the relative error carries over as $\frac{dT}{T}=\frac{1}{2}\frac{dL}{L}=\frac{1}{2}(0.03)=0.015$ , that is 1.5 percent. Hence $dT=0.015\times 2.00607\approx 0.03009$ s. Exact value: $T(1.03)=2\pi\sqrt{\frac{1.03}{9.81}}\approx 2.03594$ s, so $\Delta T\approx 0.02987$ s. The estimate is correct to within one percent of the change itself, which is more than enough for a design calculation.

b)

$E=\frac{1}{2}(1200)v^{2}=600v^{2}$ , so $dE=1200v\,dv$ . With $v=20$ and $dv=0.4$ : $dE=1200(20)(0.4)=9600$ J. Exact value: $E(20.4)-E(20)=600\left(416.16-400\right)=600(16.16)=9696$ J. The two differ by 96 J, 1 percent of the estimate: the differential slightly underestimates, and part d) explains why.

c)

The exponent of $v$ in $E=\frac{1}{2}mv^{2}$ is 2, so $\frac{dE}{E}=2\frac{dv}{v}=2(0.02)=0.04$ , that is 4 percent. No energy calculation is needed: the rule of amplification by the exponent is enough. Check with the numbers of part b): $\frac{9696}{240000}=0.0404$ , that is 4.04 percent ✓. This factor 2 is why speeding by 10 percent increases the energy to be dissipated in braking by 21 percent, and it is the central physical argument of road safety campaigns.

d)

The differential keeps only the LINEAR part of the change: it measures the rise along the TANGENT LINE, whereas the exact change follows the CURVE. The gap is the neglected second-order term, approximately $\frac{f''}{2}(dx)^{2}$ . For the energy, $E(v)=600v^{2}$ is CONCAVE UP ( $E''=1200>0$ ), so the tangent line lies below the curve and the differential UNDERESTIMATES: we do find 9600 J against 9696 J actually. We can even recover the gap exactly, since for a polynomial of degree 2 the neglected term is $600(dv)^{2}=600(0.16)=96$ J ✓, which accounts for the whole difference. For the period, $T(L)\propto\sqrt{L}$ is CONCAVE DOWN, the tangent line lies above the curve and the differential OVERESTIMATES: $0.03009$ s predicted against $0.02987$ s actually ✓. In both cases, the direction of the error is read from the sign of the second derivative, exactly as for the linear approximation.

e)

The copper fills the difference between two cylinders of the same length $h=300$ cm, with radii $2$ and $2.15$ cm: it is a change $\Delta V$ caused by $dr=0.15$ . The differential gives $dV=2\pi rh\,dr=2\pi(2)(300)(0.15)=180\pi\approx 565.5$ cm $^{3}$ . The factor $2\pi rh$ is the LATERAL AREA of the cylinder: the volume of the wall is estimated as area times thickness, like the shell of the sphere in exercise 3 a). Exact volume: $\pi h\left(2.15^{2}-2^{2}\right)=\pi(300)(0.6225)=186.75\pi\approx 586.7$ cm $^{3}$ . The estimate is too low by $6.75\pi\approx 21.2$ cm $^{3}$ , that is $3.6$ percent: the wall is not that thin compared with the radius ( $\frac{0.15}{2}=7.5$ percent), and the lateral area was taken at the INNER radius, the smaller of the two. Units pitfall: the length is written $300$ cm, not $3$ , otherwise the result is a hundred times too small.

f)

$2\pi(2.075)(300)(0.15)=186.75\pi\approx 586.7$ cm $^{3}$ : it is exactly the exact volume. The reason is a special product: $\pi h\left(R^{2}-r^{2}\right)=\pi h(R-r)(R+r)=2\pi\,\frac{R+r}{2}\,h\,(R-r)$ . The volume of a cylindrical wall IS the lateral area at the mean radius times the thickness, with no approximation at all. The differential taken at the inner radius underestimates, taken at the outer radius it would overestimate by the same amount, and the midpoint lands exactly because $r^{2}$ is a polynomial of degree two. For a sphere, where the volume goes as $r^{3}$ , the mean radius would no longer give the exact value.

Exercise 6: How the relative error is amplified

One general result governs all error propagation in science: the exponent multiplies the relative error. It takes three lines to prove and is used everywhere.

a) Prove that for $y=x^{n}$ , we have $\frac{dy}{y}=n\frac{dx}{x}$ . Does the result depend on the sign of $n$ ?
b) A 1 percent measurement error on the radius of a disk produces what error on its area? And on the radius, if it is the area that is measured to within 1 percent?
c) The period of a pendulum is $T=2\pi\sqrt{\frac{L}{g}}$ . A 2 percent error on $L$ produces what relative error on $T$ ? Comment on how comfortable this result is for the experimenter.
d) To measure $g$ with the formula $g=\frac{4\pi^{2}L}{T^{2}}$ , should you take more care over the measurement of $L$ or of $T$ ? Justify quantitatively.

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Answers

a) $\frac{dy}{y}=n\frac{dx}{x}$ , for every $n$
b) $2$ % on the area; $0.5$ % on the radius
c) $1$ %
d) Take care over $T$ : its error counts double

a)

From $y=x^{n}$ we get $dy=nx^{n-1}dx$ , so $\frac{dy}{y}=\frac{nx^{n-1}dx}{x^{n}}=n\frac{dx}{x}$ ✓. The proof never uses the sign of $n$ , nor the fact that it is an integer: the result holds for negative $n$ (reciprocals) and for fractional $n$ (roots). For negative $n$ , the relative error changes SIGN, which simply reflects that increasing $x$ decreases $y$ ; in absolute value the amplification is still $|n|$ . An elegant way to get the same result is logarithmic differentiation: $\ln y=n\ln x$ , then differentiating, $\frac{dy}{y}=n\frac{dx}{x}$ directly.

b)

Area of the disk: $A=\pi r^{2}$ , exponent $n=2$ , so $\frac{dA}{A}=2\frac{dr}{r}=2$ percent. The other way round, if it is the area that is known to within 1 percent, then $r=\sqrt{\frac{A}{\pi}}$ has exponent $n=\frac{1}{2}$ , hence $\frac{dr}{r}=\frac{1}{2}\frac{dA}{A}=0.5$ percent. The relation is inverted by inverting the exponent, which is consistent: measuring a quantity and then taking a root of it REDUCES the error, while raising it to a power amplifies it.

c)

$T=2\pi\sqrt{\frac{L}{g}}$ depends on $L$ with exponent $n=\frac{1}{2}$ , so $\frac{dT}{T}=\frac{1}{2}\frac{dL}{L}=\frac{1}{2}(2\text{ percent})=1$ percent. The error is HALVED going from the length to the period. This is very comfortable for the experimenter: the square root damps imprecision, so a mediocre length measurement still gives an acceptable period. It is one of the reasons the pendulum long served as a time standard.

d)

You must take more care over the measurement of $T$ . In $g=\frac{4\pi^{2}L}{T^{2}}$ , the length appears with exponent $+1$ and the period with exponent $-2$ . Combining the contributions: $\frac{dg}{g}=\frac{dL}{L}-2\frac{dT}{T}$ , so in absolute value a 1 percent error on $L$ gives 1 percent on $g$ , while a 1 percent error on $T$ gives 2. The time measurement weighs TWICE as much as the length measurement. The classic practical consequence in the lab: rather than timing a single oscillation, time twenty and divide, which divides the relative error on $T$ by twenty and makes its contribution much smaller (exercise 10 puts numbers on it). The error reasoning dictates the experimental procedure, not the other way round.

Exercise 7: When the linear approximation goes bad

The figure shows $y=\sqrt{x}$ and its tangent line at $x=4$ . They look alike near the point of tangency and then clearly drift apart: two distinct factors govern this drift.

a) Use the linearization of $\sqrt{x}$ at $a=4$ to estimate $\sqrt{9}$ . Compare with the exact value and give the relative error.
b) Is the estimate of part a) too large or too small? Relate the answer to the sign of $f''$ and check it on the figure.
c) Estimate $e^{0.5}$ by linearization at $a=0$ . Is it an overestimate or an underestimate? Same question of justification.
d) State the TWO factors that degrade the linear approximation, and say which one explains the error of part a).

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Answers

a) $L(9)=3.25$ : relative error $8.3$ %
b) Too large: $\sqrt{x}$ is concave down
c) $1.5$ against $1.6487$ : an underestimate
d) Distance to the base point and curvature; here the distance

a)

The linearization is $L(x)=2+\frac{x-4}{4}$ , so $L(9)=2+\frac{5}{4}=3.25$ , while $\sqrt{9}=3$ exactly. The absolute error is $0.25$ and the relative error $\frac{0.25}{3}\approx 8.3$ percent. Compare with the $0.0076$ percent obtained for $\sqrt{4.1}$ with the same linearization: the error was multiplied by more than a thousand, only because we moved away from the base point.

b)

The estimate is TOO LARGE ( $3.25>3$ ). Justification by concavity: $f(x)=x^{1/2}$ gives $f'(x)=\frac{1}{2}x^{-1/2}$ then $f''(x)=-\frac{1}{4}x^{-3/2}<0$ for every $x>0$ . The function is therefore CONCAVE DOWN, its tangent line lies ABOVE the curve at every point other than the point of tangency, and the approximation systematically OVERESTIMATES. The figure confirms it: the line is above the root everywhere except at $x=4$ , where they touch. This is not an accident of the chosen point: for a concave down function, the overestimate is guaranteed over the whole domain.

c)

$f(x)=e^{x}$ at $a=0$ : $f(0)=1$ , $f'(0)=1$ , so $L(x)=1+x$ and $L(0.5)=1.5$ . The exact value is $e^{0.5}\approx 1.64872$ , so the estimate is TOO SMALL, with an error of $0.149$ (about 9 percent). Justification: $f''(x)=e^{x}>0$ everywhere, the function is CONCAVE UP, its tangent line lies BELOW the curve, and the approximation systematically underestimates. The general rule fits in one sentence: the sign of $f''$ gives the direction of the error, concave up gives an underestimate and concave down an overestimate.

d)

The two factors. First factor, the DISTANCE $|x-a|$ to the base point: the error grows essentially like the square of this distance, which explains why it blows up quickly. Second factor, the CURVATURE, measured by $|f''|$ nearby: a strongly curved function moves away from its tangent line quickly, an almost straight function stays close for a long time. The precise result, which the chapter on Taylor series will prove, is that the error equals $\frac{f''(c)}{2}(x-a)^{2}$ for some $c$ between $a$ and $x$ : both factors appear explicitly in it, as a product. For part a), the DISTANCE is clearly to blame: the curvature of $\sqrt{x}$ is small and even decreases as $x$ increases, but we moved 5 units away instead of $0.1$ , and the square of that distance is $25$ against $0.01$ , a factor of $2500$ that accounts for most of the observed degradation.

Exercise 8: Synthesis: linearization as a first stage

The linear approximation is not an isolated trick: it is the first stage of a construction that the integral calculus course will complete, and the engine of a numerical method you have already seen.

a) Show that Newton's method, $x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$ , is nothing but the linear approximation applied in a loop. What are you looking for at each step?
b) Apply two iterations of Newton's method to $f(x)=x^{2}-2$ starting from $x_{0}=1.5$ , and compare with $\sqrt{2}$ .
c) The linear approximation is the Taylor polynomial of degree 1. Write the degree 2 polynomial of $\sqrt{x}$ at $a=4$ and estimate $\sqrt{9}$ again. By how much is the error of the previous part reduced?
d) Draw up the checklist for the chapter: given a problem, how do you recognize that it calls for a differential, a linearization or an error calculation?

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Answers

a) Newton looks for the zero of the tangent line
b) $x_{1}\approx 1.416667$ , $x_{2}\approx 1.414216$
c) $P_{2}(9)=2.859375$ : error reduced by $44$ %
d) Differential, linearization, error propagation, sign of $f''$

a)

At step $n$ , replace $f$ by its linearization at $x_{n}$ , namely $L(x)=f(x_{n})+f'(x_{n})(x-x_{n})$ , then find where this LINE crosses the $x$ -axis. Solving $L(x)=0$ : $f(x_{n})+f'(x_{n})(x-x_{n})=0$ , hence $x=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$ , which is exactly Newton's formula ✓. At each step you are therefore looking for the zero of the TANGENT LINE, hoping it is closer to the zero of the curve than the starting point. The method converges very fast because the linearization error is quadratic: each iteration roughly doubles the number of correct decimals.

b)

$f(x)=x^{2}-2$ , $f'(x)=2x$ , so $x_{n+1}=x_{n}-\frac{x_{n}^{2}-2}{2x_{n}}=\frac{x_{n}}{2}+\frac{1}{x_{n}}$ . First iteration from $x_{0}=1.5$ : $x_{1}=0.75+0.6667=1.416667$ . Second: $x_{2}=0.708333+0.705882=1.4142157$ . Now $\sqrt{2}=1.4142136$ : after only two iterations, five decimals are correct. The announced doubling is indeed observed, the error going from $8.6\times 10^{-2}$ to $2.5\times 10^{-3}$ and then to $2.1\times 10^{-6}$ .

c)

The Taylor polynomial of degree 2 is $P_{2}(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^{2}$ . For $f(x)=\sqrt{x}$ at $a=4$ : $f(4)=2$ , $f'(4)=\frac{1}{4}$ , and $f''(x)=-\frac{1}{4}x^{-3/2}$ gives $f''(4)=-\frac{1}{32}$ . So $P_{2}(x)=2+\frac{x-4}{4}-\frac{(x-4)^{2}}{64}$ . At $x=9$ : $P_{2}(9)=2+1.25-\frac{25}{64}=3.25-0.390625=2.859375$ . The error goes from $0.25$ (too high) to $0.140625$ (too low), a reduction of about 44 percent, and the direction of the error has flipped. The correction goes the right way but is still not enough here, which confirms the diagnosis of part d) of the previous exercise: at this distance from the base point, no low-degree polynomial can be enough. Adding stages never replaces a good starting point.

d)

Checklist. 1) The problem asks “by approximately how much does” a quantity change for a small change in another: that is a DIFFERENTIAL, compute $dy=f'(x)dx$ . 2) The problem asks you to ESTIMATE a numerical value without a calculator (a root, a power, a trigonometric function of an angle close to a special angle): that is a LINEARIZATION, choose the nearest convenient base point and write $L(x)=f(a)+f'(a)(x-a)$ . 3) The problem gives a measurement with an uncertainty and asks for the effect on a computed quantity: that is ERROR PROPAGATION, with two possible answers to keep apart, the absolute error $|dy|$ and the relative error $\left|\frac{dy}{y}\right|$ as a percentage; remember that exponents multiply the relative error. 4) The problem asks whether the estimate is too high or too low: look at the sign of $f''$ . And in every case, check that $dx$ really is SMALL compared with $x$ , otherwise the method guarantees nothing.

Exercise 9: The tangent line against linear interpolation

Two lines can approximate a curve: the TANGENT LINE, which touches it at one point, and the SECANT LINE, which cuts it at two. Which one is better depends entirely on where you are estimating.

a) Write the linearization of $f(x)=\sqrt{x}$ at $a=4$ , then the equation of the secant line through the points $(4,2)$ and $(9,3)$ .
b) Estimate $\sqrt{4.5}$ with both lines and compare the errors. Which one wins?
c) Estimate $\sqrt{8}$ with both lines. Which one wins this time, and why has the ranking flipped?
d) The errors of the two methods have OPPOSITE signs. Explain why, and say what this lets you assert about the true value.

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Answers

a) $L(x)=2+\frac{x-4}{4}$ ; $S(x)=2+0.2(x-4)$
b) $2.125$ against $2.100$ : the tangent line wins
c) $3$ against $2.8$ : the secant line wins
d) Bounds: $S(x)<\sqrt{x}<L(x)$

a)

TANGENT LINE: $f'(x)=\frac{1}{2\sqrt{x}}$ gives $f'(4)=\frac{1}{4}$ , hence $L(x)=2+\frac{x-4}{4}$ . SECANT LINE: it passes through $(4,2)$ and $(9,3)$ , its slope is $\frac{3-2}{9-4}=\frac{1}{5}=0.2$ , hence $S(x)=2+0.2(x-4)$ . Note that the secant requires knowing TWO exact values of the function, whereas the tangent needs only one, but requires the derivative instead.

b)

$L(4.5)=2+\frac{0.5}{4}=2.125$ and $S(4.5)=2+0.2(0.5)=2.100$ . Exact value: $\sqrt{4.5}\approx 2.121320$ . Errors: the tangent gives $+0.00368$ , the secant $-0.02132$ . The TANGENT LINE wins clearly, by a factor of nearly 6. That is expected: $4.5$ is very close to the point of tangency $a=4$ , where the tangent hugs the curve most tightly.

c)

$L(8)=2+\frac{4}{4}=3$ and $S(8)=2+0.2(4)=2.800$ . Exact value: $\sqrt{8}\approx 2.828427$ . Errors: the tangent gives $+0.17157$ , the secant $-0.02843$ . The SECANT LINE wins this time, also by a factor of 6. The ranking has flipped because $8$ is FAR from the point of tangency $a=4$ but close to the middle of the interval $[4,9]$ on which the secant was built. The error of the tangent grows like the square of the distance to $a$ and so becomes disastrous as you move away, whereas the error of the secant is BOUNDED over the whole interval and smallest near its centre. Practical rule: the tangent line to extrapolate very close to a known point, the secant line (interpolation) to estimate between two known points.

d)

The signs are opposite because $\sqrt{x}$ is CONCAVE DOWN ( $f''(x)=-\frac{1}{4}x^{-3/2}<0$ ). For a concave down function, the tangent line lies entirely ABOVE the curve, so it always overestimates: both tangent errors are indeed positive. And the chord joining two points of a concave down curve lies entirely BELOW the arc, so the secant always underestimates: both secant errors are indeed negative. What this lets you assert is valuable: the true value necessarily lies BETWEEN the two estimates. You get a pair of BOUNDS for free, without knowing the answer. For $\sqrt{8}$ : $2.800<\sqrt{8}<3.000$ , which is true ✓, and for $\sqrt{4.5}$ : $2.100<\sqrt{4.5}<2.125$ ✓. A pair of bounds is always worth more than an isolated estimate, because it carries its own guarantee on the error; and for a concave up function the two roles simply swap, the tangent lying below and the chord above.

Exercise 10: The inverse problem: how precise must the measurement be?

So far we started from the uncertainty to find the error. The engineer's real problem is the reverse: the precision of the RESULT is imposed, and you must deduce the precision required of the INSTRUMENT.

a) We want the volume of a sphere to within 1 percent. What relative precision is needed on the radius? If $r\approx 10$ cm, what absolute precision does that impose?
b) We want the period of a pendulum to within 0.1 percent. What relative precision is needed on the length? For $L\approx 1$ m, express it in millimetres.
c) We measure $g$ with $g=\frac{4\pi^{2}L}{T^{2}}$ and aim for 0.5 percent on $g$ . Sharing the requirement equally between the two measurements, what relative precision should be asked of $L$ and of $T$ ?
d) Is this equal sharing the best choice? Which measurement should be given priority, and what simple experimental move achieves it?

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Answers

a) $0.33$ % on the radius, that is $0.33$ mm
b) $0.2$ % on the length, that is $2$ mm
c) $0.25$ % on $L$ , $0.125$ % on $T$
d) Take care over the period: time twenty oscillations

a)

For $V=\frac{4}{3}\pi r^{3}$ , the exponent is 3, so $\frac{dV}{V}=3\frac{dr}{r}$ . Require $\left|\frac{dV}{V}\right|\leq 0.01$ , hence $\frac{dr}{r}\leq\frac{0.01}{3}\approx 0.00333$ , about 0.33 percent. With $r\approx 10$ cm: $dr\leq 10\times 0.00333=0.0333$ cm, that is about 0.33 mm. The requirement on the instrument is therefore THREE TIMES stricter than the one on the result: it is the amplification by the exponent, read backward.

b)

For $T=2\pi\sqrt{\frac{L}{g}}$ , the exponent of $L$ is $\frac{1}{2}$ , so $\frac{dT}{T}=\frac{1}{2}\frac{dL}{L}$ . Require $\left|\frac{dT}{T}\right|\leq 0.001$ , hence $\frac{dL}{L}\leq 0.002$ , that is 0.2 percent. For $L\approx 1$ m: $dL\leq 0.002$ m $=2$ mm. Here the requirement on the instrument is TWICE AS LENIENT as the one on the result, because the exponent is less than 1: the square root damps imprecision instead of amplifying it. Measuring a length to the nearest millimetre is therefore amply enough, which makes the pendulum experiment doable with a simple tape measure.

c)

In $g=\frac{4\pi^{2}L}{T^{2}}$ , the length appears with exponent $+1$ and the period with exponent $-2$ , so bounding the contributions: $\left|\frac{dg}{g}\right|\leq\frac{dL}{L}+2\frac{dT}{T}$ . Sharing the 0.5 percent equally, each term gets 0.25 percent. For the length: $\frac{dL}{L}\leq 0.0025$ , that is 0.25 percent. For the period: $2\frac{dT}{T}\leq 0.0025$ , so $\frac{dT}{T}\leq 0.00125$ , that is 0.125 percent. The constraint on TIME is therefore twice as strict as the one on length, even though the requirement was shared equally: it is the factor 2 of the exponent that imposes it.

d)

Equal sharing is not optimal, because it ignores the COST of each measurement. The right principle is to relax the requirement where it is hard to meet and tighten it where it is easy. Here, the measurement to give priority is TIME, for two combined reasons: its exponent is 2, so its error is doubled in the result; and manual timing suffers from a reaction time of about 0.2 s, which, over a 2 s period, is already 10 percent, eighty times the requirement. The experimental move that solves everything is to time NOT one oscillation but twenty, then divide by twenty: the reaction error occurs only once at the start and once at the end, so the relative error on the period is divided by twenty and drops to 0.5 percent. That is still four times the 0.125 percent allowed: timing eighty oscillations instead of twenty, or averaging about sixteen repeated trials (a random error shrinks like the square root of the number of trials, and the square root of sixteen is four), closes the gap. The length, for its part, is measured to 0.25 percent without difficulty with a tape measure. This is how error analysis DICTATES the experimental procedure, instead of merely commenting on it afterward.

Exercise 11: Linearizing around a special angle, in radians

We want to estimate $\cos 62°$ and $\tan 46°$ without a calculator, from the known values at $60°$ and at $45°$ .

a) Write the linearization of $\cos x$ at $a=\frac{\pi}{3}$ and estimate $\cos 62°$ to four decimal places.
b) The exact value is $\cos 62°\approx 0.46947$ . Is the estimate too high or too low? Justify with concavity.
c) In the same way, estimate $\tan 46°$ by linearization at $\frac{\pi}{4}$ , and say whether the estimate is too high or too low.
d) What would the calculation of a) give if you took $dx=2$ , in degrees? Why is that absurd?
e) Estimate each error with $\frac{|f''(a)|}{2}(dx)^{2}$ and explain why the error on the tangent is the larger one.

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Answers

a) $\cos 62°\approx 0.4698$
b) Too high: $\cos$ is concave down
c) $\tan 46°\approx 1.0349$ : too low
d) $\approx-1.23$ : degrees are not allowed
e) $3.0\times 10^{-4}$ and $6.1\times 10^{-4}$

a)

$L(x)=\cos\frac{\pi}{3}-\sin\frac{\pi}{3}\left(x-\frac{\pi}{3}\right)=\frac{1}{2}-\frac{\sqrt{3}}{2}\left(x-\frac{\pi}{3}\right)$ . The $2°$ step is $dx=\frac{2\pi}{180}=\frac{\pi}{90}\approx 0.034907$ rad. So $\cos 62°\approx 0.5-0.866025\times 0.034907\approx 0.5-0.030230=0.4698$ .

b)

$0.46977>0.46947$ : the estimate is TOO HIGH, by $3.0\times 10^{-4}$ . On $\left[0,\frac{\pi}{2}\right]$ , $\cos''(x)=-\cos x<0$ : the function is concave down, its tangent line lies above, and the approximation overestimates.

c)

$(\tan x)'=\sec^{2}x$ equals $2$ at $\frac{\pi}{4}$ : $\tan 46°\approx 1+2\times\frac{\pi}{180}\approx 1+0.034907=1.0349$ . The exact value is $1.03553$ : the estimate is TOO LOW, by $6.2\times 10^{-4}$ . On $\left(0,\frac{\pi}{2}\right)$ , $\tan''(x)=2\sec^{2}x\tan x>0$ : concave up, tangent line below.

d)

$0.5-0.866\times 2\approx-1.23$ , a cosine less than $-1$ : impossible. The formula $(\cos x)'=-\sin x$ only holds in radians; in degrees, the derivative would carry the factor $\frac{\pi}{180}$ . A step of $2°$ is not “small” if it is written $2$ : it only is when written $0.0349$ .

e)

For the cosine: $\frac{|-\cos 60°|}{2}(0.034907)^{2}=0.25\times 0.001218\approx 3.0\times 10^{-4}$ , the observed error. For the tangent: $\tan''\left(\frac{\pi}{4}\right)=2\times 2\times 1=4$ , and $\frac{4}{2}(0.017453)^{2}\approx 6.1\times 10^{-4}$ , the observed error too. The angle step is half as large, but the curvature is eight times stronger: $\frac{4}{0.5}=8$ , and $8\times\frac{1}{4}=2$ . The curvature wins.

Exercise 12: Measuring a height by sighting: the angle error

To measure the height $h$ of a tower, you stand at a distance $d=50$ m, measured very precisely, and sight its top at an angle $\theta=35°$ above the horizontal, measured to within $\pm 0.5°$ . Neglect the height of the eye.

a) Express $h$ in terms of $\theta$ and compute it.
b) Estimate the absolute error on $h$ with the differential.
c) Show that the relative error is $\frac{dh}{h}=\frac{2\,d\theta}{\sin 2\theta}$ and compute it.
d) For what sighting angle is the relative error smallest? How far from a tower about $35$ m high should you therefore stand?
e) Compute the relative error for a sighting at $80°$ . What do you conclude?

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Answers

a) $h=50\tan 35°\approx 35.01$ m
b) $dh\approx 0.65$ m
c) $\frac{2\,d\theta}{\sin 2\theta}\approx 1.86$ %
d) $45°$ , about $35$ m away
e) $5.10$ % at $80°$

a)

In the right triangle, $\tan\theta=\frac{h}{d}$ , so $h=50\tan\theta=50\tan 35°\approx 35.01$ m.

b)

$dh=50\sec^{2}\theta\,d\theta$ , with $d\theta$ IN RADIANS: $0.5°=\frac{\pi}{360}\approx 0.008727$ rad. $\sec^{2}35°=\frac{1}{\cos^{2}35°}\approx 1.4903$ . So $dh\approx 50\times 1.4903\times 0.008727\approx 0.65$ m. We write $h=35.0\pm 0.7$ m.

c)

$\frac{dh}{h}=\frac{50\sec^{2}\theta\,d\theta}{50\tan\theta}=\frac{d\theta}{\cos^{2}\theta}\cdot\frac{\cos\theta}{\sin\theta}=\frac{d\theta}{\sin\theta\cos\theta}=\frac{2\,d\theta}{\sin 2\theta}$ . At $35°$ : $\frac{0.017453}{\sin 70°}\approx\frac{0.017453}{0.93969}\approx 1.86$ percent.

d)

The relative error is smallest when $\sin 2\theta$ is largest, that is $2\theta=90°$ and $\theta=45°$ : it is then $2\,d\theta\approx 1.75$ percent. A $45°$ sighting means $h=d$ : you should stand about $35$ m from the tower.

e)

At $80°$ : $\frac{0.017453}{\sin 160°}\approx\frac{0.017453}{0.34202}\approx 5.10$ percent, nearly three times as much. Too close to the tower, the line of sight is almost vertical and half a degree of error moves the sighted point a great deal: the tangent function is very steep there. Choosing where to stand is part of the measurement.

Exercise 13: The common logarithm and small changes: decibels and pH

The sound level is $L=10\log_{10}\frac{I}{I_{0}}$ , in decibels, where $I$ is the intensity of the sound. The pH of a solution is $\text{pH}=-\log_{10}[\text{H}^{+}]$ .

a) Show that $d\left(\log_{10}u\right)=\frac{1}{\ln 10}\cdot\frac{du}{u}\approx 0.4343\,\frac{du}{u}$ .
b) The intensity of a sound increases by $10$ percent. Estimate the change in sound level, then compute it exactly.
c) The intensity doubles. What do the estimate and the exact calculation give? Why does the estimate fail?
d) The concentration of $\text{H}^{+}$ ions is known to within $5$ percent. What is the uncertainty on the pH?
e) With what relative precision must $[\text{H}^{+}]$ be known to get the pH to within $\pm 0.01$ ?

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a) $d\left(\log_{10}u\right)=\frac{0.4343\,du}{u}$
b) $0.434$ dB against $0.414$ dB
c) $4.34$ dB against $3.01$ dB: not a small change
d) $\pm 0.022$
e) $2.3$ %

a)

$\log_{10}u=\frac{\ln u}{\ln 10}$ , so $\frac{d}{du}\log_{10}u=\frac{1}{u\ln 10}$ and $d\left(\log_{10}u\right)=\frac{1}{\ln 10}\cdot\frac{du}{u}$ , with $\frac{1}{\ln 10}\approx 0.4343$ . A RELATIVE change in $u$ gives an ABSOLUTE change in the logarithm: that is why logarithmic scales add what the quantities multiply.

b)

$dL=10\times 0.4343\times 0.10\approx 0.434$ dB. Exactly: $10\log_{10}1.1\approx 0.414$ dB. The estimate is good to within $0.02$ dB, and too high, because the logarithm is concave down.

c)

Estimate: $10\times 0.4343\times 1=4.34$ dB. Exact: $10\log_{10}2\approx 3.01$ dB. The gap is $44$ percent: a relative change of $100$ percent is not small, and the differential is only valid for small $\frac{du}{u}$ . Remember the exact value, “doubling the intensity adds $3$ dB”.

d)

$|d\,\text{pH}|=0.4343\times 0.05\approx 0.022$ : the pH is known to within $\pm 0.02$ . A $5$ percent imprecision on the concentration only affects the second decimal of the pH, which explains why a pH can be given to two decimals with approximate concentrations.

e)

$0.4343\,\frac{du}{u}\le 0.01$ gives $\frac{du}{u}\le\frac{0.01}{0.4343}\approx 0.023$ , that is $2.3$ percent. This is the inverse problem: the precision required of the instrument follows from the precision wanted on the result.

Exercise 14: Problem: thermal expansion of a bridge and a ring

A steel part of length $L_{0}$ at the reference temperature measures $L=L_{0}(1+\alpha\,\Delta T)$ after a temperature rise $\Delta T$ , with $\alpha=1.2\times 10^{-5}$ per degree Celsius. Every volume of the part is multiplied by $(1+\alpha\,\Delta T)^{3}$ .

a) A steel bridge is $800$ m long. The deck temperature varies from $-30$ °C to $35$ °C over the year. By how much does its length vary?
b) A steel cube of $1$ m $^{3}$ is heated by $100$ °C. Compute the exact increase in volume, then its estimate with $(1+x)^{3}\approx 1+3x$ . What is the relative error of the estimate?
c) Deduce the approximate coefficients of area and volume expansion, in terms of $\alpha$ .
d) A steel ring with inner diameter $49.95$ mm must be slipped onto a $50.00$ mm shaft. By how much at least must it be heated?
e) The density of steel is $7850$ kg/m $^{3}$ at the reference temperature. Estimate it after a temperature rise of $100$ °C.

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a) $0.624$ m
b) $3604.3$ cm³ against $3600$ cm³: $0.12$ %
c) $2\alpha$ and $3\alpha$
d) At least $83.4$ °C
e) $\approx 7821.7$ kg/m³

a)

$\Delta T=65$ °C and $\Delta L=L_{0}\alpha\,\Delta T=800\times 1.2\times 10^{-5}\times 65=0.624$ m. The bridge gains more than $60$ cm between winter and summer: that is what the expansion joints must absorb.

b)

$x=\alpha\,\Delta T=1.2\times 10^{-3}$ . Exactly: $(1.0012)^{3}-1\approx 0.0036043$ m $^{3}$ , that is $3604.3$ cm $^{3}$ . Estimate: $3x=0.0036$ m $^{3}$ , that is $3600$ cm $^{3}$ . The gap, $4.3$ cm $^{3}$ , is $0.12$ percent of the change: the neglected terms $3x^{2}+x^{3}$ are about a thousand times smaller than $3x$ , since $x$ is about one thousandth.

c)

An area is multiplied by $(1+x)^{2}\approx 1+2x$ and a volume by $(1+x)^{3}\approx 1+3x$ : area coefficient $2\alpha$ , volume coefficient $3\alpha$ . It is the rule of the exponent multiplying relative changes, applied to expansion.

d)

We need $49.95(1+\alpha\,\Delta T)\ge 50.00$ , that is $\Delta T\ge\frac{0.05}{49.95\times 1.2\times 10^{-5}}\approx 83.4$ °C. Heat the ring by at least $84$ °C: as it cools, it grips the shaft with no weld or screw; this is a shrink fit.

e)

The mass does not change and the volume is multiplied by $1+3x$ , so $\rho=\frac{\rho_{0}}{(1+x)^{3}}\approx\rho_{0}(1-3x)$ , using $(1+u)^{-1}\approx 1-u$ . $\rho\approx 7850(1-0.0036)\approx 7821.7$ kg/m $^{3}$ , a drop of $28$ kg/m $^{3}$ . Exact calculation: $\frac{7850}{1.0036043}\approx 7821.8$ kg/m $^{3}$ .

Exercise 15: Problem: the distance to the horizon

An observer at a height $h$ above the sea sees the horizon at the distance $d=\sqrt{2Rh+h^{2}}$ , where $R=6\,371$ km is the radius of the Earth. The figure, which is not to scale, shows the situation. Work in metres.

a) Justify the formula with Pythagoras, then show that $d\approx\sqrt{2Rh}$ as long as $h$ is small compared with $R$ . Compute $d$ for $h=100$ m.
b) Using the differential, by how many metres does the distance to the horizon increase when you climb one metre starting from $100$ m?
c) Show that $\frac{dd}{d}=\frac{1}{2}\frac{dh}{h}$ . Estimate the gain in distance when climbing from $100$ m to $110$ m, then compute it exactly.
d) The altitude of $100$ m is known to within $\pm 5$ m. What is the uncertainty on the distance to the horizon?
e) What altitude must you reach to see twice as far as at $100$ m?

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a) $d\approx 35.7$ km
b) $\approx 178.5$ m per metre
c) $1785$ m estimated against $1742$ m
d) $\pm 892$ m
e) $400$ m

a)

The line of sight is tangent to the Earth at the horizon point, so it is perpendicular to the radius of the Earth there. The right triangle has legs $R$ and $d$ and hypotenuse $R+h$ : $(R+h)^{2}=R^{2}+d^{2}$ , that is $d^{2}=2Rh+h^{2}$ . The ratio $\frac{h^{2}}{2Rh}=\frac{h}{2R}$ is less than $10^{-5}$ for $h=100$ m: we neglect it. $d\approx\sqrt{2\times 6\,371\,000\times 100}\approx 35\,696$ m, about $35.7$ km.

b)

$d=\sqrt{2R}\,h^{1/2}$ , so $\frac{dd}{dh}=\frac{\sqrt{2R}}{2\sqrt{h}}=\frac{3569.6}{20}\approx 178.5$ . Each metre gained in altitude pushes the horizon back by about $178$ m.

c)

Exponent $\frac{1}{2}$ : $\frac{dd}{d}=\frac{1}{2}\frac{dh}{h}$ . Climbing by $10$ percent gives about $5$ percent more distance, that is $0.05\times 35\,696\approx 1785$ m. Exactly: $\sqrt{2R\times 110}\approx 37\,438$ m, a gain of $1742$ m. The estimate is $43$ m too high, because the square root is concave down.

d)

$dd=178.5\times 5\approx 892$ m, about $\pm 0.9$ km, $2.5$ percent of the distance, half of the $5$ percent uncertainty on the altitude.

e)

$d$ is proportional to $\sqrt{h}$ : to double $d$ , you must QUADRUPLE $h$ , that is $400$ m. The differential is no use here, the change is not small: it is the exact square-root form that answers, and it answers without any calculation.

Corrected exercises: differentials and linear approximation (201-SN2-RE)

Course recap

Calculations: differentials and estimates to compute

Exercise 0.1: Compute a differential and estimate with the tangent line: ten direct calculations

Part A: the differential, linearization and errors (/50)

Exercise 1: The differential and the actual change

Exercise 2: Linear approximation: computing without a calculator

Exercise 3: Propagation of measurement errors

Exercise 4: The table of standard approximations and their range of validity

Exercise 5: Estimating a change in a formula

Part B: reliability, bounds and the inverse problem (/50)

Exercise 6: How the relative error is amplified

Exercise 7: When the linear approximation goes bad

Exercise 8: Synthesis: linearization as a first stage

Exercise 9: The tangent line against linear interpolation

Exercise 10: The inverse problem: how precise must the measurement be?

Part C: the classics (/50)

Exercise 11: Linearizing around a special angle, in radians

Exercise 12: Measuring a height by sighting: the angle error

Exercise 13: The common logarithm and small changes: decibels and pH

Exercise 14: Problem: thermal expansion of a bridge and a ring

Exercise 15: Problem: the distance to the horizon

See also

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