Approximate calculation by differentials (linearization)

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There is a circular sector with radius R=100R=100 cm and central angle α=60°\alpha =60°. If (1) its radius RR increases by 11 cm; (2) the angle α\alpha decreases by 30′30', what is the change in the area of the sector? Find both the exact and the approximate solutions.

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(a)
(b)

The period of oscillation of a simple pendulum (measured in seconds) is determined by the formula:

T=2πlg, T=2\pi \sqrt{\frac{l}{g}},

where ll is the pendulum length measured in centimeters, and g=981g=981 cm/sec2^{2} is the acceleration due to gravity.

In order to increase the period TT by 0.050.05 seconds, how much modification is needed to the pendulum length l=20l=20 cm?

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Replacing the increment of the function by its differential, find the approximate value of the following:1.023\sqrt[3]{1.02}.

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Replacing the increment of the function by its differential, find the approximate value of the following:sin⁡29°\sin 29°.

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Replacing the increment of the function by its differential, find the approximate value of the following:cos⁡151°\cos 151°.

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Replacing the increment of the function by its differential, find the approximate value of the following:arctg⁡1.05\arctg 1.05.

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Replacing the increment of the function by its differential, find the approximate value of the following:lg⁡11\lg 11.

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Prove the approximate formula:

a2+x≈a+x2a (a>0), \sqrt{a^{2}+x} \approx a+\frac{x}{2a} \ (a>0),

where ∣x∣≪a\left|x\right| \ll a (the relation A≪BA \ll B between positive numbers AA and BB means that when AA is compared with BB, AA is a higher-order infinitesimal).

Use this formula to approximately compute:

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(a)
(b)

34\sqrt{34};

(c)

120\sqrt{120}

and compare with the values in the table.

Prove the approximate formula:

an+xn≈a+xnan−1 (a>0). \sqrt[n]{a^{n}+x} \approx a+\frac{x}{na^{n-1}} \ (a>0) .

where ∣x∣≪a\left|x\right| \ll a. Use this formula to approximately compute:

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(a)

93\sqrt[3]{9};

(b)

804\sqrt[4]{80};

(c)

1007\sqrt[7]{100};

(d)

100010\sqrt[10]{1000}.

The side of a square is x=2.4x=2.4 m ±0.05\pm 0.05 m; what are the relative error and absolute error of the resulting computed area of the square?

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In order to compute the volume of a sphere accurate to 1%1\%, what relative error is permitted when measuring the radius RR of the sphere?

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Using the oscillation of a simple pendulum, the formula g=4π2lT2g=\dfrac {4\pi^{2}l}{T^{2}} (where ll is the pendulum length and TT is the full period of the pendulum's oscillation) is used to find the acceleration due to gravity. How does the relative error δ\delta when measuring (1) the pendulum length ll, (2) the period TT, affect the value gg?

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(a)
(b)

Find the absolute error of the common logarithm of a number x (x>0)x \ (x>0), given that the relative error of this number equals δ\delta.

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Prove: an angle found from a table of logarithms of tangents is more precise than an angle found from a table of logarithms of sines having the same number of decimal places.

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Prove the approximate formula

an+xn≈a+xnan−1 \sqrt[n]{a^{n}+x} \approx a+\frac{x}{na^{n-1}}

and use it to compute the roots: 2455\sqrt[5]{245}, 1297\sqrt[7]{129}, 5159\sqrt[9]{515}, 102710\sqrt[10]{1027}.

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Prove that the tabular difference of ordinary five-digit logarithms of numbers, i.e. lg⁡(n+1)−lg⁡n\lg (n+1)-\lg n, for 1000<n<100001000 < n < 10000 is approximately expressed by the formula Δ=105⋅Mn=43429n\Delta = 10^{5} \cdot \dfrac {M}{n} = \dfrac {43429}{n}; M=0,43429M = 0{,}43429 is the modulus of common logarithms, Δ\Delta being expressed in hundred-thousandths.

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Prove that the tabular difference of logarithms of sines given one minute apart, expressed in hundred-thousandths, can be given by the approximate formula

Δlg⁡sin⁡x≈12,6ctg⁡x. \Delta \lg \sin x \approx 12{,}6\ctg x.
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Prove that formulas of the same kind for sines and for logarithms of tangents have the form:

Δsin⁡x≈29,1cos⁡x,Δlg⁡tg⁡x≈25,2sin⁡2x. \Delta \sin x \approx 29{,}1\cos x, \qquad \Delta \lg \tg x \approx \frac{25{,}2}{\sin 2x}.
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The approximate equalities sin⁡x≈x\sin x \approx x, tg⁡x≈x\tg x \approx x have high accuracy for small angles (in some cases the more accurate formulas sin⁡x≈x−x36\sin x \approx x-\dfrac {x^{3}}{6}, tg⁡x≈x+x33\tg x \approx x+\dfrac {x^{3}}{3} are needed). At what visual angle is the diameter of a factory chimney of radius 2 m seen from a distance of 10 km?

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The disk of the Sun is seen at an angle of 30′30'. How many times greater is the distance to the Sun than the diameter of the Sun?

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A railway embankment has a grade of 0,0120{,}012. What angle does it make with the horizontal?

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Prove the approximate formula: l=13hl = \sqrt{13h}, where hh is the height of the observer above the horizon in meters, and ll is the distance to the horizon in kilometers.

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Replacing the increment of the function by its differential, find the approximate value of the function y=y(x)y = y(x) at the indicated points:

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(1)

y=x3y = \sqrt[3]{x} a) x=65x = 65 b) x=125,1324x = 125,1324

(2)

y=x4y = \sqrt[4]{x} a) x=90x = 90 b) x=15,8x = 15,8

(3)

y=sin⁡xy = \sin x a) x=29∘x = 29^{\circ } b) x=359∘x = 359^{\circ }

(4)

y=tg⁡x,x=44∘50′y = \operatorname {tg} x, x = 44^{\circ } 50^{\prime }

(5)

y=arcsin⁡x,x=0,51y = \arcsin x, x = 0,51;

(6)

y=arctg⁡x,x=1,05y = \operatorname {arctg} x, x = 1,05;

(7)

y=ln⁡tg⁡x,x=47∘15′y = \ln \operatorname {tg} x, x = 47^{\circ } 15^{\prime };

(8)

y=2−x2+x,x=0,15y = \sqrt{\frac{2-x}{2+x}}, x = 0,15.

Prove that for all values of Δx\Delta x small compared with x0x_{0} the approximate formula holds

x0+Δxn≈x0n+x0nnx0Δx,x0>0 \sqrt[n]{x_{0}+\Delta x} \approx \sqrt[n]{x_{0}}+\frac{\sqrt[n]{x_{0}}}{n x_{0}} \Delta x, \quad x_{0} > 0

Using this formula, approximately compute:

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(1)

640\sqrt{640}

(2)

2003\sqrt[3]{200}

(3)

243,455\sqrt[5]{243,45}

(4)

100010\sqrt[10]{1000}

Determine by approximately how much the volume of a ball increases if its radius R=15R = 15 cm is increased by 0.2 cm.

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Determine approximately the relative error in computing the surface area of a sphere if the relative error in determining its radius was 1%1\%.

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By approximately how much (in percent) will the current in a conductor change if its resistance increases by 1%1\%?

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By approximately how much should the length of a pendulum l=20 cml = 20 \mathrm{~ cm} be changed so that the period of oscillation of the pendulum increases by 0.05s0.05 \mathrm{s}? The period TT is given by the formula T=2πl/gT = 2 \pi \sqrt{l / g}.

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References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Предел. Непрерывность. Дифференцируемость. 2-е изд., ФИЗМАТЛИТ, 2003.