Advanced Mathematics (Seventh Edition) Tongji University exercises 2-5 personal solutions

Advanced mathematics （ The seventh edition ） Tongji University exercises 2-5

 

$\begin{array}{rlr}& 1.It\; is\; known\; thaty=x^3-x,\; Calculated\; atx=2Chudang\Delta xRespectively\; equal\; to1,0.1,0.01At\; the\; time\; of\; the\Delta yAnddy.& \end{array}$

Explain ：

$\begin{array}{rlr}& \Delta y=f(x+\Delta x)-f(x)=(x+\Delta x{)}^3-(x+\Delta x)-x^3+x=3x(\Delta x{)}^2+3x^2\Delta x+(\Delta x{)}^3-\Delta x& \\ & & \\ & stayx=2It\text{'}s\; about\; ,\Delta x=1when\; ,\Delta y=18,\Delta x=0.1when\; ,\Delta y=1.161,\Delta x=0.01when\; ,\Delta y=0.110601& \\ & & \\ & dy=f^{\prime }(x)\Delta x=(3x^2-1)\Delta x& \\ & & \\ & stayx=2It\text{'}s\; about\; ,\Delta x=1when\; ,dy=11,\Delta x=0.1when\; ,dy=1.1,\Delta x=0.01when\; ,dy=0.11& \end{array}$

---

$\begin{array}{rlr}& 2.Let\text{'}s\; set\; the\; functiony=f(x)The\; figure\; of\; is\; shown\; in\; the\; figure2-12,\; Try\; in\; Figure2-12(a)、(b)、(c)、(d)Mark\; the\; points\; respectively\; in{x}_{0}Ofdy、\Delta y& \\ & & \\ & And\Delta y-dy,\; And\; explain\; its\; positive\; and\; negative\; .& \end{array}$

Explain ：

$\begin{array}{rlr}& (a)\Delta y>0,dy>0,\Delta y-dy>0& \\ & & \\ & (b)\Delta y>0,dy>0,\Delta y-dy<0& \\ & & \\ & (c)\Delta y<0,dy<0,\Delta y-dy<0& \\ & & \\ & (d)\Delta y<0,dy<0,\Delta y-dy>0& \end{array}$

---

$\begin{array}{rlr}& 3.Find\; the\; differential\; of\; the\; following\; function\; ：& \end{array}$

$\begin{array}{rlr}& (1)y=\frac{1}{x}+2\sqrt{x};(2)y=xsin2x;& \\ & & \\ & (3)y=\frac{x}{\sqrt{x^2+1}};(4)y=l{n}^2(1-x);& \\ & & \\ & (5)y=x^2{e}^{2x};(6)y=e^{-x}cos(3-x);& \\ & & \\ & (7)y=arcsin\sqrt{1-x^2};(8)y=ta{n}^2(1+2x^2);& \\ & & \\ & (9)y=arctan\frac{1-x^2}{1+x^2};(10)s=Asin(\omega t+\phi )(A、\omega 、\phi Is\; constant).& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)dy=y^{\prime }dx=\left(-\frac{1}{x^2}+\frac{1}{\sqrt{x}}\right)dx& \\ & & \\ & (2)dy=y^{\prime }dx=(sin2x+2xcos2x)dx& \\ & & \\ & (3)dy=y^{\prime }dx=\frac{x^{\prime }\sqrt{x^2+1}-x(\sqrt{x^2+1}{)}^{\prime }}{x^2+1}dx=\frac{1}{\sqrt{(x^2+1{)}^3}}dx& \\ & & \\ & (4)dy=y^{\prime }dx=(l{n}^2(1-x){)}^{\prime }dx=2ln(1-x)\frac{-1}{1-x}dx=\frac{2ln(1-x)}{x-1}dx& \\ & & \\ & (5)dy=y^{\prime }dx=(x^2{e}^{2x}{)}^{\prime }dx=(2x{e}^{2x}+2x^2{e}^{2x})dx=2x{e}^{2x}(1+x)dx& \\ & & \\ & (6)dy=y^{\prime }dx=(e^{-x}cos(3-x){)}^{\prime }dx=(-e^{-x}cos(3-x)+e^{-x}sin(3-x))dx=e^{-x}[sin(3-x)-cos(3-x)]dx& \\ & & \\ & (7)dy=y^{\prime }dx=(arcsin\sqrt{1-x^2}{)}^{\prime }dx=-\frac{x}{|x|\sqrt{1-x^2}}dx=\{\begin{array}{l}\frac{dx}{\sqrt{1-x^2}},-1<x<0,\\ \\ -\frac{dx}{\sqrt{1-x^2}},0<x<1\end{array}& \\ & & \\ & (8)dy=y^{\prime }dx=(ta{n}^2(1+2x^2){)}^{\prime }dx=\frac{8xsin(1+2x^2)}{co{s}^3(1+2x^2)}dx& \\ & & \\ & (9)dy=y^{\prime }dx={\left(arctan\frac{1-x^2}{1+x^2}\right)}^{\prime }dx=\left(\frac{1}{1+\frac{(1-x^2{)}^2}{(1+x^2{)}^2}}\cdot \frac{(1-x^2{)}^{\prime }(1+x^2)-(1-x^2)(1+x^2{)}^{\prime }}{(1+x^2{)}^2}\right)dx=-\frac{2x}{x^4+1}dx& \\ & & \\ & (10)ds=s^{\prime }dt=(Asin(\omega t+\phi ){)}^{\prime }dt=A\omega cos(\omega t+\phi )dt& \end{array}$

---

$\begin{array}{rlr}& 4.Fill\; the\; appropriate\; function\; in\; the\; following\; brackets\; ,\; Make\; the\; equation\; hold\; ：& \end{array}$

$\begin{array}{rlr}& (1)d()=2dx;(2)d()=3xdx;& \\ & & \\ & (3)d()=costdt;(4)d()=sin\omega xdx(\omega \ne 0);& \\ & & \\ & (5)d()=\frac{1}{1+x}dx;(6)d()=e^{-2x}dx;& \\ & & \\ & (7)d()=\frac{1}{\sqrt{x}}dx;(8)d()=se{c}^{2}3xdx.& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)d(2x+C)=2dx,CIs\; an\; arbitrary\; constant& \\ & & \\ & (2)d(\frac{3}{2}{x}^2+C)=3xdx,CIs\; an\; arbitrary\; constant& \\ & & \\ & (3)d(sint+C)=costdt,CIs\; an\; arbitrary\; constant& \\ & & \\ & (4)d(-\frac{1}{\omega }cos\omega x+C)=sin\omega xdx,CIs\; an\; arbitrary\; constant& \\ & & \\ & (5)d(ln(1+x)+C)=\frac{1}{1+x}dx,CIs\; an\; arbitrary\; constant& \\ & & \\ & (6)d(-\frac{1}{2}{e}^{-2x}+C)=e^{-2x}dx,CIs\; an\; arbitrary\; constant& \\ & & \\ & (7)d(2\sqrt{x}+C)=\frac{1}{\sqrt{x}}dx,CIs\; an\; arbitrary\; constant& \\ & & \\ & (8)d(\frac{1}{3}tan3x+C)=se{c}^{2}3xdx,CIs\; an\; arbitrary\; constant& \end{array}$

---

$\begin{array}{rlr}& 5.Pictured2-13Cable\; shown\stackrel{⌢}{\mathrm{AOB}}The\; length\; ofs,\; The\; span\; is2l,\; The\; lowest\; point\; of\; the\; cableOConnect\; with\; the\; rod\; topABThe\; distance\; tof,& \\ & & \\ & Then\; the\; cable\; length\; can\; be\; calculated\; according\; to\; the\; following\; formulas=2l\left(1+\frac{2f^2}{3l^2}\right),\; WhenfChanged\Delta fwhen\; ,\; How\; much\; does\; the\; cable\; length\; change\; ？& \end{array}$

Explain ：

$\begin{array}{rlr}& s=2l\left(1+\frac{2f^2}{3l^2}\right),\Delta s\approx ds=2l\cdot \frac{4f}{3l^2}\Delta f=\frac{8f}{3l}\Delta f& \end{array}$

---

$\begin{array}{rlr}& 6.Set\; the\; center\; angle\; of\; the\; fan\alpha =60^{\circ },\; radiusR=100cm（\; chart2-14）.\; IfRunchanged\; ,\alpha Reduce30^{\prime },& \\ & & \\ & Ask\; about\; the\; change\; of\; sector\; area\; ？\; And\; if\alpha unchanged\; ,RAdded1cm,\; Ask\; about\; the\; change\; of\; sector\; area\; ？& \end{array}$

Explain ：

$\begin{array}{rlr}& Sector\; area\; formulaS=\frac{R^2}{2}\alpha ,\Delta S\approx dS=\frac{R^2}{2}\Delta \alpha & \\ & & \\ & WhenR=100,\Delta \alpha =-30^{\prime }=-\frac{\pi }{360},\alpha =\frac{\pi }{3}when\; ,\Delta S\approx \frac{1}{2}\cdot 100^2\cdot \left(-\frac{\pi }{360}\right)\approx -43.63c{m}^2& \\ & & \\ & WhenR=100,\Delta R=1,\alpha =\frac{\pi }{3}when\; ,\Delta S\approx \frac{\pi }{3}\cdot 100\cdot 1\approx 104.72c{m}^2& \end{array}$

---

$\begin{array}{rlr}& 7.Calculate\; the\; approximate\; value\; of\; the\; following\; trigonometric\; function\; ：& \end{array}$

$\begin{array}{rlr}& (1)cos29^{\circ };(2)tan136^{\circ }.& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)cos29^{\circ }=cos\left(\frac{\pi }{6}-\frac{\pi }{180}\right)\approx cos\frac{\pi }{6}+(-sin\frac{\pi }{6})\cdot \left(-\frac{\pi }{180}\right)\approx \frac{\sqrt{3}}{2}+\frac{\pi }{360}\approx \mathrm{0.8748.}& \\ & & \\ & (2)tan136^{\circ }=tan\left(\frac{3}{4}\pi +\frac{\pi }{180}\right)\approx tan\frac{3}{4}\pi +se{c}^2\frac{3}{4}\pi \cdot \frac{\pi }{180}\approx -0.9651& \end{array}$

---

$\begin{array}{rlr}& 8.Calculate\; the\; approximate\; value\; of\; the\; following\; inverse\; trigonometric\; function\; ：& \end{array}$

$\begin{array}{rlr}& (1)arcsin0.5002;(2)arccos\mathrm{0.4995.}& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)arcsin0.5002\approx arcsin0.5+\frac{1}{\sqrt{1-(0.5{)}^2}}\cdot 0.0002\approx 30^{\circ }47^{\prime \prime }& \\ & & \\ & (2)arccos0.4995\approx arccos0.5-\frac{1}{\sqrt{1-(0.5{)}^2}}\cdot (-0.0005)\approx 60^{\circ }{2}^{\prime }& \end{array}$

---

$\begin{array}{rlr}& 9.When|x|More\; hours\; ,\; Prove\; the\; following\; approximate\; formula\; ：& \end{array}$

$\begin{array}{rlr}& (1)tanx\approx x（xIs\; the\; radian\; value\; of\; the\; angle\; ）;(2)ln(1+x)\approx x;& \\ & & \\ & (3)\sqrt[n]{1+x}\approx 1+\frac{1}{n}x;(4)e^x\approx 1+x.& \\ & & \\ & And\; calculatetan45^{\prime }andln1.002Approximate\; value\; .& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)tanx\approx tan0+(tan0{)}^{\prime }\cdot x=0+se{c}^{2}0\cdot x=x& \\ & & \\ & (2)ln(1+x)\approx ln(1+0)+[ln(1+x){]}^{\prime }\cdot x=0+\frac{1}{1+0}x=x& \\ & & \\ & (3)\sqrt[n]{1+x}\approx \sqrt[n]{1+0}+(\sqrt[n]{1+0}{)}^{\prime }\cdot x=1+\frac{1}{n}(1+0{)}^{\frac{1}{n}-1}\cdot x=1+\frac{1}{n}x& \\ & & \\ & (4)e^x\approx e^0+(e^0{)}^{\prime }\cdot x=1+e^{0}x=1+x.& \\ & & \\ & tan45^{\prime }=tan0.0131\approx 0.0131,ln(1.002)\approx \mathrm{0.002.}& \end{array}$

---

$\begin{array}{rlr}& 10.Calculate\; the\; approximate\; values\; of\; the\; following\; roots\; ：& \end{array}$

$\begin{array}{rlr}& (1)\sqrt[3]{996};(2)\sqrt[6]{65}.& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)because\sqrt[n]{1+x}\approx 1+\frac{x}{n},\; therefore\sqrt[3]{996}=\sqrt[3]{1000-4}=\sqrt[3]{1-\frac{4}{1000}}\approx 10\left[1+\frac{1}{3}\left(-\frac{4}{1000}\right)\right]\approx 9.987& \\ & & \\ & (2)\sqrt[6]{65}=\sqrt[6]{64+1}=2\sqrt[6]{1+\frac{1}{64}}\approx 2\left(1+\frac{1}{6}\cdot \frac{1}{64}\right)\approx 2.0052& \end{array}$

---

$\begin{array}{rlr}& 11.When\; calculating\; the\; volume\; of\; the\; sphere\; ,\; The\; accuracy\; is\; required\; to\; be2\%within\; .\; Ask\; at\; this\; time\; to\; measure\; the\; diameterDThe\; relative\; error\; of\; cannot\; exceed\; much\; ？& \end{array}$

Explain ：

$\begin{array}{rlr}& The\; volume\; formula\; of\; the\; sphere\; isV=\frac{1}{6}\pi R^3,\; thereforedV=\frac{1}{6}\pi D^3,dV=\frac{\pi }{2}{D}^2\Delta D,& \\ & & \\ & because\left|\frac{dV}{V}\right|=\left|\frac{\frac{\pi }{2}{D}^2\Delta D}{\frac{1}{6}\pi D^3}\right|=3\left|\frac{\Delta D}{D}\right|\le 2\%,\; obtain\left|\frac{\Delta D}{D}\right|\le \frac{0.02}{3}\approx 0.667\%& \end{array}$

---

$\begin{array}{rlr}& 12.The\; production\; of\; a\; factory\; is\; shown\; in\; the\; figure2-15Sector\; plate\; shown\; ,\; radiusR=200mm,\; Center\; angle\; required\alpha by55^{\circ }.\; During\; product\; inspection\; ,& \\ & & \\ & Generally,\; the\; chord\; length\; is\; measuredlTo\; indirectly\; measure\; the\; central\; angle\alpha .\; If\; you\; measure\; the\; chord\; lengthlError\; in{\delta }_l=0.1mm,& \\ & & \\ & Ask\; about\; the\; central\; angle\; measurement\; error\; caused\; by\; this{\delta }_{\alpha }How\; much\; is\; the\; ？& \end{array}$

Explain ：

$\begin{array}{rlr}& from\frac{l}{2}=Rsin\frac{\alpha }{2}have\; to\; ,\alpha =2arcsin\frac{l}{2R}=2arcsin\frac{l}{400},{\delta }_{\alpha }=|{\alpha }_l^{\prime }|{\delta }_l=\frac{2}{\sqrt{1-{\left(\frac{l}{400}\right)}^2}}\cdot \frac{1}{400}\cdot {\delta }_l.& \\ & & \\ & When\alpha =55^{\circ }when\; ,l=2Rsin\frac{\alpha }{2}=400sin(27.5^{\circ })\approx 184.7.\; because{\delta }_l=0.1,\; therefore{\delta }_{\alpha }\approx \frac{2}{\sqrt{1-{\left(\frac{184.7}{400}\right)}^2}}\cdot \frac{1}{400}\cdot 0.1\approx 0.00056=1^{\prime }55^{\prime \prime }.& \end{array}$