Advanced Mathematics (Seventh Edition) Tongji University General exercises two person solution

Advanced mathematics （ The seventh edition ） Tongji University Total exercise 2

 

$\begin{array}{rlr}& 1.stay“to\; the\; full”“necessary”and“It\; is\; necessary\; and\; sufficient”Choose\; the\; correct\; one\; of\; the\; three\; and\; fill\; in\; the\; space\; below\; ：& \end{array}$

$\begin{array}{rlr}& (1)f(x)At\; point{x}_{0}Derivable\; yesf(x)At\; point{x}_{0}Successive\_\_\_\_\_\_Conditions\; .f(x)At\; point{x}_{0}Continuous\; isf(x)At\; point{x}_{0}To\; guide\_\_\_\_\_\_Conditions\; ;& \\ & & \\ & (2)f(x)At\; point{x}_{0}The\; left\; derivative\; of{f}_{-}^{\prime }(x_0)And\; right\; derivative{f}_{+}^{\prime }(x_0)All\; exist\; and\; are\; equalf(x)At\; point{x}_{0}To\; guide\_\_\_\_\_\_Conditions\; ;& \\ & & \\ & (3)f(x)At\; point{x}_{0}Derivable\; yesf(x)At\; point{x}_{0}Differentiable\_\_\_\_\_\_Conditions\; .& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)to\; the\; full\; ,\; necessary& \\ & & \\ & (2)It\; is\; necessary\; and\; sufficient& \\ & & \\ & (3)It\; is\; necessary\; and\; sufficient& \end{array}$

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$\begin{array}{rlr}& 2.set\; upf(x)=x(x+1)(x+2)\cdot \cdot \cdot (x+n)(n\ge 2),\; be{f}^{\prime }(0)=\_\_\_\_\_\_& \end{array}$

Explain ：

$\begin{array}{rlr}& f^{\prime }(0)=\underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0}{\lim }[(x+1)(x+2)\cdot \cdot \cdot (x+n)]=n!& \end{array}$

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$\begin{array}{rlr}& 3.Four\; conclusions\; are\; given\; in\; the\; following\; questions\; ,\; Choose\; a\; correct\; conclusion\; from\; it\; ：& \end{array}$

$\begin{array}{rlr}& set\; upf(x)stayx=aThere\; is\; a\; definition\; in\; a\; neighborhood\; of\; ,\; bef(x)stayx=aA\; sufficient\; condition\; for\; local\; differentiability\; is()& \\ & & \\ & (A)\underset{h\to +\infty }{\lim }h\left[f\left(a+\frac{1}{h}\right)-f(a)\right]There\; is& \\ & & \\ & (B)\underset{h\to 0}{\lim }\frac{f(a+2h)-f(a+h)}{h}There\; is& \\ & & \\ & (C)\underset{h\to 0}{\lim }\frac{f(a+h)-f(a-h)}{2h}There\; is& \\ & & \\ & (D)\underset{h\to 0}{\lim }\frac{f(a)-f(a-h)}{h}There\; is\; .& \end{array}$

Explain ：

$\begin{array}{rlr}& ACan\; only\; know{f}_{+}^{\prime }(a)There\; is.& \\ & & \\ & B,\; takef(x)=\{\begin{array}{l}1,x\ne 0,\\ \\ 0,x=0.\end{array}f(x)stayx=0You\; can\text{'}t\; lead.& \\ & & \\ & C,\; takef(x)=|x|,f(x)stayx=0You\; can\text{'}t\; lead.& \\ & & \\ & Dcorrect\; .& \end{array}$

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$\begin{array}{rlr}& 4.There\; is\; a\; thin\; rod\; ,\; Take\; one\; end\; of\; the\; bar\; as\; the\; origin\; ,\; The\; coordinates\; of\; any\; point\; on\; the\; bar\; arex,\; So\; it\; is\; distributed\; in\; the\; interval[0,x]Top\; thin\; bar& \\ & & \\ & qualitymAndxThere\; is\; a\; functional\; relationshipm=m(x).\; How\; to\; determine\; the\; thin\; bar\; at\; the\; point{x}_{0}Linear\; density\; at\; （\; For\; uniform\; thin\; rods\; ,& \\ & & \\ & The\; mass\; of\; a\; thin\; rod\; per\; unit\; length\; is\; called\; its\; linear\; density\; ）？& \end{array}$

Explain ：

$\begin{array}{rlr}& In\; the\; interval[x_0,x_0+\Delta x]The\; average\; linear\; density\; on\; is\overline{\rho }=\frac{\Delta m}{\Delta x}=\frac{m(x_0+\Delta x)-m(x_0)}{\Delta x}& \\ & & \\ & At\; point{x}_{0}The\; linear\; density\; at\; is\rho (x_0)=\underset{\Delta x\to 0}{\lim }\frac{m(x_0+\Delta x)-m(x_0)}{\Delta x}=\frac{dm}{dx}{|}_{x=x_0}& \end{array}$

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$\begin{array}{rlr}& 5.According\; to\; the\; definition\; of\; derivative\; ,\; seekf(x)=\frac{1}{x}The\; derivative\; of\; .& \end{array}$

Explain ：

$\begin{array}{rlr}& According\; to\; the\; definition\; of\; derivative\; ,\; Whenx\ne 0when\; ,{\left(\frac{1}{x}\right)}^{\prime }=\underset{\Delta x\to 0}{\lim }\frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}=\underset{\Delta x\to 0}{\lim }\frac{-1}{x(x+\Delta x)}=-\frac{1}{x^2}& \end{array}$

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$\begin{array}{rlr}& 6.Find\; the\; following\; functionf(x)Of{f}_{-}^{\prime }(0)And{f}_{+}^{\prime }(0),\; also{f}^{\prime }(0)Whether\; there\; is\; ：& \end{array}$

$\begin{array}{rlr}& (1)f(x)=\{\begin{array}{l}sinx,x<0,\\ \\ ln(1+x),x\ge 0;\end{array}& \\ & & \\ & (2)f(x)=\{\begin{array}{l}\frac{x}{1+e^{\frac{1}{x}}},x\ne 0,\\ \\ 0,x=0.\end{array}& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)f_{-}^{\prime }(0)=\underset{x\to 0^{-}}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0^{-}}{\lim }\frac{sinx}{x}=1,f_{+}^{\prime }(0)=\underset{x\to 0^{+}}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0^{+}}{\lim }\frac{ln(1+x)}{x}=1.& \\ & & \\ & because{f}_{-}^{\prime }(0)=f_{+}^{\prime }(0)=1,\; therefore{f}^{\prime }(0)There\; is\; .& \\ & & \\ & (2)f_{-}^{\prime }(0)=\underset{x\to 0^{-}}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0^{-}}{\lim }\frac{\frac{x}{1+e^{\frac{1}{x}}}-0}{x}=\underset{x\to 0^{-}}{\lim }\frac{1}{1+e^{\frac{1}{x}}}=1,& \\ & & \\ & f_{+}^{\prime }(0)=\underset{x\to 0^{+}}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0^{+}}{\lim }\frac{\frac{x}{1+e^{\frac{1}{x}}}-0}{x}=\underset{x\to 0^{-}}{\lim }\frac{1}{1+e^{\frac{1}{x}}}=0.& \\ & & \\ & because{f}_{-}^{\prime }(0)\ne f_{+}^{\prime }(0),\; therefore{f}^{\prime }(0)non-existent\; .& \end{array}$

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$\begin{array}{rlr}& 7.Discuss\; functionsf(x)=\{\begin{array}{l}xsin\frac{1}{x},x\ne 0,\\ \\ 0,x=0.\end{array}stayx=0Continuity\; and\; derivability\; of\; .& \end{array}$

Explain ：

$\begin{array}{rlr}& \underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }xsin\frac{1}{x}=0=f(0),\; thereforef(x)stayx=0Continuous.& \\ & & \\ & f^{\prime }(0)=\underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0}{\lim }\frac{xsin\frac{1}{x}}{x}=\underset{x\to 0}{\lim }sin\frac{1}{x},\; The\; limit\; does\; not\; exist\; ,\; thereforef(x)stayx=0You\; can\text{'}t\; lead.& \end{array}$

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$\begin{array}{rlr}& 8.Find\; the\; derivative\; of\; the\; following\; function\; ：& \end{array}$

$\begin{array}{rlr}& (1)y=arcsin(sinx);(2)y=arctan\frac{1+x}{1-x};& \\ & & \\ & (3)y=lntan\frac{x}{2}-cosx\cdot lntanx;(4)y=ln(e^x+\sqrt{1+e^{2x}});& \\ & & \\ & (5)y=x^{\frac{1}{x}}(x>0).& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)y^{\prime }=\frac{1}{\sqrt{1-si{n}^{2}x}}\cdot cosx=\frac{cosx}{|cosx|}& \\ & & \\ & (2)y^{\prime }=\frac{1}{1+{\left(\frac{1+x}{1-x}\right)}^2}\cdot {\left(\frac{1+x}{1-x}\right)}^{\prime }=\frac{(1-x{)}^2}{2+2x^2}\cdot \frac{2}{(1-x{)}^2}=\frac{1}{1+x^2}& \\ & & \\ & (3)y^{\prime }=(lntan\frac{x}{2}{)}^{\prime }-(cosx\cdot lntanx{)}^{\prime }=\frac{1}{tan\frac{x}{2}}\cdot se{c}^2\frac{x}{2}\cdot \frac{1}{2}-(-sinx\cdot ln\tan x+cosx\cdot \frac{1}{tanx}\cdot se{c}^{2}x)=& \\ & & \\ & \frac{1}{sinx}+sinx\cdot ln\tan x-\frac{1}{sinx}=sinx\cdot lntanx.& \\ & & \\ & (4)y^{\prime }=\frac{1}{e^x+\sqrt{1+e^{2x}}}\cdot (e^x+\sqrt{1+e^{2x}}{)}^{\prime }=\frac{1}{e^x+\sqrt{1+e^{2x}}}\cdot (e^x+\frac{1}{2\sqrt{1+e^{2x}}}\cdot e^{2x}\cdot 2)=& \\ & & \\ & \frac{1}{e^x+\sqrt{1+e^{2x}}}\cdot \frac{e^x(e^x+\sqrt{1+e^{2x}})}{\sqrt{1+e^{2x}}}=\frac{e^x}{\sqrt{1+e^{2x}}}& \\ & & \\ & (5)lny=\frac{lnx}{x},\; Seek\; guidance\; on\; both\; sides\; ,\frac{1}{y}{y}^{\prime }=\frac{1-lnx}{x^2},y^{\prime }=x^{\frac{1}{x}-2}(1-lnx)& \end{array}$

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$\begin{array}{rlr}& 9.Find\; the\; derivative\; of\; the\; following\; function\; ：& \end{array}$

$\begin{array}{rlr}& (1)y=co{s}^{2}x\cdot lnx;(2)y=\frac{x}{\sqrt{1-x^2}}.& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)y^{\prime }=(co{s}^{2}x{)}^{\prime }lnx+co{s}^{2}x(lnx{)}^{\prime }=-sin2x\cdot \ln x+\frac{co{s}^{2}x}{x}& \\ & & \\ & y^{\prime \prime }=(-sin2x\cdot lnx{)}^{\prime }+{\left(\frac{co{s}^{2}x}{x}\right)}^{\prime }=-2cos2x\cdot lnx-\frac{sin2x}{x}-\frac{xsin2x+co{s}^{2}x}{x^2}=& \\ & & \\ & -2cos2x\cdot lnx-\frac{2sin2x}{x}-\frac{co{s}^{2}x}{x^2}& \\ & & \\ & (2)y^{\prime }=\frac{\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}}{1-x^2}=\frac{1}{\sqrt{(1-x^2{)}^3}}& \\ & & \\ & y^{\prime \prime }=\frac{-[(1-x^2{)}^{\frac{3}{2}}{]}^{\prime }}{(1-x^2{)}^3}=\frac{3x\sqrt{1-x^2}}{(1-x^2{)}^3}=\frac{3x}{(1-x^2{)}^{\frac{5}{2}}}& \end{array}$

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$\begin{array}{rlr}& 10.Find\; thenDerivative,\; ：& \end{array}$

$\begin{array}{rlr}& (1)y=\sqrt[m]{1+x};(2)y=\frac{1-x}{1+x}.& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)y^{\prime }=\frac{1}{m}(1+x{)}^{\frac{1}{m}-1},y^{\prime \prime }=\frac{1}{m}\left(\frac{1}{m}-1\right)(1+x{)}^{\frac{1}{m}-2},\cdot \cdot \cdot ,y^{(n)}=\frac{1}{m}\left(\frac{1}{m}-1\right)\cdot \cdot \cdot \left(\frac{1}{m}-n+1\right)(1+x{)}^{\frac{1}{m}-n}& \\ & & \\ & (2)because{\left(\frac{1}{1+x}\right)}^{(n)}=\frac{(-1{)}^{n}n!}{(1+x{)}^{n+1}},\; therefore{y}^{(n)}={\left(\frac{1-x}{1+x}\right)}^{(n)}={\left(-1+\frac{2}{x+1}\right)}^{(n)}=\frac{2\cdot (-1{)}^{n}n!}{(1+x{)}^{n+1}}& \end{array}$

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$\begin{array}{rlr}& 11.Let\text{'}s\; set\; the\; functiony=y(x)By\; the\; equation{e}^y+xy=eDetermined\; ,\; seek{y}^{\prime \prime }(0).& \end{array}$

Explain ：

$\begin{array}{rlr}& The\; two\; sides\; of\; the\; equation\; are\; rightxDerivation\; ,\; have\; to{e}^y{y}^{\prime }+y+x{y}^{\prime }=0.\; becausex=0,\; Put\; it\; into\; the\; equation{e}^y+xy=e,\; have\; toy=1,& \\ & & \\ & thenx=0,y=1Put\; it\; into\; the\; equation{e}^y{y}^{\prime }+y+x{y}^{\prime }=0,\; have\; toy‘=-\frac{1}{e},\; For\; the\; equation{e}^y{y}^{\prime }+y+x{y}^{\prime }=0Seek\; guidance\; on\; both\; sides\; ,& \\ & & \\ & have\; to{e}^y{y}^{\prime 2}+e^y{y}^{\prime \prime }+y^{\prime }+y^{\prime }+x{y}^{\prime \prime }=0.\; takex=0,y=1,y’=-\frac{1}{e}Put\; it\; into\; the\; equation{e}^y{y}^{\prime 2}+e^y{y}^{\prime \prime }+y^{\prime }+y^{\prime }+x{y}^{\prime \prime }=0,& \\ & & \\ & have\; to{y}^{\prime \prime }(0)=\frac{1}{e^2}& \\ & & \\ & & \end{array}$

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$\begin{array}{rlr}& 12.Find\; the\; first\; derivative\; of\; the\; following\; function\; determined\; by\; the\; parametric\; equation\frac{dy}{dx}And\; the\; second\; derivative\frac{d^{2}y}{d{x}^2}:& \end{array}$

$\begin{array}{rlr}& (1)\{\begin{array}{l}x=aco{s}^3\theta ,\\ \\ y=asi{n}^3\theta ;\end{array}(2)\{\begin{array}{l}x=ln\sqrt{1+t^2},\\ \\ y=arctant.\end{array}& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3asi{n}^2\theta cos\theta }{3aco{s}^2\theta (-sin\theta )}=-tan\theta & \\ & & \\ & \frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{d\theta }\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta }}=\frac{-se{c}^2\theta }{-3aco{s}^2\theta sin\theta }=\frac{1}{3a}se{c}^4\theta csc\theta & \\ & & \\ & (2)\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{1}{1+t^2}}{\frac{t}{1+t^2}}=\frac{1}{t}& \\ & & \\ & \frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{-\frac{1}{t^2}}{\frac{t}{1+t^2}}=-\frac{1+t^2}{t^3}& \end{array}$

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$\begin{array}{rlr}& 13.Find\; the\; curve\{\begin{array}{l}x=2e^t,\\ \\ y=e^{-t}\end{array}stayt=0Tangent\; equation\; and\; normal\; equation\; at\; corresponding\; points\; .& \end{array}$

Explain ：

$\begin{array}{rlr}& \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-e^{-t}}{2e^t}=-\frac{1}{2e^{2t}},\; Whent=0when\; ,y^{\prime }=-\frac{1}{2},t=0The\; corresponding\; point\; is(2,1),\; So\; the\; curve\; is\; at\; point(2,1)The\; tangent\; equation\; at\; is& \\ & & \\ & y-1=-\frac{1}{2}(x-2),\; namelyx+2y-4=0.\; The\; normal\; equation\; isy-1=2(x-2),\; namely2x-y-3=0.& \end{array}$

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$\begin{array}{rlr}& 14.It\; is\; known\; thatf(x)Yes,\; the\; period\; is5The\; continuous\; function\; of\; ,\; It\text{'}s\; inx=0Satisfy\; the\; relation\; in\; a\; neighborhood\; off(1+sinx)-3f(1-sinx)=8x+o(x),& \\ & & \\ & Andf(x)stayx=1Place\; can\; lead\; ,\; Find\; the\; curvey=f(x)At\; point(6,f(6))The\; tangent\; equation\; at\; .& \end{array}$

Explain ：

$\begin{array}{rlr}& becausef(x)continuity\; ,\; The\; two\; ends\; of\; the\; relationship\; arex\to 0when\; ,\; Take\; the\; limitf(1)-3f(1)=0,f(1)=0.& \\ & & \\ & because\underset{x\to 0}{\lim }\frac{f(1+sinx)-3f(1-sinx)}{x}=8,& \\ & & \\ & and\underset{x\to 0}{\lim }\frac{f(1+sinx)-3f(1-sinx)}{x}=\underset{x\to 0}{\lim }\frac{f(1+sinx)-3f(1-sinx)}{sinx}\cdot \underset{x\to 0}{\lim }\frac{sinx}{x}& \\ & & \\ & Maket=sinx,\underset{t\to 0}{\lim }\frac{f(1+t)-3f(1-t)}{t}=\underset{t\to 0}{\lim }\frac{f(1+t)-f(1)}{t}+3\underset{t\to 0}{\lim }\frac{f(1-t)-f(1)}{-t}=4f^{\prime }(1),\; therefore{f}^{\prime }(1)=2.& \\ & & \\ & becausef(x+5)=f(x),\; thereforef(6)=f(1)=0.f^{\prime }(6)=\underset{x\to 0}{\lim }\frac{f(6+x)-f(6)}{x}=\underset{x\to 0}{\lim }\frac{f(1+x)-f(1)}{x}=f^{\prime }(1)=2.& \\ & & \\ & curvey=f(x)At\; point(6,f(6))The\; tangent\; equation\; at\; isy-0=2(x-6),\; namely2x-y-12=0.& \end{array}$

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$\begin{array}{rlr}& 15.When\; heightHWhen\; the\; plane\; flying\; horizontally\; begins\; to\; descend\; towards\; the\; airport\; runway\; ,\; Pictured2-16The\; horizontal\; ground\; distance\; shown\; from\; the\; aircraft\; to\; the\; airport\; isL.& \\ & & \\ & Suppose\; the\; path\; of\; the\; aircraft\; descending\; is\; a\; cubic\; functiony=a{x}^3+b{x}^2+cx+dThe\; graphic\; ,\; amongy{|}_{x=-L}=H,y{|}_{x=0}=0.& \\ & & \\ & Try\; to\; determine\; the\; landing\; path\; of\; the\; aircraft\; .& \end{array}$

Explain ：

$\begin{array}{rlr}& becausey{|}_{x=0}=0\Rightarrow d=0,y{|}_{x=-L}=H\Rightarrow -a{L}^3+b{L}^2-cL=H.& \\ & & \\ & To\; make\; the\; plane\; land\; smoothly\; ,\; Need\; to\; meet{y}^{\prime }{|}_{x=0}=0\Rightarrow c=0,y^{\prime }{|}_{x=-L}=0\Rightarrow 3a{L}^2-2bL=0.& \\ & & \\ & To\; solve\; the\; equation\; ,a=\frac{2H}{L^3},b=\frac{3H}{L^2},\; So\; the\; landing\; path\; of\; the\; plane\; isy=H\left[2{\left(\frac{x}{L}\right)}^3+3{\left(\frac{x}{L}\right)}^2\right]& \end{array}$

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$\begin{array}{rlr}& 16.Jia\; Chuanyi6km/hDrive\; east\; at\; a\; rate\; of\; ,\; B\; ship\; with8km/hDriving\; southward\; at\; a\; rate\; of\; .\; At\; twelve\; o\text{'}clock\; sharp\; at\; noon\; ,& \\ & & \\ & Ship\; B\; is\; located\; to\; the\; north\; of\; ship\; a16kmIt\text{'}s\; about\; .\; Ask\; the\; speed\; at\; which\; the\; two\; ships\; leave\; at\; exactly\; 1:00\; p.m\; ？& \end{array}$

Explain ：

$\begin{array}{rlr}& Set\; from\; noon12It\; starts\; at\; o\text{'}clock\; sharp\; ,\; aftertHours\; ,\; The\; distance\; between\; the\; two\; ships\; iss=\sqrt{(16-8t{)}^2+(6t{)}^2},\; ratev=\frac{ds}{dt}=\frac{2(16-8t)\cdot (-8)+72t}{2\sqrt{(16-8t{)}^2+(6t{)}^2}}& \\ & & \\ & Whent=1when\; ,\; The\; rate\; at\; which\; two\; ships\; separatev=\frac{-128+72}{20}=-2.8km/h& \end{array}$

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$\begin{array}{rlr}& 17.Use\; the\; differential\; of\; the\; function\; to\; replace\; the\; increment\; of\; the\; function\; to\; find\sqrt[3]{1.02}Approximate\; value\; .& \end{array}$

Explain ：

$\begin{array}{rlr}& because\sqrt[3]{1+x}\approx 1+\frac{1}{3}x,\; takex=0.02,\; be\sqrt[3]{1.02}\approx 1+\frac{1}{3}\times (0.02)=1.007& \end{array}$

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$\begin{array}{rlr}& 18.The\; vibration\; period\; of\; a\; simple\; pendulum\; is\; knownT=2\pi \sqrt{\frac{l}{g}},\; amongg=980cm/s^2,lFor\; pendulum\; length\; （\; Unit\; iscm）.& \\ & & \\ & Let\; the\; length\; of\; the\; original\; pendulum\; be20cm,\; To\; make\; the\; cycleTincrease0.05s,\; The\; length\; of\; the\; pendulum\; needs\; to\; be\; lengthened\; ？& \end{array}$

Explain ：

$\begin{array}{rlr}& from\Delta T\approx dT=\frac{\pi }{\sqrt{gl}}\Delta l,\; have\; to\Delta l=\frac{\sqrt{gl}}{\pi }dT\approx \frac{\sqrt{gl}}{\pi }\Delta T,\; Whenl=20when\; ,\Delta l\approx \frac{\sqrt{980\times 20}}{3.14}\times 0.05\approx 2.23cm,& \\ & & \\ & The\; length\; of\; the\; pendulum\; needs\; to\; be\; lengthened2.23cm.& \end{array}$