1.7E: Exercises
- Page ID
- 18544
This page is a draft and is under active development.
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Calculus of the Hyperbolic Functions
Exercise \(\PageIndex{1}\)
Find expressions for \(\displaystyle oshx+sinhx\) and \(\displaystyle coshx−sinhx.\) Use a calculator to graph these functions and ensure your expression is correct.
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\(\displaystyle e^x\) and \(\displaystyle e^{−x}\)
Exercise \(\PageIndex{2}\)
From the definitions of \(\displaystyle cosh(x)\) and \(\displaystyle sinh(x)\), find their antiderivatives.
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Exercise \(\PageIndex{3}\)
Show that \(\displaystyle cosh(x)\) and \(\displaystyle sinh(x)\) satisfy \(\displaystyle y''=y\).
- Answer
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Exercise \(\PageIndex{4}\)
Use the quotient rule to verify that \(\displaystyle tanh(x)'=sech^2(x).\)
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Exercise \(\PageIndex{5}\)
Derive \(\displaystyle cosh^2(x)+sinh^2(x)=cosh(2x)\) from the definition.
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Exercise \(\PageIndex{6}\)
Take the derivative of the previous expression to find an expression for \(\displaystyle sinh(2x)\).
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Exercise \(\PageIndex{7}\)
Prove \(\displaystyle sinh(x+y)=sinh(x)cosh(y)+cosh(x)sinh(y)\) by changing the expression to exponentials.
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Exercise \(\PageIndex{8}\)
Take the derivative of the previous expression to find an expression for \(\displaystyle cosh(x+y).\)
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For the following exercises, find the derivatives of the given functions and graph along with the function to ensure your answer is correct.
Exercise \(\PageIndex{9}\)
\(\displaystyle cosh(3x+1)\)
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\(\displaystyle 3sinh(3x+1)\)
Exercise \(\PageIndex{10}\)
\(\displaystyle sinh(x^2)\)
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Exercise \(\PageIndex{11}\)
\(\displaystyle \frac{1}{cosh(x)}\)
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\(\displaystyle −tanh(x)sech(x)\)
Exercise \(\PageIndex{12}\)
\(\displaystyle sinh(ln(x))\)
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Exercise \(\PageIndex{13}\)
\(\displaystyle cosh^2(x)+sinh^2(x)\)
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\(\displaystyle 4cosh(x)sinh(x)\)
Exercise \(\PageIndex{14}\)
\(\displaystyle cosh^2(x)−sinh^2(x)\)
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Exercise \(\PageIndex{15}\)
\(\displaystyle tanh(\sqrt{x^2+1})\)
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\(\displaystyle \frac{xsech^2(\sqrt{x^2+1})}{\sqrt{x^2+1}}\)
Exercise \(\PageIndex{16}\)
\(\displaystyle \frac{1+tanh(x)}{1−tanh(x)}\)
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Exercise \(\PageIndex{17}\)
\(\displaystyle sinh^6(x)\)
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\(\displaystyle 6sinh^5(x)cosh(x)\)
Exercise \(\PageIndex{18}\)
\(\displaystyle ln(sech(x)+tanh(x))\)
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For the following exercises, find the antiderivatives for the given functions.
Exercise \(\PageIndex{19}\)
\(\displaystyle cosh(2x+1)\)
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\(\displaystyle \frac{1}{2}sinh(2x+1)+C\)
Exercise \(\PageIndex{20}\)
\(\displaystyle tanh(3x+2)\)
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Exercise \(\PageIndex{21}\)
\(\displaystyle xcosh(x^2)\)
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\(\displaystyle \frac{1}{2}sinh^2(x^2)+C\)
Exercise \(\PageIndex{22}\)
\(\displaystyle 3x^3tanh(x^4)\)
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Exercise \(\PageIndex{23}\)
\(\displaystyle cosh^2(x)sinh(x)\)
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\(\displaystyle \frac{1}{3}cosh^3(x)+C\)
Exercise \(\PageIndex{24}\)
\(\displaystyle tanh^2(x)sech^2(x)\)
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Exercise \(\PageIndex{25}\)
\(\displaystyle \frac{sinh(x)}{1+cosh(x)}\)
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\(\displaystyle ln(1+cosh(x))+C\)
Exercise \(\PageIndex{26}\)
\(\displaystyle coth(x)\)
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Exercise \(\PageIndex{27}\)
\(\displaystyle cosh(x)+sinh(x)\)
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\(\displaystyle cosh(x)+sinh(x)+C\)
Exercise \(\PageIndex{28}\)
\(\displaystyle (cosh(x)+sinh(x))^n\)
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For the following exercises, find the derivatives for the functions.
Exercise \(\PageIndex{29}\)
\(\displaystyle tanh^{−1}(4x)\)
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\(\displaystyle \frac{4}{1−16x^2}\)
Exercise \(\PageIndex{30}\)
\(\displaystyle sinh^{−1}(x^2)\)
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Exercise \(\PageIndex{31}\)
\(\displaystyle sinh^{−1}(cosh(x))\)
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\(\displaystyle \frac{sinh(x)}{\sqrt{cosh^2(x)+1}}\)
Exercise \(\PageIndex{32}\)
\(\displaystyle cosh^{−1}(x^3)\)
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Exercise \(\PageIndex{33}\)
\(\displaystyle tanh^{−1}(cos(x))\)
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\(\displaystyle −csc(x)\)
Exercise \(\PageIndex{34}\)
\(\displaystyle e^{sinh^{−1}(x)}\)
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Exercise \(\PageIndex{35}\)
\(\displaystyle ln(tanh^{−1}(x))\)
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\(\displaystyle −\frac{1}{(x^2−1)tanh^{−1}(x)}\)
For the following exercises, find the antiderivatives for the functions.
Exercise \(\PageIndex{36}\)
\(\displaystyle ∫\frac{dx}{4−x^2}\)
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Exercise \(\PageIndex{37}\)
\(\displaystyle ∫\frac{dx}{a^2−x^2}\)
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\(\displaystyle \frac{1}{a}tanh^{−1}(\frac{x}{a})+C\)
Exercise \(\PageIndex{38}\)
\(\displaystyle ∫\frac{dx}{\sqrt{x^2+1}}\)
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Exercise \(\PageIndex{39}\)
\(\displaystyle ∫\frac{xdx}{\sqrt{x^2+1}}\)
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\(\displaystyle \sqrt{x^2+1}+C\)
Exercise \(\PageIndex{40}\)
\(\displaystyle ∫−\frac{dx}{x\sqrt{1−x^2}}\)
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Exercise \(\PageIndex{41}\)
\(\displaystyle ∫\frac{e^x}{\sqrt{e^{2x}−1}}\)
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\(\displaystyle cosh^{−1}(e^x)+C\)
Exercise \(\PageIndex{42}\)
\(\displaystyle ∫−\frac{2x}{x^4−1}\)
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For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation \(\displaystyle dv/dt=g−v^2\).
Exercise \(\PageIndex{43}\)
Show that \(\displaystyle v(t)=\sqrt{g}tanh(\sqrt{gt})\) satisfies this equation.
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Exercise \(\PageIndex{44}\)
Derive the previous expression for \(\displaystyle v(t)\) by integrating \(\displaystyle \frac{dv}{g−v^2}=dt\).
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Exercise \(\PageIndex{45}\)
Estimate how far a body has fallen in \(\displaystyle 12\)seconds by finding the area underneath the curve of \(\displaystyle v(t)\).
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\(\displaystyle 37.30\)
For the following exercises, use this scenario: A cable hanging under its own weight has a slope \(\displaystyle S=dy/dx\) that satisfies \(\displaystyle dS/dx=c\sqrt{1+S^2}\). The constant \(\displaystyle c\) is the ratio of cable density to tension.
Exercise \(\PageIndex{46}\)
Show that \(\displaystyle S=sinh(cx)\) satisfies this equation.
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Exercise \(\PageIndex{47}\)
Integrate \(\displaystyle dy/dx=sinh(cx)\) to find the cable height \(\displaystyle y(x)\) if \(\displaystyle y(0)=1/c\).
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\(\displaystyle y=\frac{1}{c}cosh(cx)\)
Exercise \(\PageIndex{48}\)
Sketch the cable and determine how far down it sags at \(\displaystyle x=0\).
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For the following exercises, solve each problem.
Exercise \(\PageIndex{49}\)
A chain hangs from two posts \(\displaystyle 2\)m apart to form a catenary described by the equation \(\displaystyle y=2cosh(x/2)−1\). Find the slope of the catenary at the left fence post.
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\(\displaystyle −0.521095\)
Exercise \(\PageIndex{50}\)
A chain hangs from two posts four meters apart to form a catenary described by the equation \(\displaystyle y=4cosh(x/4)−3.\) Find the total length of the catenary (arc length).
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Exercise \(\PageIndex{51}\)
A high-voltage power line is a catenary described by \(\displaystyle y=10cosh(x/10)\). Find the ratio of the area under the catenary to its arc length. What do you notice?
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\(\displaystyle 10\)
Exercise \(\PageIndex{52}\)
A telephone line is a catenary described by \(\displaystyle y=acosh(x/a).\) Find the ratio of the area under the catenary to its arc length. Does this confirm your answer for the previous question?
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Exercise \(\PageIndex{53}\)
Prove the formula for the derivative of \(\displaystyle y=sinh^{−1}(x)\) by differentiating \(\displaystyle x=sinh(y).\)
(Hint: Use hyperbolic trigonometric identities.)
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Exercise \(\PageIndex{54}\)
Prove the formula for the derivative of \(\displaystyle y=cosh^{−1}(x)\) by differentiating \(\displaystyle x=cosh(y).\)
(Hint: Use hyperbolic trigonometric identities.)
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Exercise \(\PageIndex{55}\)
Prove the formula for the derivative of \(\displaystyle y=sech^{−1}(x)\) by differentiating \(\displaystyle x=sech(y).\)
(Hint: Use hyperbolic trigonometric identities.)
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Exercise \(\PageIndex{56}\)
Prove that \(\displaystyle cosh(x)+sinh(x))^n=cosh(nx)+sinh(nx).\)
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Exercise \(\PageIndex{57}\)
Prove the expression for \(\displaystyle sinh^{−1}(x).\) Multiply \(\displaystyle x=sinh(y)=(1/2)(e^y−e^{−y})\) by \(\displaystyle 2e^y\) and solve for \(\displaystyle y\). Does your expression match the textbook?
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Exercise \(\PageIndex{58}\)
Prove the expression for \(\displaystyle cosh^{−1}(x).\) Multiply \(\displaystyle x=cosh(y)=(1/2)(e^y−e^{−y})\) by \(\displaystyle 2e^y\) and solve for \(\displaystyle y\). Does your expression match the textbook?
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