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Assume a hanging cable has the shape $15\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}15\right)$ for $\mathrm{-20}\le x\le 20.$ Determine the length of the cable (in feet).
$52.95\phantom{\rule{0.2em}{0ex}}\text{ft}$
[T] Find expressions for $\text{cosh}\phantom{\rule{0.2em}{0ex}}x+\text{sinh}\phantom{\rule{0.2em}{0ex}}x$ and $\text{cosh}\phantom{\rule{0.2em}{0ex}}x-\text{sinh}\phantom{\rule{0.2em}{0ex}}x.$ Use a calculator to graph these functions and ensure your expression is correct.
${e}^{x}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{e}^{\text{\u2212}x}$
From the definitions of $\text{cosh}\left(x\right)$ and $\text{sinh}\left(x\right),$ find their antiderivatives.
Show that $\text{cosh}\left(x\right)$ and $\text{sinh}\left(x\right)$ satisfy $y\text{\u2033}=y.$
Answers may vary
Use the quotient rule to verify that $\text{tanh}\left(x\right)\prime ={\text{sech}}^{2}\left(x\right).$
Derive ${\text{cosh}}^{2}\left(x\right)+{\text{sinh}}^{2}\left(x\right)=\text{cosh}\left(2x\right)$ from the definition.
Answers may vary
Take the derivative of the previous expression to find an expression for $\text{sinh}\left(2x\right).$
Prove $\text{sinh}\left(x+y\right)=\text{sinh}\left(x\right)\text{cosh}\left(y\right)+\text{cosh}\left(x\right)\text{sinh}\left(y\right)$ by changing the expression to exponentials.
Answers may vary
Take the derivative of the previous expression to find an expression for $\text{cosh}\left(x+y\right).$
For the following exercises, find the derivatives of the given functions and graph along with the function to ensure your answer is correct.
[T] $\text{cosh}\left(3x+1\right)$
$3\phantom{\rule{0.2em}{0ex}}\text{sinh}\left(3x+1\right)$
[T] $\text{sinh}\left({x}^{2}\right)$
[T] $\frac{1}{\text{cosh}\left(x\right)}$
$\text{\u2212}\text{tanh}\left(x\right)\text{sech}\left(x\right)$
[T] $\text{sinh}\left(\text{ln}\left(x\right)\right)$
[T] ${\text{cosh}}^{2}\left(x\right)+{\text{sinh}}^{2}\left(x\right)$
$4\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\right)\text{sinh}\left(x\right)$
[T] ${\text{cosh}}^{2}\left(x\right)-{\text{sinh}}^{2}\left(x\right)$
[T] $\text{tanh}\left(\sqrt{{x}^{2}+1}\right)$
$\frac{x\phantom{\rule{0.2em}{0ex}}{\text{sech}}^{2}\left(\sqrt{{x}^{2}+1}\right)}{\sqrt{{x}^{2}+1}}$
[T] $\frac{1+\text{tanh}\left(x\right)}{1-\text{tanh}\left(x\right)}$
[T] ${\text{sinh}}^{6}\left(x\right)$
$6\phantom{\rule{0.2em}{0ex}}{\text{sinh}}^{5}\left(x\right)\text{cosh}\left(x\right)$
[T] $\text{ln}\left(\text{sech}\left(x\right)+\text{tanh}\left(x\right)\right)$
For the following exercises, find the antiderivatives for the given functions.
$\text{cosh}\left(2x+1\right)$
$\frac{1}{2}\text{sinh}\left(2x+1\right)+C$
$\text{tanh}\left(3x+2\right)$
$x\phantom{\rule{0.2em}{0ex}}\text{cosh}\left({x}^{2}\right)$
$\frac{1}{2}{\text{sinh}}^{2}\left({x}^{2}\right)+C$
$3{x}^{3}\text{tanh}\left({x}^{4}\right)$
${\text{cosh}}^{2}\left(x\right)\text{sinh}\left(x\right)$
$\frac{1}{3}{\text{cosh}}^{3}\left(x\right)+C$
${\text{tanh}}^{2}\left(x\right){\text{sech}}^{2}\left(x\right)$
$\frac{\text{sinh}\left(x\right)}{1+\text{cosh}\left(x\right)}$
$\text{ln}\left(1+\text{cosh}(x)\right)+C$
$\text{coth}\left(x\right)$
$\text{cosh}\left(x\right)+\text{sinh}\left(x\right)$
$\text{cosh}\left(x\right)+\text{sinh}\left(x\right)+C$
${\left(\text{cosh}\left(x\right)+\text{sinh}\left(x\right)\right)}^{n}$
For the following exercises, find the derivatives for the functions.
${\text{tanh}}^{\mathrm{-1}}\left(4x\right)$
$\frac{4}{1-16{x}^{2}}$
${\text{sinh}}^{\mathrm{-1}}\left({x}^{2}\right)$
${\text{sinh}}^{\mathrm{-1}}\left(\text{cosh}\left(x\right)\right)$
$\frac{\text{sinh}\left(x\right)}{\sqrt{{\text{cosh}}^{2}\left(x\right)+1}}$
${\text{cosh}}^{\mathrm{-1}}\left({x}^{3}\right)$
${\text{tanh}}^{\mathrm{-1}}\left(\text{cos}\left(x\right)\right)$
$\text{\u2212}\text{csc}\left(x\right)$
${e}^{{\text{sinh}}^{\mathrm{-1}}\left(x\right)}$
$\text{ln}\left({\text{tanh}}^{\mathrm{-1}}\left(x\right)\right)$
$-\frac{1}{\left({x}^{2}-1\right){\text{tanh}}^{\mathrm{-1}}\left(x\right)}$
For the following exercises, find the antiderivatives for the functions.
$\int \frac{dx}{4-{x}^{2}}$
$\int \frac{dx}{{a}^{2}-{x}^{2}}$
$\frac{1}{a}{\text{tanh}}^{\mathrm{-1}}\left(\frac{x}{a}\right)+C$
$\int \frac{dx}{\sqrt{{x}^{2}+1}}$
$\int \frac{x\phantom{\rule{0.2em}{0ex}}dx}{\sqrt{{x}^{2}+1}}$
$\sqrt{{x}^{2}+1}+C$
$\int -\frac{dx}{x\sqrt{1-{x}^{2}}}$
$\int \frac{{e}^{x}}{\sqrt{{e}^{2x}-1}}$
${\text{cosh}}^{\mathrm{-1}}\left({e}^{x}\right)+C$
$\int -\frac{2x}{{x}^{4}-1}$
For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation $dv\text{/}dt=g-{v}^{2}.$
Show that $v(t)=\sqrt{g}\phantom{\rule{0.2em}{0ex}}\text{tanh}\left(\sqrt{gt}\right)$ satisfies this equation.
Answers may vary
Derive the previous expression for $v(t)$ by integrating $\frac{dv}{g-{v}^{2}}=dt.$
[T] Estimate how far a body has fallen in $12$ seconds by finding the area underneath the curve of $v(t).$
$37.30$
For the following exercises, use this scenario: A cable hanging under its own weight has a slope $S=dy\text{/}dx$ that satisfies $dS\text{/}dx=c\sqrt{1+{S}^{2}}.$ The constant $c$ is the ratio of cable density to tension.
Show that $S=\text{sinh}(cx)$ satisfies this equation.
Integrate $dy\text{/}dx=\text{sinh}(cx)$ to find the cable height $y(x)$ if $y(0)=1\text{/}c.$
$y=\frac{1}{c}\text{cosh}(cx)$
Sketch the cable and determine how far down it sags at $x=0.$
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