We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Guias de estudo > Calculus Volume 1

Calculus of the Hyperbolic Functions

Learning Objectives

  • Apply the formulas for derivatives and integrals of the hyperbolic functions.
  • Apply the formulas for the derivatives of the inverse hyperbolic functions and their associated integrals.
  • Describe the common applied conditions of a catenary curve.

We were introduced to hyperbolic functions in Introduction to Functions and Graphs, along with some of their basic properties. In this section, we look at differentiation and integration formulas for the hyperbolic functions and their inverses.

Derivatives and Integrals of the Hyperbolic Functions

Recall that the hyperbolic sine and hyperbolic cosine are defined as

sinhx=ex−e−x2andcoshx=ex+e−x2.\text{sinh}\phantom{\rule{0.2em}{0ex}}x=\frac{{e}^{x}-{e}^{\text{−}x}}{2}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{cosh}\phantom{\rule{0.2em}{0ex}}x=\frac{{e}^{x}+{e}^{\text{−}x}}{2}.

The other hyperbolic functions are then defined in terms of sinhx\text{sinh}\phantom{\rule{0.2em}{0ex}}x and coshx.\text{cosh}\phantom{\rule{0.2em}{0ex}}x. The graphs of the hyperbolic functions are shown in the following figure.

Graphs of the hyperbolic functions.
This figure has six graphs. The first graph labeled “a” is of the function y=sinh(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled “b” and is of the function y=cosh(x). It decreases in the second quadrant to the intercept y=1, then becomes an increasing function. The third graph labeled “c” is of the function y=tanh(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled “d” and is of the function y=coth(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled “e” and is of the function y=sech(x). It is a curve above the x-axis, increasing in the second quadrant, to the y-axis at y=1 and then decreases. The sixth graph is labeled “f” and is of the function y=csch(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.

It is easy to develop differentiation formulas for the hyperbolic functions. For example, looking at sinhx\text{sinh}\phantom{\rule{0.2em}{0ex}}x we have

ddx(sinhx)=ddx(ex−e−x2)=12[ddx(ex)−ddx(e−x)]=12[ex+e−x]=coshx.\begin{array}{cc}\hfill \frac{d}{dx}\left(\text{sinh}\phantom{\rule{0.2em}{0ex}}x\right)& =\frac{d}{dx}\left(\frac{{e}^{x}-{e}^{\text{−}x}}{2}\right)\hfill \\ & =\frac{1}{2}\left[\frac{d}{dx}\left({e}^{x}\right)-\frac{d}{dx}\left({e}^{\text{−}x}\right)\right]\hfill \\ & =\frac{1}{2}\left[{e}^{x}+{e}^{\text{−}x}\right]=\text{cosh}\phantom{\rule{0.2em}{0ex}}x.\hfill \end{array}

Similarly, (d/dx)coshx=sinhx.\left(d\text{/}dx\right)\text{cosh}\phantom{\rule{0.2em}{0ex}}x=\text{sinh}\phantom{\rule{0.2em}{0ex}}x. We summarize the differentiation formulas for the hyperbolic functions in the following table.

Derivatives of the Hyperbolic Functions
f(x)f\left(x\right)ddxf(x)\frac{d}{dx}f\left(x\right)
sinhx\text{sinh}\phantom{\rule{0.2em}{0ex}}xcoshx\text{cosh}\phantom{\rule{0.2em}{0ex}}x
coshx\text{cosh}\phantom{\rule{0.2em}{0ex}}xsinhx\text{sinh}\phantom{\rule{0.2em}{0ex}}x
tanhx\text{tanh}\phantom{\rule{0.2em}{0ex}}xsech2x{\text{sech}}^{2}\phantom{\rule{0.2em}{0ex}}x
cothx\text{coth}\phantom{\rule{0.2em}{0ex}}x−csch2x\text{−}{\text{csch}}^{2}\phantom{\rule{0.2em}{0ex}}x
sechx\text{sech}\phantom{\rule{0.2em}{0ex}}x−sechxtanhx\text{−}\text{sech}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{tanh}\phantom{\rule{0.2em}{0ex}}x
cschx\text{csch}\phantom{\rule{0.2em}{0ex}}x−cschxcothx\text{−}\text{csch}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{coth}\phantom{\rule{0.2em}{0ex}}x

Let’s take a moment to compare the derivatives of the hyperbolic functions with the derivatives of the standard trigonometric functions. There are a lot of similarities, but differences as well. For example, the derivatives of the sine functions match: (d/dx)sinx=cosx\left(d\text{/}dx\right)\text{sin}\phantom{\rule{0.2em}{0ex}}x=\text{cos}\phantom{\rule{0.2em}{0ex}}x and (d/dx)sinhx=coshx.\left(d\text{/}dx\right)\text{sinh}\phantom{\rule{0.2em}{0ex}}x=\text{cosh}\phantom{\rule{0.2em}{0ex}}x. The derivatives of the cosine functions, however, differ in sign: (d/dx)cosx=−sinx,\left(d\text{/}dx\right)\text{cos}\phantom{\rule{0.2em}{0ex}}x=\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}x, but (d/dx)coshx=sinhx.\left(d\text{/}dx\right)\text{cosh}\phantom{\rule{0.2em}{0ex}}x=\text{sinh}\phantom{\rule{0.2em}{0ex}}x. As we continue our examination of the hyperbolic functions, we must be mindful of their similarities and differences to the standard trigonometric functions.

These differentiation formulas for the hyperbolic functions lead directly to the following integral formulas.

∫sinhudu=coshu+C∫csch2udu=−cothu+C∫coshudu=sinhu+C∫sechutanhudu=−sechu+C∫sech2udu=tanhu+C∫cschucothudu=−cschu+C\begin{array}{cccccccc}\hfill \int \text{sinh}\phantom{\rule{0.2em}{0ex}}u\phantom{\rule{0.2em}{0ex}}du& =\hfill & \text{cosh}\phantom{\rule{0.2em}{0ex}}u+C\hfill & & & \hfill \int {\text{csch}}^{2}\phantom{\rule{0.2em}{0ex}}u\phantom{\rule{0.2em}{0ex}}du& =\hfill & \text{−}\text{coth}\phantom{\rule{0.2em}{0ex}}u+C\hfill \\ \hfill \int \text{cosh}\phantom{\rule{0.2em}{0ex}}u\phantom{\rule{0.2em}{0ex}}du& =\hfill & \text{sinh}\phantom{\rule{0.2em}{0ex}}u+C\hfill & & & \hfill \int \text{sech}\phantom{\rule{0.2em}{0ex}}u\phantom{\rule{0.2em}{0ex}}\text{tanh}\phantom{\rule{0.2em}{0ex}}u\phantom{\rule{0.2em}{0ex}}du& =\hfill & \text{−}\text{sech}\phantom{\rule{0.2em}{0ex}}u+C\hfill \\ \hfill \int {\text{sech}}^{2}u\phantom{\rule{0.2em}{0ex}}du& =\hfill & \text{tanh}\phantom{\rule{0.2em}{0ex}}u+C\hfill & & & \hfill \int \text{csch}\phantom{\rule{0.2em}{0ex}}u\phantom{\rule{0.2em}{0ex}}\text{coth}\phantom{\rule{0.2em}{0ex}}u\phantom{\rule{0.2em}{0ex}}du& =\hfill & \text{−}\text{csch}\phantom{\rule{0.2em}{0ex}}u+C\hfill \end{array}
Differentiating Hyperbolic Functions

Evaluate the following derivatives:

  1. ddx(sinh(x2))\frac{d}{dx}\left(\text{sinh}\left({x}^{2}\right)\right)
  2. ddx(coshx)2\frac{d}{dx}{\left(\text{cosh}\phantom{\rule{0.2em}{0ex}}x\right)}^{2}

Using the formulas in [link] and the chain rule, we get

  1. ddx(sinh(x2))=cosh(x2)⋅2x\frac{d}{dx}\left(\text{sinh}\left({x}^{2}\right)\right)=\text{cosh}\left({x}^{2}\right)·2x
  2. ddx(coshx)2=2coshxsinhx\frac{d}{dx}{\left(\text{cosh}\phantom{\rule{0.2em}{0ex}}x\right)}^{2}=2\phantom{\rule{0.2em}{0ex}}\text{cosh}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{sinh}\phantom{\rule{0.2em}{0ex}}x

Evaluate the following derivatives:

  1. ddx(tanh(x2+3x))\frac{d}{dx}\left(\text{tanh}\left({x}^{2}+3x\right)\right)
  2. ddx(1(sinhx)2)\frac{d}{dx}\left(\frac{1}{{\left(\text{sinh}\phantom{\rule{0.2em}{0ex}}x\right)}^{2}}\right)
  1. ddx(tanh(x2+3x))=(sech2(x2+3x))(2x+3)\frac{d}{dx}\left(\text{tanh}\left({x}^{2}+3x\right)\right)=\left({\text{sech}}^{2}\left({x}^{2}+3x\right)\right)\left(2x+3\right)
  2. ddx(1(sinhx)2)=ddx(sinhx)−2=−2(sinhx)−3coshx\frac{d}{dx}\left(\frac{1}{{\left(\text{sinh}\phantom{\rule{0.2em}{0ex}}x\right)}^{2}}\right)=\frac{d}{dx}{\left(\text{sinh}\phantom{\rule{0.2em}{0ex}}x\right)}^{-2}=-2{\left(\text{sinh}\phantom{\rule{0.2em}{0ex}}x\right)}^{-3}\text{cosh}\phantom{\rule{0.2em}{0ex}}x
Hint

Use the formulas in [link] and apply the chain rule as necessary.

Integrals Involving Hyperbolic Functions

Evaluate the following integrals:

  1. ∫xcosh(x2)dx\int x\phantom{\rule{0.2em}{0ex}}\text{cosh}\left({x}^{2}\right)dx
  2. ∫tanhxdx\int \text{tanh}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx

We can use u-substitution in both cases.

  1. Let u=x2.u={x}^{2}. Then, du=2xdxdu=2x\phantom{\rule{0.2em}{0ex}}dx and

    ∫xcosh(x2)dx=∫12coshudu=12sinhu+C=12sinh(x2)+C.\int x\phantom{\rule{0.2em}{0ex}}\text{cosh}\left({x}^{2}\right)dx=\int \frac{1}{2}\text{cosh}\phantom{\rule{0.2em}{0ex}}u\phantom{\rule{0.2em}{0ex}}du=\frac{1}{2}\text{sinh}\phantom{\rule{0.2em}{0ex}}u+C=\frac{1}{2}\text{sinh}\left({x}^{2}\right)+C.
  2. Let u=coshx.u=\text{cosh}\phantom{\rule{0.2em}{0ex}}x. Then, du=sinhxdxdu=\text{sinh}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx and

    ∫tanhxdx=∫sinhxcoshxdx=∫1udu=ln∣u∣+C=ln∣coshx∣+C.\int \text{tanh}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx=\int \frac{\text{sinh}\phantom{\rule{0.2em}{0ex}}x}{\text{cosh}\phantom{\rule{0.2em}{0ex}}x}dx=\int \frac{1}{u}du=\text{ln}|u|+C=\text{ln}|\text{cosh}\phantom{\rule{0.2em}{0ex}}x|+C.

    Note that coshx>0\text{cosh}\phantom{\rule{0.2em}{0ex}}x>0 for all x,x, so we can eliminate the absolute value signs and obtain

    ∫tanhxdx=ln(coshx)+C.\int \text{tanh}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx=\text{ln}\left(\text{cosh}\phantom{\rule{0.2em}{0ex}}x\right)+C.

Evaluate the following integrals:

  1. ∫sinh3xcoshxdx\int {\text{sinh}}^{3}x\phantom{\rule{0.2em}{0ex}}\text{cosh}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx
  2. ∫sech2(3x)dx\int {\text{sech}}^{2}\left(3x\right)dx
  1. ∫sinh3xcoshxdx=sinh4x4+C\int {\text{sinh}}^{3}x\phantom{\rule{0.2em}{0ex}}\text{cosh}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx=\frac{{\text{sinh}}^{4}x}{4}+C
  2. ∫sech2(3x)dx=tanh(3x)3+C\int {\text{sech}}^{2}\left(3x\right)dx=\frac{\text{tanh}\left(3x\right)}{3}+C
Hint

Use the formulas above and apply u-substitution as necessary.

Calculus of Inverse Hyperbolic Functions

Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. Most of the necessary range restrictions can be discerned by close examination of the graphs. The domains and ranges of the inverse hyperbolic functions are summarized in the following table.

Domains and Ranges of the Inverse Hyperbolic Functions
FunctionDomainRange
sinh−1x{\text{sinh}}^{-1}x(−∞,∞)\left(\text{−}\infty ,\infty \right)(−∞,∞)\left(\text{−}\infty ,\infty \right)
cosh−1x{\text{cosh}}^{-1}x(1,∞)\left(1,\infty \right)[0,∞)\left[0,\infty \right)
tanh−1x{\text{tanh}}^{-1}x(−1,1)\left(-1,1\right)(−∞,∞)\left(\text{−}\infty ,\infty \right)
coth−1x{\text{coth}}^{-1}x(−∞,−1)∪(1,∞)\left(\text{−}\infty ,-1\right)\cup \left(1,\infty \right)(−∞,0)∪(0,∞)\left(\text{−}\infty ,0\right)\cup \left(0,\infty \right)
sech−1x{\text{sech}}^{-1}x(0, 1)\left(0\text{, 1}\right)[0,∞)\left[0,\infty \right)
csch−1x{\text{csch}}^{-1}x(−∞,0)∪(0,∞)\left(\text{−}\infty ,0\right)\cup \left(0,\infty \right)(−∞,0)∪(0,∞)\left(\text{−}\infty ,0\right)\cup \left(0,\infty \right)

The graphs of the inverse hyperbolic functions are shown in the following figure.

Graphs of the inverse hyperbolic functions.
This figure has six graphs. The first graph labeled “a” is of the function y=sinh^-1(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled “b” and is of the function y=cosh^-1(x). It is in the first quadrant, beginning on the x-axis at 2 and increasing. The third graph labeled “c” is of the function y=tanh^-1(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled “d” and is of the function y=coth^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled “e” and is of the function y=sech^-1(x). It is a curve decreasing in the first quadrant and stopping on the x-axis at x=1. The sixth graph is labeled “f” and is of the function y=csch^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.

To find the derivatives of the inverse functions, we use implicit differentiation. We have

y=sinh−1xsinhy=xddxsinhy=ddxxcoshydydx=1.\begin{array}{ccc}\hfill y& =\hfill & {\text{sinh}}^{-1}\phantom{\rule{0.2em}{0ex}}x\hfill \\ \hfill \text{sinh}\phantom{\rule{0.2em}{0ex}}y& =\hfill & x\hfill \\ \hfill \frac{d}{dx}\text{sinh}\phantom{\rule{0.2em}{0ex}}y& =\hfill & \frac{d}{dx}x\hfill \\ \hfill \text{cosh}\phantom{\rule{0.2em}{0ex}}y\frac{dy}{dx}& =\hfill & 1.\hfill \end{array}

Recall that cosh2y−sinh2y=1,{\text{cosh}}^{2}y-{\text{sinh}}^{2}y=1, so coshy=1+sinh2y.\text{cosh}\phantom{\rule{0.2em}{0ex}}y=\sqrt{1+{\text{sinh}}^{2}y}. Then,

dydx=1coshy=11+sinh2y=11+x2.\frac{dy}{dx}=\frac{1}{\text{cosh}\phantom{\rule{0.2em}{0ex}}y}=\frac{1}{\sqrt{1+{\text{sinh}}^{2}y}}=\frac{1}{\sqrt{1+{x}^{2}}}.

We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.

Derivatives of the Inverse Hyperbolic Functions
f(x)f\left(x\right)ddxf(x)\frac{d}{dx}f\left(x\right)
sinh−1x{\text{sinh}}^{-1}x11+x2\frac{1}{\sqrt{1+{x}^{2}}}
cosh−1x{\text{cosh}}^{-1}x1x2−1\frac{1}{\sqrt{{x}^{2}-1}}
tanh−1x{\text{tanh}}^{-1}x11−x2\frac{1}{1-{x}^{2}}
coth−1x{\text{coth}}^{-1}x11−x2\frac{1}{1-{x}^{2}}
sech−1x{\text{sech}}^{-1}x−1x1−x2\frac{-1}{x\sqrt{1-{x}^{2}}}
csch−1x{\text{csch}}^{-1}x−1∣x∣1+x2\frac{-1}{|x|\sqrt{1+{x}^{2}}}

Note that the derivatives of tanh−1x{\text{tanh}}^{-1}\phantom{\rule{0.2em}{0ex}}x and coth−1x{\text{coth}}^{-1}\phantom{\rule{0.2em}{0ex}}x are the same. Thus, when we integrate 1/(1−x2),1\text{/}\left(1-{x}^{2}\right), we need to select the proper antiderivative based on the domain of the functions and the values of x.x. Integration formulas involving the inverse hyperbolic functions are summarized as follows.

\begin{array}{cccccccc}\hfill \int \frac{1}{\sqrt{1+{u}^{2}}}du& =\hfill & {\text{sinh}}^{-1}u+C\hfill & & & \hfill \int \frac{1}{u\sqrt{1-{u}^{2}}}du& =\hfill & \text{−}{\text{sech}}^{-1}|u|+C\hfill \\ \hfill \int \frac{1}{\sqrt{{u}^{2}-1}}du& =\hfill & {\text{cosh}}^{-1}u+C\hfill & & & \hfill \int \frac{1}{u\sqrt{1+{u}^{2}}}du& =\hfill & \text{−}{\text{csch}}^{-1}|u|+C\hfill \\ \hfill \int \frac{1}{1-{u}^{2}}du& =\hfill & \left\{\begin{array}{c}{\text{tanh}}^{-1}u+C\phantom{\rule{0.2em}{0ex}}\text{if}\phantom{\rule{0.2em}{0ex}}|u|<1\hfill \\ {\text{coth}}^{-1}u+C\phantom{\rule{0.2em}{0ex}}\text{if}\phantom{\rule{0.2em}{0ex}}|u|>1\hfill \end{array}\hfill & & & & & \end{array}
Differentiating Inverse Hyperbolic Functions

Evaluate the following derivatives:

  1. ddx(sinh−1(x3))\frac{d}{dx}\left({\text{sinh}}^{-1}\left(\frac{x}{3}\right)\right)
  2. ddx(tanh−1x)2\frac{d}{dx}{\left({\text{tanh}}^{-1}x\right)}^{2}

Using the formulas in [link] and the chain rule, we obtain the following results:

  1. ddx(sinh−1(x3))=131+x29=19+x2\frac{d}{dx}\left({\text{sinh}}^{-1}\left(\frac{x}{3}\right)\right)=\frac{1}{3\sqrt{1+\frac{{x}^{2}}{9}}}=\frac{1}{\sqrt{9+{x}^{2}}}
  2. ddx(tanh−1x)2=2(tanh−1x)1−x2\frac{d}{dx}{\left({\text{tanh}}^{-1}x\right)}^{2}=\frac{2\left({\text{tanh}}^{-1}x\right)}{1-{x}^{2}}

Evaluate the following derivatives:

  1. ddx(cosh−1(3x))\frac{d}{dx}\left({\text{cosh}}^{-1}\left(3x\right)\right)
  2. ddx(coth−1x)3\frac{d}{dx}{\left({\text{coth}}^{-1}x\right)}^{3}
  1. ddx(cosh−1(3x))=39x2−1\frac{d}{dx}\left({\text{cosh}}^{-1}\left(3x\right)\right)=\frac{3}{\sqrt{9{x}^{2}-1}}
  2. ddx(coth−1x)3=3(coth−1x)21−x2\frac{d}{dx}{\left({\text{coth}}^{-1}x\right)}^{3}=\frac{3{\left({\text{coth}}^{-1}x\right)}^{2}}{1-{x}^{2}}
Hint

Use the formulas in [link] and apply the chain rule as necessary.

Integrals Involving Inverse Hyperbolic Functions

Evaluate the following integrals:

  1. ∫14x2−1dx\int \frac{1}{\sqrt{4{x}^{2}-1}}dx
  2. ∫12x1−9x2dx\int \frac{1}{2x\sqrt{1-9{x}^{2}}}dx

We can use u-substitutionu\text{-substitution} in both cases.

  1. Let u=2x.u=2x. Then, du=2dxdu=2dx and we have

    ∫14x2−1dx=∫12u2−1du=12cosh−1u+C=12cosh−1(2x)+C.\int \frac{1}{\sqrt{4{x}^{2}-1}}dx=\int \frac{1}{2\sqrt{{u}^{2}-1}}du=\frac{1}{2}{\text{cosh}}^{-1}u+C=\frac{1}{2}{\text{cosh}}^{-1}\left(2x\right)+C.
  2. Let u=3x.u=3x. Then, du=3dxdu=3dx and we obtain

    ∫12x1−9x2dx=12∫1u1−u2du=−12sech−1∣u∣+C=−12sech−1∣3x∣+C.\int \frac{1}{2x\sqrt{1-9{x}^{2}}}dx=\frac{1}{2}\int \frac{1}{u\sqrt{1-{u}^{2}}}du=-\frac{1}{2}{\text{sech}}^{-1}|u|+C=-\frac{1}{2}{\text{sech}}^{-1}|3x|+C.

Evaluate the following integrals:

  1. ∫1x2−4dx,x>2\int \frac{1}{\sqrt{{x}^{2}-4}}dx,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}x>2
  2. ∫11−e2xdx\int \frac{1}{\sqrt{1-{e}^{2x}}}dx
  1. ∫1x2−4dx=cosh−1(x2)+C\int \frac{1}{\sqrt{{x}^{2}-4}}dx={\text{cosh}}^{-1}\left(\frac{x}{2}\right)+C
  2. ∫11−e2xdx=−sech−1(ex)+C\int \frac{1}{\sqrt{1-{e}^{2x}}}dx=\text{−}{\text{sech}}^{-1}\left({e}^{x}\right)+C
Hint

Use the formulas above and apply u-substitutionu\text{-substitution} as necessary.

Applications

One physical application of hyperbolic functions involves hanging cables. If a cable of uniform density is suspended between two supports without any load other than its own weight, the cable forms a curve called a catenary. High-voltage power lines, chains hanging between two posts, and strands of a spider’s web all form catenaries. The following figure shows chains hanging from a row of posts.

Chains between these posts take the shape of a catenary. (credit: modification of work by OKFoundryCompany, Flickr)
An image of chains hanging between posts that all take the shape of a catenary.

Hyperbolic functions can be used to model catenaries. Specifically, functions of the form y=acosh(x/a)y=a\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}a\right) are catenaries. [link] shows the graph of y=2cosh(x/2).y=2\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}2\right).

A hyperbolic cosine function forms the shape of a catenary.
This figure is a graph. It is of the function f(x)=2cosh(x/2). The curve decreases in the second quadrant to the y-axis. It intersects the y-axis at y=2. Then the curve becomes increasing.
Using a Catenary to Find the Length of a Cable

Assume a hanging cable has the shape 10cosh(x/10)10\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}10\right) for −15≤x≤15,-15\le x\le 15, where xx is measured in feet. Determine the length of the cable (in feet).

Recall from Section 6.46.4 that the formula for arc length is

Arc Length=∫ab1+[f′(x)]2dx.\text{Arc Length}={\int }_{a}^{b}\sqrt{1+{\left[{f}^{\prime }\left(x\right)\right]}^{2}}\phantom{\rule{0.2em}{0ex}}dx.

We have f(x)=10cosh(x/10),f\left(x\right)=10\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}10\right), so f′(x)=sinh(x/10).{f}^{\prime }\left(x\right)=\text{sinh}\left(x\text{/}10\right). Then

Arc Length=∫ab1+[f′(x)]2dx=∫−15151+sinh2(x10)dx.\begin{array}{cc}\hfill \text{Arc Length}& ={\int }_{a}^{b}\sqrt{1+{\left[{f}^{\prime }\left(x\right)\right]}^{2}}\phantom{\rule{0.2em}{0ex}}dx\hfill \\ & ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}\left(\frac{x}{10}\right)}\phantom{\rule{0.2em}{0ex}}dx.\hfill \end{array}

Now recall that 1+sinh2x=cosh2x,1+{\text{sinh}}^{2}x={\text{cosh}}^{2}x, so we have

Arc Length=∫−15151+sinh2(x10)dx=∫−1515cosh(x10)dx=10sinh(x10)∣−1515=10[sinh(32)−sinh(−32)]=20sinh(32)≈42.586ft.\begin{array}{cc}\hfill \text{Arc Length}& ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}\left(\frac{x}{10}\right)}\phantom{\rule{0.2em}{0ex}}dx\hfill \\ & ={\int }_{-15}^{15}\text{cosh}\left(\frac{x}{10}\right)dx\hfill \\ & =10\phantom{\rule{0.2em}{0ex}}\text{sinh}{\left(\frac{x}{10}\right)|}_{-15}^{15}=10\left[\text{sinh}\left(\frac{3}{2}\right)-\text{sinh}\left(-\frac{3}{2}\right)\right]=20\phantom{\rule{0.2em}{0ex}}\text{sinh}\left(\frac{3}{2}\right)\hfill \\ & \approx 42.586\phantom{\rule{0.2em}{0ex}}\text{ft}\text{.}\hfill \end{array}

Assume a hanging cable has the shape 15cosh(x/15)15\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}15\right) for −20≤x≤20.-20\le x\le 20. Determine the length of the cable (in feet).

52.95ft52.95\phantom{\rule{0.2em}{0ex}}\text{ft}

Hint

Use the procedure from the previous example.

Key Concepts

  • Hyperbolic functions are defined in terms of exponential functions.
  • Term-by-term differentiation yields differentiation formulas for the hyperbolic functions. These differentiation formulas give rise, in turn, to integration formulas.
  • With appropriate range restrictions, the hyperbolic functions all have inverses.
  • Implicit differentiation yields differentiation formulas for the inverse hyperbolic functions, which in turn give rise to integration formulas.
  • The most common physical applications of hyperbolic functions are calculations involving catenaries.

[T] Find expressions for coshx+sinhx\text{cosh}\phantom{\rule{0.2em}{0ex}}x+\text{sinh}\phantom{\rule{0.2em}{0ex}}x and coshx−sinhx.\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.

exande−x{e}^{x}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{e}^{\text{−}x}

From the definitions of cosh(x)\text{cosh}\left(x\right) and sinh(x),\text{sinh}\left(x\right), find their antiderivatives.

Show that cosh(x)\text{cosh}\left(x\right) and sinh(x)\text{sinh}\left(x\right) satisfy y″=y.y\text{″}=y.

Answers may vary

Use the quotient rule to verify that tanh(x)′=sech2(x).\text{tanh}\left(x\right)\prime ={\text{sech}}^{2}\left(x\right).

Derive cosh2(x)+sinh2(x)=cosh(2x){\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 sinh(2x).\text{sinh}\left(2x\right).

Prove sinh(x+y)=sinh(x)cosh(y)+cosh(x)sinh(y)\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 cosh(x+y).\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]cosh(3x+1)\text{cosh}\left(3x+1\right)

3sinh(3x+1)3\phantom{\rule{0.2em}{0ex}}\text{sinh}\left(3x+1\right)

[T]sinh(x2)\text{sinh}\left({x}^{2}\right)

[T]1cosh(x)\frac{1}{\text{cosh}\left(x\right)}

−tanh(x)sech(x)\text{−}\text{tanh}\left(x\right)\text{sech}\left(x\right)

[T]sinh(ln(x))\text{sinh}\left(\text{ln}\left(x\right)\right)

[T]cosh2(x)+sinh2(x){\text{cosh}}^{2}\left(x\right)+{\text{sinh}}^{2}\left(x\right)

4cosh(x)sinh(x)4\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\right)\text{sinh}\left(x\right)

[T]cosh2(x)−sinh2(x){\text{cosh}}^{2}\left(x\right)-{\text{sinh}}^{2}\left(x\right)

[T]tanh(x2+1)\text{tanh}\left(\sqrt{{x}^{2}+1}\right)

xsech2(x2+1)x2+1\frac{x\phantom{\rule{0.2em}{0ex}}{\text{sech}}^{2}\left(\sqrt{{x}^{2}+1}\right)}{\sqrt{{x}^{2}+1}}

[T]1+tanh(x)1−tanh(x)\frac{1+\text{tanh}\left(x\right)}{1-\text{tanh}\left(x\right)}

[T]sinh6(x){\text{sinh}}^{6}\left(x\right)

6sinh5(x)cosh(x)6\phantom{\rule{0.2em}{0ex}}{\text{sinh}}^{5}\left(x\right)\text{cosh}\left(x\right)

[T]ln(sech(x)+tanh(x))\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.

cosh(2x+1)\text{cosh}\left(2x+1\right)

12sinh(2x+1)+C\frac{1}{2}\text{sinh}\left(2x+1\right)+C

tanh(3x+2)\text{tanh}\left(3x+2\right)

xcosh(x2)x\phantom{\rule{0.2em}{0ex}}\text{cosh}\left({x}^{2}\right)

12sinh2(x2)+C\frac{1}{2}{\text{sinh}}^{2}\left({x}^{2}\right)+C

3x3tanh(x4)3{x}^{3}\text{tanh}\left({x}^{4}\right)

cosh2(x)sinh(x){\text{cosh}}^{2}\left(x\right)\text{sinh}\left(x\right)

13cosh3(x)+C\frac{1}{3}{\text{cosh}}^{3}\left(x\right)+C

tanh2(x)sech2(x){\text{tanh}}^{2}\left(x\right){\text{sech}}^{2}\left(x\right)

sinh(x)1+cosh(x)\frac{\text{sinh}\left(x\right)}{1+\text{cosh}\left(x\right)}

ln(1+cosh(x))+C\text{ln}\left(1+\text{cosh}\left(x\right)\right)+C

coth(x)\text{coth}\left(x\right)

cosh(x)+sinh(x)\text{cosh}\left(x\right)+\text{sinh}\left(x\right)

cosh(x)+sinh(x)+C\text{cosh}\left(x\right)+\text{sinh}\left(x\right)+C

(cosh(x)+sinh(x))n{\left(\text{cosh}\left(x\right)+\text{sinh}\left(x\right)\right)}^{n}

For the following exercises, find the derivatives for the functions.

tanh−1(4x){\text{tanh}}^{-1}\left(4x\right)

41−16x2\frac{4}{1-16{x}^{2}}

sinh−1(x2){\text{sinh}}^{-1}\left({x}^{2}\right)

sinh−1(cosh(x)){\text{sinh}}^{-1}\left(\text{cosh}\left(x\right)\right)

sinh(x)cosh2(x)+1\frac{\text{sinh}\left(x\right)}{\sqrt{{\text{cosh}}^{2}\left(x\right)+1}}

cosh−1(x3){\text{cosh}}^{-1}\left({x}^{3}\right)

tanh−1(cos(x)){\text{tanh}}^{-1}\left(\text{cos}\left(x\right)\right)

−csc(x)\text{−}\text{csc}\left(x\right)

esinh−1(x){e}^{{\text{sinh}}^{-1}\left(x\right)}

ln(tanh−1(x))\text{ln}\left({\text{tanh}}^{-1}\left(x\right)\right)

−1(x2−1)tanh−1(x)-\frac{1}{\left({x}^{2}-1\right){\text{tanh}}^{-1}\left(x\right)}

For the following exercises, find the antiderivatives for the functions.

∫dx4−x2\int \frac{dx}{4-{x}^{2}}

∫dxa2−x2\int \frac{dx}{{a}^{2}-{x}^{2}}

1atanh−1(xa)+C\frac{1}{a}{\text{tanh}}^{-1}\left(\frac{x}{a}\right)+C

∫dxx2+1\int \frac{dx}{\sqrt{{x}^{2}+1}}

∫xdxx2+1\int \frac{x\phantom{\rule{0.2em}{0ex}}dx}{\sqrt{{x}^{2}+1}}

x2+1+C\sqrt{{x}^{2}+1}+C

∫−dxx1−x2\int -\frac{dx}{x\sqrt{1-{x}^{2}}}

∫exe2x−1\int \frac{{e}^{x}}{\sqrt{{e}^{2x}-1}}

cosh−1(ex)+C{\text{cosh}}^{-1}\left({e}^{x}\right)+C

∫−2xx4−1\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/dt=g−v2.dv\text{/}dt=g-{v}^{2}.

Show that v(t)=gtanh(gt)v\left(t\right)=\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)v\left(t\right) by integrating dvg−v2=dt.\frac{dv}{g-{v}^{2}}=dt.

[T] Estimate how far a body has fallen in 1212 seconds by finding the area underneath the curve of v(t).v\left(t\right).

37.3037.30

For the following exercises, use this scenario: A cable hanging under its own weight has a slope S=dy/dxS=dy\text{/}dx that satisfies dS/dx=c1+S2.dS\text{/}dx=c\sqrt{1+{S}^{2}}. The constant cc is the ratio of cable density to tension.

Show that S=sinh(cx)S=\text{sinh}\left(cx\right) satisfies this equation.

Integrate dy/dx=sinh(cx)dy\text{/}dx=\text{sinh}\left(cx\right) to find the cable height y(x)y\left(x\right) if y(0)=1/c.y\left(0\right)=1\text{/}c.

y=1ccosh(cx)y=\frac{1}{c}\text{cosh}\left(cx\right)

Sketch the cable and determine how far down it sags at x=0.x=0.

For the following exercises, solve each problem.

[T] A chain hangs from two posts 22 m apart to form a catenary described by the equation y=2cosh(x/2)−1.y=2\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}2\right)-1. Find the slope of the catenary at the left fence post.

−0.521095-0.521095

[T] A chain hangs from two posts four meters apart to form a catenary described by the equation y=4cosh(x/4)−3.y=4\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}4\right)-3. Find the total length of the catenary (arc length).

[T] A high-voltage power line is a catenary described by y=10cosh(x/10).y=10\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}10\right). Find the ratio of the area under the catenary to its arc length. What do you notice?

1010

A telephone line is a catenary described by y=acosh(x/a).y=a\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}a\right). Find the ratio of the area under the catenary to its arc length. Does this confirm your answer for the previous question?

Prove the formula for the derivative of y=sinh−1(x)y={\text{sinh}}^{-1}\left(x\right) by differentiating x=sinh(y).x=\text{sinh}\left(y\right). (Hint: Use hyperbolic trigonometric identities.)

Prove the formula for the derivative of y=cosh−1(x)y={\text{cosh}}^{-1}\left(x\right) by differentiating x=cosh(y).x=\text{cosh}\left(y\right).

(Hint: Use hyperbolic trigonometric identities.)

Prove the formula for the derivative of y=sech−1(x)y={\text{sech}}^{-1}\left(x\right) by differentiating x=sech(y).x=\text{sech}\left(y\right). (Hint: Use hyperbolic trigonometric identities.)

Prove that (cosh(x)+sinh(x))n=cosh(nx)+sinh(nx).{\left(\text{cosh}\left(x\right)+\text{sinh}\left(x\right)\right)}^{n}=\text{cosh}\left(nx\right)+\text{sinh}\left(nx\right).

Prove the expression for sinh−1(x).{\text{sinh}}^{-1}\left(x\right). Multiply x=sinh(y)=(1/2)(ey−e−y)x=\text{sinh}\left(y\right)=\left(1\text{/}2\right)\left({e}^{y}-{e}^{\text{−}y}\right) by 2ey2{e}^{y} and solve for y.y. Does your expression match the textbook?

Prove the expression for cosh−1(x).{\text{cosh}}^{-1}\left(x\right). Multiply x=cosh(y)=(1/2)(ey−e−y)x=\text{cosh}\left(y\right)=\left(1\text{/}2\right)\left({e}^{y}-{e}^{\text{−}y}\right) by 2ey2{e}^{y} and solve for y.y. Does your expression match the textbook?

Chapter Review Exercises

True or False? Justify your answer with a proof or a counterexample.

The amount of work to pump the water out of a half-full cylinder is half the amount of work to pump the water out of the full cylinder.

False

If the force is constant, the amount of work to move an object from x=ax=a to x=bx=b is F(b−a).F\left(b-a\right).

The disk method can be used in any situation in which the washer method is successful at finding the volume of a solid of revolution.

False

If the half-life of seaborgium-266\text{seaborgium-}266 is 360360 ms, then k=(ln(2))/360.k=\left(\text{ln}\left(2\right)\right)\text{/}360.

For the following exercises, use the requested method to determine the volume of the solid.

The volume that has a base of the ellipse x2/4+y2/9=1{x}^{2}\text{/}4+{y}^{2}\text{/}9=1 and cross-sections of an equilateral triangle perpendicular to the y-axis.y\text{-axis}\text{.} Use the method of slicing.

32332\sqrt{3}

y=x2−x,y={x}^{2}-x, from x=1tox=4,x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=4, rotated around they-axis using the washer method

x=y2x={y}^{2} and x=3yx=3y rotated around the y-axis using the washer method

162π5\frac{162\pi }{5}

x=2y2−y3,x=0,andy=0x=2{y}^{2}-{y}^{3},x=0,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y=0 rotated around the x-axis using cylindrical shells

For the following exercises, find

  1. the area of the region,
  2. the volume of the solid when rotated around the x-axis, and
  3. the volume of the solid when rotated around the y-axis. Use whichever method seems most appropriate to you.

y=x3,x=0,y=0,andx=2y={x}^{3},x=0,y=0,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}x=2

a. 4,4, b. 128π7,\frac{128\pi }{7}, c. 64π5\frac{64\pi }{5}

y=x2−xandx=0y={x}^{2}-x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}x=0

[T]y=ln(x)+2andy=xy=\text{ln}\left(x\right)+2\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y=x

a. 1.949,1.949, b. 21.952,21.952, c. 17.09917.099

y=x2y={x}^{2} and y=xy=\sqrt{x}

y=5+x,y=5+x,y=x2,y={x}^{2},x=0,x=0, and x=1x=1

a. 316,\frac{31}{6}, b. 452π15,\frac{452\pi }{15}, c. 31π6\frac{31\pi }{6}

Below x2+y2=1{x}^{2}+{y}^{2}=1 and above y=1−xy=1-x

Find the mass of ρ=e−x\rho ={e}^{\text{−}x} on a disk centered at the origin with radius 4.4.

245.282245.282

Find the center of mass for ρ=tan2x\rho ={\text{tan}}^{2}x on x∈(−π4,π4).x\in \left(-\frac{\pi }{4},\frac{\pi }{4}\right).

Find the mass and the center of mass of ρ=1\rho =1 on the region bounded by y=x5y={x}^{5} and y=x.y=\sqrt{x}.

Mass: 12,\frac{1}{2}, center of mass: (1835,911)\left(\frac{18}{35},\frac{9}{11}\right)

For the following exercises, find the requested arc lengths.

The length of xx for y=cosh(x)y=\text{cosh}\left(x\right) from x=0tox=2.x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=2.

The length of yy for x=3−yx=3-\sqrt{y} from y=0y=0 to y=4y=4

17+18ln(33+817)\sqrt{17}+\frac{1}{8}\text{ln}\left(33+8\sqrt{17}\right)

For the following exercises, find the surface area and volume when the given curves are revolved around the specified axis.

The shape created by revolving the region between y=4+x,y=4+x, y=3−x,y=3-x, x=0,x=0, and x=2x=2 rotated around the y-axis.

The loudspeaker created by revolving y=1/xy=1\text{/}x from x=1x=1 to x=4x=4 around the x-axis.

Volume: 3π4,\frac{3\pi }{4}, surface area: π(2−sinh−1(1)+sinh−1(16)−25716)\pi \left(\sqrt{2}-{\text{sinh}}^{-1}\left(1\right)+{\text{sinh}}^{-1}\left(16\right)-\frac{\sqrt{257}}{16}\right)

For the following exercises, consider the Karun-3 dam in Iran. Its shape can be approximated as an isosceles triangle with height 205205 m and width 388388 m. Assume the current depth of the water is 180180 m. The density of water is 10001000 kg/m 3.{}^{3}.

Find the total force on the wall of the dam.

You are a crime scene investigator attempting to determine the time of death of a victim. It is noon and 45°F45\text{°}\text{F} outside and the temperature of the body is 78°F.78\text{°}\text{F}. You know the cooling constant is k=0.00824°F/min.k=0.00824\text{°}\text{F/min}\text{.} When did the victim die, assuming that a human’s temperature is 98°F98\text{°}\text{F} ?

11:02 a.m.

For the following exercise, consider the stock market crash in 19291929 in the United States. The table lists the Dow Jones industrial average per year leading up to the crash.

Source: http://stockcharts.com/freecharts/historical/djia19201940.html
Years after 1920Value (💲)
1163.9063.90
33100100
55110110
77160160
99381.17381.17

[T] The best-fit exponential curve to these data is given by y=40.71+1.224x.y=40.71+{1.224}^{x}. Why do you think the gains of the market were unsustainable? Use first and second derivatives to help justify your answer. What would this model predict the Dow Jones industrial average to be in 20142014 ?

For the following exercises, consider the catenoid, the only solid of revolution that has a minimal surface, or zero mean curvature. A catenoid in nature can be found when stretching soap between two rings.

Find the volume of the catenoid y=cosh(x)y=\text{cosh}\left(x\right) from x=−1tox=1x=-1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=1 that is created by rotating this curve around the x-axis,x\text{-axis}, as shown here.

This figure is an image of a catenoid. It has been formed by rotating a catenary curve about a vertical axis.

π(1+sinh(1)cosh(1))\pi \left(1+\text{sinh}\left(1\right)\text{cosh}\left(1\right)\right)

Find surface area of the catenoid y=cosh(x)y=\text{cosh}\left(x\right) from x=−1x=-1 to x=1x=1 that is created by rotating this curve around the x-axis.x\phantom{\rule{0.2em}{0ex}}\text{-axis.}

Glossary

catenary
a curve in the shape of the function y=acosh(x/a)y=a\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(x\text{/}a\right) is a catenary; a cable of uniform density suspended between two supports assumes the shape of a catenary