Calculus of the Hyperbolic Functions: Fresh Take

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Calculus of the Hyperbolic Functions: Fresh Take

Differentiate and integrate hyperbolic functions and their inverse forms
Understand the practical situations where the catenary curve appears

Derivatives and Integrals of the Hyperbolic Functions

The Main Idea

Definitions of Hyperbolic Functions:
- $\sinh x = \frac{e^x - e^{-x}}{2}$
- $\cosh x = \frac{e^x + e^{-x}}{2}$
Key Derivatives:
- $\frac{d}{dx}(\sinh x) = \cosh x$
- $\frac{d}{dx}(\cosh x) = \sinh x$
- $\frac{d}{dx}(\tanh x) = \text{sech}^2 x$
Key Integrals:
- $\int \sinh x , dx = \cosh x + C$
- $\int \cosh x , dx = \sinh x + C$
- $\int \text{sech}^2 x , dx = \tanh x + C$
Similarities with Trigonometric Functions:
- Derivatives of sinh and sin are similar
- But $\frac{d}{dx}(\cos x) = -\sin x$ , while $\frac{d}{dx}(\cosh x) = \sinh x$
U-substitution is often useful for integrals involving hyperbolic functions

Evaluate the following derivatives:

$\frac{d}{dx}(\text{tanh}({x}^{2}+3x))$
$\frac{d}{dx}(\dfrac{1}{{(\text{sinh}x)}^{2}})$

$\frac{d}{dx}(\text{tanh}({x}^{2}+3x))=({\text{sech}}^{2}({x}^{2}+3x))(2x+3)$
$\frac{d}{dx}(\frac{1}{{(\text{sinh}x)}^{2}})=\frac{d}{dx}{(\text{sinh}x)}^{-2}=-2{(\text{sinh}x)}^{-3}\text{cosh}x$

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “2.9 Calculus of Hyperbolic Functions” here (opens in new window).

Evaluate the following integrals:

$\displaystyle\int {\text{sinh}}^{3}x\text{cosh}xdx$
$\displaystyle\int {\text{sech}}^{2}(3x)dx$

$\displaystyle\int {\text{sinh}}^{3}x\text{cosh}xdx=\frac{{\text{sinh}}^{4}x}{4}+C$
$\displaystyle\int {\text{sech}}^{2}(3x)dx=\frac{\text{tanh}(3x)}{3}+C$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.9 Calculus of Hyperbolic Functions” here (opens in new window).

Evaluate the following integral:

$\int x \sinh(x^2) , dx$

This integral suggests a u-substitution. Let $u = x^2$ .

Then $du = 2x , dx$ , or $x , dx = \frac{1}{2} du$ .

Rewrite the integral in terms of u:

$\int x \sinh(x^2) , dx = \int \frac{1}{2} \sinh(u) , du$

Now we can use the integration formula for sinh:

$\frac{1}{2} \int \sinh(u) , du = \frac{1}{2} \cosh(u) + C$

Substitute back $x^2$ for $u$ :

$\frac{1}{2} \cosh(x^2) + C$

Therefore, $\int x \sinh(x^2) , dx = \frac{1}{2} \cosh(x^2) + C$ .

Calculus of Inverse Hyperbolic Functions

The Main Idea

All hyperbolic functions have inverses with appropriate range restrictions
Key Derivatives:
- $\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{1+x^2}}$
- $\frac{d}{dx}(\cosh^{-1} x) = \frac{1}{\sqrt{x^2-1}}$
- $\frac{d}{dx}(\tanh^{-1} x) = \frac{1}{1-x^2}$
Key Integrals:
- $\int \frac{1}{\sqrt{1+u^2}} du = \sinh^{-1} u + C$
- $\int \frac{1}{\sqrt{u^2-1}} du = \cosh^{-1} u + C$
- $\int \frac{1}{1-u^2} du = \begin{cases} \tanh^{-1} u + C & \text{if } |u| < 1 \ \coth^{-1} u + C & \text{if } |u| > 1 \end{cases}$
Each inverse hyperbolic function has specific domain and range restrictions
Derivatives of inverse hyperbolic functions are found using implicit differentiation

Evaluate the following derivatives:

$\frac{d}{dx}({\text{cosh}}^{-1}(3x))$
$\frac{d}{dx}{({\text{coth}}^{-1}x)}^{3}$

$\frac{d}{dx}({\text{cosh}}^{-1}(3x))=\frac{3}{\sqrt{9{x}^{2}-1}}$
$\frac{d}{dx}{({\text{coth}}^{-1}x)}^{3}=\frac{3{({\text{coth}}^{-1}x)}^{2}}{1-{x}^{2}}$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.9 Calculus of Hyperbolic Functions” here (opens in new window).

Evaluate the following integrals:

$\displaystyle\int \frac{1}{\sqrt{4{x}^{2}-1}}dx$
$\displaystyle\int \frac{1}{2x\sqrt{1-9{x}^{2}}}dx$

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

Let $u=2x.$ Then, $du=2dx$ and we have
$\displaystyle\int \frac{1}{\sqrt{4{x}^{2}-1}}dx=\displaystyle\int \frac{1}{2\sqrt{{u}^{2}-1}}du=\frac{1}{2}{\text{cosh}}^{-1}u+C=\frac{1}{2}{\text{cosh}}^{-1}(2x)+C.$
Let $u=3x.$ Then, $du=3dx$ and we obtain
$\displaystyle\int \frac{1}{2x\sqrt{1-9{x}^{2}}}dx=\frac{1}{2}\displaystyle\int \frac{1}{u\sqrt{1-{u}^{2}}}du=-\frac{1}{2}{\text{sech}}^{-1}|u|+C=-\frac{1}{2}{\text{sech}}^{-1}|3x|+C.$

Evaluate the following integral:

$\int \frac{x}{\sqrt{x^2+9}} dx$

This integral suggests a u-substitution. Let $u = x^2 + 9$ .

Then $du = 2x dx$ , or $x dx = \frac{1}{2} du$ .

Rewrite the integral in terms of $u$ :

$\int \frac{x}{\sqrt{x^2+9}} dx = \frac{1}{2} \int \frac{1}{\sqrt{u}} du$

This is a standard integral. We get:

$\frac{1}{2} \int \frac{1}{\sqrt{u}} du = \frac{1}{2} \cdot 2\sqrt{u} + C = \sqrt{u} + C$

Substitute back $x^2 + 9$ for $u$ :

$\sqrt{x^2 + 9} + C$

However, we can express this result using an inverse hyperbolic function:

$\sqrt{x^2 + 9} = 3\sqrt{1 + (\frac{x}{3})^2} = 3\sinh(\sinh^{-1}(\frac{x}{3}))$

Therefore, $\int \frac{x}{\sqrt{x^2+9}} dx = 3\sinh(\sinh^{-1}(\frac{x}{3})) + C$ .

Applications of Hyperbolic Functions

The Main Idea

Catenary Curves:
- A catenary is the curve formed by a hanging cable or chain under uniform gravity
- Mathematically modeled by: $y = a \cosh(\frac{x}{a})$
Arc Length Formula:
- Used to calculate the length of a catenary
- $\text{Arc Length} = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx$
Hyperbolic Function Properties:
- $1 + \sinh^2 x = \cosh^2 x$
- This identity is useful in simplifying arc length calculations

Assume a hanging cable has the shape $15\text{cosh}\left(\frac{x}{15}\right)$ for $-20\le x\le 20.$ Determine the length of the cable (in feet).

$52.95\text{ft}$

A suspension bridge cable forms a catenary with the equation $y = 50 \cosh(\frac{x}{50})$ , where $x$ and $y$ are measured in meters. The cable is anchored at $x = -100$ and $x = 100$ . Calculate the length of the cable.

We use the arc length formula: $\text{Arc Length} = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx$

First, find $f'(x)$ :

$f'(x) = 50 \cdot \frac{1}{50} \sinh(\frac{x}{50}) = \sinh(\frac{x}{50})$

Substitute into the arc length formula:

$\text{Arc Length} = \int_{-100}^{100} \sqrt{1 + [\sinh(\frac{x}{50})]^2} dx$

Use the identity $1 + \sinh^2 x = \cosh^2 x$ :

$\text{Arc Length} = \int_{-100}^{100} \cosh(\frac{x}{50}) dx$

Integrate:

$\begin{array}{rcl} \text{Arc Length} &=& 50 \sinh(\frac{x}{50}) \Big|_{-100}^{100} \\ &=& 50 [\sinh(2) - \sinh(-2)] \\ &=& 100 \sinh(2) \\ &\approx& 364.1 \text{ meters} \end{array}$

Therefore, the length of the cable is approximately $364.1$ meters.