Wednesday, 30 July 2014

CHAPTER 11 - Inverse Hyperbolic Functions

 Inverse Hyperbolic Functions

The hyperbolic sine function is a one-to-one function, and thus has an inverse. As usual, we obtain the graph of the inverse hyperbolic sine function tex2html_wrap_inline53 (also denoted by tex2html_wrap_inline55 ) by reflecting the graph oftex2html_wrap_inline57 about the line y=x:

Since tex2html_wrap_inline61 is defined in terms of the exponential function, you should not be surprised that its inverse function can be expressed in terms of the logarithmic function:
Let's set tex2html_wrap_inline63 , and try to solve for x:
This is a quadratic equation with tex2html_wrap_inline67 instead of x as the variable. y will be considered a constant.
So using the quadratic formula, we obtain
Since tex2html_wrap_inline73 for all x, and since tex2html_wrap_inline77 for all y, we have to discard the solution with the minus sign, so
and consequently
Read that last sentence again slowly!
We have found out that
  • tex2html_wrap_inline81

Try it yourself!

You know what's coming up, don't you? Here's the graph. Note that the hyperbolic cosine function is not one-to-one, so let's restrict the domain to tex2html_wrap_inline83 .

Here it is: Express the inverse hyperbolic cosine functions in terms of the logarithmic function!


Hyperbolic Functions

The hyperbolic functions enjoy properties similar to the trigonometric functions; their definitions, though, are much more straightforward:

Here are their graphs: the tex2html_wrap_inline125 (pronounce: "kosh") is pictured in red, the tex2html_wrap_inline127 function (rhymes with the "Grinch") is depicted in blue.

As their trigonometric counterparts, the tex2html_wrap_inline125 function is even, while the tex2html_wrap_inline127 function is odd.
Their most important property is their version of the Pythagorean Theorem.

  • tex2html_wrap_inline133
The verification is straightforward:eqnarray18
While tex2html_wrap_inline135 , tex2html_wrap_inline137 , parametrizes the unit circle, the hyperbolic functions tex2html_wrap_inline139 , tex2html_wrap_inline141 , parametrize the standard hyperbola tex2html_wrap_inline143 , x>1.
In the picture below, the standard hyperbola is depicted in red, while the point tex2html_wrap_inline139 for various values of the parameter t is pictured in blue.

The other hyperbolic functions are defined the same way, the rest of the trigonometric functions is defined:

tanh x
coth x
sech x
csch x

For every formula for the trigonometric functions, there is a similar (not necessary identical) formula for the hyperbolic functions:
Let's consider for example the addition formula for the hyperbolic cosine function:
  • tex2html_wrap_inline151
Start with the right side and multiply out:eqnarray61

Try it yourself!

Prove the addition formula for the hyperbolic sine function:Show that tex2html_wrap_inline153 .


The Derivatives of Trigonometric Functions

\begin{displaymath}\begin{array}{cc}\sin^\prime x=\cos x&\cos^\prime x=-\cos x\\...
...\prime x=\sec x \tan x&\csc^\prime x=-\csc x \cot x
Trigonometric functions are useful in our practical lives in diverse areas such as astronomy, physics, surveying, carpentry etc. How can we find the derivatives of the trigonometric functions?
Our starting point is the following limit:

\begin{displaymath}\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1\cdot\end{displaymath}

Using the derivative language, this limit means that $\sin'(0) = 1$. This limit may also be used to give a related one which is of equal importance:

\begin{displaymath}\lim_{x \rightarrow 0} \frac{\cos(x)-1}{x} = 0\end{displaymath}

To see why, it is enough to rewrite the expression involving the cosine as

\begin{displaymath}\frac{\cos(x)-1}{x} = \frac{(\cos(x)-1)(\cos(x) + 1)}{x(\cos(x) + 1)} = \frac{(\cos^2(x)-1)}{x(\cos(x) + 1)}\end{displaymath}

But $\cos^2(x)-1 = -\sin^2(x)$, so we have

\begin{displaymath}\lim_{x \rightarrow 0} \frac{\cos(x)-1}{x} = \lim_{x \rightar...
... \rightarrow 0} x \frac{-\sin^2(x)}{x^2(\cos(x) + 1)} = 0 \cdot\end{displaymath}

This limit equals $\cos'(0)$ and thus $\cos'(0) = 0$.

In fact, we may use these limits to find the derivative of $\sin(x)$ and $\cos(x)$ at any point x=a. Indeed, using the addition formula for the sine function, we have

\begin{displaymath}\sin(a + h) = \sin(a) \cos(h) + \sin(h) \cos(a) \cdot\end{displaymath}


\begin{displaymath}\frac{\sin(a + h) - \sin(a)}{h} = \sin(a)\frac{1 - \cos(h)}{h} + \cos(a) \frac{\sin(h)}{h}\end{displaymath}

which implies

\begin{displaymath}\lim_{h \rightarrow 0} \frac{\sin(a + h) - \sin(a)}{h} = \cos(a) \cdot\end{displaymath}

So we have proved that $\sin'(a)$ exists and $\sin'(a) =
\cos(a)$.Similarly, we obtain that $\cos'(a)$ exists and that $\cos'(a) =
Since $\tan(x)$$\cot(x)$$\sec(x)$, and $\csc(x)$ are all quotients of the functions $\sin(x)$ and $\cos(x)$, we can compute their derivatives with the help of the quotient rule:

\displaystyle \frac{d}{dx} (\tan(x)) = \s... \frac{d}{dx} (\csc(x)) = -\csc(x) \cot(x) \\

It is quite interesting to see the close relationship between $\tan(x)$ and $\sec(x)$ (and also between $\cot(x)$ and $\csc(x)$).
From the above results we get

\begin{displaymath}\sin''(x) = - \sin(x)\;\;\mbox{and}\;\; \cos''(x) = - \cos(x)\cdot\end{displaymath}

These two results are very useful in solving some differential equations.Example 1. Let $f(x) = \sin(2 x)$. Using the double angle formula for the sine function, we can rewrite

\begin{displaymath}\sin(2 x) = 2 \sin(x) \cos(x)\cdot\end{displaymath}

So using the product rule, we get

\begin{displaymath}\frac{d}{dx}\Big(\sin(2x)\Big) = 2 \Big( \cos(x) \cos(x) - \sin(x) \sin(x) \Big) = 2 \Big( \cos^2(x) - \sin^2(x) \Big)\end{displaymath}

which implies, using trigonometric identities,

\begin{displaymath}\frac{d}{dx}\Big(\sin(2x)\Big) = 2 \cos(2x)\cdot\end{displaymath}

Exercise 1. Find the equations of the tangent line and the normal line to the graph of $f(x) = \sec(x) + \tan(x)$ at the point $\left(\displaystyle \frac{\pi}{4},
Answer. First we need to find the derivative of f(x) so we can get the slope of the tangent line and the normal line. We have

\begin{displaymath}f'(x) = \sec(x) \tan(x) + \sec^2(x)\;.\end{displaymath}

So we have

\begin{displaymath}f'\left(\frac{\pi}{4}\right) = \sqrt{2} + 2\end{displaymath}

knowing that $\sec(\pi/4) = \sqrt{2}$ and $\tan(\pi/4) = 1$. Note that

\begin{displaymath}f\left(\frac{\pi}{4}\right) = \sqrt{2} + 1\;.\end{displaymath}

So the equation of the tangent line at the point $(\pi/4,
\sqrt{2} + 1)$ is

\begin{displaymath}y - \sqrt{2} - 1 = (\sqrt{2} + 2) \left(x- \frac{\pi}{4}\right) \;.\end{displaymath}

the slope of the normal line to the graph at the point $(\pi/4,
\sqrt{2} + 1)$ is

\begin{displaymath}- \frac{1}{\sqrt{2} + 2}\end{displaymath}

which helps us find the equation of the normal line as

\begin{displaymath}y - \sqrt{2} - 1 = - \frac{1}{\sqrt{2} + 2} \left(x- \frac{\pi}{4}\right) \;.\end{displaymath}

Exercise 2. Find the x-coordinates of all points on the graph of $f(x) = x +2\cos(x)$ in the interval $[0,\pi]$ at which the tangent line is horizontal.
Answer. The points (x,f(x)) at which the tangent line is horizontal are the ones for which f'(x) = 0. Let us first find f'(x). We have

\begin{displaymath}f'(x) = 1 - 2\sin(x)\;.\end{displaymath}

So we have to solve $1 - 2\sin(x) = 0$ which gives $\sin(x) =
1/2$. We get

\begin{displaymath}x = \frac{\pi}{6}\;\;\mbox{or}\;\; x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \cdot\end{displaymath}