Tangent to hyperbolic function

**klooless** · June 12th 2009, 10:58 AM

Question:
At what point of the curve y=cosh (x) does the tangent have slope 1?

It seems so simple but I'm having a hard time working through this one. I know I need to find X when sinh (x)=1. I looked at the answer however and it is;

(ln(1+sqr(2)), sqr(2))

The X term in this point is similar to the arcsinh (x)= ln(x+sqr(x^2 + 1)). I don't understand how it got there? Any help walking through this one??

Thanks all,
cheers

**Random Variable** · June 12th 2009, 11:14 AM

let $y = \sinh(x) = \frac {e^{x}-e^{-x}}{2}$

If we can find x in terms of y, then we have found $\sinh^{-1}x$

$2y = e^{x}-e^{-x}$

multlipy both sides by $e^{x}$

$2ye^{x} = e^{2x} -1$

so ${(e^{x})}^{2} - 2y(e^{x}) - 1 = 0$

now apply the quadratic forumula to find $e^{x}$

we only need the positive root because $e^{x}$ is always positive

and then take the ln of both sides to arrive at the answer

**Plato** · June 12th 2009, 11:23 AM

Solve this equation: $e^x - e^{-x}=2$ .
That is the value of arcsinh(1).
You should get $e^x=1+\sqrt(2)$ so $x=?$

**klooless** · June 13th 2009, 03:01 PM

Thank you all,

Plato - I'm still getting stuck after e^x - e^-x = 2. I multiplied both sides by e^x, then used the quadratic function but I'm gettin x= ln( (-2+sqr(8))/2 )....

cheers

**mr fantastic** · June 13th 2009, 04:38 PM

Originally Posted by klooless

Thank you all,

Plato - I'm still getting stuck after e^x - e^-x = 2. I multiplied both sides by e^x, then used the quadratic function but I'm gettin x= ln( (-2+sqr(8))/2 )....

cheers

$(e^x)^2 - 2 e^x - 1 = 0$

a = 1, b = -2, c = -1:

$e^x = \frac{2 \pm \sqrt{4 + 4}}{2}$ .

Note that $\sqrt{8} = 2 \sqrt{2}$ , divide by the common factor of 2, reject the -ve root etc.

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