Finding the Minimum Value of x in Hyperbolic Calculus

[synkk]

q: http://gyazo.com/297417b9665206ae8e38cb8b5d930a83

I'm stuck trying to find the value of x when TN is a minimum

here's what I've tried so far:

Let T be the point (a,0) and N be the point (b,0)

line of tangent through P:

$y = sinh(x)(x-a)$
line of normal through P $y = \dfrac{-1}{sinh(x)}(x-b)$

my plan was to rearrange for a and b and find b-a and try differentiate that and set the derivative to be 0 and solve for x:

$a = x - \dfrac{y}{sinh(x)}$
$b = y(sinh(x)) + x$
$b - a = y(sinh(x) + \dfrac{1}{sinh(x)})$
$\dfrac{d(b-a)}{dx} = \dfrac{dy}{dx} (sinh(x) + \dfrac{1}{sinh(x)}) + y(cosh(x) - coth(x)cosech(x)) = sinh^2(x) + 1 + y(cosh(x) - coth(x)cosech(x)) = 0$ but I can't seem to solve that

any ideas on where I went wrong?

 

[tiny-tim]

hi synkk!

$a = x - \dfrac{y}{sinh(x)}$
$b = y(sinh(x)) + x$
$b - a = y(sinh(x) + \dfrac{1}{sinh(x)})$
$\dfrac{d(b-a)}{dx} = \dfrac{dy}{dx} (sinh(x) + \dfrac{1}{sinh(x)}) + y(cosh(x) - coth(x)cosech(x)) = sinh^2(x) + 1 + y(cosh(x) - coth(x)cosech(x)) = 0$ but I can't seem to solve that

erm …

y isn't a variable, y is coshx ! ​

 

[synkk]

hi synkk! erm …

y isn't a variable, y is coshx ! ​

oh thank you, but is my method correct?

following on from this I get $x = arsinh(\dfrac{1}{\sqrt{2}})$ is this correct?

 

[tiny-tim]

yup!

 

[FeDeX_LaTeX]

You could have saved yourself differentiating $\sinh(x) + \frac{1}{\sinh(x)}$ by instead adding the two terms and noting the hyperbolic trig identity -- otherwise, all correct.