- #1
synkk
- 216
- 0
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?
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?