f-x-e-x-g-x-ln-x-h-x-x-A-line-L-is-perpendicular-to-h-x-at-point-P-x-y-and-extends-and-disects-f-x-and-g-x-The-length-of-L-between-f-x-and-g-x-is-r-When-is-r-minimum-

Question Number 5692 by FilupSmith last updated on 24/May/16

f (x) = e^{x}

g (x) = \ln x

h (x) = x

A line L is perpendicular to h (x) at point

P (x, y) and extends and disects f (x) and g (x) .

The length of L between f (x) a n d g (x)

is r . When is r minimum ?

Commented by FilupSmith last updated on 24/May/16

Commented by Yozzii last updated on 25/May/16

(1) y = e^{x}, (2) y = x (x \in R)

T h e p e r p e n d i c u l a r d i s t a n c e b e t w e e n

(1) a n d (2) i s n e e d e d . T h e p o i n t s o f

i n t e r e c t i o n c a n b e f o u n d u s i n g t h e

l i n e y = c - x w h i c h i s n o r m a l t o (2)

b u t h a s a n u n k n o w n y - i n t e r c e p t .

F o r (2) m e e t i n g y = c - x

c - x = x \Rightarrow x = 0.5 c \Rightarrow (0.5 c, 0.5 c)

F o r (1) m e e t i n g y = c - x

c - x = e^{x} \Rightarrow (c - W (e^{c}), W (e^{c}))

I u s e d W o l f r a m A l p h a t o t e l l m e

a b o u t t h e p r o p e r t i e s o f t h e p r o d u c t - l o g

f u n c t i o n W (x) . T h e c a l c u l a t i o n t h a t

f o l l o w s f i n d s h a l f t h e v a l u e o f r .

U s i n g l = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} (l = 0.5 r)

\Rightarrow l^{2} = {(W (e^{c}) - 0.5 c)}^{2} + {(W (e^{c}) - 0.5 c)}^{2}

l = \sqrt{2} (W (e^{c}) - 0.5 c)

\frac{d l}{d c} = \sqrt{2} (\frac{d W (e^{c})}{d c} - 0.5)

\frac{d W (e^{c})}{d c} = \frac{d W (u)}{d u} \times \frac{d u}{d c} = \frac{W (u) e^{c}}{u (1 + W (u))} = \frac{W (e^{c})}{W (e^{c}) + 1} (c h a i n r u l e)

\frac{d l}{d c} = \sqrt{2} (\frac{W (e^{c}) e^{c}}{e^{u} (1 + W (e^{c}))} - 0.5)

\frac{d l}{d c} = \sqrt{2} (\frac{0.5 (W (e^{c}) - 1)}{1 + W (e^{c})})

\frac{d l}{d c} = \frac{\sqrt{2}}{2} (\frac{W (e^{c}) - 1}{W (e^{c}) + 1})

A t s t a t i o n a r y v a l u e o f l, \frac{d l}{d c} = 0 .

\Rightarrow W (e^{c}) = 1

∴ W (e^{c}) = 1 \Rightarrow e^{c} = e \Rightarrow c = 1

\frac{d^{2} l}{{d c}^{2}} = \frac{\sqrt{2}}{2} (\frac{(W (e^{c}) + 1) (\frac{W (e^{c})}{1 + W (e^{c})}) - (W (e^{c}) - 1) \frac{W (e^{c})}{1 + W (e^{c})}}{{(1 + W (e^{c}))}^{2}})

\frac{d^{2} l}{{d c}^{2}} = \frac{W (e^{c}) \sqrt{2}}{{(1 + W (e^{c}))}^{3}}

S i n c e W (e^{c}) = W (e^{1}) = 1 \Rightarrow \frac{d^{2} l}{{d c}^{2}} = \frac{\sqrt{2}}{8} > 0 \Rightarrow m i n i m u m a t c = 1 .

∴ m i n (l) = \sqrt{2} (1 - 0.5) = \frac{\sqrt{2}}{2}

T h e r e q u i r e d d i s t a n c e i s 2 m i n (l) = \sqrt{2}

s i n c e y = x i s t h e r e f l e c t i o n l i n e .

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