The-shortest-distance-between-the-point-3-2-0-and-the-curve-y-x-x-gt-0-is-

Question Number 56914 by rahul 19 last updated on 26/Mar/19

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

(\frac{3}{2}, 0) a n d t h e c u r v e y = \sqrt{x}, (x > 0) i s ?

Answered by Smail last updated on 26/Mar/19

d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

= \sqrt{{(x - \frac{3}{2})}^{2} + {(\sqrt{x} - 0)}^{2}} = \sqrt{x^{2} - 2 x + \frac{9}{4}}

d^{'} = \frac{2 x - 2}{2 \sqrt{x^{2} - 2 x + \frac{9}{2}}}

C V i s x = 1

d_{m i n} = \sqrt{1^{2} - 2 \times 1 + \frac{9}{4}} = \frac{\sqrt{5}}{2}

Commented by rahul 19 last updated on 26/Mar/19

thank you sir!

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Mar/19

l e t (x, \sqrt{x}) b e a p o i n t o n t h e c u r v e .

d i s t a n c e l = \sqrt{{(x - \frac{3}{2})}^{2} + {(\sqrt{x} - 0)}^{2}}

l^{2} = x^{2} - 2 \times x \times \frac{3}{2} + \frac{9}{4} + x

l^{2} = x^{2} - 2 x + \frac{9}{4}

l = \sqrt{x^{2} - 2 x + \frac{9}{4}}

\frac{d l}{d x} = \frac{1}{2 \sqrt{x^{2} - 2 x + \frac{9}{4}}} \times (2 x - 2)

\frac{d l}{d x} = \frac{x - 1}{\sqrt{x^{2} - 2 x + \frac{9}{4}}}

f o r m a x / m i n \frac{d l}{d x} = 0 s o x = 1

\frac{d^{2} l}{{d x}^{2}} = \frac{\sqrt{x^{2} - 2 x + \frac{9}{4}} \times 1 - (x - 1) \times \frac{1}{2 \sqrt{x^{2} - 2 x + \frac{9}{4}}} \times (2 x - 2)}{(x^{2} - 2 x + \frac{9}{4})}

{(\frac{d^{2} l}{{d x}^{2}})}_{x = 1} = \frac{\sqrt{\frac{5}{4}} \times 1}{(\frac{5}{4})} = \sqrt{\frac{4}{5}} > 0

s o m i n i m u m d i s t a n c e i s b e t w e e n p o i n t (x, \sqrt{x})

t h a t m e a n s (1, 1) a n d (\frac{3}{2}, 0) i s

l = \sqrt{{(\frac{3}{2} - 1)}^{2} + {(1 - 0)}^{2}}

= \sqrt{\frac{1}{4} + 1}

= \sqrt{\frac{5}{4}}

= \frac{\sqrt{5}}{2} \leftarrow t h i s i s t h e s n s w e r

Commented by rahul 19 last updated on 26/Mar/19

thank you sir!

Answered by mr W last updated on 26/Mar/19

d = \sqrt{{(x - \frac{3}{2})}^{2} + {(\sqrt{x} - 0)}^{2}} = \sqrt{x^{2} - 2 x + \frac{9}{4}}

= \sqrt{{(x - 1)}^{2} + \frac{5}{4}} ⩾ \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}

\Rightarrow d_{m i n} = \frac{\sqrt{5}}{2} a t x = 1

Commented by rahul 19 last updated on 26/Mar/19

thank you sir!