Respuesta :
Answer:
Step-by-step explanation:
d(x)=√((x-2)2+(y-0)²)
=√((x-2)²+y²)
=√((x-2)²+(x-1)²)
=√(x²-4x+4+x²-2x+1)
=√(2x²-6x+5)
D=d²(x)=2x²-6x+5
[tex]b.\\\frac{dD}{dx}=4x-6\\\frac{dD}{dx}=0,gives \\4x-6=0\\x=\frac{3}{2}\\\frac{d^2D}{dx^2}=4 >0 at x=\frac{3}{2}\\so~D~or~d^2~or~d~is~minimum~at~x=\frac{3}{2}\\so~y=1(\frac{3}{2} )-1=1/2=0.5\\so~nearest~point~is~(1.5,0.5)[/tex]
For the line [tex]y=x-1[/tex] and the point [tex]P=(2,0)[/tex]
- The distance function is [tex]D(x)=\sqrt{2x^2-6x+5}[/tex]
- The critical number is [tex]x=\frac{3}{2}[/tex]
- The closest point on the line to P is [tex](\frac{3}{2},\frac{1}{2})[/tex]
1. Let [tex]D(x)[/tex] be the distance function for the distance between the point
[tex]P=(2,0)[/tex] and the line [tex]y=x-1[/tex]. The distance function will be
[tex]D(x)=\sqrt{(x-2)^2+y^2}\\\implies D(x)=\sqrt{2x^2-6x+5}[/tex]
2. To find the critical number, first find [tex]D'(x)[/tex]. Then, find the points at
which
[tex]D'(x)=0 \text{ or } D'(x) \text{ does not exist}[/tex]
[tex]D'(x)=\frac{d}{dx}\sqrt{2x^2-6x+5}=\frac{2x-3}{\sqrt{2x^2-6x+5}}[/tex]
[tex]D'(x)=0\text{ when }2x-3=0\\\implies x=\frac{3}{2}[/tex]
[tex]D'(x)\text{ does not exist when }\sqrt{2x^2-6x+5}=0\\[/tex]
but, [tex]2x^2-6x+5 =0[/tex] only has complex solutions for [tex]x[/tex]. Therefore, the
only critical point is [tex]x=\frac{3}{2}[/tex]
3. To find the closest point on the line to P, find out if the critical point
[tex]x=\frac{3}{2}[/tex] is a minimum point
[tex]D''(x)=\frac{1}{\sqrt{(2x^2-6x+5)^3}}[/tex]
[tex]D''(\frac{3}{2})=\frac{1}{\sqrt{(2(\frac{3}{2})^2-6(\frac{3}{2})+5)^3}}=2\sqrt{2}\\\\\therefore x=\frac{3}{2} \text{ is a minimum point since } D''(\frac{3}{2})>0[/tex]
[tex]y=x-1=(\frac{3}{2})-1=\frac{1}{2}[/tex]
The closest point is [tex](\frac{3}{2},\frac{1}{2})[/tex]
Another solved problem on distance between a point and a line can be found in the link: https://brainly.com/question/13942941
