Question

Consider the line y=4 x -1 and the point P=(2,0). (a) Write the formula for a function d(x) that describes the distance between the point P and a point (x,y) on the line. You final answer should only involve the variable x. Then d(x) = √(4−x)2(4x−1)2 (b) d'(x)= (c) The critical number is x= . (d) The closest point on the line to P is ( , ).

Answer 1

Answer:

a) d(x)=$\sqrt{17x^{2} -12x+5}$

b)d'(x)=$\frac{17x-6}{\sqrt{17x^{2} -12x+5} }$

c)The critical point is x=$\frac{6}{17}$

d)Closest point is ($\frac{6}{17}$,$\frac{7}{17}$

Step-by-step explanation:

We are given the line

$y=4x-1$

Let a point Q($x,y$) lie on the line.

Point P is given as P(2,0)

By distance formula, we have the distance D between any two points

A($x_{1},y_{1}$) and B($x_{2},y_{2}$) as

D=$\sqrt{(x_{1}-x_{2})^2 + (y_{1}-y_2)^2}$

Thus,

d(x)=$\sqrt{(x-2)^2+(y-0)^2}$

But we have, $y=4x-1$

So,

d(x)=$\sqrt{(x-2)^2+(4x-1)^2}$

Expanding,

d(x)=$\sqrt{17x^2-12x+5}$  - - - (a)

Now,

d'(x)= $\frac{\frac{d}{dx} (17x^2-12x+5)}{2(\sqrt{17x^2-12x+5}) }$

i.e.

d'(x)=$\frac{17x-6}{\sqrt{17x^{2} -12x+5} }$ - - - (b)

Now, the critical point is where d'(x)=0

⇒ $\frac{17x-6}{\sqrt{17x^{2} -12x+5} }$ =0

⇒$x=\frac{6}{17}$    - - - (c)

Now,

The closest point on the given line to point P is the one for which d(x) is minimum i.e. d'(x)=0

⇒$x=\frac{6}{17}$

as $y=4x-1$

⇒y=$\frac{7}{17}$

So, closest point is ($\frac{6}{17},\frac{7}{17}$)   - - -(d)