Q. 34 - Calculus

Question

find the point on the graph of the function $f$ that is closest to the point $(a,b)$ by minimizing the square of the distance from the graph to the point. 

      $f\left(x\right)=\sqrt{x^2+1}$  and the point $(2,0)$

Step-by-Step Solution

Answer

Ans: The point is, $( 1,\sqrt{2} )$

Step 1. Given information.

given,  $f\left(x\right)=\sqrt{x^2+1}$

and the point is, (2,0)

Step 2. The objective is to find a point closest to the function that is closest to the point by minimizing the graph from the distance to the point.

Let, the point be $(x,y)$.

The distance between $(2,0)$ and $(x,y)$ is,

    $$\begin{gathered} D\left(x\right) = \sqrt{{(0-y)}^2+{(2-x)}^2} \\ D\left(x\right) =\sqrt{y^2+4+x^2-4x} \\ D\left(x\right) =\sqrt{{\left(\sqrt{x^2+1}\right)}^2+x^2+4-4x} \\ D\left(x\right) =\sqrt{2x^2-4x+5} \end{gathered}$$

Step 3. Now,

$$\begin{gathered} D\left(x\right) =\sqrt{2x^2-4x+5} \\ D\text{'}\left(x\right)=\frac{2x-2}{\sqrt{2x^2-4x+5}} \end{gathered}$$

Again,

       $D\text{'}\left(x\right)=0$

      $$\begin{gathered} \frac{2x-2}{\sqrt{2x^2-4x+5}}=0 \\ 2x-2=0 \\ x=1 \end{gathered}$$

Step 4. Putting x = 1 in f ( x ) = x 2 + 1 .

      $f\left(x\right)=\sqrt{2}$

Therefore, the point is $( 1,\sqrt{2} )$ .