Chapter 5.4, Problem 41E

To determine

To find the minimum distance of the curve.

Answer to Problem 41E

The minimum distance is $d=\frac{\sqrt{5}}{2}.$

Explanation of Solution

Given information: Given point $\left(\frac{3}{2},0\right).$

Calculation:

Distance of the point $\left(\frac{3}{2},0\right)$ from the point $\left(x,\sqrt{x}\right)$ is $d=\sqrt{{\left(x-\frac{3}{2}\right)}^2+{\left(\sqrt{x}-0\right)}^2}$

Square of the distance is ,

  $\begin{array}{l}D={\left(x-\frac{3}{2}\right)}^2+{\left(\sqrt{x}-0\right)}^2\\ =x^2+\frac{9}{4}-3x+x\\ =x^2+\frac{9}{4}-2x\end{array}$

For any function $f\left(x\right)$ the condition of minimum and maximum is obtained by find $f^{\prime }\left(x\right)=0$ ,

Critical points and end points

  $\begin{array}{l}D=x^2+\frac{9}{4}-2x\\ \frac{dD}{dx}=2x-2\end{array}$

Critical points occur at,

  $\begin{array}{l}{D}^{\prime }\left(x\right)=0\\ 2x-2=0\\ x=1\end{array}$

Here $\frac{dD}{dx}<0\text{ for }x<1\text{ and}\frac{dD}{dx}>0\text{ for }x>1$ this corresponds to minimum distance.

For $x=1$

  $\begin{array}{l}D=x^2+\frac{9}{4}-2x\\ =1+\frac{9}{4}-2\\ =\frac{5}{4}\end{array}$

Thus, the minimum distance is $d=\frac{\sqrt{5}}{2}.$