Chapter 4.4, Problem 41E

To determine

To find: The minimum distance such that the curve $y=\sqrt{x}$ come to the point $\left(3/2,0\right)$ .

Answer to Problem 41E

: The minimum distance is $\frac{\sqrt{5}}{2}\text{ units}$.

Explanation of Solution

Given information: The equation of the curve $y=\sqrt{x}$ and its plot along with the point $\left(3/2,0\right)$ is given.

Formula used:

Distance d between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by the distance formula, that is,

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The minimum value of the function or the closest distance will be at the critical point of that function.

Calculation:

Determine the distance between the points $\left(3/2,0\right)$ and $\left(x,\sqrt{x}\right)$ as shown below:

  $\begin{array}{c}d=\sqrt{{\left(x-\frac{3}{2}\right)}^2+{\left(\sqrt{x}-0\right)}^2}\\ =\sqrt{{\left(x-\frac{3}{2}\right)}^2+{\left(\sqrt{x}-0\right)}^2}\end{array}$

Square on both sides of the obtained equation and simplify the resulting equation.

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

Differentiate the obtained equation on both sides with respect to x.

  $\begin{array}{c}\frac{d}{dx}\left(D\right)=\frac{d}{dx}\left(x^2+\frac{9}{4}-2x\right)\\ D^{\prime }\left(x\right)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{9}{4}\right)-2\frac{d}{dx}\left(x\right)\\ =2x+0-2\left(1\right)\\ =2x-2\end{array}$

Equate the obtained derivative to zero to get the critical point.

  $\begin{array}{c}2x-2=0\\ 2x=2\\ x=1\end{array}$

Now, determine the second derivative of the obtained distance by differentiating the obtained first derivative.

  $\begin{array}{c}\frac{d}{dx}\left(D^{\prime }\left(x\right)\right)=\frac{d}{dx}\left(2x-2\right)\\ D^{″}\left(x\right)=\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(2\right)\\ =2\frac{d}{dx}\left(x\right)-0\\ =2\left(1\right)\\ =2\end{array}$

Since it is evident that the second derivative is positive or greater than zero at x=1, the distance between the given points will be minimum at x=1.

Determine the minimum distance by substituting 1 for x into the obtained distance function.

  $\begin{array}{c}{\left.D\right|}_{x=1}=1^2+\frac{9}{4}-2\left(1\right)\\ =1+\frac{9}{4}-2\\ =\frac{4+9-8}{4}\\ =\frac{5}{4}\end{array}$

Substitute $d^2$ for D into the obtained equation and solve for d to obtain the required result.

  $\begin{array}{c}{d}^2=\frac{5}{4}\\ d=\sqrt{\frac{5}{4}}\\ =\frac{\sqrt{5}}{2}\text{ units}\end{array}$

As a result, the curve would come to close the point (3/2, 0) at a distance of $\frac{\sqrt{5}}{2}\text{ units}$ .