Chapter 4.4, Problem 42E

To determine

To find: The closest or minimum distance such that the semicircle $y=\sqrt{16-x^2}$ come to the point $\left(1,\sqrt{3}\right)$ .

Answer to Problem 42E

The minimum distance is $2\text{ units}$.

Explanation of Solution

Given information: The equation of the semicircle $y=\sqrt{16-x^2}$ is given along with the point $\left(1,\sqrt{3}\right)$ .

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:

Let the point on the given semicircle be $\left(x,y\right)$ , such that the point will become $\left(x,\sqrt{16-x^2}\right)$ .

Consider the distance between the points $\left(1,\sqrt{3}\right)$ and $\left(x,\sqrt{16-x^2}\right)$ to be d.

Determine the distance d between the points $\left(1,\sqrt{3}\right)$ and $\left(x,\sqrt{16-x^2}\right)$ as shown below:

  $d=\sqrt{{\left(x-1\right)}^2+{\left(\sqrt{16-x^2}-\sqrt{3}\right)}^2}$

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

  $\begin{array}{c}{d}^2={\left(\sqrt{{\left(x-1\right)}^2+{\left(\sqrt{16-x^2}-\sqrt{3}\right)}^2}\right)}^2\\ D={\left(x-1\right)}^2+{\left(\sqrt{16-x^2}-\sqrt{3}\right)}^2\\ =x^2+1-2x+{\left(\sqrt{16-x^2}\right)}^2+{\left(\sqrt{3}\right)}^2-2\sqrt{3}\sqrt{16-x^2}\\ =x^2+1-2x+16-x^2+3-2\sqrt{3}\sqrt{16-x^2}\\ =-2x+20-2\sqrt{48-3x^2}\end{array}$

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

  $\begin{array}{c}\frac{d}{dx}\left(D\right)=\frac{d}{dx}\left(-2x+20-2\sqrt{48-3x^2}\right)\\ D^{\prime }\left(x\right)=-2\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(20\right)-2\frac{d}{dx}\left(\sqrt{48-3x^2}\right)\\ =-2\left(1\right)+0-2\left(\frac{1}{2}{\left(48-3x^2\right)}^{\frac{1}{2}-1}\frac{d}{dx}\left(48-3x^2\right)\right)\\ =-2-\left({\left(48-3x^2\right)}^{-\frac{1}{2}}\left(\frac{d}{dx}\left(48\right)-3\frac{d}{dx}\left(x^2\right)\right)\right)\\ =-2-\left({\left(48-3x^2\right)}^{-\frac{1}{2}}\left(0-3\left(2x\right)\right)\right)\\ =-2-\left({\left(48-3x^2\right)}^{-\frac{1}{2}}\left(-6x\right)\right)\\ =-2+\frac{6x}{\sqrt{48-3x^2}}\end{array}$

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

  $\begin{array}{c}-2+\frac{6x}{\sqrt{48-3x^2}}=0\\ \frac{6x}{\sqrt{48-3x^2}}=2\\ 6x=2\sqrt{48-3x^2}\\ 3x=\sqrt{48-3x^2}\\ 9x^2=48-3x^2\\ 12x^2=48\\ x^2=\frac{48}{12}\\ =4\end{array}$

This implies that $x=\pm 2$ .

Since the distance along with the point $\left(1,\sqrt{3}\right)$ has to be determined, consider only the positive value of x. This implies that $x=2$ .

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

  $\begin{array}{c}\frac{d}{dx}\left(D^{\prime }\left(x\right)\right)=\frac{d}{dx}\left(2+\frac{6x}{\sqrt{48-3x^2}}\right)\\ D^{″}\left(x\right)=\frac{d}{dx}\left(2\right)+\frac{d}{dx}\left(\frac{6x}{\sqrt{48-3x^2}}\right)\\ =0+\left(\frac{\sqrt{48-3x^2}\frac{d}{dx}\left(6x\right)-6x\frac{d}{dx}\left(\sqrt{48-3x^2}\right)}{{\left(\sqrt{48-3x^2}\right)}^2}\right)\\ =\frac{6\sqrt{48-3x^2}-6x\left(\frac{1}{2}{\left(48-3x^2\right)}^{\frac{1}{2}-1}\frac{d}{dx}\left(48-3x^2\right)\right)}{48-3x^2}\\ =\frac{6\sqrt{48-3x^2}-3x\left(\frac{-6x}{\sqrt{48-3x^2}}\right)}{48-3x^2}\\ =\frac{6\sqrt{48-3x^2}+\frac{18x^2}{\sqrt{48-3x^2}}}{48-3x^2}\\ =\frac{6\left(48-3x^2\right)+18x^2}{{\left(48-3x^2\right)}^{\frac{3}{2}}}\\ =\frac{288-18x^2+18x^2}{{\left(48-3x^2\right)}^{\frac{3}{2}}}\\ =\frac{288}{{\left(48-3x^2\right)}^{\frac{3}{2}}}\end{array}$

Determine the value of the second derivative at x=2.

  $\begin{array}{c}{D}^{″}\left(2\right)=\frac{288}{{\left(48-3{\left(2\right)}^2\right)}^{\frac{3}{2}}}\\ =\frac{288}{{\left(48-3\left(4\right)\right)}^{\frac{3}{2}}}\\ =\frac{288}{{\left(48-12\right)}^{\frac{3}{2}}}\\ =\frac{288}{{\left(36\right)}^{\frac{3}{2}}}\\ =\frac{288}{{\left(6^2\right)}^{\frac{3}{2}}}\\ =\frac{288}{216}\\ >0\end{array}$

Since the second derivative at x=2 is positive, the distance will be closest or minimum at x=2.

Determine the minimum distance by substituting 2 for x into the obtained distance function.

  $\begin{array}{c}{\left.D\right|}_{x=2}=-2\left(2\right)+20-2\sqrt{48-3{\left(2\right)}^2}\\ =-4+20-2\sqrt{48-12}\\ =16-2\sqrt{36}\\ =16-2\left(6\right)\\ =16-12\\ =4\end{array}$

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

  $\begin{array}{c}{d}^2=4\\ d=\pm 2\text{ units}\end{array}$

As a result, the curve would come to close the point $\left(1,\sqrt{3}\right)$ at a distance of $2\text{ units}$ .