Q. 32 - Calculus

Question

In Exercises 31–34, 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.

$f\left(x\right)=x^2$ and the point $(0,3)$

Step-by-Step Solution

Answer

The point on the graph of the function that is close to the point is $(\sqrt{\frac{5}{2}},\frac{5}{2}),(-\sqrt{\frac{5}{2}},\frac{5}{2})$

Step 1. Given Information.

The function:

$f\left(x\right)=x^2$

The point:

$(0,3)$

Step 2. Write the distance between thee function and the point.

The distance between $(0,3)$ and $(x,y)$ is:

$$\begin{gathered} D\left(x\right)=\sqrt{{(3-y)}^2+{(0-x)}^2} \\ =\sqrt{{(3-x^2)}^2+{(-x)}^2} \\ =\sqrt{x^4-5x^2+9} \end{gathered}$$

Step 3. Find the derivative of the function.

$$\begin{gathered} D\left(x\right)=\sqrt{x^4-5x^2+9} \\ D\text{'}\left(x\right)=\frac{1}{2}{(x^4-5x^2+9)}^{\frac{-1}{2}}.(4x^3-10x) \\ =\frac{2(2x^3-5x)}{2(x^4-5x^2+9)} \\ =\frac{(2x^3-5x)}{x^4-5x^2+9} \end{gathered}$$

Step 4. Find the point close to the function and the point.

To find the point that is closest to the function and the point, 

$$\begin{gathered} D\text{'}\left(x\right)=0 \\ \frac{(2x^3-5x)}{x^4-5x^2+9}=0 \\ 2x^3-5x=0 \\ 2x^2-5=0 \\ 2x^2=5 \\ x^2=\frac{5}{2} \\ x=\pm \sqrt{\frac{5}{2}} \end{gathered}$$

Step 5. Find y.

Substitute the value of x in the function to get y, 

$$\begin{gathered} y=x^2 \\ ={(\pm \sqrt{\frac{5}{2}})}^2 \\ =\frac{5}{2} \end{gathered}$$

So the closest point is $(\sqrt{\frac{5}{2}},\frac{5}{2}),(-\sqrt{\frac{5}{2}},\frac{5}{2})$