give data:
The currve moving along with point P is,
[tex]y=x^2-1[/tex]The line moving along the point Q is,
[tex]y=x-3[/tex]Let us now graph the given curve and line,
The closest point for the curve moving point P is,
[tex]\begin{gathered} \frac{d(x^2-1)}{dx}=1 \\ 2x=1 \\ x=\frac{1}{2} \end{gathered}[/tex]thus,
[tex]\begin{gathered} y=x^2-1 \\ =(\frac{1}{2})^2-1 \\ y=-\frac{3}{4} \end{gathered}[/tex]The distance between the two given paths is the perpendicular distance from the line y=x-3 to the point (1/2,-3/4).
The equation of the line in point and slope form is,
[tex]\begin{gathered} y-(\frac{3}{4})=-1(x-\frac{1}{2}) \\ y=-x+\frac{1}{2}-\frac{3}{4}=-x-\frac{1}{4} \\ y=-x-\frac{1}{4} \end{gathered}[/tex]at the coordinates of point Q is the distance between the two points have minimum value.,
[tex]\begin{gathered} -x-\frac{1}{4}=x-3 \\ 2-x=3-\frac{3}{4}=2\frac{1}{4}=\frac{9}{4} \\ x=\frac{9}{8} \\ y=x-3 \\ =\frac{9}{8}-3 \\ y=-\frac{15}{8} \end{gathered}[/tex]The point is (9/8,-15/8)
now we find the distance between the point by using the formula,'
[tex]\begin{gathered} \text{length}=\sqrt[]{(y_2-y_1)^2+(^{_{}}x}_2-x_1)^2 \\ \end{gathered}[/tex]thus, the distance is,
[tex]\begin{gathered} \sqrt[]{(\frac{9}{8}-\frac{1}{2})^2+(-\frac{15}{8}-(-\frac{3}{4}))^2} \\ =\sqrt[]{\frac{53}{32}} \\ length=1.28 \end{gathered}[/tex]
