Chapter 3 - Applications of Differentiation - 3.7 Exercises - Page 220: 13

$$\left( {1,1} \right)$$

$$\eqalign{ & {\text{Let the function be }}f\left( x \right) = {x^2} \cr & y = {x^2}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{The distance between two points }}\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right){\text{ is given}} \cr & {\text{by : }}d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \cr & {\text{We know the point }}\left( {{x_1},{y_1}} \right) = \left( {2,\frac{1}{2}} \right),{\text{ and }}\left( {x,y} \right) = \left( {{x_2},{y_2}} \right) \cr & d = \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - \frac{1}{2}} \right)}^2}} {\text{ }}\left( {\bf{2}} \right) \cr & {\text{Substitute }}{x^2}{\text{ for }}y{\text{ into equation }}\left( {\bf{2}} \right) \cr & d = \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {{x^2} - \frac{1}{2}} \right)}^2}} \cr & d = \sqrt {{x^2} - 4x + 4 + {x^4} - {x^2} + \frac{1}{4}} \cr & d = \sqrt {{x^4} - 4x + \frac{{17}}{4}} \cr & {\text{If the radicand is the smallest, then }}d{\text{ is the smallest, so we need the}} \cr & {\text{minimum value of the radicand to be }}r\left( x \right) = {x^4} - 4x + \frac{{17}}{4} \cr & {\text{Differentiate}} \cr & \frac{{dr}}{{dx}} = \frac{d}{{dx}}\left[ {{x^4} - 4x + \frac{{17}}{4}} \right] \cr & \frac{{dr}}{{dx}} = 4{x^3} - 4 \cr & \frac{{dr}}{{dx}} = 0 \cr & 4{x^3} - 4 = 0 \cr & x = 1 \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}r}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {4{x^3} - 4} \right] \cr & \frac{{{d^2}r}}{{d{x^2}}} = 12{x^2} \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}r}}{{d{x^2}}}} \right|_{x = 1}} = 12{\left( 1 \right)^2} > 0{\text{ Relative minimum}} \cr & {\text{Let }}x = 1 \cr & {\text{Calculate }}y \cr & y = {x^2} \to y = {\left( 1 \right)^2} = 1 \cr & {\text{The point }}\left( {x,y} \right){\text{ nearest to }}\left( {2,\frac{1}{2}} \right){\text{ is }}\left( {x,y} \right) = \left( {1,1} \right) \cr} $$