Minimize distance from curve to fixed point

Find the point of the parabola   $\displaystyle{ x =y^2 }$   nearest to the point   $(3,-3)\;$ .   This is illustrated by the illustrated diagram below.   Note that this example is algebraically somewhat more involved than the preceding two.


Experimentation suggests that the distance is minimized when   $y$   is about   $-1.83\,$ , and this minimal distance is approximately   $1.22\;$ .   There is also an unusual dip in the distance function when   $y$   is approximately   $1\;$ .   As   $y$   increases from   $0\,$ , when the distance increases to about   $4.473\,$ , hovers there for a bit, then dips slightly when   $y$   is about   $1\,$ , and then increases rapidly as   $y$   continues to get larger.

Letting the second name ( $\,y\,$ ) of a point on the curve serve as our variable, we wish to minimize

$\displaystyle{ \text{dist}=\sqrt{(x-3)^2+(y-(-3))^2}=\sqrt{\left(y^2-3\right)^2+(y-(-3))^2}\; {\rm .} }$

In fact, as this function is strictly positive, and squaring is an increasing function for positive real numbers, the minimum of this occurs for precisely the same   $y$   value as the minimum of its square.   Thus we instead minimize

$\displaystyle{ \text{dist}^2=(x-3)^2+(y-(-3))^2=\left(y^2-3\right)^2+(y-(-3))^2 \; {\rm .} }$

This is a common simplification of a problem   –   optimize a related but easier function.

This new function, $\displaystyle{ \, \text{dist}^2(y)\, }$ , is defined for all   $y$   values.   We cannot directly apply the Extreme Value Theorem.   Nevertheless, let us consider the derivative of this function.

$\displaystyle{ \frac{\mathrm{d}\phantom{y}}{\mathrm{d}y}\,\left(\text{dist}^2(y)\right) =2\, \left(2\, y\, \left(y^2-3\right) +(y+3)\right) \; {\rm .} }$

This exists for all   $y\in \mathbb{R}\; $ .   So if we have a minimum, it will be at a place where the derivative   $=0\;$ .   We proceed to solve this equation.

As   $\displaystyle{ \frac{\mathrm{d}\phantom{y}}{\mathrm{d}y}\,\left(\text{dist}^2(y)\right) =2\, \left(2\, y\, \left(y^2-3\right) +(y+3)\right) =0 }$   when   $\displaystyle{ 2\, y\, \left(y^2-3\right) +(y+3) =0\, }$ , or   $\displaystyle{ 2\, y^3 -5\, y +3 =0\, }$ , we study this cubic equation.   It factors to   $\displaystyle{ (y-1)\,\left( 2\, y^2 +2\, y -3 \right) =0\, }$ , and thus   $\displaystyle{ \frac{\mathrm{d}\phantom{y}}{\mathrm{d}y}\,\left(\text{dist}^2(y)\right) =0 }$   when either   $y=1\,$ , or   $\displaystyle{ 2\, y^2 +2\, y -3 =0\; }$ .   This last quadratic equation has the two solutions   $\displaystyle{ y =-\frac{1}{2} +\frac{1}{2}\,\sqrt{7} \sim .8229 }$   and   $\displaystyle{ y =-\frac{1}{2} -\frac{1}{2}\,\sqrt{7} \sim -1.8229\; }$ .

Consider the derivative   $\displaystyle{ \frac{\mathrm{d}\phantom{y}}{\mathrm{d}y}\,\left(\text{dist}^2(y)\right) =2\, \left(2\, y\, \left(y^2-3\right) +(y+3)\right) = 4\, y^3 -10\, y +6\, }$ , for   $y\in \mathbb{R}\;$ .   We have seen that it   $=0$   for   $y=1\,$ , $\displaystyle{ \, y =-\frac{1}{2} +\frac{1}{2}\,\sqrt{7}\, }$ , and   $\displaystyle{ y =-\frac{1}{2} -\frac{1}{2}\,\sqrt{7}\; }$ .   On the remaining intervals of   $\mathbb{R}\,$ , that is on   $\displaystyle{ \left( -\infty, -\frac{1}{2} -\frac{1}{2}\,\sqrt{7} \right)\, }$ , $\displaystyle{ \, \left( -\frac{1}{2} -\frac{1}{2}\,\sqrt{7} , -\frac{1}{2} +\frac{1}{2}\,\sqrt{7} \right)\, }$ , $\displaystyle{ \, \left( -\frac{1}{2} +\frac{1}{2}\,\sqrt{7} , 1 \right)\, }$ , and $(1,\infty)\,$ , this derivative is negative, positive, negative and positive respectively.   We see that   $\displaystyle{ y =-\frac{1}{2} -\frac{1}{2}\,\sqrt{7} }$   and   $y=1$   are local minima, and   $\displaystyle{ y =-\frac{1}{2} +\frac{1}{2}\,\sqrt{7} }$   is a local maximum, of the function   $\displaystyle{ \text{dist}^2=\left(y^2-3\right)^2+(y+3)^2 \; }$ .   The function has a minimum and it must occur at one of   $\displaystyle{ y =-\frac{1}{2} -\frac{1}{2}\,\sqrt{7} }$   and   $y=1\;$ .

Since   $\displaystyle{ \text{dist}^2\left( y =-\frac{1}{2} -\frac{1}{2}\,\sqrt{7} \right) }$   is   $\displaystyle{ \frac{43}{4} -\frac{7}{2}\,\sqrt{7} \sim 1.4897\, }$ , so that   $\displaystyle{ \text{dist}\left( y =-\frac{1}{2} -\frac{1}{2}\,\sqrt{7} \right) = \sqrt{ \frac{43}{4} -\frac{7}{2}\,\sqrt{7} } \sim 1.2205\, }$ , and   $\displaystyle{ \text{dist}^2 ( y =1 ) }$   is   $20\,$ , so that   $\displaystyle{ \text{dist}(y=1) =\sqrt{2\,\sqrt{7} } \sim 4.4722\, }$ , the distance   $\text{dist}(y)$   is minimized when   $\displaystyle{ y =-\frac{1}{2} -\frac{1}{2}\,\sqrt{7} \sim -1.8229\, }$ , and the minimum distance is

$\displaystyle{ \text{dist}\left( y =-\frac{1}{2} -\frac{1}{2}\,\sqrt{7} \right) = \sqrt{ \frac{43}{4} -\frac{7}{2}\,\sqrt{7} } \sim 1.2205\; }$ .

This analysis gives firm support to our experimental observations.