minimum distance between two curves parabola and any function

Exercise:

Find the minimum distance between the curves $\left( C \right)\,\,:\,\,\,f\left( x \right)=\sqrt{x}\,\,\,\And \,\,\,\left( C' \right)\,\,:\,\,g\left( x \right)={{x}^{2}}+1$

Note that the distance is not horizontal or vertical line ( see the figure )

Solution: Let $P$ be a point on the parabola $\left( C \right)$

So the coordinates of $P$ will be $P\left( \alpha ,\sqrt{\alpha } \right)$

And $Q$ be a point on the curve $\left( C' \right)$

So the coordinate of $Q$ are $Q\left( x,{{x}^{2}}+1 \right)$

So the slopes of the tangent lines to $\left( C \right)\,\,\And \,\left( C' \right)$ are:

${{\vec{v}}_{t}}={{\left. \frac{d}{dx}f\left( x \right) \right|}_{x=\alpha }}={{\left. \frac{1}{2\sqrt{x}} \right|}_{x=\alpha }}=\frac{1}{2\sqrt{\alpha }}$ and ${{\vec{u}}_{t}}=\frac{d}{dx}g\left( x \right)=2x$

Note that the tangent slopes are parallel iff the tangent slopes are equal

So $\frac{1}{2\sqrt{\alpha }}=2x\Leftrightarrow 4x\sqrt{\alpha }=1\Leftrightarrow 16{{x}^{2}}\alpha =1\Leftrightarrow \alpha =\frac{1}{16{{x}^{2}}}$     (*)

But the slope of $PQ$ is $m=\frac{{{y}_{Q}}-{{y}_{P}}}{{{x}_{Q}}-{{x}_{P}}}=\frac{{{x}^{2}}+1-\sqrt{\alpha }}{x-\alpha }$

Now to have $\left( QP \right)\bot \left( C \right)$ we need to find the normal slopes

So the normal slope of $\left( C \right)$ is $2xm'=-1\Leftrightarrow m'=\frac{-1}{2x}$

Observe that the normal slope of $\left( C \right)$ is parallel to $\left( PQ \right)$

Hence $m=m'\Leftrightarrow \frac{{{x}^{2}}+1-\sqrt{\alpha }}{x-\alpha }=\frac{-1}{2x}\Leftrightarrow \frac{{{x}^{2}}+1-\frac{1}{4x}}{x-\frac{1}{16{{x}^{2}}}}+\frac{1}{2x}=0$

$\Leftrightarrow \frac{\frac{4{{x}^{3}}+4x-1}{4x}}{\frac{16{{x}^{3}}-1}{16{{x}^{2}}}}+\frac{1}{2x}=0\Leftrightarrow \frac{16{{x}^{2}}\left( 4{{x}^{3}}+4x-1 \right)}{4x\left( 16{{x}^{3}}-1 \right)}+\frac{1}{2x}=0$

$\Leftrightarrow \frac{8{{x}^{2}}\left( 4{{x}^{3}}+4x-1 \right)+16{{x}^{3}}-1}{2x\left( 16{{x}^{3}}-1 \right)}=0$

$\Leftrightarrow 32{{x}^{5}}+32{{x}^{3}}-8{{x}^{2}}+16{{x}^{3}}-1=0\Leftrightarrow 32{{x}^{5}}+48{{x}^{3}}-8{{x}^{2}}-1=0$

Solving to $x=0.331695\Leftrightarrow \alpha =\frac{1}{16}{{x}^{-2}}=\frac{1}{16}{{\left( 0.331695 \right)}^{-2}}=1.46797$

Hence the minimum distance from $P$ to $Q$ is $PQ=\sqrt{{{\left( {{x}_{Q}}-{{x}_{P}} \right)}^{2}}+{{\left( {{y}_{Q}}-{{y}_{P}} \right)}^{2}}}$

$=\sqrt{{{\left( 1.46797-0.331695 \right)}^{2}}+{{\left( \sqrt{1.46797}-{{\left( 0.331695 \right)}^{2}}-1 \right)}^{2}}}=1.14081$