# Find the shortest path from a point to curve

## Problem

Find the shortest length from the point $$(2,8)$$ to the curve $$C=\{(x,y)|y=x^{2/3}+8, x \ge 0\}$$

## Attempt to solve

Here is the associated plot of the situation: By drawing a triangle to this image we can form a length function of $$x$$ by Pythagora's theorem. This function $$L(x)$$ is:

$$L^2(x)=(2-x)^2+(x^{2/3}+8)^2$$

Now since $$\frac{d}{dx}L^2(x)=0 \iff \frac{d}{dx}L(x)=0$$

$$L'(x)=2x+\frac{4 \sqrt{x}}{3}-4$$

Now I want to solve when

$$L'(x)=0 \implies 2x + \frac{4\sqrt{x}}{3}-4=0$$

$$\implies 2x+\frac{4\sqrt{x}}{3}=4$$

$$\implies 6x+4\sqrt{x}=12$$ $$\implies \sqrt{x}=3-\frac{6x}{4}$$ $$\implies x = (3-\frac{6x}{4})^3$$ $$\implies 27x^3-162x^2+332x-216=0$$ $$\implies x= \frac{2}{9}(9-\frac{2}{\sqrt{\sqrt{737}-27}})+\sqrt{\sqrt{737}- 27} \approx 1.2768$$

Now if you take a look at the image this is probably wrong since I would approximate just by looking at the image that the shortest length is in $$x \in [1.5,2]$$.

• @ParclyTaxel Why do you think i would be using wrong image ? This might be the case but i would like to know how did you come up with this ? – Tuki Oct 7 '18 at 15:27

Hint: $$(i)$$ Find the equation of normal of the curve and put the point $$(2,8)$$. Then you will get the equation of normal passing through the point. Normals will be of the form $$y=mx+c$$, with $$m=-\frac{dx}{dy}$$. As, tangent and normal are perpendicular to each other, slope of tangent,i.e $$dy/dx$$, times slope of normal is $$-1$$.
$$(ii)$$ Find the point on the curve where that line intersects.
$$(iii)$$ Then find the distance of that point(on curve) to $$(2,8)$$.
That distance is the shortest path, as it is perpendicular from $$(2,8)$$.