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 $\frac{d}{dx}L^2(x)=0 \iff \frac{d}{dx}L(x)=0$
$$ L'(x)=2x+\frac{4 \sqrt[3]{x}}{3}-4 $$
Now I want to solve when
$$ L'(x)=0 \implies 2x + \frac{4\sqrt[3]{x}}{3}-4=0 $$
$$ \implies 2x+\frac{4\sqrt[3]{x}}{3}=4 $$
$$ \implies 6x+4\sqrt[3]{x}=12 $$ $$ \implies \sqrt[3]{x}=3-\frac{6x}{4} $$ $$ \implies x = (3-\frac{6x}{4})^3 $$ $$ \implies 27x^3-162x^2+332x-216=0 $$ $$ \implies x= \frac{2}{9}(9-\frac{2}{\sqrt[3]{\sqrt{737}-27}})+\sqrt[3]{\sqrt{737}- 27} \approx 1.2768 $$
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 $x \in [1.5,2]$.
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 $(2,8)$.