Find the point(s) on the curve $y^3=x^2$ closest to the point $P=(0,4).$
I understand that there is a way to solve this, using the distance formula, however this turns out to seem rather complicated. I am also aware that there is a calculus method to solving this question, however am unsure as to what that method is, exactly.
Any help is appreciated. Thanks 🙂
[Math] Finding the points on a curve, closest to a specific point
calculus
Best Answer
The square of the distance between the point $P$ and another point with coordinates $(x,y)$ is $$d^2=x^2+(y-4)^2.$$
Since the curve is defined as a set of points $(x,y)$ related to each other by the relation $y^3=x^2$, we have the square of the distance from $P$ to the curve equal to: $$d^2=y^3+(y-4)^2$$ Thus, the task is to find the smallest $d$ possible, when $d$ is actually a function of $y$. For this, we employ the derivative of $d(y)$ to find extremums of the function: $$d(y)=\sqrt{y^3+(y-4)^2}$$ $$d'(y)=\frac{3y^2+2(y-4)}{2\sqrt{y^3+(y-4)^2}}$$ Now, to find extremums, solve $d'(y)=0$, or in our case $3y^2+2(y-4)=0$. This quadratic equation has two solutions $y=-2$ and $y=\frac{4}{3}$. But the solution $y=-2$ is not acceptable, since $y^3=x^2$, and that would mean $-8=x^2$. Hence, we are left with only one solution and two pairs (do you know why?) of points:
$(\frac{8}{\sqrt{28}},\frac{4}{3})$ and $(-\frac{8}{\sqrt{28}},\frac{4}{3})$.
P.S. I did not check that obtained solution is actually a minimum. Can you validate that?
Edit: Dear reviewer, I want to edit the typo in "suare" to "square". Stupid StackExchange doesn't let me because it's less then 6 symbols!