Finding the points on a curve, closest to a specific point
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 :)
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$\begingroup$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!
$\endgroup$ 1 $\begingroup$Hint:
The distance is $ \sqrt{x^2+(y-4)^2}$
replace $x^2$ with $y^3$, the distance become $ \sqrt{y^3+(y-4)^2}$
Let $u(y) = y^3+(y-4)^2$ find the stationary point of u(y):
$u'(y) = 3y^2+2y-8 = 0 $ => $y = -2$ or $4/3$
substitute the roots into distance function and choose the minimal one.
References:correlation between stationary points and extrema
Basic identities of derivatives
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