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10

How about this? Say the two curves are $A$ and $B$, with the min distance achieved at point $a \in A$ and $b \in B$. Let me assume both curves are smooth, or at least have 1st derivatives everywhere. Claim: the segment $ab$ is orthogonal to $A$ at $a$ and to $B$ at $b$. Suppose to the contrary that $ab$ is not orthogonal to $A$ at $a$. Then one can move $a$ ...

3

I would be more careful with the phrasing of the question. Here are some points to explain why. The first point is an answer under certain conditions. When looking for the distance between a point and a curve (in two dimensions), do you prove that for a point $P$ not on the curve when $B$ is the closest point on the curve, then $\overleftrightarrow{PB}$ is ...

2

[T]o justify that it should occur when the slopes of the two curves are parallel: If $ab$ is a minimum, then the curve through $b$ has no points in the interior of the circle centered at $a$ with radius $ab$. (Any point inside is closer to $a$ than $b$.) Likewise, the curve through $a$ has no points in the interior of the circle centered at $b$ with ...

1

So, you want to find the minimum distance between two curves in the most general form using calculus and optimization. The curves you are interested at this particular example are: \begin{equation} xy=1 ~~and~~ y=-x \tag{01} \end{equation} The very first step is to write them in parametric form. i.e, \begin{equation} xy=1 \Rightarrow \begin{cases} x_1 &...

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