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## Understand the problem

Consider the following subsets of the plane: $$\displaystyle { C_1 = \{ (x, y) : x > 0, y = \frac{1}{x} \} }$$ and $$\displaystyle { C_2 = \{ (x, y) : x < 0, y = -1 + \frac{1}{x} \} }$$ Given any two points P = (x, y) and Q = (u, v) of the plane, their distance d(P, Q) is defined by $$\displaystyle { d(P, Q) = \sqrt{(x-u)^2 + (y-v)^2} }$$ Show that there exists a unique choice of points $P_0 \in C_1$ and $Q_0 \in C_2$ such that $$d(P_0, Q_0) \leq d(P, Q)$$ for all $P \in C_1$ and $Q \in C_2$.

##### Source of the problem

I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 8 from 2019

##### Topic

Differential Calculus

8 out 10

##### Suggested Book

Problems in Calculus by I.A. Maron

## Start with hints

Do you really need a hint? Try it first!

Shortest distance from a point to a line is the length of the perpendicular drawn from the point to the line. This is the key idea. Suppose f(t) and g(s) are two (smooth) curves. Then the shortest distance between them is along the common normal (perpendicular) on the two curves. How to draw perpendiculars to curves? Pick a point on the curve (suppose A in the picture).   Next, draw a tangent at A (since the curve is smooth, that infinitely differentiable at every point, we can do this). Locally (near A), this tangent is the approximation of the curve.  Finally draw a perpendicular to this tangent line, at the point A. This is regarded as perpendicular to the curve (normal to the curve) at A.

# How to find the shortest path between two (smooth curves)?

Draw all possible normals to both curves. That draws all ‘perpendiculars’ to both curves erected at all points on both curves.  If any normal (perpendicular) is common between the two curves then that is possibly the shortest path. (You still need to check some other details. But that is part of a calculus course, not this discussion).  Can you show there is a common normal between f(x) = 1/x (x > 0) and $g(x) = -1 + \frac{1}{x}, (x < 0)$?

Consider $f(x) = \frac{1}{x}, (x > 0)$. The slope of the tangent line at any point is the derivative of the function. $\frac {d} {dx} f(x) = \frac {-1}{x^2}$ Slope of normal which is perpendicular to the tangent is negative reciprocal of it. Hence it is $x^2$. Parametrize the curve $f(x) = \frac{1}{ x}$ as $(t, \frac{1}{ t})$ Hence the equation of the normal through a point $(t, \frac{1}{ t})$ is $$y – \frac{1}{t} = t^2 (x – t)$$ Similarly, parametrize the other curve and find the equation of the normal. Use some other variable for this.

Consider $g(x) = -1 + \frac{1}{x}, (x > 0)$. The slope of the tangent line at any point is the derivative of the function. $\frac {d} {dx} g(x) = \frac {-1}{x^2}$ Slope of normal which is perpendicular to the tangent is negative reciprocal of it. Hence it is $x^2$. Parametrize the curve $g(x) = -1 + \frac{1}{ x}$ as $(r, -1 + \frac{1}{ r})$ Hence the equation of the normal through a point $(r, -1 + \frac{1}{ r})$ is $$y + 1 – \frac{1}{r} = r^2 (x – r)$$ We have generic equations for family of normal for each curve:  $y + 1 – \frac{1}{r} = r^2 (x – r)$  $y – \frac{1}{t} = t^2 (x – t)$ Can you find (or show the existence of) r and t such that these two equations are equal? This would show that there is a common normal and hence prove the existence of shortest path.

Let us write the equations in the slope-intercept form:  $y = r^2 x + \frac{1}{r} – r^3 – 1 \\ y = t^2 x + \frac{1}{t} – t^3$ Comparing the coefficients we have: $r^2 = t^2$ hence r = t or r = -t  We also have $$\frac {1}{r} – r^3 – 1 = \frac {1}{t} – t^3$$ Plugging in r = t we get -1 = 0 from second equation which is contradiction.  Plug in r = – t $$-\frac {1}{t} + t^3 – 1 = \frac {1}{t} – t^3$$ (Assume t is not 0 (why?)) Simplifying we have  $$2t^4 – t – 2 = 0$$ We want to show that all the roots of this equation are not complex. (If we have a real root, then we get a common normal, hence shortest part; this common normal will not be the longest path because the length of the longest path between these two curves easily goes to infinity). It is enough to show the existence of real roots.  Consider the function $2t^4 – t – 2 = h(t)$ Notice that h(0) = -2 < 0 and h(2) > 0  Hence by intermediate value property theorem there is at least one real root between 0 and 2.  Hence we are done.

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