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 June 23rd, 2012, 03:48 PM #1 Senior Member   Joined: Jan 2012 Posts: 159 Thanks: 0 Finding The Shortest Distance Between Two Points... that also is tangent to a given function. No idea about this one. Seems not that tough just do not know where to begin.
 June 23rd, 2012, 08:02 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Finding The Shortest Distance Between Two Points... Let point P be $$$x_P,0$$$ and point Q be $$$0,y_Q$$$ and using the two-intercept formula for a line, the line passing through P and Q is given by: $\frac{x}{x_P}+\frac{y}{y_Q}=1$ Now substitute for y using $y=\frac{2}{x}$: $\frac{x}{x_P}+\frac{2}{xy_Q}=1$ Multiply through by x and arrange in standard quadratic form: $\frac{1}{x_P}x^2-x+\frac{2}{y_Q}=0$ We require this equation to have 1 real repeated root, hence the discriminant must be zero: $(-1)^2-4$$\frac{1}{x_P}$$$$\frac{2}{y_Q}$$=0$ $y_Q=\frac{8}{x_P}$ Now, the square of the distance L from P to Q is: $L^2=x_P^2+y_Q^2=x_P^2+$$\frac{8}{x_P}$$^2=x_P^2+64 x_P^{-2}$ Implicitly differentiate with respect to $x_P$: $2L\cdot\frac{dL}{dx_P}=2x_P-128x_P^{-3}$ Simplify and equate $\frac{dL}{dx_P}$ to zero: $\frac{dL}{dx_P}=\frac{x_P^4-64}{Lx_P^3}=0$ ${x_P^4-64=0$ $x_P^2=8$ Hence: $L^2=8+\frac{64}{8}=16$ $L=4$
 June 24th, 2012, 07:11 AM #3 Senior Member   Joined: Jan 2012 Posts: 159 Thanks: 0 Re: Finding The Shortest Distance Between Two Points... thanks mark going through your answer now!
 June 24th, 2012, 12:16 PM #4 Senior Member   Joined: Jan 2012 Posts: 159 Thanks: 0 Re: Finding The Shortest Distance Between Two Points... I am surprised I am actually understanding your solution. Just did a review of the what you can tell from the value of a discriminant, cool stuff. So I am down to where you set dL/xp equal to zero. This is the rate of change of the distance over the rate of change of xp. So I know we are looking for the min L (distance) but could you explain a bit why we set this ratio to 0?
June 24th, 2012, 12:55 PM   #5
Senior Member

Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
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Math Focus: Calculus/ODEs
Re: Finding The Shortest Distance Between Two Points...

Here is a graph of $y=L^2$. This should make it obvious why we set the derivative equal to zero:

[attachment=0:37i9uwqm]lengthsquared.jpg[/attachment:37i9uwqm]
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 lengthsquared.jpg (9.4 KB, 200 views)

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