Given $$y=2\sqrt{\ln(x+3) }$$ How do I determine a (x,y) pair satisfying the above relation which is the closest to the origin (0,0)?
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1$x^2 + y^2$ is minimized – Will Jagy Jun 10 '13 at 01:46
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Using Lagrange multiplier under this constraint. – Shuhao Cao Jun 10 '13 at 01:46
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Well, I have no idea how to begin. – klaufir Jun 10 '13 at 02:04
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A related problem. – Mhenni Benghorbal Jun 10 '13 at 02:24
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Minimize the square of the distance from the curve to the origin (and hence, the distance from the curve to the origin), namely: $$x^2+y^2=x^2+4\ln(x+3).$$
Cameron Buie
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To minimize the distance from the origin, minimize the square of the distance from the origin, given by $d^2 = M = x^2+y^2$,
$$M = x^2+y^2=x^2+(2\sqrt{\ln(x+3)})^2 = x^2 + 4\ln(x+3)$$
$$\frac{dM}{dx} = 2x + \frac{4}{x+3} = \frac{2x(x+3) + 4}{x+3} = \frac{2x^2 + 6x + 4}{x+3}.$$
This is only (possibly) equal to zero when the numerator is equal to zero,
$$2x^2 + 6x + 4 = 0 \implies x=-2 \ \ \text{and} \ \ x=-1$$
We know that at least one of these must be where the minimum occurs, so just plug the values of $x$ into the given equation and compute the distance from the origin.
$y(-1) = 2\sqrt{\ln(-1+3)} = 2 \sqrt{\ln(2)} = 1.665 \implies d^2 = (-1)^2 + (1.665)^2 = 2.386$
$y(-2) = 2\sqrt{\ln(-2+3)} = 2 \sqrt{\ln(1)} = 0 \implies d^2 = (-1)^2 + (0)^2 = 1$
So the point nearest to the origin is $(-2,0)$.
Gamma Function
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