Find the point on the curve y =√ln x, x > 1 which is closest to the point (2, 0).
I'm not sure if it is because i tried solving for x with the derivative of the distance function but it's undefined
Find the point on the curve y =√ln x, x > 1 which is closest to the point (2, 0).
I'm not sure if it is because i tried solving for x with the derivative of the distance function but it's undefined
minimize the distance between $(x,\sqrt{\ln{x}})$ and $(2,0)$ for $x>1$...
$D = \sqrt{(x-2)^2 + \ln{x}}$
minimizing the radicand, $(x-2)^2 + \ln{x}$, will minimize $D$ ...
$2(x-2) + \dfrac{1}{x} = 0$
$2x^2 - 4x + 1 = 0$
$x = 1 \pm \dfrac{\sqrt{2}}{2}$
since $x > 1$, $x = 1 + \dfrac{\sqrt{2}}{2}$
second derivative, $2 - \dfrac{1}{x^2} > 0$ for $x = 1 + \dfrac{\sqrt{2}}{2}$ which confirms that value of $x$ yields a minimum.
The function is $y= \sqrt{ln(x)}= (ln(x))^{1/2}$?
If so then the distance from (x, y) to the point (2, 0) is $D(x)= \sqrt{(x- 2)^2+ ln(x)= x^2- 2x+ 1+ ln(x)}$ but it is sufficient to minimize the square of that, $(x- 2)^2+ ln(x)= x^2- 2x+ 1+ ln(x)$. The derivative of that certainly does exist, it is $2x- 2+ \frac{1}{x}$.
Setting that equal to 0, $2x- 2+ \frac{1}{x}= 0$. Multiplying by x, $2x^2- 2x+ 1= 0$. That is a quadratic equation with solutions $x= \frac{2\pm\sqrt{4- 4(2)}}{4}$ which are not real. If the condition is actually x> 1, as you ive, then there is no point closest to (2, 0)- the closer you get to 1, the smaller that distance is. If the condition were $x\ge 1$ then the point closest to (2, 0) is (1, 0).