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Thread: Not sure if i learned this (is this a optimization problem for distance?)

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    Not sure if i learned this (is this a optimization problem for distance?)

    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
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    Re: Not sure if i learned this (is this a optimization problem for distance?)

    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.
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    Re: Not sure if i learned this (is this a optimization problem for distance?)

    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).
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    Re: Not sure if i learned this (is this a optimization problem for distance?)

    i do have the answer key here. the answer is suppose to be (2+sqrt(2) , sqrt(ln(2+sqrt(2))))
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    Re: Not sure if i learned this (is this a optimization problem for distance?)

    Quote Originally Posted by HallsofIvy View Post
    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, $(x- 2)^2+ ln(x)= \color{red}{x^2- 2x+ 1+ ln(x)}$.
    $(x-2)^2 + \ln{x} = \color{red}{x^2 - 4x + 4 + \ln{x}}$
    Thanks from HallsofIvy
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    Re: Not sure if i learned this (is this a optimization problem for distance?)

    I never was any good at arithmetic!
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    Re: Not sure if i learned this (is this a optimization problem for distance?)

    Quote Originally Posted by lc99 View Post
    i do have the answer key here. the answer is suppose to be (2+sqrt(2) , sqrt(ln(2+sqrt(2))))
    Quote Originally Posted by skeeter View Post
    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 $D$ ...
    $2(x-2) + \dfrac{1}{x} = 0$
    $x = 1 \pm \dfrac{\sqrt{2}}{2}$
    I did it an entirely different way (using the tangent at the point $P$ where $\overline{P(2,0)}$ is perpendicular to the tangent. Got the same answer as skeeter. Thus that given answer must be mistaken/
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