Question #ec325

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Question #ec325

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Answer 1 · 2016-07-28T12:53:15

The perpendicular from origin $O$ $O$ to a normal to $y = {log}_{e} (x)$ $y=log_e(x)$ at point $P$ $P$ is the segment $O P$ $OP$ .
It's length is $\sqrt{X_{P}^{2} + Y_{P}^{2}}$ $sqrt(X_P^2+Y_P^2)$ , where $X_{P}$ $X_P$ is the abscissa of point $P$ $P$ and $Y_{P} = {log}_{e} (X_{P})$ $Y_P=log_e(X_P)$ .

Explanation:

Normal to a curve at point $P$ $P$ lying on this curve, by definition, is a line perpendicular to a tangent to a curve at this point (presuming the curve is smooth and has a tangent, otherwise a normal is undefined).

Since $O P$ $OP$ is a tangent to curve $y = {log}_{e} (x)$ $y=log_e(x)$ from origin $O$ $O$ to point $P$ $P$ lying on this curve, it is also a perpendicular to a normal to a curve at point $P$ $P$ .

So, our task is to measure the length of $O P$ $OP$ .
This is done by Pythagorean Theorem as the distance between two points:
origin $O$ $O$ with coordinates $(0, 0)$ $(0,0)$ and
point $P$ $P$ on a curve with coordinates $(X_{P}, Y_{P})$ $(X_P,Y_P)$ , where
$Y_{P} = {log}_{e} (X_{P})$ $Y_P = log_e(X_P)$ .

$O P = \sqrt{X_{P}^{2} + Y_{P}^{2}} = \sqrt{X_{P}^{2} + {log}_{e}^{2} (X_{P})}$ $OP = sqrt(X_P^2+Y_P^2)=sqrt(X_P^2+log_e^2(X_P))$

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Question #ec325

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