The equation of the normal to the curve y = c | vedclass

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Question

(C) Given the curve $y = \cosh x$. To find the point nearest to the origin $(0, 0)$,we minimize the squared distance $D^2 = x^2 + y^2 = x^2 + (\cosh x

Vedclass · Accepted Answer

$x = 0$

Vedclass · Answer

(C) Given the curve $y = \cosh x$. To find the point nearest to the origin $(0, 0)$,we minimize the squared distance $D^2 = x^2 + y^2 = x^2 + (\cosh x)^2$. Let $f(x) = x^2 + \cosh^2 x$. Taking the derivative,$f'(x) = 2x + 2 \cosh x \sinh x = 2x + \sinh(2x)$. Setting $f'(x) = 0$,we get $2x + \sinh(2x) = 0$. The only solution to this equation is $x = 0$. At $x = 0$,$y = \cosh(0) = 1$. The slope of the tangent at $(0, 1)$ is $y' = \sinh(0) = 0$. The slope of the normal $m$ is given by $m = -\frac{1}{y'} = -\frac{1}{0}$,which is undefined (vertical line). The equation of a vertical line passing through $(0, 1)$ is $x = 0$.

The equation of the normal to the curve $y = \cosh x$ drawn at the point nearest to the origin is

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