If x^2 + y^2 = R^2 (R 0),then k = frac{y''} | vedclass

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(B) Given the equation of the circle: $x^2 + y^2 = R^2$.Differentiating with respect to $x$: $2x + 2yy' = 0$,which simplifies to $x + yy' = 0$,so $y'

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$-\frac{1}{R}$

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(B) Given the equation of the circle: $x^2 + y^2 = R^2$.Differentiating with respect to $x$: $2x + 2yy' = 0$,which simplifies to $x + yy' = 0$,so $y' = -\frac{x}{y}$.Differentiating again with respect to $x$: $1 + y'y' + yy'' = 0$,which gives $1 + (y')^2 + yy'' = 0$.Therefore,$y'' = -\frac{1 + (y')^2}{y}$.The expression for $k$ is given by $k = \frac{y''}{(1 + (y')^2)^{3/2}}$.Substituting $y''$: $k = \frac{-(1 + (y')^2)}{y(1 + (y')^2)^{3/2}} = -\frac{1}{y\sqrt{1 + (y')^2}}$.Since $y' = -\frac{x}{y}$,we have $1 + (y')^2 = 1 + \frac{x^2}{y^2} = \frac{y^2 + x^2}{y^2} = \frac{R^2}{y^2}$.Thus,$\sqrt{1 + (y')^2} = \sqrt{\frac{R^2}{y^2}} = \frac{R}{|y|}$.Substituting this back into $k$: $k = -\frac{1}{y \cdot (R/|y|)} = -\frac{1}{R} \cdot \frac{|y|}{y}$.Assuming the upper semicircle where $y > 0$,we get $k = -\frac{1}{R}$.

If $x^2 + y^2 = R^2$ $(R > 0)$,then $k = \frac{y''}{\sqrt{(1 + (y')^2)^3}}$. The value of $k$ in terms of $R$ alone is equal to:

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