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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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'

Vedclass · Accepted Answer

$-\frac{1}{R}$

Vedclass · Answer

(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:

Similar Questions

If the tangent to the curve $y^{2} = x^{3}$ at $(m^{2}, m^{3})$ is also a normal to the curve at $(M^{2}, M^{3})$, then the value of $mM$ is

If the tangent drawn at the point $(\alpha, \beta)$ on the curve $x^{2/3} + y^{2/3} = 4$ is parallel to the line $\sqrt{3}x + y = 1$,then $\alpha^2 + \beta^2 =$

The equation of the tangent to the curve $(1+x^2)y = 2-x$,where it crosses the $X$-axis,is

If $P$ and $Q$ are two different points on the curve $y = x^3 - x$ such that the tangent at $P$ intersects the curve again at $Q$,then $\frac{m_{OQ} + 1}{m_{OP} + 1}$ is equal to,where $O$ is the origin and $m_{AB}$ represents the slope of the line segment $AB$.

If $\theta$ denotes the acute angle between the curves $y = 10 - x^2$ and $y = 2 + x^2$ at a point of their intersection,then $|\tan \theta |$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module