Question

the radius of curvature of the curve y = ex at the point (0,1) is answer

Answer 1

$\textbf{Concept used:}$

$\text{The radius of curvature of the given curve y=f(x) is}$

$\bf\;R=\displaystyle\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}$

$\text{Given curve is}\;y=e^x$

$\text{Then,}$

$\displaystyle\frac{dy}{dx}=e^x$

$\displaystyle\frac{d^2y}{dx^2}=e^x$

$\text{At the point (0,1)}$

$\displaystyle\frac{dy}{dx}=e^0=1$

$\displaystyle\frac{d^2y}{dx^2}=e^0=1$

$\text{Radius of curvature}$

$\bf\;R=\displaystyle\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}$

$R=\displaystyle\frac{[1+(1)^2]^{\frac{3}{2}}}{1}$

$R=\displaystyle\;2^{\frac{3}{2}}$

$R=\displaystyle\;2\sqrt{2}$

$\therefore\textbf{Radius of curvature is }\bf\;2\sqrt{2}$

Answer 2

Your answer is Right

2 sqrt 2 is The Right Answer