Question

The radius of curvature of the curve y = e^x at x = 0 is

Answer 1

Answer:

The radius of curvature is $2\sqrt{2}$

Explanation:

The radius of curvature for y=f(x) is,

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

From the question, we have

$y= e^x$ at (0,1)

At point (0,1)

$\frac{dy}{dx} = 1$

Radius of curvature is,

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

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