Question

the radius of curvature of the curve y=e^theta at any point on it is​

Answer 1

Answer:

y = e^x

The graph is a continuous curve that passes through (0, 1).

Differentiate: y’ = e^x

Differentiate again: y’’ = e^x

radius of curvature = | (1 + y’ ²)^(3/2) / y’’ |

= | (1 + e^2x)^(3/2) / e^x |

When x = 0 the radius of curvature is

| (1 + e^0)^(3/2) / e^0 | = | (1+1)^(3/2) / 1 | = 2^(3/2)

or 2√2