Question

the radius of curvature of the curve y=e^x at origine is​

Answer 1

Answer:

Mark, and now Robert , having found the radius or curvature to be r =2√2 ,it would be interesting to find where the centre of curvature of that circle is .

h = x - (dy/dx )[1 + (dy/dx )^2]/ d^2y/dx^2 , at (x,y) = (0,1)

h = 0 - (1) [ 1+ (1)^2] /1 = - 2. A

k = y + [1 + (dy/dx )^2] /( d^2 y / dx^2) = 1. + (1+1)/1 =3 B

Putting A and B together together the centre of curvature is

(h, k ) = C ( - 2 ,3 )

Check ; find the distance from P ( 0 , 1 ) to C ( -2 , 3)

r = √[(- 2 -0)^2 +(3 - 1)^2] = √8 = 2 √ 2 = radius of curvature.

So (h , k) = ( - 2 , 3 ) is the centre.