Find the radius of curvature of the curve y=e®x at (0.1)?
Answers
Answer:
What is the radius of curvature of the curve y=ex at the point where it crosses the y -axis?
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
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.