Find the radius of curvature at x = 0 on y = ex
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Question:-
Find the radius of curvature at x = 0 on y = ex
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
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