Question

Consider the function.
y = 13 e^( x)
At what point does the curve have maximum curvature?

Answer 1

you want i guess i would say at 0

Answer 2

\[\frac{\sqrt{(1+f')^3}}{f''}\] i think but i could be wrong

Answer 3

well I used the formula |f"(x)|/[1+(f'(x)^2]^
3/2
and got 13e^x/[1+(13e^x)^2]^3/2
then differentiated
quotient rule
[1+169^2x]^3/2.13e^x-13e^x.[3/2(1+169e^2x)^1/2.169e^2x]/[1+169e^2x]^3
now I have hard time setting the numerator to 0

Answer 4

maybe i am wrong but i think your formula is upside down. let me check

Answer 5

yeah your formula is upside down

Answer 6

http://en.wikipedia.org/wiki/Radius_of_curvature_%28applications%29

Answer 7

i cheated and took derivative here
http://www.wolframalpha.com/input/?i=sqrt%28%281%2B%2813e^x%29^2%29%29^3%2F%2813e^x%29

Answer 8

oh i see you want max and i am finding min, but min of radius of curvature is max of curvature, so it amount to the same thing. we set
\[338e^{2x}-1=0\] solve for x to get it

Answer 9

since all that other stuff in the derivative is never zero

Answer 10

Thank you satellite73!

Answer 11

A solution and plot from Mathematica is attached.

Answer 12

Thank you as well robtobey!

Answer 13

let me see if it is the same..
\[338e^{2x}=1\]
\[e^{2x}=\frac{1}{338}\]
\[2x=\ln(\frac{1}{338})=-\ln(338)\]
\[x=-\frac{\ln(338)}{2}\]\]