Find the curvature of the curve y=((e)^x)^2 at the point P=(0,1)

Question

anonymous · Answer

$$y=e^{x ^{2}}$$

if that is more clear

amistre64 · Answer

i always gotta relook up the k when we deal with functions; can never quite remember it

amistre64 · Answer

http://tutorial.math.lamar.edu/Classes/CalcIII/Curvature.aspx is always a good sourse for me

amistre64 · Answer

$$k=\frac{f'}{(1+f'^2 )^{3/2}}$$

amistre64 · Answer

those are spose to be f's :)

amistre64 · Answer

$$k=\frac{e'^{x^2}}{(1+(e'^{x^2})^2 )^{3/2}}$$

$$k=\frac{2xe^{x^2}}{(1+(2xe^{x^2})^2 )^{3/2}}$$

$$k=\frac{2xe^{x^2}}{(1+4x^2 e^{2x^2} )^{3/2}}$$

amistre64 · Answer

when x=0, this appears to = 0 as well ... just not too confident about that tho

anonymous · Answer

Sam does this make sense...?

anonymous · Answer

I follow the stuff up the the last k=... but am not sure where you get that it equals zero

anonymous · Answer

When you plug in x=0, you get 0/(1+0)=0

amistre64 · Answer

http://www.wolframalpha.com/input/?i=%28d+e%5E%28x%5E2%29%2Fdx%29%2F%281%2B%28d+e%5E%28x%5E2%29%2Fdx%29%5E2%29%5E%283%2F2%29 its got a neat looking graph i think :)