Find the point(s) on the graph of y=e^(-x^2) at which the curvature is zero.

Question

idealist10 · Answer

@hartnn @Compassionate @ParthKohli @.Sam.

anonymous · Answer

I may not have a solution for you, but may I ask what math class this is from?

anonymous · Answer

also is it (-x)^2  or -(x^2)

idealist10 · Answer

It's Calculus 3. $$y=e ^{-x^2}$$

idealist10 · Answer

@zepdrix @thomaster

campbell_st · Answer

why not find the 1st derivative... and then find the stationary points.

irishboy123 · Answer

curvature is not the same as gradient.  it measures the change of direction (a vector) wrt to arc length.  

parameterising is easier, i think, which gives me this:

|dw:1426367405718:dw|

that means curvature = 0 @ x = 1.

the workings could be wrong - have done this quite quickly - but the idea behind it is right.