Asked by C
                Find the point where the curve y =e^-x has maximum curvature.
            
            
        Answers
                    Answered by
            oobleck
            
    k = |y"|/(1+y'^2)^(3/2)
so,
k = e^-x/(1+e^2x)^(3/2)
find where k' = 0 for max curvature. Just thinking of the curve, where do you estimate it has max curvature? I'd guess somewhere near x = -3/2
Let's see what it really is.
k' = e^-x (e^2(2x+3) + 2) / 2(1+e^2x)^(5/2)
So, we want
e^2(2x+3) + 2 = 0
x = (-2/e^2 - 3)/2 = -(1/e^2 + 3/2) ≈ -1.635
    
so,
k = e^-x/(1+e^2x)^(3/2)
find where k' = 0 for max curvature. Just thinking of the curve, where do you estimate it has max curvature? I'd guess somewhere near x = -3/2
Let's see what it really is.
k' = e^-x (e^2(2x+3) + 2) / 2(1+e^2x)^(5/2)
So, we want
e^2(2x+3) + 2 = 0
x = (-2/e^2 - 3)/2 = -(1/e^2 + 3/2) ≈ -1.635
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.