Find the point where the curve y =e^-x has maximum curvature.

To find the point where the curve y = e^(-x) has maximum curvature, we need to find the maximum of the curvature function.

The curvature of a curve is given by the formula K = |(d^2y/dx^2)| / (1 + (dy/dx)^2)^(3/2), where dy/dx represents the derivative of y with respect to x and d^2y/dx^2 represents the second derivative of y with respect to x.

First, let's find the derivatives of y = e^(-x):
dy/dx = -e^(-x) (using the chain rule)
d^2y/dx^2 = e^(-x) (since the derivative of -e^(-x) is just e^(-x))

Next, let's calculate the curvature function K(x):
K(x) = |e^(-x)| / (1 + (-e^(-x))^2)^(3/2)
= e^(-x) / (1 + e^(-2x))^(3/2)

To find the maximum curvature of K(x), we need to find the critical points of K(x) by setting its derivative equal to zero.

dK(x)/dx = 0

Let's calculate the derivative of K(x):

dK(x)/dx = (e^(-x) * (1 + e^(-2x))^(3/2) - 3/2 * e^(-x) * e^(-2x) * (1 + e^(-2x))^(1/2)) / (1 + e^(-2x))^3

Setting dK(x)/dx equal to zero and solving for x:
(e^(-x) * (1 + e^(-2x))^(3/2) - 3/2 * e^(-x) * e^(-2x) * (1 + e^(-2x))^(1/2)) / (1 + e^(-2x))^3 = 0

Since e^(-x) is always positive, we can simplify the equation by canceling out e^(-x) from both sides:

(1 + e^(-2x))^(3/2) - 3/2 * e^(-2x) * (1 + e^(-2x))^(1/2) = 0

Now, let's solve this equation to find the critical points for the maximum curvature.

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