find the points on the given curve at which the curvature is a maximum?
a)y=e^-x
b)y=lnX
I'll do it for lnx. By symmetry it should then be easy to figure it for e^-x
curvature is k=|y''|/(1+y'^2)^3/2
y=ln(x)
y'=1/x
y''=1/x^2
1+y'^2=1+1/x^2=(x^2+1)/x^2
1/(1+y'^2)=x^2/(x^2+1)
1/(1+y'^2)^3/2=x^3/(x^2+1)^3/2
k=(1/x^2)x^3/(x^2+1)^3/2=x/(x^2+1)^3/2
this is a maximum when
dk/dx = { (x^2+1)^3/2-x(3/2)(2x)(x^2+1)^1/2 } / (x^2+1)^3 = 0
(x^2+1)^3/2-x(3/2)(2x)(x^2+1)^1/2 = 0
(x^2+1)^3/2 = x(3/2)(2x)(x^2+1)^1/2
(x^2+1) = 3x^2
1 = 2x^2
x=+sqrt(1/2)=1/sqrt(2)=(2)^(-1/2) (we must have x>0)
and y = ln(x) = -(1/2)ln(2) or 1/2 ln(1/2)
that's brilliant!Thx!
To find the points on a curve where the curvature is a maximum, you need to follow these steps:
1. Differentiate the given equation twice to find the second derivative, which represents the curvature of the curve.
2. Set the second derivative equal to zero and solve for x to find the critical points.
3. Determine the corresponding y-values for each critical point by substituting the x-values back into the original equation.
4. Evaluate the second derivative at each critical point to determine whether the curvature is a maximum or minimum.
5. If the second derivative is negative at a critical point, it represents a maximum curvature. If positive, it represents a minimum curvature.
6. The points where the curvature is a maximum correspond to the x-values of the critical points.
Now let's apply these steps to the given curves:
a) y = e^-x
1. Differentiating y = e^-x, we get dy/dx = -e^-x.
2. Differentiating again, we get d^2y/dx^2 = e^-x.
3. Setting d^2y/dx^2 = 0, we find e^-x = 0, which has no real solutions. Therefore, there are no critical points for this curve.
4. Since there are no critical points, there are no points where the curvature is a maximum.
b) y = ln(x)
1. Differentiating y = ln(x), we get dy/dx = 1/x.
2. Differentiating again, we get d^2y/dx^2 = -1/x^2.
3. Setting d^2y/dx^2 = 0, we find -1/x^2 = 0, which has no solutions. Therefore, there are no critical points for this curve.
4. Since there are no critical points, there are no points where the curvature is a maximum.
In conclusion, for both curves, there are no points where the curvature is a maximum.