I am trying to find the point of maximum curvature of y=(abx)/(b-x)?

Question

I have forgotten how to find dy/dx and d2y/dx2 of the function.

Answer 1

The curvature κ is given by

κ = y''/(1 + y'²)^(3/2)

To maximize κ, you will need to calculate dκ/dx and set that to zero.

y' = ab[(b-x)(1) - x(-1)]/(b-x)^2 = ab(b-x+x)/(b-x)^2 = ab²/(b-x)² = ab² (b-x)^(-2)

y'' = ab² [(-2)(b-x)^(-3) (-1)] = 2ab² (b-x)^(-3)

y''' = 2ab² [(-3)(b-x)^(-4) (-1)] = 6ab² (b-x)^(-4)

You will need y''' to calculate dκ/dx

Note: in each of the derivatives calculated above, the final factor of (-1) in the square brackets is the derivative of (b-x) with respect to x.

Answer 2

ok. so. umm see the maximum will be when dy/dx = 0.

now, to solve for dy/dx, you will nexd to use the quotient rule

if y = (f(x))/(g(x))
then dy/dx= (f dash of (x)g(x) - g dash of (x)f(x))/((g(x))^2)

ie the derivative of the first times the second unchanged minus the derivative of the second times the first unchanged all divided by the 2nd squared

...helpful at all?

Answer 3

* It's where the slope goes to zero.

Answer 4

If i remember right y=-b/2a