What is the curvature of x 2 plus y2-4x-6y plus 10 equals 0 at any point on it?

Answer 1

First, we can see that this curve is a circle with a radius of √3 and a center point of [2, 3] by converting it to a more readable format. This is done by "completing the squares":

x2 + y2 - 4x - 6y + 10 = 0

∴(x2 - 4x + 4) + (y2 - 6y + 9) + 10 - 13 = 0

∴(x - 2)2 + (y - 3)2 = 3

So the curve at any point will be a simple arc.

Now, if by "what is the curvature ... at any point", you mean "what is the slope of the curve", then we would need to find the rate of change of y with respect to x. This is done by first solving for y and then taking it's derivative:

(x - 2)2 + (y - 3)2 = 3

∴(y - 3)2 = 3 - (x - 2)2

∴y - 3 = [3 - (x - 2)2]1/2

∴y = [3 - (x - 2)2]1/2 + 3

∴y = (3 - x2 + 4x - 4)1/2 + 3

∴y = (x2 + 4x - 1)1/2 + 3

We now have our curve solved for y, and we can take it's derivative with respect to x to find it's slope:

y = (x2 + 4x - 1)1/2 + 3

∴dy/dx = (1/2)(x2 + 4x - 1)-1/2(2x + 4)

∴dy/dx = (x + 2) / (x2 + 4x - 1)1/2

And to put it in proper form with radicals in the numerator:

dy/dx = (x + 2)(x2 + 4x - 1)1/2 / x2 + 4x - 1