First, let's draw the graph and see what the question means.

y = 2x3 x + 3


I have used equal scaling along the 2 axes (so that later, when I draw the circle, it will not have an elliptical shape).

Now, to find the radius of curvature, we need:


And then



So now we are ready to substitute into the formula to give us the radius at any point x:




Now to find the radius of curvature at the required point x = 1, we substitute:

`[[36x^4-12x^2+2]^(3//2)/{|12x|}]_(x=1)` `=11.04787562`

To show what we have done, let's look at the graph of the curve (blue) with the approximating circle (dark red) overlaid. The circle is a good approximation for the curve at (1, 4).

We can show that the center of the approximating circle is `(−9.8, 6.17)`.

radius curvature graph

How did I find that center?

We know the length of the radius shown in the diagram (`11.05` units).

We know 1 point on that radius line, `(1,4)`, and we need to find the one at the other end, the center. Let's call it (x1, y1).

We're going to set up 2 equations using these 2 unknowns.

We know that


So the slope of the tangent at `x = 1` is

6(1)2 − 1 = 5.

Now, the slope of the normal (the line at right angles to the tangent at the point of contact) is −1/5. (See Tangents and Normals.)

So we can use the formula yy1 = m(xx1) and the known slope `-1/5` and point `(1, 4)` to find the equation of the line containing the radius as follows:

y − 4 = (−1/5) (x − 1), which gives:

`y = −x/5 + 21/5`

The unknown point (x1, y1) lies on this line, so we can say


Next, we use the distance formula:


We know the distance r (since it is the radius), and one of our points is (1,4), and using our expression connecting x1 and y1, we can solve these simultaneously as:

`sqrt((1-x_1)^2+[4-(21/5-1/5x_1)]^2)` `=11.04787562`

This is one equation in one unknown, which after some algebra gives us

x1 = − 9.833 (and another positive solution which doesn't apply here)

Substituting this back into `y_1=-(x_1)/5+21/5` gives

y1 = 6.167

So the center of the circle is `(−9.8, 6.17)`.