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:
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:
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)`.
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 y − y1 = m(x − x1) 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:
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)`.