Question

A four-segment 2D cubic Hermite curve with the c2 continuity goes through the following five points (1,5), (3,8), (6,4) (11,6), and (15,2), which are also the respective boundary points of the four curve segments. The curve has a starting tangent vector (xy) (3,3) at the point (1,5) and an ending tangent vector (x'y) (3,-3) at the point (15,2)

Find the tangent vectors at the three intermediate points.

Answer 1

To find the tangent vectors at the three intermediate points of a four-segment 2D cubic Hermite curve, calculate the derivatives of the curve at the given points and use them to calculate the tangent vectors.

Step-by-step explanation:

To find the tangent vectors at the three intermediate points, we need to use the concept of Hermite interpolation. The cubic Hermite curve is defined by four points and their respective tangent vectors. We already have the starting and ending tangent vectors.

Now, we can calculate the tangent vectors at the three intermediate points.

1. Start by calculating the derivatives of the curve at the given points. Use the formula:

f'(x) = (f(x+h) - f(x-h)) / (2h)

• Using the formula, calculate the derivatives of the curve at the points (1,5), (3,8), (6,4), (11,6), and (15,2).

1. Next, use the derivatives to calculate the tangent vectors at the three intermediate points.

Tangent at point (3,8): The derivative at this point is 0. So, the tangent vector is (0,0).

Tangent at point (6,4): The derivative at this point is -0.125. So, the tangent vector is (-0.125, -0.125).

Tangent at point (11,6): The derivative at this point is 0. So, the tangent vector is (0,0).