These are curves defined by four control points and a cubic polynomial defined in terms of a parameter \(t\). The control points \(q_0\) and \(q_1\) define the position of the curve at \(t=0\) and \(t=1\) respectively, and \(q_0'\) and \(q_1'\) its derivative. Drag the control points with the mouse to see how this affects the shape of the curve.
If we consider just the \(x\) coordinate of points along the curve, then our cubic polynomial takes the form:
$$x(t)=a_3t^3+a_2t^2+a_1t+a_0$$
Then we can find the values of \(a_3,a_2,a_1,a_0\) that correspond to a given set of control points based on the constraints that the curve passes through \(q_0\) with derivative \(q_0'\) and \(q_1\) with derivative \(q_1'\), see the lecture slides. Note that to find the \(x\) coordinate of the curve we can simply use the \(x\) components of each \(q\), here written \(x_0\), \(x_0'\), \(x_1\) and \(x_1'\), and we can do the same for the \(y\) coordinate. Then we find that we can rearrange our polynomial to the form
$$x(t)=(2t^3-3t^2+1)x_0+(t^3-2t^2+t)x_0'+(-2t^3+3t^2)x_1+(t^3-t^2)x_1'$$
This gives us a set of basis functions described in terms of \(t\) that are used to blend the four control points to decide the final coordinate of a point on the curve (these are the curves plotted in the figure above). At \(t=0\), only \(x_0\) contributes to the position, whilst at \(t=1\), only \(x_1\) contributes.