Section 13.1: Vector Functions and Space Curves

SOLUTION In Section 12.5 we found a vector equation for the line segment that joins the tip of the vector \(\mathbf{r}_0\) to the tip of the vector \(\mathbf{r}_1\): \[ \mathbf{r}(t) = (1-t)\mathbf{r}_0 + t\mathbf{r}_1 \qquad 0 \le t \le 1 \] (See Equation 12.5.4.) Here we take \(\mathbf{r}_0 = \langle 1, 3, -2 \rangle\) and \(\mathbf{r}_1 = \langle 2, -1, 3 \rangle\) to obtain a vector equation of the line segment from \(P\) to \(Q\): \[ \mathbf{r}(t) = (1-t)\langle 1, 3, -2 \rangle + t\langle 2, -1, 3 \rangle \qquad 0 \le t \le 1 \] or \[ \mathbf{r}(t) = \langle 1+t, 3-4t, -2+5t \rangle \qquad 0 \le t \le 1 \] The corresponding parametric equations are \[ x = 1+t \qquad y = 3-4t \qquad z = -2+5t \qquad 0 \le t \le 1 \]

SOLUTION

Figure 5 shows how the plane and the cylinder intersect, and Figure 6 shows the curve of intersection \(C\), which is an ellipse. The projection of \(C\) onto the xy-plane is the circle \(x^2 + y^2 = 1, z=0\). So we know from Example 10.1.2 that we can write \[ x = \cos t \qquad y = \sin t \qquad 0 \le t \le 2\pi \] From the equation of the plane, we have \[ z = 2-y = 2 - \sin t \] So we can write parametric equations for \(C\) as \[ x = \cos t \qquad y = \sin t \qquad z = 2 - \sin t \qquad 0 \le t \le 2\pi \] The corresponding vector equation is \[ \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + (2-\sin t)\mathbf{k} \qquad 0 \le t \le 2\pi \] This equation is called a parametrization of the curve \(C\). The arrows in Figure 6 indicate the direction in which \(C\) is traced as the parameter \(t\) increases.