Question

Sketch the space curve and find its length over the given interval. $$ \mathbf{r}(t)=2 t \mathbf{i}-3 t \mathbf{j}+t \mathbf{k} $$ $$ [0,2] $$

Short answer

The space curve goes from (0,0,0) to (4,-6,2) and its length over the interval [0,2] is 2√14.

Step-by-step solution

Sketch the Space Curve

A simple way to draw the space curve is to examine each component function. The \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) components represent the x, y, and z coordinates of points along the curve respectively. Given \(\mathbf{r}(t)=2t\mathbf{i} - 3t\mathbf{j} + t\mathbf{k }\), x = 2t, y = -3t, and z = t. This vector's position in 3D space at any time t can be found by plugging a specific value of t into \(\mathbf{r}(t)\). For \(t = 0\) to \(2\) we get: \(\mathbf{r}(0) = 0\mathbf{i} - 0\mathbf{j} + 0\mathbf{k}\) at t=0 and \(\mathbf{r}(2) = 4\mathbf{i} - 6\mathbf{j} + 2\mathbf{k } \) at t=2. Sketch these points and connect them to get the curve.

Calculate the Length of Curve

To compute the length of the space curve over an interval, apply the formula for curve length: \(L = \int_a^b |\mathbf{r}'(t)| dt \) where \(\mathbf{r}'(t)\) is the derivative of \(\mathbf{r}(t)\). First find \(\mathbf{r}'(t)\). Given \(\mathbf{r}(t) = 2t\mathbf{i} - 3t\mathbf{j} + t\mathbf{k }\), \(\mathbf{r}'(t)\) = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k }. Next, find the magnitude of \(\mathbf{r}'(t)\), which is \(\sqrt{(2)^2+(-3)^2 + (1)^2} = \sqrt{14}\). Substitute this into the length formula and apply limits for the interval [0,2] to get \(L = \int_0^2 \sqrt{14} dt = \sqrt{14} \cdot t | _0^2 = 2\sqrt{14}\).