Question

Sketch the space curve and find its length over the given interval. $$ \mathbf{r}(t)=\langle 3 t, 2 \cos t, 2 \sin t\rangle $$ $$ \left[0, \frac{\pi}{2}\right] $$

Short answer

The curve moves from (0,2,0) to \(\frac{3\pi}{2},0,2\) in the first octant. The length of the curve is \(\frac{\pi\sqrt{17}}{2}\).

Step-by-step solution

Understanding the curve

The given vector-valued function may be rewritten in parametric form: \(x=3t\), \(y=2 \cos(t)\), and \(z=2\sin(t)\). Here \(t\) represents the parameter. Each of these equations describes a simple curve in their respective planes. \(x=3t\) is a straight line in the x-t plane, while \(y=2\cos(t)\) and \(z=2\sin(t)\) are oscillatory functions in the y-t and z-t planes. For \(t\) in \([0, \frac{\pi}{2}]\), \(y\) decreases from 2 to 0 and \(z\) increases from 0 to 2. In summary, the curve starts at (0,2,0) and finishes at \(\frac{3\pi}{2},0,2\). It lies entirely in the first octant.

Sketching the curve

In the 3D Cartesian Coordinate system, plot points according to the parametric equations. Start by marking the point (0,2,0) and as you increase \(t\) from 0 to \(\frac{\pi}{2}\), mark the new points. Connect these points smoothly to graph the space curve.

Calculating the length of the curve

The length of a curve defined by a vector-valued function \(\mathbf{r}(t)\) over the interval \([a, b]\) is given by the integral \(\int_{a}^{b}\|\mathbf{r}'(t)\|dt\), where \(\|\mathbf{r}'(t)\|\) is the magnitude of the derivative of \(\mathbf{r}(t)\). Start by finding the derivative of \(\mathbf{r}(t)\), which is \(\mathbf{r}'(t) = \langle 3, -2\sin(t), 2\cos(t)\rangle\). The magnitude of \(\mathbf{r}'(t)\) can be found using the Euclidean norm calculation: \(\|\mathbf{r}'(t)\| = \sqrt{3^2 + (-2\sin(t))^2 + (2\cos(t))^2} = \sqrt{9 + 4 + 4} = \sqrt{17}\). Then, plug into the integral formula: \(\int_{0}^{\frac{\pi}{2}} \sqrt{17}dt = \sqrt{17}t \Big|_0^\frac{\pi}{2} = \frac{\pi\sqrt{17}}{2}\).