Question

Sketch the curve represented by the vectorvalued function and give the orientation of the curve. \(\mathbf{r}(t)=\langle\cos t+t \sin t, \sin t-t \cos t, t\rangle\)

Short answer

The curve represented by the vector-valued function \(\mathbf{r}(t)=\langle\cos t+t \sin t, \sin t-t \cos t, t\rangle\) has an orientation upwards along the \(z\)-axis once it's sketched onto a 3D space.

Step-by-step solution

Understand the Vector-Valued Function

Given, the vector-valued function is \(\mathbf{r}(t)=\langle\cos t+t \sin t, \sin t-t \cos t, t\rangle\). In a vector-valued function, each scalar component is a function of the same variable, in this case, \(t\). Thus, this vector function represents a curve in three-dimensional space as \(t\) varies.

Convert it into a Parametric Form

The vector-valued function can be transformed into a set of parametric equations, where \(x\), \(y\), and \(z\) are all functions of \(t\). The given vector-valued function can therefore be written as: \(x = \cos t+t \sin t\), \(y = \sin t-t \cos t\) and \(z = t\).

Visualize and Define the Orientation

Once we have the parametric equations, we can sketch the function in a 3D plot. The orientation of the curve is defined by the direction in which \(t\) increases. By observing how the values of \(x\), \(y\), and \(z\) change as \(t\) increases, we can define the orientation of the curve. In this case, as \(t\) increases, \(x\) and \(y\) perform oscillating behaviors due to the sinusoidal components, while \(z\) increases linearly with \(t\), so we can say the orientation of the curve is upwards along the \(z\)-axis.