Question

Find \(\mathbf{r}(t) \cdot \mathbf{u}(t) .\) Is the result a vector-valued function? Explain. \(\mathbf{r}(t)=\langle 3 \cos t, 2 \sin t, t-2\rangle\) \(\mathbf{u}(t)=\left\langle 4 \sin t,-6 \cos t, t^{2}\right\rangle\)

Short answer

The result is \(\mathbf{r}(t) \cdot \mathbf{u}(t) = t^{3} - 2t^{2}\), which is a scalar function not a vector-valued function.

Step-by-step solution

Write down the given functions

Write down the given vector-valued functions \(\mathbf{r}(t)=\langle 3 \cos t, 2 \sin t, t-2\rangle\) and \(\mathbf{u}(t)=\left\langle 4 \sin t,-6 \cos t, t^{2}\right\rangle\).

Compute the dot product

Compute the dot product, which means multiplying matching components of the vectors together and sum them up: \(\mathbf{r}(t) \cdot \mathbf{u}(t) = (3 \cos t)(4 \sin t) + (2 \sin t)(-6 \cos t) + (t-2)(t^{2}).\)

Simplify the expression

Simplify the expression: \(\mathbf{r}(t) \cdot \mathbf{u}(t) = 12 \cos t \sin t - 12 \cos t \sin t + t^{3} - 2t^{2} = t^{3} - 2t^{2},\) as the first and second terms cancel each other out.

Determine the type of the result

The result of a dot product is always a scalar - a single number - not a vector. Step 3 clearly shows that the result of \(\mathbf{r}(t) \cdot \mathbf{u}(t)\) is a scalar-valued function, not a vector-valued function, because the result is a single expresssion in terms of \(t\), not a three-component vector.