Question

Find \(\mathbf{r}(t) \cdot \mathbf{u}(t) .\) Is the result a vector-valued function? Explain. \(\mathbf{r}(t)=(3 t-1) \mathbf{i}+\frac{1}{4} t^{3} \mathbf{j}+4 \mathbf{k}\) \(\mathbf{u}(t)=t^{2} \mathbf{i}-8 \mathbf{j}+t^{3} \mathbf{k}\)

Short answer

The result of the dot product \(\mathbf{r}(t) \cdot \mathbf{u}(t)\) is \(5t^3-t^2\), which is a scalar function, not a vector-valued function.

Step-by-step solution

Identify the components of the vectors

The components of the vector \(\mathbf{r}(t)\) are \(r_1(t) = 3t-1\), \(r_2(t) = \frac{1}{4}t^3\), and \(r_3(t) = 4\). For the vector \(\mathbf{u}(t)\), the components are \(u_1(t) = t^2\), \(u_2(t) = -8\), and \(u_3(t) = t^3\).

Find the dot product

Use the formula for the dot product of two vectors, \(r\cdot u = r_1u_1 + r_2u_2 + r_3u_3\), and substitute the identified components: \((3t-1)t^2 + (\frac{1}{4}t^3)(-8) + 4(t^3) = 3t^3-t^2-2t^3+4t^3 = 5t^3-t^2\), which is the dot product.

Identify the type of resulting function

The result of the dot product is \(5t^3-t^2\), which is a scalar function of \(t\). It consists of a single value for each \(t\), not a vector, so it's not a vector-valued function.