Question

Find \(\|r(t)\| .\) \(\mathbf{r}(t)=\sin 3 t \mathbf{i}+\cos 3 t \mathbf{j}+t \mathbf{k}\)

Short answer

So, the magnitude of the vector r(t) = sin(3t)i + cos(3t)j + tk is given by \|r(t)\| = \sqrt{1 + t^2}.

Step-by-step solution

Identify the components of the vector

The vector r(t) has been given in the i, j, k representation as r(t) = sin(3t)i + cos(3t)j + tk. The components of this vector are: r_i(t) = sin(3t), r_j(t) = cos(3t), and r_k(t) = t.

Calculate the squares of the components

Next, we want to square each of these components. This gives us: (r_i(t))^2 = (sin(3t))^2, (r_j(t))^2 = (cos(3t))^2, and (r_k(t))^2 = t^2.

Add up the squares and take the square root

Once the squares of the components have been computed, these need to be added together and the square root of the sum needs to be taken to get the magnitude of the vector r(t). The magnitude is then given by the formula: \|r(t)\| = \sqrt{ (sin(3t))^2 + (cos(3t))^2 + t^2 }.

Simplify the square root

Observe that our expression inside the square root becomes \((sin(3t))^2 + (cos(3t))^2 + t^2. We know that \(sin^2x + cos^2x = 1\) for any x. So, this simplifies to \sqrt{1 + t^2}.