Question

Find \(\|r(t)\| .\) \(\mathbf{r}(t)=\sqrt{t} \mathbf{i}+3 t \mathbf{j}-4 t \mathbf{k}\)

Short answer

The magnitude of the vector function is \( \|r(t)\| = \sqrt{t + 25t^{2}} \).

Step-by-step solution

Define the Vector Function Components

In this vector function, the components are \( \sqrt{t} \), \( 3t \), and \( -4t \) along the i, j, and k directions respectively.

Square Each Component

Now each of these components need to be squared: \( (\sqrt{t})^{2} = t, (3t)^{2} = 9t^{2}, \) and \( (-4t)^{2} = 16t^{2} \).

Sum of Squares

You need to sum up the squares of all these component: \( t + 9t^{2} + 16t^{2} = t + 25t^{2} \).

Calculate Square Root

Finally, the norm or magnitude of the vector function is the square root of this sum: \( \|r(t)\| = \sqrt{t + 25t^{2}} \).