91Ó°ÊÓ

In vector calculus, understanding vector components is fundamental. A vector can be broken down into its directional components, usually expressed in terms of unit vectors. For example, in the exercise provided, the vector function \(\mathbf{r}(t)\) is described as \(\sqrt{t} \mathbf{i} + 3t \mathbf{j} - 4t \mathbf{k}\). Each term represents a component of the vector:

• \(\sqrt{t}\mathbf{i}\) is the component in the direction of the unit vector \(\mathbf{i}\) (usually representing the x-axis).
• \(3t\mathbf{j}\) is the component in the direction of the unit vector \(\mathbf{j}\) (usually representing the y-axis).
• \(-4t\mathbf{k}\) is the component in the direction of the unit vector \(\mathbf{k}\) (usually representing the z-axis).

Each of these components influences the overall direction and magnitude of the vector. Recognizing and manipulating these components allows us to perform calculations such as determining the vector's magnitude.

A vector function is a special kind of mathematical function that takes a single variable input and returns a vector output. In simpler terms, it's like a formula that tells you a point in space based on a parameter, often time \(t\). For our example, the vector function is \(\mathbf{r}(t) = \sqrt{t} \mathbf{i} + 3t \mathbf{j} - 4t \mathbf{k}\).
This function describes how the vector \(\mathbf{r}\) changes as \(t\) varies. Each component of the vector is dependent on \(t\), and by plugging in different values for \(t\), you can trace how the position changes in 3-dimensional space.
Understanding vector functions is crucial in fields like physics and engineering, where they help describe the motion and position of objects.

Calculating the magnitude of a vector is like finding its 'length.' It gives a scalar value representing the size of the vector, irrespective of its direction.To find the magnitude of the vector function \(\mathbf{r}(t) = \sqrt{t} \mathbf{i} + 3t \mathbf{j} - 4t \mathbf{k}\), follow these steps:
1. **Square each component**:

• \(x = \sqrt{t}\) so \(x^2 = t\)
• \(y = 3t\) so \(y^2 = 9t^2\)
• \(z = -4t\) so \(z^2 = 16t^2\)

2. **Add the squares**: Combine these squares to get \(t + 9t^2 + 16t^2\), which simplifies to \(t + 25t^2\).3. **Take the square root**: The final step is to compute the square root, giving the magnitude \(\|\mathbf{r}(t)\| = \sqrt{t + 25t^2}\).This process adheres to the general formula for magnitude in three-dimensional space, \(\sqrt{x^2 + y^2 + z^2}\). This gives you a definitive sense of how 'big' the vector is at any point \(t\).