Question

Find the domains of the following vector-valued functions. $$\mathbf{r}(t)=\frac{2}{t-1} \mathbf{i}+\frac{3}{t+2} \mathbf{j}$$

Short answer

The domain of the function \(\mathbf{r}(t)\) is \((-\infty, -2) \cup (-2, 1) \cup (1, \infty)\).

Step-by-step solution

Find the domain for the i-component

We first need to analyze the \(\mathbf{i}\) component of the function: $$\frac{2}{t-1} \mathbf{i}$$. We must identify when it is undefined, which occurs only when there is division by zero, i.e., when \(t-1 = 0\). Solving for \(t\), we get \(t = 1\). So, we should exclude \(t = 1\) from the domain for the \(\mathbf{i}\) component.

Find the domain for the j-component

Now let's analyze the \(\mathbf{j}\) component of the function: $$\frac{3}{t+2} \mathbf{j}$$. Similar to the \(\mathbf{i}\) component, we need to identify when it is undefined, which occurs when there is division by zero, i.e., when \(t+2 = 0\). Solving for \(t\), we get \(t = -2\). So, we should exclude \(t = -2\) from the domain for the \(\mathbf{j}\) component.

Combine the restrictions to get the overall domain

Combining the restrictions for both components, the domain of the function \(\mathbf{r}(t)\) will exclude both \(t = 1\) and \(t = -2\). So the domain of \(\mathbf{r}(t)\) is $$(-\infty, -2) \cup (-2, 1) \cup (1, \infty)$$.

Key concepts

Domain Exclusion

When working with vector-valued functions like \( \mathbf{r}(t) = \frac{2}{t-1} \mathbf{i} + \frac{3}{t+2} \mathbf{j} \), determining the domain involves understanding when each component is defined.

The domain of a function is the set of all input values, here represented by \( t \), that allow the function to output a valid result. In simpler terms, the domain tells us what values \( t \) can take so that the entire expression is mathematically sound.

In our function, the presence of denominators means we need to be cautious about values that make any of them zero because division by zero is undefined in mathematics. Such values should be excluded from the domain. We'll address this exclusion by finding values of \( t \) that lead to zero in any denominator of the components, thereby guidance us to exclude these specific \( t \)-values from the domain.

Component-Wise Analysis

When analyzing a vector-valued function, it is vital to break down the problem into simpler parts by examining each vector component independently.

This approach is often called component-wise analysis and is highly effective in solving problems related to vector functions.

Take the \( \mathbf{r}(t) = \frac{2}{t-1} \mathbf{i} + \frac{3}{t+2} \mathbf{j} \) function as an example:

• We start by looking at the \( \mathbf{i} \) component: \( \frac{2}{t-1} \). To find when this part becomes undefined, resolve the denominator \( t - 1 = 0 \), which gives \( t = 1 \). Hence, exclude \( t = 1 \) from the domain.
• Next, tackle the \( \mathbf{j} \) component: \( \frac{3}{t+2} \). Similarly, solve \( t + 2 = 0 \) to obtain \( t = -2 \). So, exclude \( t = -2 \) as well.

By addressing each component separately and understanding where they become undefined, we assemble the restrictions that together help define the overall domain of the function.

Division by Zero

Division by zero is an essential concept to grasp in mathematics, especially when working with functions. It refers to any scenario in an expression where a number is divided by zero, resulting in a situation that is undefined.

When assessing vector-valued functions such as \( \mathbf{r}(t) = \frac{2}{t-1} \mathbf{i} + \frac{3}{t+2} \mathbf{j} \), we must ensure that none of the operations encompass division by zero.

For instance, examine the components:

• For the \( \mathbf{i} \) component \( \frac{2}{t-1} \), dividing by zero occurs at \( t = 1 \) because the denominator becomes zero at this point.
• Likewise, the \( \mathbf{j} \) component \( \frac{3}{t+2} \) has a division by zero issue at \( t = -2 \) when the denominator equals zero.

When constructing the domain, exclude these critical values causing the zero denominator to assure the function remains well-defined and avoids undefined mathematical errors.