Question

Find the domains of the following vector-valued functions. $$\mathbf{r}(t)=\sqrt{4-t^{2}} \mathbf{i}+\sqrt{t} \mathbf{j}-\frac{2}{\sqrt{1+t}} \mathbf{k}$$

Short answer

Answer: The domain of the vector-valued function is $$0 \leq t \leq 2$$.

Step-by-step solution

Identify the restrictions in each component function

Here are the restrictions on 't' that we need to consider in each component function to ensure they are defined. 1. In the $$\sqrt{4-t^2} \mathbf{i}$$: We must have $$4 - t^2 \geq 0$$, since the expression inside the square root sign must be non-negative for the square root to be defined. 2. In $$\sqrt{t} \mathbf{j}$$: We must have $$t \geq 0$$, since the expression inside the square root sign must be non-negative for the square root to be defined. 3. In $$-\frac{2}{\sqrt{1+t}} \mathbf{k}$$: We want to avoid the case that division by zero. Thus, we must have $$1 + t \neq 0$$, meaning $$t \neq -1$$.

Determine the domain by combining all restrictions

Now we need to combine these restrictions to find the domain of the entire vector-valued function. 1. Solve the inequality $$4 - t^2 \geq 0$$ for 't'. Factoring the expression we get: $$(2-t)(2+t) \geq 0$$ So the domain for $$\sqrt{4-t^{2}} \mathbf{i}$$ is given by $$-2 \leq t \leq 2$$. 2. The domain for $$\sqrt{t} \mathbf{j}$$ is given by: $$t \geq 0$$. 3. The restriction to avoid division by zero for $$-\frac{2}{\sqrt{1+t}} \mathbf{k}$$ is given by: $$t \neq -1$$.

Combine the restrictions to get the final domain

Now we need to find the values of 't' satisfying all three component functions: - The domain for $$\sqrt{4-t^{2}} \mathbf{i}$$ is $$-2 \leq t \leq 2$$. - The domain for $$\sqrt{t} \mathbf{j}$$ is $$t \geq 0$$. - The restriction for $$-\frac{2}{\sqrt{1+t}} \mathbf{k}$$ is $$t \neq -1$$. Combining the above restrictions, we have $$0 \leq t \leq 2$$ except $$t = -1$$. However, since $$t = -1$$ is not in the range $$0 \leq t \leq 2$$, we can conclude that the domain of the given vector-valued function is $$0 \leq t \leq 2$$.

Key concepts

Function Domain

A function domain is essentially the set of all possible input values (in this case, values of the variable \( t \)) that allow the function to work correctly without running into any mathematical problems. For vector-valued functions, each component function has its own specific domain requirements. To find the overall domain, we must consider all these requirements together.

For the vector-valued function \( \mathbf{r}(t) = \sqrt{4-t^{2}} \mathbf{i} + \sqrt{t} \mathbf{j} - \frac{2}{\sqrt{1+t}} \mathbf{k} \), understanding the domain means identifying the 't' values that allow all three component expressions to be valid. Let's break down these restrictions further to ensure a grasp on each part.

Square Root Restrictions

Square root restrictions are rules we need to follow when handling square roots in mathematics. Essentially, the expression under the square root sign must be zero or positive to ensure that the square root is real, as square roots of negative numbers will lead to complex numbers.

For the expression \( \sqrt{4 - t^2} \mathbf{i} \), we impose the constraint \( 4 - t^2 \geq 0 \). Solving this inequality helps us find the valid 't' values. Similarly, \( \sqrt{t} \mathbf{j} \) requires \( t \geq 0 \). These restrictions guarantee that we obtain real numbers for these square root expressions. Always ensure to solve such inequalities carefully to determine where the function is defined.

Inequality Solutions

In mathematical exercises involving inequalities, we're essentially finding all values that satisfy a certain condition expressed as an inequality.

For the function component \( \sqrt{4-t^2} \mathbf{i} \), solving the inequality \( 4-t^2 \geq 0 \) involves finding 't' values such that \((2-t)(2+t) \geq 0\). This is accomplished by determining the critical points and testing intervals to check where the inequality holds. The resulting solution, \(-2 \leq t \leq 2\), tells us the range where this expression within the square root is defined.
Applying these concepts ensures that we clearly define the domain of functions with square roots, resulting in precise and correct interpretations of possible values.

Component Functions

A vector-valued function is composed of multiple component functions, each impacting the overall domain. In our exercise, \( \mathbf{r}(t) = \sqrt{4-t^{2}} \mathbf{i} + \sqrt{t} \mathbf{j} - \frac{2}{\sqrt{1+t}} \mathbf{k} \), each term contributes its own set of restrictions.
By reviewing these components individually:

• \( \sqrt{4-t^2} \) constrains \( t \) to \(-2 \leq t \leq 2\).
• \( \sqrt{t} \) dictates that \( t \geq 0 \).
• The term \( \frac{-2}{\sqrt{1+t}} \) disallows \( t = -1 \) to prevent division by zero.

These separate conditions must all be satisfied simultaneously, meaning that the resulting domain of the entire vector-valued function consists of values where all components are viable. This leads to a domain of \( 0 \leq t \leq 2 \).
Understanding each component's influence on the whole function enables us to address domain questions effectively.