Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

When no domain is given in the definition of a vectorvalued function, it is to be understood that the domain is the set of all (real) scalars for which the rule for the function makes sense and gives real vectors (i.e., vectors with real components). Find the domain of each of the following vector-valued functions: (a) \(\mathbf{r}(t)=\frac{2}{t-4} \mathbf{i}+\sqrt{3-t} \mathbf{j}+\ln |4-t| \mathbf{k}\) (b) \(\mathbf{r}(t)=\left[t^{2}\right] \mathbf{i}-\sqrt{20-t} \mathbf{j}+3 \mathbf{k}([]\) denotes the greatest integer function.) (c) \(\mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+\sqrt{9-t^{2}} \mathbf{k}\)

Step-by-Step Solution

Answer

(a) Domain is \((-\infty, 3]\); (b) Domain is \((-\infty, 20]\); (c) Domain is \([-3, 3]\).

Step 1: Analyze the Domain of Function (a)

The function \(\mathbf{r}(t)=\frac{2}{t-4} \mathbf{i}+\sqrt{3-t} \mathbf{j}+\ln |4-t| \mathbf{k}\) contains three parts: \(\frac{2}{t-4}\), \(\sqrt{3-t}\), and \(\ln |4-t|\). Each of these has its own domain conditions. For the first part, \(\frac{2}{t-4}\), \(t\) cannot be 4 because of division by zero. For the second part, \(\sqrt{3-t}\), \(3-t\) must be non-negative, so \(t \leq 3\). For the third part, \(\ln |4-t|\), \(4-t\) must be positive or negative, so \(t eq 4\). Combining these constraints, the domain is \(t \in (-\infty, 3]\).

Step 2: Analyze the Domain of Function (b)

The function \(\mathbf{r}(t)=\left[t^{2}\right] \mathbf{i}-\sqrt{20-t} \mathbf{j}+3 \mathbf{k}\) contains two parts affecting the domain: \(\left[t^{2}\right]\) and \(\sqrt{20-t}\). The greatest integer function \([t^2]\) is defined for all real \(t\). However, \(\sqrt{20-t}\) requires \(20-t \geq 0\), so the domain condition here is \(t \leq 20\). Therefore, the total domain is \(t \in (-\infty, 20]\).

Step 3: Analyze the Domain of Function (c)

The function \(\mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+\sqrt{9-t^{2}} \mathbf{k}\) includes \(\cos t\) and \(\sin t\) that are defined for all real \(t\). The term \(\sqrt{9-t^2}\) imposes \(9-t^2 \geq 0\), meaning \(-3 \leq t \leq 3\). Thus, the domain is \(t \in [-3, 3]\).

Key Concepts

Domain of Functions

The domain of a function is the set of all input values (in this case, real numbers) for which the function is defined. For vector-valued functions, this means looking at each component of the vector separately and determining where all parts are valid.
The common goal is to find the intersection of these individual domains.

• When dealing with fractional expressions like \( \frac{2}{t-4} \), we need to avoid division by zero, so \( t eq 4 \).
• For square root expressions \( \sqrt{3-t} \), ensure the quantity inside is non-negative, thus \( 3-t \geq 0 \) or \( t \leq 3 \).
• Logarithmic expressions like \( \ln |4-t| \) are defined as long as the input to the logarithm is positive, which excludes \( t = 4 \).

These constraints combine to form the domain. In practice, determining the domain requires careful attention to the type of expression and the mathematical requirements for each part.

Greatest Integer Function

The greatest integer function, often denoted as \( [x] \) or \( \lfloor x \rfloor \), rounds a real number down to the nearest integer less than or equal to that number. This function is defined for all real numbers as every number has a greatest integer less than or equal to it.
It looks like a step function where the output jumps at integer values of the input.

• Example: For \( x = 3.7 \), \([3.7] = 3\).
• For a negative value like \( x = -1.2 \), \([-1.2] = -2\) because \(-2\) is the nearest integer less than \(-1.2\).

This function is particularly useful in computer applications and discrete mathematics, where aligning continuous data to discrete steps is necessary. In the context of vector functions, it can be applied to components without restricting the domain over real numbers, but impacts how the output behaves.

Real Vectors

Real vectors are ordered sets of real numbers. Vectors can be imagined as points in a space where each component represents a dimension of that space. In our case, these vectors exist in three dimensions with components often represented by \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), which are unit vectors in the direction of the x, y, and z axes respectively.
Each component can vary independently and collectively describe the vector's position or direction in space.

• The vector \( \mathbf{r}(t) = a(t) \mathbf{i} + b(t) \mathbf{j} + c(t) \mathbf{k} \) is a typical example, where \( a(t), b(t), \text{and } c(t)\) are scalar functions of \( t \).
• If these functions include operations like division or square roots, their individual constraints help define the domain of the vector as a whole.

Understanding real vectors is key to disciplines like physics and engineering, where they describe forces, velocities, and other physical phenomena in space.

Trigonometric Functions

Trigonometric functions, such as \( \sin t \), \( \cos t \), and \( \tan t \), are periodic and based on the ratios of the sides of a right triangle. They are fundamental in describing oscillatory and wave-like phenomena, crucial for fields like physics and engineering.
The primary trigonometric functions \( \sin \) and \( \cos \) have a period of \( 2\pi \), meaning they repeat their values every \( 2\pi \) units along the t-axis.

• Both \( \sin t \) and \( \cos t \) are defined for all real numbers \( t \). There is no restriction on their domain.
• However, if included in expressions like \( \sqrt{9-t^2} \), the valid range is further restricted by the requirement that the expression inside the square root is non-negative, such as limiting \( t \) to \(-3 \leq t \leq 3 \).

Such functions are not just mathematical abstractions but model real-world scenarios like sound waves and light frequencies.