Chapter 12: Problem 18
Determine whether the statement is true or false. Explain your answer. If the line \(y=x\) is parametrized by the vector-valued function \(\mathbf{r}(t)\), then \(\mathbf{r}(t)\) is smooth.
Short Answer
Expert verified
The statement is true; the vector function \( \mathbf{r}(t) = \langle t, t \rangle \) is smooth.
Step by step solution
01
Understanding Smoothness
A vector-valued function \( \mathbf{r}(t) \) is called smooth on an interval if its derivative \( \mathbf{r}'(t) \) is continuous and \( \mathbf{r}'(t) eq \mathbf{0} \) for all \( t \) in the given interval. First, we need to understand if the given parametrization leads to these conditions.
02
Define the Parametrization
The line \( y = x \) can be parametrized by \( \mathbf{r}(t) = \langle t, t \rangle \), where both the x and y components are equal to the parameter \( t \).
03
Calculate the Derivative
To check for smoothness, calculate \( \mathbf{r}'(t) \):\[ \mathbf{r}'(t) = \frac{d}{dt} \langle t, t \rangle = \langle 1, 1 \rangle. \]
04
Verify Smoothness Condition
Check if \( \mathbf{r}'(t) = \langle 1, 1 \rangle \) is non-zero and continuous. Since it is constant and non-zero everywhere, the derivative is both continuous and non-zero for all \( t \).
05
Conclude the Smoothness
Since \( \mathbf{r}'(t) \) is continuous and non-zero, the function \( \mathbf{r}(t) = \langle t, t \rangle \) is smooth on the entire real line, confirming that the parametrization is smooth.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Smoothness
The concept of smoothness in mathematics refers to a specific property of a function that makes it "smooth". When we say a function is smooth, we usually mean it lacks any sharp corners or breaks. Specifically, for vector-valued functions like \( \mathbf{r}(t) \), smoothness implies that throughout the interval in consideration, the derivative \( \mathbf{r}'(t) \) is continuous and, importantly, never zero.
- A continuous derivative means the function changes gradually without any sudden jumps or interruptions.
- Having \( \mathbf{r}'(t) eq \mathbf{0} \) ensures that the function is not just sitting still at any point, but is progressing smoothly along its path.
Vector-Valued Functions
Vector-valued functions extend the idea of single-variable functions to higher dimensions. Instead of returning a single real number as their output, vector-valued functions like \( \mathbf{r}(t) \) return a vector for every input value of \( t \). This is particularly useful for describing paths and curves in spaces like the plane or 3D space.
- Each component of the vector depends on the parameter \( t \), which describes how the point moves as \( t \) changes.
- In contexts such as physics and engineering, these functions can describe motion in terms of position, velocity, and acceleration.
- The line \( y = x \) can be described by \( \mathbf{r}(t) = \langle t, t \rangle \), showing both x and y coordinates vary identically as \( t \) changes.
Derivative Continuity
Derivative continuity is a critical aspect that ensures smoothness in vector-valued functions. For a function to be smooth, its derivative must not only be defined everywhere in the interval but also be continuous.
- Continuity of the derivative means that small changes in \( t \) lead to small changes in the value of \( \mathbf{r}'(t) \).
- This is important because jumps or discontinuities in the derivative could lead to kinks or sharp turns in the resulting graph, contradicting smoothness.
- In the example of \( \mathbf{r}(t) = \langle t, t \rangle \), the derivative \( \mathbf{r}'(t) = \langle 1, 1 \rangle \) is constant, thereby trivially continuous.