Problem 45
Question
Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=\mathbf{i}+\mathbf{j}+\mathbf{k}$$
Step-by-Step Solution
Verified Answer
The derivative is the zero vector \( \mathbf{0} \).
1Step 1: Understanding the Vector-Valued Function
The vector-valued function given is \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \). This means that it is a constant vector, not dependent on \( t \).
2Step 2: Setting Up the Derivative
To find the derivative of a vector-valued function, we take the derivative of each of the component functions. In this case, the components are the scalar multipliers of \( \mathbf{i}, \mathbf{j}, \text{ and } \mathbf{k} \), which are all constants: 1.
3Step 3: Calculating the Derivative
Differentiate each component separately with respect to \( t \). Since each component is a constant (\(1\)), their derivatives are \(0\). Thus, the derivative of \( \mathbf{r}(t) \) is \(0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} = \mathbf{0} \).
4Step 4: Interpreting the Result
The derivative of a constant vector-valued function is the zero vector, \( \mathbf{0} \), because there is no change with respect to the parameter \( t \).
Key Concepts
Understanding Vector-Valued FunctionsDerivatives of Vector-Valued FunctionsRole of Constants in DerivativesThe Zero Vector as a Derivative
Understanding Vector-Valued Functions
Vector-valued functions are functions that have vectors as their output. These functions take a variable, often denoted as \( t \), and produce vectors as a result. These vectors can be in two or three dimensions or even higher, depending on the problem.
A typical vector-valued function is represented as \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \). Here, \( x(t) \), \( y(t) \), and \( z(t) \) are the component functions that determine the vector's direction and magnitude. Each of these components is a function of \( t \).
In the problem you're exploring, the function is \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \). Notice there is no \( t \) dependence, which means it's actually a constant vector. The components are constants: 1 for each of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), and they don't change with \( t \). This is critical for solving the derivative in this case.
A typical vector-valued function is represented as \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \). Here, \( x(t) \), \( y(t) \), and \( z(t) \) are the component functions that determine the vector's direction and magnitude. Each of these components is a function of \( t \).
In the problem you're exploring, the function is \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \). Notice there is no \( t \) dependence, which means it's actually a constant vector. The components are constants: 1 for each of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), and they don't change with \( t \). This is critical for solving the derivative in this case.
Derivatives of Vector-Valued Functions
Derivatives play a crucial role in calculus, even for vector-valued functions. The derivative of a vector-valued function is found by differentiating each of its components individually.
If you have \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \), the derivative \( \mathbf{r}'(t) \) is calculated by taking the derivative of each component with respect to \( t \):
In your given problem, each component is simply the number 1. So taking the derivative of each – since the derivative of a constant is zero – you get: \( 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} = \mathbf{0} \). This shows that the derivative of a constant vector is always the zero vector.
If you have \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \), the derivative \( \mathbf{r}'(t) \) is calculated by taking the derivative of each component with respect to \( t \):
- Derivative of \( x(t) \), or \( x'(t) \)
- Derivative of \( y(t) \), or \( y'(t) \)
- Derivative of \( z(t) \), or \( z'(t) \)
In your given problem, each component is simply the number 1. So taking the derivative of each – since the derivative of a constant is zero – you get: \( 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} = \mathbf{0} \). This shows that the derivative of a constant vector is always the zero vector.
Role of Constants in Derivatives
Constants have a straightforward property in calculus: their derivative is zero. This makes sense because constants do not change; they're fixed values.
In the context of vector-valued functions, if a component of the vector is a constant – like in \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \) where each component equals 1 – the derivative of each is zero. This is true regardless of how you might try to change the variable \( t \).
To see this in practice, remember that the derivative of a constant \( c \) is \( 0 \). Therefore, \( \frac{d}{dt}(c) = 0 \). Apply this to each component of your vector function: each remains unchanged.
In the context of vector-valued functions, if a component of the vector is a constant – like in \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \) where each component equals 1 – the derivative of each is zero. This is true regardless of how you might try to change the variable \( t \).
To see this in practice, remember that the derivative of a constant \( c \) is \( 0 \). Therefore, \( \frac{d}{dt}(c) = 0 \). Apply this to each component of your vector function: each remains unchanged.
The Zero Vector as a Derivative
The concept of the zero vector arises naturally when dealing with derivatives of constant vectors. A zero vector in three dimensions looks like \( \mathbf{0} = 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} \). It is the vector form of the number zero and has no magnitude or direction.
In the exercise, when you differentiate the constant vector \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \), you end up with \( \mathbf{0} \). The zero vector as a result of this differentiation signifies that there is no change in the vector with respect to \( t \).
When you obtain a zero vector from a derivative, this implies that your initial vector was "flat" or "constant" in relation to the variable \( t \); it doesn't vary or evolve over time. Understanding this can be quite helpful in interpreting results in both physics and engineering frameworks, where constant rates or states often lead to stationary outcomes.
In the exercise, when you differentiate the constant vector \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \), you end up with \( \mathbf{0} \). The zero vector as a result of this differentiation signifies that there is no change in the vector with respect to \( t \).
When you obtain a zero vector from a derivative, this implies that your initial vector was "flat" or "constant" in relation to the variable \( t \); it doesn't vary or evolve over time. Understanding this can be quite helpful in interpreting results in both physics and engineering frameworks, where constant rates or states often lead to stationary outcomes.
Other exercises in this chapter
Problem 43
Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=e^{-t} \mathbf{i}+\sin (3 t) \mathbf{j}+10 \sqrt{t} \mathbf{k}$$ A sketch of the graph i
View solution Problem 44
Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=e^{t} \mathbf{i}+2 e^{t} \mathbf{j}+\mathbf{k}$$
View solution Problem 46
Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=t e^{t} \mathbf{i}+t \ln (t) \mathbf{j}+\sin (3 t) \mathbf{k}$$
View solution Problem 47
Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=\frac{1}{t+1} \mathbf{i}+\arctan (t) \mathbf{j}+\ln t^{3} \mathbf{k}$$
View solution