Chapter 13: Problem 13
Find the derivative of the vector function. $$ \mathbf{r}(t)=e^{t^{\prime}} \mathbf{i}-\mathbf{j}+\ln (1+3 t) \mathbf{k} $$
Short Answer
Expert verified
The derivative of \( \mathbf{r}(t) \) is \( e^{t} \mathbf{i} + \frac{3}{1+3t} \mathbf{k} \).
Step by step solution
01
Identify Components of the Vector Function
The given vector function is \( \mathbf{r}(t) = e^{t} \mathbf{i} - \mathbf{j} + \ln(1+3t) \mathbf{k} \). This function has three components: \( e^{t} \) associated with \( \mathbf{i} \), \( -1 \) associated with \( \mathbf{j} \), and \( \ln(1+3t) \) associated with \( \mathbf{k} \).
02
Differentiate Each Component
Find the derivative of each function component with respect to \( t \):1. Differentiate \( e^{t} \): The derivative is \( \frac{d}{dt}(e^{t}) = e^{t} \).2. Differentiate \( -1 \): Since \( -1 \) is a constant, its derivative is 0.3. Differentiate \( \ln(1+3t) \) using the chain rule: The derivative is \( \frac{d}{dt}(\ln(1+3t)) = \frac{3}{1+3t} \).
03
Combine the Derivatives into a Vector
Combine the derivatives into a vector to form \( \mathbf{r}'(t) \): \[ \mathbf{r}'(t) = e^{t} \mathbf{i} + 0 \mathbf{j} + \frac{3}{1+3t} \mathbf{k} \].
04
Simplify the Result
Simplify the expression for the derivative vector:\( \mathbf{r}'(t) = e^{t} \mathbf{i} + \frac{3}{1+3t} \mathbf{k} \).This is the final expression for the derivative of the vector function \( \mathbf{r}(t) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Functions
A vector function is a mathematical expression used to describe a curve in space using a vector. It assigns a vector to each point in its domain, often parameterized in terms of a variable like time or generically as \( t \). Vector functions are expressed in the form \( \mathbf{r}(t) = f_1(t) \mathbf{i} + f_2(t) \mathbf{j} + f_3(t) \mathbf{k} \). Each component function \( f_1(t) \), \( f_2(t) \), and \( f_3(t) \) represents a coordinate function in the x, y, and z directions respectively.
Here are some key points:
Here are some key points:
- Vector functions provide a way to capture paths in physics and engineering.
- They are handy for defining curves and trajectories in a multi-dimensional space.
- The components can be any differentiable function.
Chain Rule
The chain rule is a fundamental technique in calculus used to find the derivative of composite functions. When you have a function nested inside another function, the chain rule allows you to differentiate it by considering the inner function and the outer function separately.
Essential points about the chain rule include:
Essential points about the chain rule include:
- The formula is typically written as \( \frac{d}{dt} f(g(t)) = f'(g(t)) \cdot g'(t) \).
- It effectively "unzips" composite functions, differentiating each layer step by step.
- The rule uses the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
Differentiation
Differentiation is a key concept in calculus, referring to the process of finding a derivative, which shows the rate at which one quantity changes with respect to another.
Differentiating a vector function requires differentiating each of its component functions separately, treating them as regular functions:
Differentiating a vector function requires differentiating each of its component functions separately, treating them as regular functions:
- Derivatives tell us how a function behaves small changes in the input.
- It involves applying rules such as the power rule, product rule, and chain rule.
- It can describe how velocity or acceleration varies over time.
Calculus
Calculus is the broader mathematical field that encompasses both differentiation and integration, allowing for a deep understanding of change and areas under curves. Calculus has two main branches:
In the context of this exercise, calculus facilitates understanding how the given vector function \( \mathbf{r}(t) \) changes over time by differentiating it. This ability to compute the derivative is a powerful element of calculus, revealing the underlying dynamic processes in mathematical functions and physical phenomena.
- Differential Calculus: focuses on the concept of the derivative, enabling us to explore changes and rates.
- Integral Calculus: concentrates on the integral, helping us find areas and accumulate quantities.
In the context of this exercise, calculus facilitates understanding how the given vector function \( \mathbf{r}(t) \) changes over time by differentiating it. This ability to compute the derivative is a powerful element of calculus, revealing the underlying dynamic processes in mathematical functions and physical phenomena.