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Chapter 12: Problem 15

Find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=e^{-t} \mathbf{i}+4 \mathbf{j} $$

Short Answer

Expert verified
The derivative of the vector function is \(\mathbf{r}^{\prime}(t) = -e^{-t} \mathbf{i}\)

Step by step solution

01

Identify component functions

The vector function (\(\mathbf{r}(t)\)) is given and its components are \(\mathbf{i}\) and \(\mathbf{j}\), where \(f(t) = e^{-t}\) and \(g(t) = 4\) are the component functions. Mathematically, \(\mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j}\). Now, find the derivatives \(f'(t)\) and \(g'(t)\) of these component functions.
02

Differentiate component functions

The derivative of \(e^{-t}\) is \(-e^{-t}\) by chain rule, so \(f'(t) = -e^{-t}\). The derivative of the constant \(g(t) = 4\) is zero, so \(g'(t) = 0\).
03

Derive the vector function

Then derivative of the vector function \(\mathbf{r}(t)\) is obtained by replacing each component by their derivative. So, following the formula for the derivative of a vector function, \(\mathbf{r}'(t) = f'(t) \mathbf{i} + g'(t) \mathbf{j}\). Substituting the derivatives of the component functions, we find \(\mathbf{r}^{\prime}(t) = -e^{-t} \mathbf{i} + 0 \mathbf{j}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a key concept in calculus that deals with finding the rate at which a function is changing at any given point. When you differentiate a function, you are essentially finding its derivative, which is a new function describing the rate of change of the original function.

In this exercise, we differentiate a vector function to determine how each of its component functions changes with respect to time.
  • The **chain rule** is often used in differentiation to find the derivative of composite functions. For example, when differentiating the component function \( f(t) = e^{-t} \), we apply the chain rule to get \( f'(t) = -e^{-t} \).
  • For constant functions like \( g(t) = 4 \), the derivative is always zero, because the rate of change of a constant is always zero.
By understanding differentiation, we can accurately determine the changes in both scalar and vector quantities, which is essential for solving many practical mathematical problems.
Vector Function
A vector function provides a way to encode multiple quantities using vectors. In calculus, these functions map a variable, often time \( t \), to a vector that depends on that variable.

In this exercise, the vector function \( \mathbf{r}(t) = e^{-t} \mathbf{i} + 4 \mathbf{j} \) is given as an example. This vector function describes a vector with two components: one in the \( \mathbf{i} \) direction and one in the \( \mathbf{j} \) direction.
  • Each component of the vector function can change with time, according to its own function.
  • Vector functions are often used in physics to describe quantities that have both magnitude and direction, like velocity or force.
Vector functions are powerful tools for modeling real-world scenarios where several quantities interact simultaneously.
Component Functions
Component functions are the individual functions that make up a vector function. Each function corresponds to a particular vector component, such as \( \mathbf{i} \) or \( \mathbf{j} \). Understanding how to work with these components is crucial for finding the derivative of a vector function.

In the provided vector function \( \mathbf{r}(t) \), the component functions are \( f(t) = e^{-t} \) and \( g(t) = 4 \):
  • \( f(t) = e^{-t} \) is the component function in the \( \mathbf{i} \) direction. Its derivative, \( f'(t) = -e^{-t} \), tells us how it changes over time.
  • \( g(t) = 4 \) is the constant component function in the \( \mathbf{j} \) direction. Since it's constant, its derivative is \( 0 \).
By identifying and differentiating component functions, you can piece together the overall derivative of a vector function, like \( \mathbf{r}'(t) = -e^{-t} \mathbf{i} + 0 \mathbf{j} \), effectively capturing the rate of change of the vector components.

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Most popular questions from this chapter

Sketch the space curve and find its length over the given interval. Function \(\quad\) Interval \(\mathbf{r}(t)=\langle 3 t, 2 \cos t, 2 \sin t\rangle \quad\left[0, \frac{\pi}{2}\right]\)

Find the curvature \(K\) of the curve. \(\mathbf{r}(t)=2 \cos \pi t \mathbf{i}+\sin \pi t \mathbf{j}\)

Find all points on the graph of the function such that the curvature is zero. $$ y=(x-1)^{3}+3 $$

Repeat Exercise 19 for the curve represented by the vector-valued function \(\mathbf{r}(t)=\left\langle 4(\sin t-t \cos t), 4(\cos t+t \sin t), \frac{3}{2} t^{2}\right\rangle\)

Find \(\mathrm{T}(t), \mathrm{N}(t), a_{\mathrm{T}}\), and \(a_{\mathrm{N}}\) at the given time \(t\) for the space curve \(\mathbf{r}(t) .\) [Hint: Find \(\mathrm{a}(t), \mathrm{T}(t)\), and \(a_{\mathrm{N}^{*}}\) Solve for \(\mathbf{N}\) in the equation \(\left.\mathbf{a}(t)=a_{\mathrm{T}} \mathbf{T}+a_{\mathrm{N}} \mathbf{N} .\right]\) $$ \mathbf{r}(t)=e^{t} \sin t \mathbf{i}+e^{t} \cos t \mathbf{j}+e^{t} \mathbf{k} \quad t=0 $$
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