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Chapter 3: 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 is shown here. Notice the varying periodic nature of the graph.

Short Answer

Expert verified
\( \mathbf{r}'(t) = -e^{-t} \mathbf{i} + 3\cos(3t) \mathbf{j} + 5t^{-1/2} \mathbf{k} \).

Step by step solution

01

Understand the Problem

We have a vector-valued function \( \mathbf{r}(t) = e^{-t} \mathbf{i} + \sin(3t) \mathbf{j} + 10\sqrt{t} \mathbf{k} \). Our goal is to find the derivative of this function with respect to \( t \), i.e., \( \mathbf{r}'(t) \).
02

Differentiate Component-wise 鈥� x-component

For the x-component, we have \( e^{-t} \). The derivative of \( e^{-t} \) with respect to \( t \) is \( -e^{-t} \). So, the derivative of the x-component is \( -e^{-t} \mathbf{i} \).
03

Differentiate Component-wise 鈥� y-component

For the y-component, we have \( \sin(3t) \). Using the chain rule, the derivative is \( 3 \cos(3t) \) because \( \frac{d}{dt} [\sin(3t)] = 3\cos(3t) \). Therefore, the y-component of the derivative is \( 3\cos(3t) \mathbf{j} \).
04

Differentiate Component-wise 鈥� z-component

For the z-component, we have \( 10\sqrt{t} = 10t^{1/2} \). Using the power rule, the derivative is \( 10 \times \frac{1}{2} t^{-1/2} = 5t^{-1/2} \). Thus, the z-component of the derivative is \( 5t^{-1/2} \mathbf{k} \).
05

Combine the Derivatives

Combine all the component derivatives to find the overall derivative: \( \mathbf{r}'(t) = -e^{-t} \mathbf{i} + 3\cos(3t) \mathbf{j} + 5t^{-1/2} \mathbf{k}\).

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Key Concepts

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

Derivative Calculations
Calculating the derivative of a vector-valued function involves differentiating each of its component functions separately with respect to the same variable, typically time \(t\).
For the given function \(\mathbf{r}(t) = e^{-t} \mathbf{i} + \sin(3t) \mathbf{j} + 10\sqrt{t} \mathbf{k}\), we will calculate the derivative, denoted as \(\mathbf{r}'(t)\).
  • This derivative shows how the vector changes as the parameter \(t\) changes, which gives insight into the velocity of the parameterized curve represented by the vector-valued function.
  • We determine the derivative by calculating the derivative for each unit vector component: \(\mathbf{i}, \mathbf{j}, \mathbf{k}\).
The process of finding this derivative involves systematic component-wise differentiation, ensuring all parts are correctly interpreted to reveal the dynamic behavior of \(\mathbf{r}(t)\).
Component-wise Differentiation
Component-wise differentiation means taking the derivative of each component of a vector-valued function individually. This is essential because each component might involve different functions, each requiring their method of differentiation.
For \(\mathbf{r}(t)\), we have:
  • **x-component** \(e^{-t} \mathbf{i}\): The derivative is \(-e^{-t} \mathbf{i}\).
  • **y-component** \(\sin(3t) \mathbf{j}\): Here, use the chain rule to obtain \(3\cos(3t) \mathbf{j}\).
  • **z-component** \(10\sqrt{t} \mathbf{k}\): Apply the power rule to get \(5t^{-1/2} \mathbf{k}\).
This step-wise breakdown ensures each part is addressed, respecting the unique features and differentiation rules each component function entails.
Chain Rule
The Chain Rule is a crucial tool in calculus, especially for functions that are compositions of one or more other functions. It allows us to differentiate composite functions effectively.
For the y-component \(\sin(3t) \mathbf{j}\) of \(\mathbf{r}(t)\), it is composed of the base function \(\sin(x)\) and an inner function \(3t\).
  • According to the chain rule, the derivative of \(\sin(3t)\) is calculated as \(3\cos(3t)\).
  • The derivative of the outer function \(\sin\) evaluated at \(\frac{d}{dt}(3t) = 3\).
The chain rule concept is vital for handling situations where one variable relies on another, capturing the rate of change effectively in such multi-layered contexts.
Power Rule
The Power Rule simplifies finding derivatives of power functions and is one of the fundamental rules of differentiation. It applies primarily to monomials of the form \(ax^n\).
For the z-component \(10\sqrt{t} \mathbf{k}\), rewrite \(\sqrt{t}\) as \(t^{1/2}\). Utilizing the power rule, we differentiate to obtain \(5t^{-1/2} \mathbf{k}\).
  • Initially, adjust the form to \(10t^{1/2}\).
  • Apply the power rule: Derivative of \(t^n\) is \(n\cdot t^{n-1}\).
  • Resulting in \(5t^{-1/2}\), derived from \(10 \times \frac{1}{2} t^{-1/2}\).
The power rule is an essential, time-saving strategy for differentiating powers, making it invaluable in calculus and related applications.

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Most popular questions from this chapter

A particle moves along the plane curve \(\mathrm{C}\) described by \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} .\) Solve the following problems. Find the curvature of the plane curve at \(t=0,1,2\).

Evaluate the following integrals. \(\int\left(\tan (t) \sec (t) \mathbf{i}-t e^{3 t} \mathbf{j}\right) d t\)

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