Chapter 11: Problem 32
Prove that \(\mid \mathbf{r}(t) \|\) is constant if and only if \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t)=0\).
Short Answer
Expert verified
The magnitude of \( \mathbf{r}(t) \) is constant if and only if it is orthogonal to its derivative \( \mathbf{r}'(t) \).
Step by step solution
01
Understand the Problem
We are given a vector function \( \mathbf{r}(t) \) and need to prove that the magnitude (i.e., length) of this vector is constant if and only if the dot product of \( \mathbf{r}(t) \) and its derivative \( \mathbf{r}'(t) \) is zero.
02
Express the Magnitude of \( \mathbf{r}(t) \)
The magnitude of a vector \( \mathbf{r}(t) \) is given by \( ||\mathbf{r}(t)|| = \sqrt{\mathbf{r}(t) \cdot \mathbf{r}(t)} \). We need to show conditions under which this magnitude remains constant.
03
Differentiate and Apply Chain Rule
If \( ||\mathbf{r}(t)|| \) is constant, then its derivative with respect to \( t \) is zero: \( \frac{d}{dt} ||\mathbf{r}(t)|| = 0 \). Use the chain rule to find the derivative: \( \frac{d}{dt} \mathbf{r}(t) \cdot \mathbf{r}(t) = 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) \).
04
Analyze the Result
Setting the derivative to zero, we get \( 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \), which simplifies to \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \). This means that for the magnitude to be constant, \( \mathbf{r}(t) \) and \( \mathbf{r}'(t) \) must be orthogonal.
05
Reverse Implication
Assume \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \). Then \( \frac{d}{dt} ||\mathbf{r}(t)||^2 = 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \), which implies \( ||\mathbf{r}(t)|| \) is constant. This completes the proof.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnitude of a Vector
The magnitude of a vector represents its length. For a vector function \( \mathbf{r}(t) \), the magnitude is calculated using the formula:
The magnitude is a crucial concept because it provides a way to measure the size of a vector, irrespective of its direction. If the magnitude is constant, it suggests that the vector is not getting longer or shorter over time. In this context, if \( ||\mathbf{r}(t)|| \) is constant, we explore how it relates to the vector's nature and behavior over time.
- \( ||\mathbf{r}(t)|| = \sqrt{\mathbf{r}(t) \cdot \mathbf{r}(t)} \).
The magnitude is a crucial concept because it provides a way to measure the size of a vector, irrespective of its direction. If the magnitude is constant, it suggests that the vector is not getting longer or shorter over time. In this context, if \( ||\mathbf{r}(t)|| \) is constant, we explore how it relates to the vector's nature and behavior over time.
Derivative of a Vector Function
The derivative of a vector function like \( \mathbf{r}(t) \) is similar to finding the derivative of a scalar function, but it applies to each component of the vector. Symbolically, this derivative is written as \( \mathbf{r}'(t) \). It provides information on how the vector \( \mathbf{r}(t) \) changes as the parameter \( t \) varies.
In the context of the problem, if the magnitude of \( \mathbf{r}(t) \) is constant, then its rate of change, as given by its derivative, must be zero. According to the steps provided, when we differentiate \( ||\mathbf{r}(t)|| \) (following chain rule), and find that it's zero, it reinforces the condition for a constant magnitude. This leads us to look further into the properties that enable such a condition.
In the context of the problem, if the magnitude of \( \mathbf{r}(t) \) is constant, then its rate of change, as given by its derivative, must be zero. According to the steps provided, when we differentiate \( ||\mathbf{r}(t)|| \) (following chain rule), and find that it's zero, it reinforces the condition for a constant magnitude. This leads us to look further into the properties that enable such a condition.
Dot Product Properties
The dot product is a fundamental operation in vector calculus that combines two vectors to yield a scalar. For vectors \( \mathbf{a} \) and \( \mathbf{b} \), the dot product is expressed as:
In this exercise, we consider the dot product \( \mathbf{r}(t) \cdot \mathbf{r}'(t) \). If it equals zero, the derivative vector \( \mathbf{r}'(t) \) is orthogonal to the original vector function \( \mathbf{r}(t) \). This orthogonality is the key condition ensuring that the magnitude of \( \mathbf{r}(t) \) remains constant.
- \( \mathbf{a} \cdot \mathbf{b} = ||\mathbf{a}|| \, ||\mathbf{b}|| \cos(\theta) \)
In this exercise, we consider the dot product \( \mathbf{r}(t) \cdot \mathbf{r}'(t) \). If it equals zero, the derivative vector \( \mathbf{r}'(t) \) is orthogonal to the original vector function \( \mathbf{r}(t) \). This orthogonality is the key condition ensuring that the magnitude of \( \mathbf{r}(t) \) remains constant.
Chain Rule
The chain rule is a method in calculus for finding the derivative of a composition of functions. When dealing with vector-valued functions, the chain rule helps us understand the rate of change of a vector function's magnitude.
In the solution, we apply the chain rule to differentiate the square of the magnitude of \( \mathbf{r}(t) \), expressed as:
In the solution, we apply the chain rule to differentiate the square of the magnitude of \( \mathbf{r}(t) \), expressed as:
- \( \frac{d}{dt}(\mathbf{r}(t) \cdot \mathbf{r}(t)) = 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) \)