Chapter 13: Problem 54
Find an expression for \(\frac{d}{d t}[\mathbf{u}(t) \cdot(\mathbf{v}(t) \times \mathbf{w}(t))]\)
Short Answer
Expert verified
The derivative is \(\frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot (\frac{d\mathbf{v}}{dt} \times \mathbf{w}) + \mathbf{u} \cdot (\mathbf{v} \times \frac{d\mathbf{w}}{dt})\).
Step by step solution
01
Identify the Problem
We need to find the derivative of the expression \(\mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))\) with respect to \(t\). This involves a dot product and a cross product, so we will use the product rule and the properties of derivatives.
02
Use the Product Rule for Derivatives
The expression \(\mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))\) is a product of two vector functions. Let's denote \(\mathbf{a}(t) = \mathbf{u}(t)\) and \(\mathbf{b}(t) = \mathbf{v}(t) \times \mathbf{w}(t)\). According to the product rule, the derivative is \(\frac{d}{dt}[\mathbf{a}(t) \cdot \mathbf{b}(t)] = \frac{d\mathbf{a}}{dt} \cdot \mathbf{b} + \mathbf{a} \cdot \frac{d\mathbf{b}}{dt}\).
03
Apply the Derivative to Each Component
First, calculate \(\frac{d\mathbf{a}}{dt}\cdot\mathbf{b}(t) = \frac{d}{dt}[\mathbf{u}(t)] \cdot (\mathbf{v}(t) \times \mathbf{w}(t))\). This derivative follows directly from the derivative \(\mathbf{a}(t)\), i.e., \(\frac{d\mathbf{u}}{dt}\cdot(\mathbf{v} \times \mathbf{w})\).
04
Differentiate the Cross Product
Next, calculate \(\mathbf{a}(t) \cdot \frac{d\mathbf{b}}{dt}\). Here, the cross-product \(\mathbf{b}(t) = \mathbf{v}(t) \times \mathbf{w}(t)\), so its derivative \(\frac{d\mathbf{b}}{dt}\) is given by the formula \(\frac{d}{dt}[\mathbf{v} \times \mathbf{w}] = \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt}\).
05
Plug In Differentiated Components
Substitute the differentiated components back into the product rule expression: \[ \frac{d}{dt}[\mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))] = \left( \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) \right) + \left( \mathbf{u} \cdot \left( \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt} \right) \right) \].
06
Simplify the Expression
Combine the expressions from above: \[ \frac{d}{dt}[\mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))] = \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot (\frac{d\mathbf{v}}{dt} \times \mathbf{w}) + \mathbf{u} \cdot (\mathbf{v} \times \frac{d\mathbf{w}}{dt}) \].
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.
Cross Product
The cross product is a mathematical operation that takes two vectors and returns a new vector that is perpendicular to both. It is an essential tool in vector calculus and three-dimensional geometry. If you have two vectors \( \mathbf{v} \) and \( \mathbf{w} \), the cross product is expressed as \( \mathbf{v} \times \mathbf{w} \).
Here are some important characteristics of the cross product:
Here are some important characteristics of the cross product:
- Direction: The resulting vector is perpendicular to both original vectors. In a right-hand coordinate system, follow the right-hand rule to find the direction. Curl your fingers from \( \mathbf{v} \) to \( \mathbf{w} \), and your thumb points in the direction of \( \mathbf{v} \times \mathbf{w} \).
- Magnitude: The length of the cross product vector is equal to the area of the parallelogram that the original vectors span, calculated by \( ||\mathbf{v} \times \mathbf{w}|| = ||\mathbf{v}|| ||\mathbf{w}|| \sin(\theta) \), where \( \theta \) is the angle between \( \mathbf{v} \) and \( \mathbf{w} \).
- Non-commutativity: Keep in mind that the order matters: \( \mathbf{v} \times \mathbf{w} = - (\mathbf{w} \times \mathbf{v}) \).
Dot Product
The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It involves two vectors and results in a scalar value. If \( \mathbf{u} \) and \( \mathbf{v} \) are two vectors, their dot product is denoted by \( \mathbf{u} \cdot \mathbf{v} \).
Key properties of the dot product include:
Key properties of the dot product include:
- Commutation: The dot product is commutative, meaning \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \).
- Scalar Result: Unlike the cross product, the result of a dot product is a scalar, a single numerical value.
- Related to Angles: The dot product is related to the cosine of the angle between the vectors. Calculated by \( \mathbf{u} \cdot \mathbf{v} = ||\mathbf{u}|| ||\mathbf{v}|| \cos(\theta) \), where \( \theta \) is the angle between \( \mathbf{u} \) and \( \mathbf{v} \).
- Projection: It can be used to find the length of the projection of one vector onto another.
Product Rule for Derivatives
The product rule is a fundamental concept in calculus used to differentiate products of functions. When dealing with vector functions, it becomes slightly more sophisticated but follows a similar principle.
In vector calculus, suppose you have two differentiable vector functions \( \mathbf{a}(t) \) and \( \mathbf{b}(t) \), the derivative of their dot product with respect to \( t \) is given by:
Key factors in using the product rule include:
In vector calculus, suppose you have two differentiable vector functions \( \mathbf{a}(t) \) and \( \mathbf{b}(t) \), the derivative of their dot product with respect to \( t \) is given by:
- \( \frac{d}{dt}[\mathbf{a}(t) \cdot \mathbf{b}(t)] = \frac{d\mathbf{a}}{dt} \cdot \mathbf{b}(t) + \mathbf{a}(t) \cdot \frac{d\mathbf{b}}{dt} \).
Key factors in using the product rule include:
- Sequential Application: First, differentiate each vector function as if it were alone, maintaining the existing vectors untouched.
- Summation of Products: Combine expressions of the differentiated components. This involves adding each product of a differentiated component and the undifferentiated counterpart.
- Fit for Complex Functions: Applicable to expressions involving chaining of dot and cross products as seen in the complex calculus problem given.