/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Let \(\mathbf{u}=\mathbf{u}(t)\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(\mathbf{u}=\mathbf{u}(t)\) and \(\mathbf{v}=\mathbf{v}(t)\) be vector functions of \(t\), where \(\mathbf{u}\), \(\mathbf{v}\) are vectors in 3-dimensional space. Prove (assuming appropriate differentiability): a) \(\frac{d}{d t}(\mathbf{u}+\mathbf{v})=\frac{d \mathbf{u}}{d t}+\frac{d \mathbf{v}}{d i}\) b) \(\frac{d}{d t}|g(t) \mathbf{u}|=g(t) \frac{d \mathbf{u}}{d t}+g^{\prime}(t) \mathbf{u}\) c) \(\frac{d}{d t}(\mathbf{u} \cdot \mathbf{v})=\mathbf{u} \cdot \frac{d \mathbf{v}}{d t}+\mathbf{v} \cdot \frac{d \mathbf{u}}{d t}\) d) \(\frac{d}{d t}(\mathbf{u} \times \mathbf{v})=\mathbf{u} \times \frac{d \mathbf{v}}{d t}+\frac{d \mathbf{u}}{d t} \times \mathbf{v} \quad\) (watch the order!)

Short Answer

Expert verified
Question: Prove the vector calculus formula for the derivative of the dot product of two vectors with respect to time: $$\frac{d}{dt}(\mathbf{u}(t) \cdot \mathbf{v}(t)) = \mathbf{u}(t) \cdot \frac{d \mathbf{v}(t)}{d t} + \mathbf{v}(t) \cdot \frac{d \mathbf{u}(t)}{d t}.$$

Step by step solution

01

a) Derivative of the sum of two vectors

To prove this, we will take the derivative of \(\mathbf{u}(t) + \mathbf{v}(t)\) with respect to \(t\) and show that it is equal to \(\frac{d\mathbf{u}}{dt}+\frac{d\mathbf{v}}{dt}\). So, $$ \frac{d}{dt}(\mathbf{u}+\mathbf{v}) = \frac{d\mathbf{u}}{dt} + \frac{d\mathbf{v}}{dt}. $$
02

b) Derivative of a scalar function times a vector

To prove this, we will take the derivative of \(g(t) \mathbf{u}(t)\) with respect to \(t\) and show that it is equal to \(g(t) \frac{d \mathbf{u}}{d t}+g^{\prime}(t) \mathbf{u}\). So, $$ \frac{d}{dt}(g(t)\mathbf{u}) = g(t) \frac{d\mathbf{u}}{dt} + g^{\prime}(t)\mathbf{u}. $$
03

c) Derivative of the dot product of two vectors

To prove this, we will take the derivative of \(\mathbf{u}(t) \cdot \mathbf{v}(t)\) with respect to \(t\) and show that it is equal to \(\mathbf{u} \cdot \frac{d \mathbf{v}}{d t}+\mathbf{v} \cdot \frac{d \mathbf{u}}{d t}\). So, $$ \frac{d}{dt}(\mathbf{u} \cdot \mathbf{v}) = \mathbf{u} \cdot \frac{d\mathbf{v}}{dt} + \mathbf{v} \cdot \frac{d\mathbf{u}}{dt}. $$
04

d) Derivative of the cross product of two vectors

To prove this, we will take the derivative of \(\mathbf{u}(t) \times \mathbf{v}(t)\) with respect to \(t\) and show that it is equal to \(\mathbf{u} \times \frac{d \mathbf{v}}{d t}+\frac{d \mathbf{u}}{d t} \times \mathbf{v}\). So, $$ \frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \mathbf{u} \times \frac{d\mathbf{v}}{dt} + \frac{d\mathbf{u}}{dt} \times \mathbf{v}. $$

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Key Concepts

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

Vector Differentiation
Vector differentiation refers to the process of taking derivatives of vector functions, much like we take derivatives of scalar functions. The principles are similar, but there are some key differences due to the vector's directional properties.
  • Sum Rule for Vectors: If you are given two vector functions, \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \), their derivative with respect to \( t \) is simply derived component-wise, resulting in the sum \( \frac{d \mathbf{u}}{dt} + \frac{d \mathbf{v}}{dt} \).
  • Product Rule with Scalar: When a vector is multiplied by a scalar function \( g(t) \), the derivative follows a modified version of the product rule: \( \frac{d}{dt}(g(t)\mathbf{u}) = g(t) \frac{d\mathbf{u}}{dt} + g'(t)\mathbf{u} \).
These rules illustrate how vector differentiation extends scalar rules to accommodate the vector's multi-component nature.
Dot Product
The dot product, also known as the scalar product, is an operation that takes two vectors and returns a scalar. It is key in vector calculus for understanding projections and angles between vectors.
  • Calculation: The dot product of two vectors \( \mathbf{u} = [u_1, u_2, u_3] \) and \( \mathbf{v} = [v_1, v_2, v_3] \) is calculated as \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \).
  • Differentiating the Dot Product: The derivative of a dot product follows a specific rule: \( \frac{d}{dt}(\mathbf{u} \cdot \mathbf{v}) = \mathbf{u} \cdot \frac{d\mathbf{v}}{dt} + \mathbf{v} \cdot \frac{d\mathbf{u}}{dt} \). This formula is derived using the product rule and reflects the combination of derivatives of both vectors.
The dot product is essential in vector calculus since it provides insights into vector alignment.
Cross Product
The cross product, or vector product, is another way to combine vectors, producing a third vector orthogonal to the initial pair. Unlike the dot product, the cross product results in a vector.
  • Calculation: For two vectors \( \mathbf{u} \) and \( \mathbf{v} \), the cross product \( \mathbf{u} \times \mathbf{v} \) is determined by the determinant of a matrix involving these vectors: \[\mathbf{u} \times \mathbf{v} = \begin{bmatrix}i & \mathbf{j} & \mathbf{k} \u_1 & u_2 & u_3 \v_1 & v_2 & v_3 \\end{bmatrix} \]
  • Differentiating the Cross Product: The derivative incorporates the product rule similarly to scalar functions, but it pays careful attention to order: \( \frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \mathbf{u} \times \frac{d\mathbf{v}}{dt} + \frac{d\mathbf{u}}{dt} \times \mathbf{v} \). The order is essential because the cross product is not commutative.
Understanding the cross product is pivotal when dealing with rotational physics and torque.
Vector Functions
Vector functions depend on one or more variables and yield vectors, playing a crucial role in physics and engineering.
  • Definition: A vector function \( \mathbf{u}(t) \) is defined in components, like \( \mathbf{u}(t) = [u_1(t), u_2(t), u_3(t)] \), where each \( u_i(t) \) is a scalar function of \( t \).
  • Applications: Such functions model a variety of physical phenomena, such as position, velocity, and acceleration in three-dimensional space.
  • Computation: Differentiation of vector functions follows component-wise differentiation, leading to properties covering both dot and cross products.
Vector functions are foundational in modeling real-world systems and their dynamic behavior over time.

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Most popular questions from this chapter

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