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Chapter 13: Problem 42

Derivatives of triple scalar products a. Show that if \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are differentiable vector functions of \(t,\) then $$\frac{d}{d t}(\mathbf{u} \cdot \mathbf{v} \times \mathbf{w})=\frac{d \mathbf{u}}{d t} \cdot \mathbf{v} \times \mathbf{w}+\mathbf{u} \cdot \frac{d \mathbf{v}}{d t} \times \mathbf{w}+\mathbf{u} \cdot \mathbf{v} \times \frac{d \mathbf{w}}{d t}$$ b. Show that $$\frac{d}{d t}\left(\mathbf{r} \cdot \frac{d \mathbf{r}}{d t} \times \frac{d^{2} \mathbf{r}}{d t^{2}}\right)=\mathbf{r} \cdot\left(\frac{d \mathbf{r}}{d t} \times \frac{d^{3} \mathbf{r}}{d t^{3}}\right)$$ Hint: Differentiate on the left and look for vectors whose products are zero.)

Short Answer

Expert verified
Both results use vector calculus product rules for differentiation.

Step by step solution

01

Recognize the Triple Product Rule

To solve part (a), recognize that the expression involves a cross product inside a dot product, which is a triple scalar product. We need the product rule for differentiation in this vector context. Our task is to differentiate each vector function within the scalar product.
02

Apply the Derivative to the Triple Scalar Product (a)

Apply the derivative to each term in the scalar product: \[ \frac{d}{dt} (\mathbf{u} \cdot \mathbf{v} \times \mathbf{w}) = \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot \frac{d}{dt}(\mathbf{v} \times \mathbf{w}) \]For the second term, apply the product rule to \( \mathbf{v} \times \mathbf{w} \): \[ \frac{d}{dt} (\mathbf{v} \times \mathbf{w}) = \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt} \].
03

Substitute Back and Simplify (a)

Substitute the derivative of the cross product back into the expression: \[ \frac{d}{dt} (\mathbf{u} \cdot \mathbf{v} \times \mathbf{w}) = \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot \left(\frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt}\right) \]. Distribute \(\mathbf{u}\) to get the final expression: \[ \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{u} \cdot \mathbf{v} \times \frac{d\mathbf{w}}{dt} \].
04

Define Triple Scalar Products for Derivatives (b)

For part (b), consider the expression \( \mathbf{r} \cdot \frac{d \mathbf{r}}{d t} \times \frac{d^2 \mathbf{r}}{d t^2} \). This is another triple scalar product involving derivatives of the same vector \( \mathbf{r} \).
05

Differentiate using Product Rule (b)

Differentiate the given expression: \[ \frac{d}{dt}\left(\mathbf{r} \cdot \frac{d \mathbf{r}}{d t} \times \frac{d^2 \mathbf{r}}{d t^2}\right) = \frac{d\mathbf{r}}{dt} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2} \right) + \mathbf{r} \cdot \frac{d}{dt}\left(\frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2}\right) \].
06

Apply the Vector Product Rule (b)

Use the vector product rule on the derivative of the cross product: \[ \frac{d}{dt}\left(\frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2}\right) = \frac{d^2\mathbf{r}}{dt^2} \times \frac{d^2\mathbf{r}}{dt^2} + \frac{d\mathbf{r}}{dt} \times \frac{d^3\mathbf{r}}{dt^3} \]. Note that any vector crossed with itself is zero, thus simplifying to \[ \mathbf{r} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^3\mathbf{r}}{dt^3} \right) \].
07

Conclude on Simplification Zero Products (b)

The first term \( \frac{d\mathbf{r}}{dt} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2} \right) \) is zero because a vector dotted with its cross product with another vector is zero. This leads to the simplified result: \[ \mathbf{r} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^3\mathbf{r}}{dt^3} \right) \].

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Key Concepts

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

Vector Calculus
Vector calculus is a branch of mathematics that deals with vector fields and functions. In this exercise, we explore the differentiation of a triple scalar product, which involves vectors in three-dimensional space. Understanding vector calculus is essential because it provides the groundwork for physics and engineering concepts that deal with forces, acceleration, and more.

In a triple scalar product, you combine three vectors using a dot and cross product. For example, in the expression \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \), \( \mathbf{v} \times \mathbf{w} \) forms a new vector that is perpendicular to both \( \mathbf{v} \) and \( \mathbf{w} \). Then, \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \) results in a scalar value, which is the volume of the parallelepiped formed by the vectors.

Key operations in vector calculus include:
  • Dot Product: Measures the extent to which two vectors point in the same direction.
  • Cross Product: Gives a vector that is perpendicular to the plane containing the two input vectors.
These operations are crucial in manipulating and understanding how different vectors interact in space.
Product Rule in Calculus
The product rule is a fundamental concept in calculus that allows us to find the derivative of a product of two functions. When applied to vector functions, it becomes essential for dealing with complexities of vector differentiation as seen in this exercise.

For the usual product of two functions \( f(t) \) and \( g(t) \), the rule states:
\[ \frac{d}{dt}[f(t)g(t)] = f'(t)g(t) + f(t)g'(t) \]
In the context of the triple scalar product from the exercise, the product rule extends to handle vector functions:
  • Differentiate each component of the scalar triple product separately.
  • Reassemble the derivative by utilizing the results from differentiating individual parts.

This approach is particularly useful because:
  • Vectors have both magnitude and direction, thus requiring careful differentiation of each aspect.
  • Understanding this helps with complex problems in dynamics and electromagnetic theory where terms are often expressed as products of multiple functions.
Cross Product Differentiation
Cross product differentiation involves understanding how to handle derivatives of vector cross products. It's an extended application of the product rule to vectors and this concept is pivotal for operations involving vector fields. In the given exercise, the differentiation of cross products is a critical step.

When differentiating \( \mathbf{v} \times \mathbf{w} \), use: \[ \frac{d}{dt}(\mathbf{v} \times \mathbf{w}) = \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt} \]
  • First term represents the rate of change of \( \mathbf{v} \) and how it affects the cross product.
  • Second term accounts for changes in \( \mathbf{w} \).

This step-by-step method ensures that each element of the cross product is accurately differentiated, maintaining the vector properties throughout the calculation.

Important considerations include:
  • The order of vectors in a cross product affects the sign of the result (anti-commutative property).
  • A vector crossed with itself is zero, simplifying calculations where such terms arise.
Understanding these principles is vital not only in calculus but in fields like physics where motion, rotation, and other vector behaviors are analyzed.

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Most popular questions from this chapter

Volleyball A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6 -ft-high net. It leaves the point of impact with an initial velocity of 35 \(\mathrm{ft} / \mathrm{sec}\) at an angle of \(27^{\circ}\) and slips by the opposing team untouched. a. Find a vector equation for the path of the volleyball. b. How high does the volleyball go, and when does it reach maximum height? c. Find its range and flight time. d. When is the volleyball 7 ft above the ground? How far e. Suppose that the net is raised to 8 ft. Does this change things? Explain.

Osculating circle Find a parametrization of the osculating circle for the parabola \(y=x^{2}\) when \(x=1 .\)

Where trajectories crest For a projectile fired from the ground at launch angle \(\alpha\) with initial speed \(v_{0},\) consider \(\alpha\) as a variable and \(v_{0}\) as a fixed constant. For each \(\alpha, 0<\alpha<\pi / 2,\) we obtain a parabolic trajectory as shown in the accompanying figure. Show that the points in the plane that give the maximum heights of these parabolic trajectories all lie on the ellipse $$x^{2}+4\left(y-\frac{v_{0}^{2}}{4 g}\right)^{2}=\frac{v_{0}^{4}}{4 g^{2}}$$

Find \(\mathbf{B}\) and \(\tau\) for these space curves. \(\mathbf{r}(t)=(6 \sin 2 t) \mathbf{i}+(6 \cos 2 t) \mathbf{j}+5 t \mathbf{k}\)

In Exercises \(31-38\) you will use a CAS to explore the osculating circle at a point \(P\) on a plane curve where \(\kappa \neq 0 .\) Use a CAS to perform the following steps: $$ \begin{array}{l}{\text { a. Plot the plane curve given in parametric or function form over }} \\ {\text { the specified interval to see what it looks like. }} \\ {\text { b. Calculate the curvature } \kappa \text { of the curve at the given value } t_{0}} \\ {\text { using the appropriate formula from Exercise } 5 \text { or } 6 . \text { Use the }} \\ {\text { parametrization } x=t \text { and } y=f(t) \text { if the curve is given as a }} \\ {\quad \text { function } y=f(x) \text { . }}\\\\{\text { c. Find the unit normal vector } N \text { at } t_{0} . \text { Notice that the signs of }} \\ {\text { the components of } N \text { depend on whether the unit tangent vector }} \\\ {\text { T is turning clockwise or counterclockwise at } t=t_{0} \text { . (See }} \\ {\text { Exercise } 7 . )}\\\\{\text { d. If } \mathbf{C}=a \mathbf{i}+b \mathbf{j} \text { is the vector from the origin to the center }(a, b)} \\ {\text { of the osculating circle, find the center } \mathbf{C} \text { from the vector }} \\ {\text { equation }} \\\ {\mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right)} \\ {\text { The point } P\left(x_{0}, y_{0}\right) \text { on the curve is given by the position }} \\ {\text { vector } \mathbf{r}\left(t_{0}\right) .}\\\\{\text { e. Plot implicitly the equation }(x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}} \\ {\text { of the osculating circle. Then plot the curve and osculating }} \\ {\text { circle together. You may need to experiment with the size of }} \\ {\text { the viewing window, but be sure the axes are equally scaled. }}\end{array} $$ $$ \begin{array}{l}{\mathbf{r}(t)=(2 t-\sin t) \mathbf{i}+(2-2 \cos t) \mathbf{j}, \quad 0 \leq t \leq 3 \pi} \\ {t_{0}=3 \pi / 2}\end{array} $$
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