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Chapter 13: Problem 28

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 $$\frac{d}{d t}\left(\mathbf{r} \cdot \frac{d \mathbf{r}}{d t} \times \frac{d^{2} \mathbf{r}}{d t^{2}}\right)=\mathbf{r} \cdot\left(\frac{d \mathbf{r}}{d t} \times \frac{d^{3} \mathbf{r}}{d t^{3}}\right)$$

Short Answer

Expert verified
Both parts use product rules for differentiation and confirm given identities.

Step by step solution

01

Apply the Product Rule to the Scalar Triple Product

The problem states that we need to differentiate \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \) with respect to \( t \). We recognize that this is a product of a vector and a vector cross product. By applying the product rule for derivatives: \[ \frac{d}{dt}(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})) = \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot \frac{d}{dt}(\mathbf{v} \times \mathbf{w}) \]
02

Apply the Product Rule to the Cross Product

The cross product \( \mathbf{v} \times \mathbf{w} \) is also a product that needs to be differentiated. Again applying the product rule to the cross product, we get:\[ \frac{d}{dt}(\mathbf{v} \times \mathbf{w}) = \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt} \]Substitute this back into the equation derived in Step 1.
03

Substitute and Simplify

Substitute the expression obtained in Step 2 into the equation from Step 1:\[ \frac{d}{dt}(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})) = \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot \left( \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt} \right) \]Simplifying gives:\[ \frac{d\mathbf{u}}{dt} \cdot \mathbf{v} \times \mathbf{w} + \mathbf{u} \cdot \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{u} \cdot \mathbf{v} \times \frac{d\mathbf{w}}{dt} \]This expression matches the right-hand side of the formula we were required to show, confirming the relation.
04

Understand the Given Expression for Part b

To tackle part b, note that we want to differentiate \( \mathbf{r} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2} \right) \). Recognize this as a scalar triple product and apply the same derivative rules as in Step 1 and 2.
05

Apply the Product and Cross Product Rules

By applying the derivative rules, we start with:\[ \frac{d}{dt} \left( \mathbf{r} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2} \right) \right) = \frac{d\mathbf{r}}{dt} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2} \right) + \mathbf{r} \cdot \frac{d}{dt} \left( \frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2} \right) \]
06

Differentiate the Cross Product in Part b

To differentiate \[ \frac{d}{dt} \left( \frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2} \right) \],apply the product rule:\[ \frac{d^2\mathbf{r}}{dt^2} \times \frac{d^2\mathbf{r}}{dt^2} + \frac{d\mathbf{r}}{dt} \times \frac{d^3\mathbf{r}}{dt^3} \]Note that the cross product of a vector with itself is zero, allowing the simplification of the product to:\[ \frac{d\mathbf{r}}{dt} \times \frac{d^3\mathbf{r}}{dt^3} \]
07

Substitute and Simplify

Now, substitute back:\[ \frac{d}{dt} \left( \mathbf{r} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2} \right) \right) = \mathbf{r} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^3\mathbf{r}}{dt^3} \right) \]This confirms the given expression in part b, as the term with the double cross product goes to zero.

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Key Concepts

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

Product Rule
The product rule is a fundamental concept when differentiating products of functions. In calculus, it's widely used for the differentiation of functions that are multiplied together. When it comes to vector functions, this rule still applies but requires more attention due to the intricacies of vector operations.

When you differentiate a scalar triple product like \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \), where \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) are vector functions of a variable \( t \), the product rule aids in simplifying the differentiation process. The rule tells us that the derivative of a product of two or more functions is the sum of derivatives, taken one function at a time, with all the other functions held constant at each step.

In this specific case, we find:
  • The derivative of \( \mathbf{u} \) first, with \( \mathbf{v} \times \mathbf{w} \) as a constant, then
  • The derivative of \( (\mathbf{v} \times \mathbf{w}) \), keeping \( \mathbf{u} \) constant.
This approach simplifies the problem and makes the differentiation manageable. Remember, the product rule is vital because it breaks down the complex interactions in vector differentiation tasks.
Cross Product
The cross product is a key concept in vector mathematics, particularly in three-dimensional spaces. Unlike the dot product, which results in a scalar, the cross product of two vectors results in another vector. This new vector is orthogonal (perpendicular) to each of the original vectors.

For vectors \( \mathbf{v} \) and \( \mathbf{w} \), the cross product is written as \( \mathbf{v} \times \mathbf{w} \). This operation is not only crucial in many areas of physics such as torque and magnetic forces but also plays a significant role in computer graphics and geometrical computations.
  • The magnitude of the cross product vector is equal to the area of the parallelogram that the vectors span.
  • The direction of the cross product vector is determined by the right-hand rule.
In differentiation, when the cross product itself involves vector functions, you must apply differentiation rules to each component. For example, breaking down \( \frac{d}{dt}(\mathbf{v} \times \mathbf{w}) \) involves differentiating each vector while applying the product rule, making sure all terms and operations meet the requirements for correct derivative computation.
Scalar Triple Product
The scalar triple product connects three vectors in a way that results in a scalar quantity. It combines both the dot product and the cross product, expressed as \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \). This product has applications in determining volumes and is used in physics and engineering to describe phenomena like torque.

The value of the scalar triple product can show if vectors are coplanar. When the result is zero, it indicates that the vectors lie in the same plane.
  • This operation places one vector through the dot product with the area vector resulting from the cross product of the other two.
  • It follows the right-hand rule for direction but returns a scalar value.
In calculus, when differentiating such expressions, we rely on rules for differentiation of both dot and cross products. The product rule applies twice - once for the scalar multiplication and once for the underlying cross product. This layered application helps unravel the complexity, as seen by substituting piece by piece until simplifying to the desired form of the scalar product's derivative.

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Most popular questions from this chapter

Find the velocity and acceleration vectors in terms of \(\mathbf{u}_{r}\) and \(\mathbf{u}_{\theta}\) $$r=a \sin 2 \theta \quad \text { and } \quad \frac{d \theta}{d t}=2 t$$

Rounding the answers to four decimal places, use a CAS to find \(\mathbf{v}, \mathbf{a}\), speed, \(\mathbf{T}, \mathbf{N}, \mathbf{B}, \kappa, \tau,\) and the tangential and normal components of acceleration for the curves at the given values of \(t\). $$\mathbf{r}(t)=\left(e^{t} \cos t\right) \mathbf{i}+\left(e^{t} \sin t\right) \mathbf{j}+e^{t} \mathbf{k}, \quad t=\ln 2$$

\(\mathbf{r}(t)\) is the position of a particle in space at time \(t\) Find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of \(t .\) Write the particle's velocity at that time as the product of its speed and direction. $$\mathbf{r}(t)=(2 \ln (t+1)) \mathbf{i}+t^{2} \mathbf{j}+\frac{t^{2}}{2} \mathbf{k}, \quad t=1$$

Do the data in the accompanying table support Kepler's third law? Give reasons for your answer. $$\begin{array}{lcc} \hline & \text { Semimajor axis } & \\ \text { Planet } & a\left(\mathbf{1 0}^{10} \mathbf{m}\right) & \text { Period } T \text { (years) } \\ \hline \text { Mercury } & 5.79 & 0.241 \\ \text { Venus } & 10.81 & 0.615 \\ \text { Mars } & 22.78 & 1.881 \\ \text { Saturn } & 142.70 & 29.457 \\ \hline \end{array}$$

a. Show that the curve \(\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(1-\cos t) \mathbf{k}\) \(0 \leq t \leq 2 \pi,\) is an ellipse by showing that it is the intersection of a right circular cylinder and a plane. Find equations for the cylinder and plane. b. Sketch the ellipse on the cylinder. Add to your sketch the unit tangent vectors at \(t=0, \pi / 2, \pi,\) and \(3 \pi / 2\) c. Show that the acceleration vector always lies parallel to the plane (orthogonal to a vector normal to the plane). Thus, if you draw the acceleration as a vector attached to the ellipse, it will lie in the plane of the ellipse. Add the acceleration vectors for \(t=0, \pi / 2, \pi,\) and \(3 \pi / 2\) to your sketch. d. Write an integral for the length of the ellipse. Do not try to evaluate the integral; it is nonelementary. e. Numerical integrator Estimate the length of the ellipse to two decimal places.
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