Question

Ask you to verify parts of Theorem \(11.2 .4 .\) In each let \(f(t)=t^{3}, \vec{r}(t)=\left\langle t^{2}, t-1,1\right\rangle\) and \(\vec{s}(t)=\) \(\left\langle\sin t, e^{t}, t\right\rangle .\) Compute the various derivatives as indicated. Simplify \(\vec{r}(t) \times \vec{s}(t),\) then find its derivative; show this is the same as \(\vec{r}^{\prime}(t) \times \vec{s}(t)+\vec{r}(t) \times \vec{s}^{\prime}(t)\)

Short answer

Both methods yield \( \langle 2t - 1 - e^t, \cos t - 3t^2, t^2 e^t + 2te^t - (t-1) \cos t - \sin t \rangle \).

Step-by-step solution

Compute \( \vec{r}(t) \times \vec{s}(t) \)

First, calculate the cross product \( \vec{r}(t) \times \vec{s}(t) \). Recall that for vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), the cross product is given by:\[ \vec{a} \times \vec{b} = \langle a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1 \rangle \]Applying this to our vectors \( \vec{r}(t) = \langle t^2, t-1, 1 \rangle \) and \( \vec{s}(t) = \langle \sin t, e^t, t \rangle \), we compute:\[ \vec{r}(t) \times \vec{s}(t) = \langle (t-1)t - 1e^t, 1\sin t - t^2t, t^2e^t - (t-1)\sin t \rangle \]\[ = \langle t^2 - t - e^t, \sin t - t^3, t^2 e^t - (t-1)\sin t \rangle \].

Differentiate \( \vec{r}(t) \times \vec{s}(t) \) with respect to \( t \)

Now differentiate the result from Step 1 with respect to \( t \). Let's denote the cross product \( \vec{u}(t) = \langle u_1(t), u_2(t), u_3(t) \rangle \equiv \langle t^2 - t - e^t, \sin t - t^3, t^2 e^t - (t-1)\sin t \rangle \).Differentiate each component:\[ \frac{d}{dt}(u_1) = 2t - 1 - e^t \]\[ \frac{d}{dt}(u_2) = \cos t - 3t^2 \]\[ \frac{d}{dt}(u_3) = (2te^t + t^2 e^t) - ((t-1)\cos t + \sin t) \]\[ = t^2 e^t + 2te^t - (t-1) \cos t - \sin t \]So, \[ \frac{d}{dt}(\vec{r}(t) \times \vec{s}(t)) = \langle 2t - 1 - e^t, \cos t - 3t^2, t^2 e^t + 2te^t - (t-1) \cos t - \sin t \rangle \]

Compute \( \vec{r}^{\prime}(t) \times \vec{s}(t) + \vec{r}(t) \times \vec{s}^{\prime}(t) \)

First, calculate \( \vec{r}^{\prime}(t) \) and \( \vec{s}^{\prime}(t) \).\( \vec{r}^{\prime}(t) = \langle 2t, 1, 0 \rangle \)\( \vec{s}^{\prime}(t) = \langle \cos t, e^t, 1 \rangle \)Calculate \( \vec{r}^{\prime}(t) \times \vec{s}(t) \):\[ \vec{r}^{\prime}(t) \times \vec{s}(t) = \langle 1t - 0e^t, 0\sin t - 2t t, 2t e^t - 1\sin t \rangle \]\[ = \langle t, -2t^2, 2t e^t - \sin t \rangle \]Next, calculate \( \vec{r}(t) \times \vec{s}^{\prime}(t) \):\[ \vec{r}(t) \times \vec{s}^{\prime}(t) = \langle (t-1)1 - 1e^t, 1\cos t - t^2 \cdot 1, t^2 e^t - (t-1) \cdot \cos t \rangle \]\[ = \langle t - 1 - e^t, \cos t - t^2, t^2 e^t - (t-1) \cos t \rangle \]Finally, add these two results:\[ \vec{r}^{\prime}(t) \times \vec{s}(t) + \vec{r}(t) \times \vec{s}^{\prime}(t) = \langle t + t - 1 - e^t, -2t^2 + \cos t - t^2, (2t e^t - \sin t) + (t^2 e^t - (t-1) \cos t) \rangle \]\[ = \langle 2t - 1 - e^t, \cos t - 3t^2, t^2 e^t + 2t e^t - (t-1) \cos t - \sin t \rangle \],which is the same as the derivative of the cross product from Step 2.

Key concepts

Cross Product

The cross product is an essential operation in vector calculus. It results in a vector that is orthogonal to the plane containing the two input vectors. This operation is only applicable in three-dimensional space. Given vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), the cross product \( \vec{a} \times \vec{b} \) is calculated as follows:

• The first component is \( a_2 b_3 - a_3 b_2 \).
• The second component is \( a_3 b_1 - a_1 b_3 \).
• The third component is \( a_1 b_2 - a_2 b_1 \).

This computation results in a new vector \( \langle c_1, c_2, c_3 \rangle \), perpendicular to both original vectors. The magnitude of the cross product vector represents the area of the parallelogram formed by the two vectors. It is a key component in physics and engineering for determining torque and rotational force.

Derivative

The derivative is a fundamental tool in calculus that measures the rate of change or the slope of a function at any point. For a function \( f(t) \), the derivative, denoted as \( f'(t) \) or \( \frac{df}{dt} \), provides insights into how \( f(t) \) varies with respect to the variable \( t \). To calculate the derivative:

• Use the power rule: If \( f(t) = t^n \), then \( f'(t) = nt^{n-1} \).
• Apply the chain rule when dealing with composite functions.
• Utilize the product rule when differentiating products of functions: \( (uv)' = u'v + uv' \).

In the context of vectors, such as in the given problem, finding the derivative involves calculating the derivative of each component of the vector function individually. This allows us to understand not just the direction of change, but also its magnitude.

Vector Differentiation

Vector differentiation extends the concept of derivatives to vector functions. For a vector function \( \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \), vector differentiation involves finding the derivative of each of the vector's components. In essence:

• The derivative \( \vec{r}'(t) \) becomes \( \langle x'(t), y'(t), z'(t) \rangle \).

When working with vector products, such as cross products, the differentiation follows similar rules to single-variable derivatives, including:

• The product rule for vectors: \( (\vec{u} \times \vec{v})' = \vec{u}' \times \vec{v} + \vec{u} \times \vec{v}' \).

This principle is evident in the problem steps where the derivative of the cross product was demonstrated to be equal to the sum of two cross products involving the derivatives of the component vectors. Vector differentiation is pivotal in dynamics and electromagnetism where it helps describe motions and forces.

Calculus Theorems

Calculus theorems provide the foundational rules and truths that govern calculus. A key result applied in vector calculus is the theorem involving the derivative of a cross product, as seen in the exercise. This theorem states that for vectors \( \vec{u}(t) \) and \( \vec{v}(t) \):\[(\vec{u} \times \vec{v})' = \vec{u}' \times \vec{v} + \vec{u} \times \vec{v}'\]This theorem simplifies complex vector calculations by breaking them down into simpler components. Understanding these basic principles not only helps in verifying computations but also in solving problems efficiently.In studying such theorems, always pay attention to:

• The conditions under which the theorem applies.
• Geometric interpretations that can provide intuition.
• Applications in physics and engineering, like calculating rotational dynamics.

Grasping these fundamental theorems enriches comprehension and application of vector calculus concepts in both theoretical and practical scenarios.