Problem 95 - Calculus 3

Question

Given \(\mathbf{r}(t)=t \mathbf{i}+2 \sin t \mathbf{j}+2 \cos t \mathbf{k}\) and \(\mathbf{u}(t)=\frac{1}{t} \mathbf{i}+2 \sin t \mathbf{j}+2 \cos t \mathbf{k}\) find the following: $$\frac{d}{d t}(\mathbf{r}(t) \times \mathbf{u}(t))$$

Step-by-Step Solution

Answer

\(\frac{d}{dt}(\mathbf{r}(t) \times \mathbf{u}(t)) = (2\sin t + 2(t-1)\cos t) \mathbf{k}\).

Step 1: Understand the Cross Product

To solve this exercise, we need to find the derivative of the cross product of the vectors \(\mathbf{r}(t)\) and \(\mathbf{u}(t)\). First, recall that the cross product of two vectors \(\mathbf{a}=(a_1, a_2, a_3)\) and \(\mathbf{b}=(b_1, b_2, b_3)\) is given by: \[ \mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}. \]

Step 2: Identify Components of \(\mathbf{r}(t)\) and \(\mathbf{u}(t)\)

Given \(\mathbf{r}(t)=t \mathbf{i}+2 \sin t \mathbf{j}+2 \cos t \mathbf{k}\) and \(\mathbf{u}(t)=\frac{1}{t} \mathbf{i}+2 \sin t \mathbf{j}+2 \cos t \mathbf{k}\), identify the components: \(\mathbf{r}(t) = (t, 2\sin t, 2\cos t)\) and \(\mathbf{u}(t) = \left(\frac{1}{t}, 2\sin t, 2\cos t \right)\).

Step 3: Compute the Cross Product \(\mathbf{r}(t) \times \mathbf{u}(t)\)

Using the formula for the cross product, compute: \[ \mathbf{r}(t) \times \mathbf{u}(t) = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ t & 2\sin t & 2\cos t \ \frac{1}{t} & 2\sin t & 2\cos t \end{array} \right|. \]Calculating the determinant gives us: \[ \mathbf{r}(t) \times \mathbf{u}(t) = (0)\mathbf{i} - \left( 2t\cos t - 2t\cos t \right)\mathbf{j} + \left( 2t \sin t - 2\sin t \right)\mathbf{k}. \]Simplifying, we get: \[ \mathbf{r}(t) \times \mathbf{u}(t) = (0)\mathbf{i} - (0)\mathbf{j} + (2t\sin t - 2\sin t)\mathbf{k}. \]

Step 4: Simplify the Expression

Simplified, the cross product is: \[ \mathbf{r}(t) \times \mathbf{u}(t) = 2(t-1)\sin t \mathbf{k}. \]

Step 5: Differentiate with Respect to \(t\)

Now, differentiate \(\mathbf{r}(t) \times \mathbf{u}(t) = 2(t-1)\sin t \mathbf{k}\) with respect to \(t\):Let \(g(t) = 2(t-1)\sin t\).Using the product rule, \(\frac{d}{dt}g(t) = \frac{d}{dt}[2(t-1)] \cdot \sin t + 2(t-1) \cdot \frac{d}{dt}[\sin t].\)Differentiating each:\(\frac{d}{dt}[2(t-1)] = 2\), and \(\frac{d}{dt}[\sin t] = \cos t\).Thus, \[ \frac{d}{dt}g(t) = 2 \cdot \sin t + 2(t-1)\cos t. \]

Step 6: Final Expression for the Derivative

The derivative of the cross-product is then: \[ \frac{d}{dt}(\mathbf{r}(t) \times \mathbf{u}(t)) = (2\sin t + 2(t-1)\cos t) \mathbf{k}. \]

Key Concepts

Cross Product

The cross product is a mathematical operation used in vector calculus to find a vector that is perpendicular to two given vectors. If you have two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), their cross product, represented as \(\mathbf{a} \times \mathbf{b}\), results in a new vector. This vector is orthogonal, meaning it forms a right angle with both \(\mathbf{a}\) and \(\mathbf{b}\).

• Physically, the cross product is significant because it allows us to determine the orientation of a plane formed by \(\mathbf{a}\) and \(\mathbf{b}\).
• It also has applications in physics, such as calculating torque and angular momentum.

The cross product involves the determinant of a 3x3 matrix where the first row is the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\), and the subsequent two rows are filled with the components of \(\mathbf{a}\) and \(\mathbf{b}\). Calculating this determinant gives you the i, j, and k components of the resulting cross product vector.

Derivatives of Vector Functions

In calculus, derivatives allow us to understand how a function changes as its inputs change. When dealing with vector functions, which are functions that output vectors rather than a single value, the concept of derivatives extends to multiple dimensions.

• A vector function usually can be expressed in the form \(\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}\), where \(t\) is a parameter.
• Taking the derivative of this vector function involves differentiating each of the component functions \(f(t)\), \(g(t)\), and \(h(t)\) separately with respect to \(t\).

Understanding the derivative of a vector function is crucial in providing insight into the curve's velocity, direction, and shape by examining the rate of change of each coordinate.

Product Rule in Calculus

The product rule is a fundamental principle in calculus used to differentiate the product of two functions. This rule is just as essential when working with vector functions as it is with single-variable functions.

• The standard product rule states: if \(u(t)\) and \(v(t)\) are functions of \(t\), then the derivative of their product is \(u'(t)v(t) + u(t)v'(t)\).
• For vector functions, applying the product rule means differentiating both the scalar coefficients and the vector components.

In the context of our problem, when differentiating \(\mathbf{r}(t) \times \mathbf{u}(t)\), the product rule is necessary. This is because each component of the vectors, as well as the vector itself, must be treated as a product and differentiated accordingly.