Question

In Exercises \(\mathbf{3 7}\) and \(\mathbf{3 8 ,}\) find (a) \(\quad D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)] \quad\) and (b) \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)]\) by differentiating the product, then applying the properties of Theorem 10.2. $$ \mathbf{r}(t)=t \mathbf{i}+2 t^{2} \mathbf{j}+t^{3} \mathbf{k}, \quad \mathbf{u}(t)=t^{4} \mathbf{k} $$

Short answer

(a) \( D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)]=5 t^{4} + 4 t^{7}\). (b) \( D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)] = 5 t^{4} \mathbf{i} -12 t^{5} \mathbf{j} + 8 t^{2} \mathbf{k}.\)

Step-by-step solution

Find the derivative of \( \mathbf{r}(t) \) and \( \mathbf{u}(t) \)

The derivative of \( \mathbf{r}(t) \) is \( \mathbf{r}'(t)=\mathbf{i}+4 t \mathbf{j}+3 t^{2} \mathbf{k} \), and the derivative of \( \mathbf{u}(t) \) is \( \mathbf{u}'(t)=4 t^{3} \mathbf{k} \).

Compute \( D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)] \)

Using \( \mathbf{r}(t) \cdot \mathbf{u}(t)=t^{5}\) and the formula mentioned above, we get \( D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)]=\mathbf{r}'(t) \cdot \mathbf{u}(t) + \mathbf{r}(t) \cdot \mathbf{u}'(t) =5 t^{4} + 4 t^{7}\).

Compute \( D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)] \)

Using \( \mathbf{r}(t) \times \mathbf{u}(t)=t^{5} \mathbf{i} - 2 t^{6} \mathbf{j} + 2 t^{2} \mathbf{k}\) and the formula mentioned above, we get \( D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)] = \mathbf{r}'(t) \times \mathbf{u}(t) + \mathbf{r}(t) \times \mathbf{u}'(t) = 5 t^{4} \mathbf{i} -12 t^{5} \mathbf{j} + 8 t^{2} \mathbf{k}.\)

Key concepts

Vector Differentiation

Vector differentiation is a powerful tool in calculus, particularly when dealing with physical quantities that have both magnitude and direction. When we differentiate a vector function, such as \(\mathbf{r}(t)\), we're finding how the vector changes with respect to a variable, typically time. The derivative of a vector function is another vector that points in the direction of the greatest rate of increase of the function.

In the given exercise, we differentiated the vector functions \(\mathbf{r}(t)=t\mathbf{i}+2t^{2}\mathbf{j}+t^{3}\mathbf{k}\) and \(\mathbf{u}(t)=t^{4}\mathbf{k}\). The resulting derivatives, \(\mathbf{r}'(t)\) and \(\mathbf{u}'(t)\), give us the instantaneous rate of change of each vector at any time \(t\). It's important to differentiate each component of the vector with respect to \(t\), and make sure to apply the power rule appropriately, just as with scalar functions.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors that results in a scalar. It is calculated by multiplying corresponding components of the vectors and then summing those products. The dot product formula can be seen as \(\mathbf{A} \cdot \mathbf{B} = A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z}\).

In our exercise, the dot product of \(\mathbf{r}(t)\) and \(\mathbf{u}(t)\) is \(t^{5}\), since all components are orthogonal except for the \(\mathbf{k}\) components. Calculating the time derivative of the dot product \(D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)]\) involves using the product rule, resulting in the sum of each vector dotted with the derivative of the other vector. This step is crucial in understanding how the dot product of two vectors changes over time.

Cross Product

The cross product is another way of multiplying two vectors, but unlike the dot product, the cross product returns a vector that is perpendicular to both original vectors. The cross product, represented by \(\mathbf{A} \times \mathbf{B}\), has the magnitude equal to the area of the parallelogram formed by \(\mathbf{A}\) and \(\mathbf{B}\) and follows the right-hand rule for direction.

In the example provided, we computed \(\mathbf{r}(t) \times \mathbf{u}(t)\), which gave us a new vector function. When differentiating this cross product with respect to time \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)]\), we used the product rule analogous to that of the dot product, but kept in mind that the result will again be a vector. This provides insight into how the vector perpendicular to \(\mathbf{r}(t)\) and \(\mathbf{u}(t)\) changes as time progresses.

Derivatives of Vector Functions

Derivatives of vector functions follow rules similar to those of scalar functions, but it's essential to differentiate each component of the vector independently. The derivative rules including sum, constant multiple, and power rules translate directly to vector components.

In the step-by-step solution, after computing derivatives of \(\mathbf{r}(t)\) and \(\mathbf{u}(t)\), we applied the derivatives to the dot and cross products. Understanding the derivative of each component is key to correctly working out these products. Remember that each component acts somewhat independently, and they must be combined correctly to form the final differentiated vector.

Theorems in Vector Calculus

Theorems in vector calculus, such as Theorem 10.2 referenced in the exercise, lay the groundwork for operations on vector functions. These theorems often extend the familiar rules from scalar calculus to three dimensions and beyond. A common example is the product rule, which we use when computing the derivatives of the dot and cross products of vector functions.

Applying theorems correctly is crucial for finding accurate results; they guide us through the computation process ensuring that we don't lose the integrity of the vector's properties. In solving the given exercise, understanding and employing the product rule allowed us to differentiate the dot and cross products effectively, ensuring that the scalar and vector nature of the results are preserved.