Question

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. $$ \mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+t \mathbf{k}, \quad \mathbf{u}(t)=\mathbf{j}+t \mathbf{k} $$

Short answer

The derivatives of the dot and cross products are given by: (a) \(D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)] = \cos t + 2t\), (b) \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)] = (\cos t - 1) \mathbf{i} + \sin t \mathbf{j}\)

Step-by-step solution

Compute dot product of the given vectors

Use the formula for the dot product of two vectors: \( \mathbf{a} \cdot \mathbf{b} = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3} \). Here, \( \mathbf{r}(t) \cdot \mathbf{u}(t) = (\cos t)(0) + (\sin t)(1) + t(t) = \sin t + t^{2} \).

Differentiate the result of the dot product

Apply the chain rule to compute the derivative \( D_t[\mathbf{r}(t) \cdot \mathbf{u}(t)] \). Using this rule, the derivative is given by \( \cos t + 2t \).

Compute cross product of the given vectors

Use the formula for the cross product of two vectors. \( \mathbf{c} = \mathbf{r} \times \mathbf{u} \) will have the components \( c_1 = r_2 u_3 - r_3 u_2 \), \( c_2 = r_3 u_1 - r_1 u_3 \), and \( c_3 = r_1 u_2 - r_2 u_1 \). So, the cross product \( \mathbf{r}(t) \times \mathbf{u}(t) = (\sin t - t) \mathbf{i} - \cos t \mathbf{j} + \mathbf{k} \).

Differentiate the result of the cross product

Use the distributive property to take the derivative of each of the vector components. The result is \( D_t[\mathbf{r}(t) \times \mathbf{u}(t)] = (\cos t - 1) \mathbf{i} + \sin t \mathbf{j} \).

Key concepts

Dot Product

The dot product, also known as scalar product, measures the magnitude of one vector in the direction of another. When two vectors \textbf{a} and \textbf{b} are multiplied dot-wise, the result is a scalar: \( \mathbf{a} \cdot \mathbf{b} = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3} \). This operation is crucial in physics and engineering for it relates to work done by a force or projections of vectors.

For instance, in the given exercise, the dot product of \(\textbf{r}(t)\) and \(\textbf{u}(t)\) was calculated, and then differentiated. A common misconception is that the dot product and the derivative are interchangeable in their order of operations. However, this is not the case. The differentiation must happen after computing the dot product, as was correctly done in the step by step solution. This is a subtle but impactful detail for understanding vector calculus operations.

Cross Product

The cross product is a binary operation on two vectors in three-dimensional space, producing a third vector that is perpendicular to the plane containing the first two. Its magnitude is proportional to the area of the parallelogram that the vectors span. The cross product \( \mathbf{a} \times \mathbf{b} \) has components \( c_1 = a_2 b_3 - a_3 b_2 \), \( c_2 = a_3 b_1 - a_1 b_3 \), and \( c_3 = a_1 b_2 - a_2 b_1 \).

In our exercise, the vectors produced a cross product before differentiation. It's important to arrange the vectors correctly and adhere to the right-hand rule to determine the direction of the resulting vector. The exercise shows a proper application by finding the cross product first and then differentiating, acknowledging that, similar to the dot product, the order of these operations is not interchangeable.

Differentiation of Vectors

Differentiation of vectors extends the concept of derivative from scalars to vectors. When a vector function like \( \mathbf{r}(t) \) varies with a parameter (in this case, time \( t \)), its derivative is the vector that represents the rate of change of \( \mathbf{r}(t) \) with respect to \( t \). The derivative is found by differentiating each component of the vector function individually.

As part of improving understanding, it is valuable to recognize that differentiation of a vector maintains the vector's nature; it yields a vector, not a scalar, except in the dot product scenario where it follows the product's nature. The step by step solution illustrates this principle by differentiating each component after finding the dot and cross products.

Chain Rule

The chain rule is a formula for computing the derivative of the composition of two or more functions. In vector calculus, it serves the same purpose. If a vector function \( \mathbf{r}(t) \) is composed of functions that depend on \( t \) (e.g., \( \cos(t) \), \( \sin(t) \)), then the derivative of \( \mathbf{r}(t) \) with respect to \( t \) involves applying the chain rule to each component function.

Understanding and applying the chain rule correctly is vital, especially when dealing with product operations between two vector functions as seen in this exercise. The step by step solution differentiation is evidence of the chain rule being appropriately used to unpack the derivatives of the respective compositions and ensure accuracy in the dynamics of vector functions.