Question

Use the properties of the derivative to find the following. (a) \(\mathbf{r}^{\prime}(t)\) (b) \(\mathbf{r}^{\prime \prime}(t)\) (c) \(D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)]\) (d) \(D_{t}[3 \mathbf{r}(t)-\mathbf{u}(t)]\) (e) \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)]\) (f) \(D_{t}[\|\mathbf{r}(t)\|], \quad t>0\) $$ \begin{array}{l} \mathbf{r}(t)=t \mathbf{i}+2 \sin t \mathbf{j}+2 \cos t \mathbf{k} \\ \mathbf{u}(t)=\frac{1}{t} \mathbf{i}+2 \sin t \mathbf{j}+2 \cos t \mathbf{k} \end{array} $$

Short answer

The derived functions are \(\mathbf{r}^{\prime}(t) = \mathbf{i} + 2\cos (t) \mathbf{j} + 2\sin (t) \mathbf{k}\), \(\mathbf{r}^{\prime\prime}(t) = -2\sin (t) \mathbf{j} + 2 \cos (t) \mathbf{k}\), \(D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)]\), \(D_{t}[3 \mathbf{r}(t)-\mathbf{u}(t)]\), \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)]\) and \(D_{t}[\|\mathbf{r}(t)\|]\)

Step-by-step solution

Find \(\mathbf{r}^{\prime}(t)\)

The derivative of \(\mathbf{r}(t)\) is found by taking the derivative of each of its components. Using the chain rule, \(\mathbf{r}^{\prime}(t) = \mathbf{i} + 2\cos (t) \mathbf{j} + 2\sin (t) \mathbf{k}\).

Find \(\mathbf{r}^{\prime\prime}(t)\)

The second derivative of \(\mathbf{r}(t)\) can be found by taking the derivative of \(\mathbf{r}^{\prime}(t)\). This gives \(\mathbf{r}^{\prime\prime}(t) = -2\sin (t) \mathbf{j} + 2 \cos (t) \mathbf{k}\).

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

To find the derivative of the dot product of \(\mathbf{r}(t)\) and \(\mathbf{u}(t)\), use the product rule. This results in \(D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)] = \mathbf{r}^{\prime}(t) \cdot \mathbf{u}(t) + \mathbf{r}(t) \cdot \mathbf{u}^{\prime}(t)\) assuming \(\mathbf{u}^{\prime}(t)\) is the derivative of \(\mathbf{u}(t)\).

Derive \(D_{t}[3 \mathbf{r}(t)-\mathbf{u}(t)]\)

The derivative of a linear combination of functions can be found by taking the derivative of each term individually. Therefore, \(D_{t}[3 \mathbf{r}(t)-\mathbf{u}(t)] = 3\mathbf{r}^{\prime}(t) - \mathbf{u}^{\prime}(t)\), assuming \(\mathbf{u}^{\prime}(t)\) is the derivative of \(\mathbf{u}(t)\).

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

To derive \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)]\), it's important to note that the derivative of a cross product of two functions \(\mathbf{f}(t)\) and \(\mathbf{g}(t)\) is given by \(\mathbf{f}^{\prime}(t) \times \mathbf{g}(t) + \mathbf{f}(t) \times \mathbf{g}^{\prime}(t)\). Therefore, \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)] = \mathbf{r}^{\prime}(t) \times \mathbf{u}(t) + \mathbf{r}(t) \times \mathbf{u}^{\prime}(t)\), assuming \(\mathbf{u}^{\prime}(t)\) is the derivative of \(\mathbf{u}(t)\).

Derive \(D_{t}[\|\mathbf{r}(t)\|]\)

To find \(D_{t}[\|\mathbf{r}(t)\|]\), where \(\|\mathbf{r}(t)\|\) is the magnitude (or norm) of \(\mathbf{r}(t)\), apply the chain rule and then normalize by the magnitude of the vector after differentiation. This gives \(D_{t}[\|\mathbf{r}(t)\|] = \frac{\mathbf{r}^{\prime}(t) \cdot \mathbf{r}(t)}{\|\mathbf{r}(t)\|}\).

Key concepts

Properties of Derivatives

Understanding the properties of derivatives is essential to calculus, especially when dealing with more complex functions. The derivative represents the rate at which a function changes at any point, commonly known as the slope of the tangent line to the function at that point. There are a few key properties that make working with derivatives much more manageable.

Firstly, the sum rule tells us that the derivative of the sum of two functions is the sum of their derivatives. This property is exemplified when we differentiate functions that are added together; we simply differentiate each function individually and then add the results. Secondly, the constant multiple rule allows us to take a constant out of the derivative. If a function is multiplied by a constant, its derivative will also be multiplied by that same constant. Finally, power rules make it easy to differentiate polynomial functions by reducing the exponent by one and multiplying by the original exponent.

When it comes to more complex operations, such as dot products or cross products in vector calculus, these properties extend and are combined with the product and chain rules to successfully compute the derivatives of vector-valued functions.

Vector-Valued Functions

A vector-valued function is a function that outputs a vector rather than a single scalar value. In calculus, these functions are often denoted as \( \mathbf{r}(t) \) for a vector function dependent on the parameter \( t \). These functions can represent various things, such as the position of an object in space as a function of time.

Taking derivatives of vector-valued functions, like \( \mathbf{r}(t) \) or \( \mathbf{u}(t) \) from our exercise, involves taking derivatives component-wise. That means we differentiate each part of the vector—in the \( i \), \( j \), and \( k \) directions—separately. This process is similar to differentiating scalar functions, but it's important to apply it to each component. The concept becomes particularly interesting and more complex when these vector functions interact with each other through dot or cross products.

Product Rule

The product rule is a fundamental rule in calculus for finding the derivative of the product of two functions. It tells us that the derivative of a product \( f(t)g(t) \) is not simply the product of the derivatives, but rather \( f'(t)g(t) + f(t)g'(t) \). This rule is crucial when functions are multiplied together, and it becomes even more vital when dealing with vector-valued functions.

In vector calculus, the product rule is extended to include dot and cross products. For the dot product, \( \mathbf{r}(t) \cdot \mathbf{u}(t) \), the derivative is found by doing \( \mathbf{r}'(t) \cdot \mathbf{u}(t) + \mathbf{r}(t) \cdot \mathbf{u}'(t) \), where \( \mathbf{r}'(t) \) and \( \mathbf{u}'(t) \) are the derivatives of \( \mathbf{r}(t) \) and \( \mathbf{u}(t) \), respectively. Similarly, for the cross product, the derivative of \( \mathbf{r}(t) \times \mathbf{u}(t) \) follows the analogous pattern.

Chain Rule

The chain rule is a powerful tool in calculus that allows us to find the derivative of a composite function. When one function is nested inside another, the chain rule enables us to differentiate the entire expression by taking the derivative of the outer function and multiplying it by the derivative of the inner function. It's expressed as \( \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \).

This rule is pivotal when considering functions that are not simply a product of two other functions but are dependent on each other. In the context of vector calculus and particularly with vector norms, as in \( \Vert \mathbf{r}(t) \Vert \), the chain rule helps us find the derivative by involving both the derivative of the norm function and the derivative of the vector function inside. This approach allows us to handle more complex derivative tasks and understand how rates of change occur within nested operations.