Problem 16 - Calculus Early Transcendentals

Question

Find the derivative of the vector function. $$ \mathbf{r}(t)=t \mathbf{a} \times(\mathbf{b}+t \mathbf{c}) $$

Step-by-Step Solution

Answer

The derivative is \( \mathbf{a} \times \mathbf{b} + 2t \mathbf{a} \times \mathbf{c} \).

Step 1: Express the vector function

The given vector function is \(\mathbf{r}(t) = t \mathbf{a} \times (\mathbf{b} + t \mathbf{c})\). Here, \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\) are constant vectors.

Step 2: Simplify the expression inside the function

Simplify the expression by distributing the variables: \(\mathbf{b} + t \mathbf{c}\) becomes \(\mathbf{b} + t \mathbf{c} \). This expression is already simplified.

Step 3: Apply the product rule for derivatives

We need to find \(\frac{d\mathbf{r}}{dt}\). We notice that the function \(\mathbf{r}(t)\) is a product of \(t\) and \(\mathbf{a} \times (\mathbf{b} + t \mathbf{c})\). Use the product rule: if \(f(t) = u(t) v(t)\), then \(f'(t) = u'(t)v(t) + u(t)v'(t)\).

Step 4: Derive each component of the product

Let \(u(t) = t\) and \(v(t) = \mathbf{a} \times (\mathbf{b} + t \mathbf{c})\).- The derivative of \(u(t) = t\) is \(u'(t) = 1\).- To derive \(v(t)\), apply the derivative to \(\mathbf{a} \times (\mathbf{b} + t \mathbf{c})\). Since \(\mathbf{a} \times \mathbf{b}\) is constant with respect to \(t\), its derivative is 0. The derivative of \(\mathbf{a} \times (t \mathbf{c})\) is \(\mathbf{a} \times \mathbf{c}\).

Step 5: Apply the derivatives to the product rule

With \(u'(t) = 1\) and \(v'(t) = \mathbf{a} \times \mathbf{c}\), substitute into the product rule formula:\[ \frac{d\mathbf{r}}{dt} = 1 \cdot (\mathbf{a} \times (\mathbf{b} + t \mathbf{c})) + t (\mathbf{a} \times \mathbf{c}) \]

Step 6: Simplify the derivative

Simplify the expression:- \(1 \cdot (\mathbf{a} \times (\mathbf{b} + t \mathbf{c})) \) is \(\mathbf{a} \times \mathbf{b} + t(\mathbf{a} \times \mathbf{c})\).- Combine terms with \(t\): \(\mathbf{a} \times \mathbf{b} + t \mathbf{a} \times \mathbf{c} + t \mathbf{a} \times \mathbf{c} = \mathbf{a} \times \mathbf{b} + 2t(\mathbf{a} \times \mathbf{c})\).

Step 7: Final derivative of the vector function

The derivative of the vector function \(\mathbf{r}(t)\) is:\[ \frac{d\mathbf{r}}{dt} = \mathbf{a} \times \mathbf{b} + 2t \mathbf{a} \times \mathbf{c} \].

Key Concepts

Vector Functions

Vector functions are powerful tools in vector calculus that describe how vectors depend on one or more variables. In many applications, these variables are time-related, which makes vector functions ideal for modeling motion and change over time. For instance, when dealing with the trajectory of an object in space, we can use a vector function to represent its path.
The structure of a vector function can vary, but typically it is defined as \( \mathbf{r}(t) = \langle f_1(t), f_2(t), f_3(t) \rangle \), where each component \( f_i(t) \) is a function of the variable \( t \). In our specific example, \( \mathbf{r}(t) = t \mathbf{a} \times (\mathbf{b} + t \mathbf{c}) \), the function is defined using vector quantities.

• Here, \( t \) is a scalar variable, often representing time.
• \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) are vectors that do not depend on \( t \).

Understanding how to manipulate and differentiate these functions is crucial for interpreting physical phenomena.

Cross Product

The cross product is a mathematical operation performed on two vectors in three-dimensional space. It results in a third vector that is perpendicular to both input vectors. The cross product provides a way to calculate vectors representing directions and areas in space.
Given two vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), their cross product \( \mathbf{u} \times \mathbf{v} \) is calculated as:\[ \mathbf{u} \times \mathbf{v} = \left( u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1\right)\]

• This operation is only defined in three dimensions and results in a vector perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \).
• The magnitude of the cross product is equal to the area of the parallelogram formed by the two vectors.

In the original exercise, the cross product \( \mathbf{a} \times (\mathbf{b} + t \mathbf{c}) \) is crucial for deriving the final result of the vector function's derivative.

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. For vector functions, the process of differentiation applies to each component of the vector separately.
The derivative of a scalar function \( f(t) \) is given by:\[ f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h}\]For the derivative of a vector function \( \mathbf{r}(t) = \langle f_1(t), f_2(t), f_3(t) \rangle \), it is taken as:\[ \frac{d\mathbf{r}}{dt} = \langle f_1'(t), f_2'(t), f_3'(t) \rangle\]

• Understand that each vector component is differentiated individually, similarly to scalar functions.
• In the exercise, applying the derivative to \( \mathbf{r}(t) = t \mathbf{a} \times (\mathbf{b} + t \mathbf{c}) \) involves understanding how to manage vector derivatives.

The skill of finding derivatives of vector functions like this is essential in physics and engineering, where such functions are often used to describe changing systems.

Product Rule

The product rule is a differentiation rule used to find the derivative of the product of two functions. This rule applies to both scalar and vector functions, making it particularly useful in complex equations involving products.
For two functions \( u(t) \) and \( v(t) \), the product rule states:\[ (uv)'=u'v+uv'\]

• The rule instructs us to differentiate both functions separately, then combine them using the formula.
• For vector functions, each component is dealt with individually, but the cross product must be taken into account.

In our exercise with \( \mathbf{r}(t) = t \mathbf{a} \times (\mathbf{b} + t \mathbf{c}) \), we identified \( u(t) = t \) and \( v(t) = \mathbf{a} \times (\mathbf{b} + t \mathbf{c}) \).
We applied the product rule to find the derivative, which included recognizing the differentiation of the cross product as well as applying the derivative to a scalar multiplied by a vector expression.