Problem 44 - Essential Calculus Early Transcendentals

Question

\(39-44=\) 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 \( \mathbf{r}'(t) = \mathbf{a} \times \mathbf{b} + 2t \mathbf{a} \times \mathbf{c} \).

Step 1: Understanding the Vector Function

We have the vector function \( \mathbf{r}(t) = t \mathbf{a} \times (\mathbf{b} + t \mathbf{c}) \). Here, \( \mathbf{a}, \mathbf{b}, \text{ and } \mathbf{c} \) are constant vectors, and \( \times \) denotes the cross product between two vectors.

Step 2: Apply the Cross Product Identity

Use the identity for the cross product of a vector with a sum: \( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} \). For our function, it becomes \( t \mathbf{a} \times (\mathbf{b} + t \mathbf{c}) = t (\mathbf{a} \times \mathbf{b} + t \mathbf{a} \times \mathbf{c}) \).

Step 3: Simplify the Expression

The expression simplifies to \( t \mathbf{a} \times \mathbf{b} + t^2 \mathbf{a} \times \mathbf{c} \). This new function, \( \mathbf{r}(t) = t \mathbf{a} \times \mathbf{b} + t^2 \mathbf{a} \times \mathbf{c} \), is expressed as a linear combination of the vectors \( \mathbf{a} \times \mathbf{b} \) and \( \mathbf{a} \times \mathbf{c} \).

Step 4: Differentiate the Vector Function

Differentiate \( \mathbf{r}(t) = t \mathbf{a} \times \mathbf{b} + t^2 \mathbf{a} \times \mathbf{c} \) with respect to \( t \). Use the power rule for differentiation: \( \frac{d}{dt}(t^n) = n t^{n-1} \). The derivative is \( \mathbf{r}'(t) = \frac{d}{dt}(t \mathbf{a} \times \mathbf{b}) + \frac{d}{dt}(t^2 \mathbf{a} \times \mathbf{c}) \).

Step 5: Compute Each Term's Derivative

Compute \( \frac{d}{dt}(t \mathbf{a} \times \mathbf{b}) = \mathbf{a} \times \mathbf{b} \) using the power rule, and \( \frac{d}{dt}(t^2 \mathbf{a} \times \mathbf{c}) = 2t \mathbf{a} \times \mathbf{c} \) also using the power rule. Combine these results to find the complete derivative.

Step 6: Final Expression for the Derivative

The complete derivative is \( \mathbf{r}'(t) = \mathbf{a} \times \mathbf{b} + 2t \mathbf{a} \times \mathbf{c} \). This expression gives the rate of change of the vector function \( \mathbf{r}(t) \) with respect to \( t \).

Key Concepts

Vector Function

A vector function is an important concept in vector calculus. It is a function that takes a variable, often time *t*, and returns a vector. The vector function \( \mathbf{r}(t) = t \mathbf{a} \times (\mathbf{b} + t \mathbf{c}) \) is expressed in terms of the variable \( t \) and vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \). These vectors are considered constants in the function. This means the expression evolves as you change \( t \), allowing you to model paths or curves in three-dimensional space.

• For instance, in physics, vector functions can represent the position of a particle over time.
• They can also be utilized in engineering to describe fields and forces.

The function above includes a cross product within it, an operation which combines two vectors to form another vector. This means that as \( t \) changes, this cross product modifies the direction and nature of \( \mathbf{r}(t) \). Understanding how to manipulate and differentiate vector functions is essential in solving real-world problems that involve vector quantities.

Cross Product

The cross product is a vector operation that is used to find a vector that is perpendicular to two given vectors. If you have two vectors \( \mathbf{u} \) and \( \mathbf{v} \), their cross product, denoted \( \mathbf{u} \times \mathbf{v} \), is a new vector.

• The magnitude of the resulting vector is equal to the area of the parallelogram that the vectors span.
• Its direction is determined by the right-hand rule, which means if you point your index finger in the direction of \( \mathbf{u} \) and your middle finger in the direction of \( \mathbf{v} \), your thumb will point in the direction of \( \mathbf{u} \times \mathbf{v} \).

In the exercise, the expression \( t \mathbf{a} \times (\mathbf{b} + t \mathbf{c}) \) uses the cross product identity: \( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} \).This identity helps in simplifying the problem by breaking down the cross product into manageable components. Applying this identity translates our vector function into \( t (\mathbf{a} \times \mathbf{b} + t \mathbf{a} \times \mathbf{c}) \), which makes it easier to differentiate.

Differentiation of Vector Functions

Differentiating vector functions involves applying calculus operations to each component of a vector function. The derivative provides a new vector that describes how the vector function changes with respect to a variable, most commonly time \( t \).

• When differentiating a vector function, you treat each operation (like addition or multiplication by a scalar) in a similar manner as you would for scalar functions.
• The differentiation rules, including the power rule in our example, still apply but are done separately for each vector component.

In this exercise, the given vector function is \( \mathbf{r}(t) = t \mathbf{a} \times \mathbf{b} + t^2 \mathbf{a} \times \mathbf{c} \).Using calculus, the power rule gives us the derivatives: \( \frac{d}{dt}(t \mathbf{a} \times \mathbf{b}) = \mathbf{a} \times \mathbf{b} \) and \( \frac{d}{dt}(t^2 \mathbf{a} \times \mathbf{c}) = 2t \mathbf{a} \times \mathbf{c} \).This results in the derivative vector function: \( \mathbf{r}'(t) = \mathbf{a} \times \mathbf{b} + 2t \mathbf{a} \times \mathbf{c} \).Differentiating vector functions is vital for analyzing changes and motions represented by vectors, such as velocity and acceleration in physics.