Problem 14 - Calculus Early Transcendentals

Question

Find the derivative of the vector function. $$ \mathbf{r}(t)=a t \cos 3 t \mathbf{i}+b \sin ^{3} t \mathbf{j}+c \cos ^{3} t \mathbf{k} $$

Step-by-Step Solution

Answer

The derivative is \((a \cos 3t - 3at \sin 3t) \mathbf{i} + 3b \sin^2 t \cos t \mathbf{j} - 3c \cos^2 t \sin t \mathbf{k}.\)

Step 1: Identify the Vector Function Components

The given vector function is \( \mathbf{r}(t) = a t \cos 3t \, \mathbf{i} + b \sin^3 t \, \mathbf{j} + c \cos^3 t \, \mathbf{k} \). It has three components: \( x(t) = a t \cos 3t \), \( y(t) = b \sin^3 t \), and \( z(t) = c \cos^3 t \). We need to differentiate each component with respect to \( t \).

Step 2: Differentiate the First Component

Differentiate \( x(t) = a t \cos 3t \) using the product rule for derivatives which requires \( u = at \) and \( v = \cos 3t \). Thus, \( x'(t) = u'v + uv' = a\cos 3t - 3at \sin 3t \).

Step 3: Differentiate the Second Component

The second component is \( y(t) = b \sin^3 t \). Differentiate using the chain rule: \( y'(t) = 3b \sin^2 t \cos t \).

Step 4: Differentiate the Third Component

For the third component, \( z(t) = c \cos^3 t \), use the chain rule: \( z'(t) = -3c \cos^2 t \sin t \).

Step 5: Combine the Derivatives

The derivative of the vector function \( \mathbf{r}(t) \) is the combination of the three differentiated components: \( \mathbf{r}'(t) = (a \cos 3t - 3a t \sin 3t) \mathbf{i} + (3b \sin^2 t \cos t) \mathbf{j} - (3c \cos^2 t \sin t) \mathbf{k} \).

Key Concepts

Product Rule

When differentiating products of functions, the product rule becomes essential. This rule helps you differentiate two functions that are multiplied together.
It can be particularly useful in vector calculus. The product rule states that, for two functions \( u(t) \) and \( v(t) \), the derivative of their product \( u(t)v(t) \) is given by:

• \((u(t)v(t))' = u'(t)v(t) + u(t)v'(t)\)

In our exercise, the component \( x(t) = a t \cos 3t \) needs this rule. Here, \( u(t) = a t \) and \( v(t) = \cos 3t \).
First, differentiate each component:

• \( u'(t) = a \)
• \( v'(t) = -3 \sin 3t \)

Then, substitute back into the product rule:

• \( x'(t) = a \cos 3t - 3a t \sin 3t \)

The result efficiently gives the derivative of the product, breaking a complex problem into smaller, manageable parts.

Chain Rule

The chain rule is a fundamental technique in calculus for finding derivatives of composite functions.
It's essential when functions are nested within others. The chain rule states that if a function \( y = f(g(t)) \), then the derivative is:

• \( \frac{dy}{dt} = f'(g(t)) \cdot g'(t) \)

In the context of vector functions, you can observe its application in both \( y(t) = b \sin^3 t \) and \( z(t) = c \cos^3 t \).
Here, we differentiate powers of trigonometric functions:

• In the case of \( \sin^3 t \), let \( g(t) = \sin t \), and \( f(u) = u^3 \).
  Then differentiate:\( f'(u) = 3u^2 \), \( g'(t) = \cos t \)
• This leads to: \( y'(t) = 3b \sin^2 t \cos t \)

This method simplifies higher power functions, making them manageable by gradually unfolding each function layer by using derivatives.

Differentiation

Differentiation is a crucial concept used to compute the rate of change of a function relative to a variable.
In vector calculus, this often means finding the derivative of a vector function with respect to one of its parameters, such as time \( t \).
In the given exercise, we work with the vector function \( \mathbf{r}(t) = a t \cos 3t \ \mathbf{i} + b \sin^3 t \ \mathbf{j} + c \cos^3 t \ \mathbf{k} \).
Differentiation breaks down into calculating each component separately:

• \( x(t) = a t \cos 3t \)
• \( y(t) = b \sin^3 t \)
• \( z(t) = c \cos^3 t \)

Each of these components requires different rules or methods (such as the product or chain rule) for differentiation.
By treating each segment individually, you simplify the process, making it easier to manage and understand.
Ultimately, this method aids in comprehending how changes in \( t \) influence the overall behavior of vector functions.

Vector Calculus

Vector calculus extends the concepts of single-variable calculus to vector fields.
It's used to handle functions that involve vectors, which are quantities with both magnitude and direction, such as position vectors in three dimensions.
In our specific problem, \( \mathbf{r}(t) = a t \cos 3t \ \mathbf{i} + b \sin^3 t \ \mathbf{j} + c \cos^3 t \ \mathbf{k} \), represents a vector-valued function of time \( t \).

Differentiation of vector functions involves computing derivatives of each component vector function.
This might involve:

• Breaking down complex terms into simpler parts
• Using rules like the product or chain rule

By differentiating each component individually, we can analyze movement and understand how each axis independently affects the vector's behavior.
Vector calculus thus becomes a vital tool in physics, engineering, and computer graphics, where understanding vector fields and motion paths is crucial.