Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

\mathbf{F}(t)=\mathbf{f}(u(t)) .\( Find \)\mathbf{F}^{\prime}(t)\( in terms of \)t$ $$ \mathbf{f}(u)=\cos u \mathbf{i}+e^{3 \mu} \mathbf{j} \text { and } u(t)=3 t^{2}-4 $$

Step-by-Step Solution

Answer

\( \mathbf{F}^{\prime}(t) = -6t \sin(3t^2 - 4) \mathbf{i} + 18t e^{3(3t^2 - 4)} \mathbf{j} \).

Step 1: Identify Given Functions

You are given the vector function \( \mathbf{F}(t) = \mathbf{f}(u(t)) \) where \( \mathbf{f}(u) = \cos u \mathbf{i} + e^{3u} \mathbf{j} \) and \( u(t) = 3t^2 - 4 \). You need to find \( \mathbf{F}^{\prime}(t) \).

Step 2: Apply the Chain Rule

To find \( \mathbf{F}^{\prime}(t) \), you need to apply the chain rule. The derivative of \( \mathbf{F}(t) \) with respect to \( t \) is \( \frac{d}{dt}(\mathbf{f}(u(t))) = \mathbf{f}'(u(t)) \cdot u'(t) \).

Step 3: Differentiate \( \mathbf{f}(u(t)) \)

Find \( \mathbf{f}'(u) \):- For \( \cos u \mathbf{i} \), the derivative with respect to \( u \) is \( -\sin u \mathbf{i} \).- For \( e^{3u} \mathbf{j} \), the derivative with respect to \( u \) is \( 3e^{3u} \mathbf{j} \).Thus, \( \mathbf{f}'(u) = -\sin u \mathbf{i} + 3e^{3u} \mathbf{j} \).

Step 4: Differentiate \( u(t) \)

Compute \( u'(t) \) from \( u(t) = 3t^2 - 4 \). This gives \( u'(t) = 6t \).

Step 5: Calculate \( \mathbf{F}^{\prime}(t) \)

Substitute \( u(t) \) and \( u'(t) \) into the chain rule expression:\[ \mathbf{F}^{\prime}(t) = \left(-\sin(3t^2 - 4) \mathbf{i} + 3e^{3(3t^2 - 4)} \mathbf{j}\right) \cdot 6t \]Resulting in:\[ \mathbf{F}^{\prime}(t) = -6t \sin(3t^2 - 4) \mathbf{i} + 18t e^{3(3t^2 - 4)} \mathbf{j} \]

Key Concepts

Vector Functions

Vector functions are extremely useful in calculus, especially when we deal with multiple dimensions. A vector function is a function that has one or more variables and returns a vector as output. For example, the vector function \( \mathbf{F}(t) \) given in the exercise has the form \( \mathbf{f}(u(t)) \), which involves both vector components and a parameter \( t \).
- The vector \( \mathbf{F} \) consists of different components, here represented by \( \mathbf{i} \) and \( \mathbf{j} \). These components are affected by the input function \( u(t) \).
- The importance of vector functions lies in their capability to represent physical quantities like velocity and force in physics and engineering.
Understanding how vector functions operate is important for differentiating them, especially when each component depends on another function.

Differentiation

Differentiation is the process used to find the derivative of a function. In vector calculus, differentiation applies similarly to vector functions, meaning we differentiate each individual component of the vector separately.
- Here, the important point is to realize how differentiation works with composite functions as well. In step 3 of the solution, you notice that the derivatives are computed for both \( \cos u \mathbf{i} \) and \( e^{3u} \mathbf{j} \).
- Differentiation in this vector context requires using the standard rules of derivatives for each part. For example, the derivative of \( \cos u \) is \( -\sin u \), while for \( e^{3u} \), we apply the chain rule to get \( 3e^{3u} \).
To differentiate vector functions effectively:

• Differentiate each vector component individually.
• Apply the chain rule where necessary.
• Understand how the derivative affects your function holistically, not just component-wise.

Composite Functions

Composite functions are functions where one function's output becomes the input for another function. In the given exercise, \( \mathbf{f}(u) = \cos u \mathbf{i} + e^{3u} \mathbf{j} \) is made a composite function by the introduction of \( u(t) \), where \( u(t) = 3t^2 - 4 \).
- This means for the derivative \( \mathbf{F}'(t) \), you must respect both the outer function and inner functions (\( \mathbf{f} \) and \( u \), respectively).
- The chain rule is a critical calculus principle that helps in evaluating derivatives of composite functions by making the differentiation process systematic.
Steps to approach composite functions:

• Differentiate the outer function first, treating the inner function as a single variable.
• Differentiate the inner function next, as shown where \( u(t) \) turns into \( u'(t) = 6t \).
• Finally, multiply the derivatives together to find the overall derivative as illustrated in the solution.