Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \mathbf{f}(u)=\cos u \mathbf{i}+e^{3 u} \mathbf{j} \text { and } u(t)=3 t^{2}-4 $$

Step-by-Step Solution

Answer

Replace \( u \) with \( 3t^2 - 4 \) in \( \mathbf{f}(u) \) to get \( \mathbf{f}(t)=\cos(3t^2-4)\mathbf{i}+e^{9t^2-12}\mathbf{j} \).

Step 1: Substitute the expression for u

First, we will substitute the given expression for \( u(t) \), which is \( u(t) = 3t^2 - 4 \), into the vector function \( \mathbf{f}(u) = \cos u \mathbf{i} + e^{3u} \mathbf{j} \). Thus, replacing \( u \) with \( 3t^2 - 4 \) gives us: \[ \mathbf{f}(t) = \cos(3t^2 - 4) \mathbf{i} + e^{3(3t^2 - 4)} \mathbf{j}. \]

Step 2: Simplify the expression

Next, simplify the expression for the exponential part. We have \( 3(3t^2 - 4) = 9t^2 - 12 \). Substitute this back into the equation to get: \[ \mathbf{f}(t) = \cos(3t^2 - 4) \mathbf{i} + e^{9t^2 - 12} \mathbf{j}. \] This is the simplified form of the vector function \( \mathbf{f}(t) \).

Key Concepts

Parametric Equations

In mathematics, parametric equations are a powerful tool that describe a set of related quantities as functions of an independent variable, called a parameter. Rather than expressing one variable directly in terms of another, parametric equations allow us to define complex curves by expressing each coordinate in terms of a separate parameter. This approach is especially useful in vector calculus and physics.

• A common use of parametric equations is in describing the trajectory of objects. Rather than calculating the position directly with Cartesian coordinates, we define variables such as time as the parameter.


• Parametric equations also play a key role in defining curves in three dimensions, such as the path of an aircraft or a roller coaster.

In the context of our problem, we use a parametric equation where the parameter is given by the expression for \( u(t) = 3t^2 - 4 \), allowing us to generate a vector function \( \mathbf{f}(t) \). By substituting this expression into \( \mathbf{f}(u) = \cos u \mathbf{i} + e^{3u} \mathbf{j} \), we successfully map the parameter \( t \) to the vector functions that define a path.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in mathematics due to their ability to describe oscillatory motion and wave patterns. These functions are defined based on the ratios of sides in a right triangle and are periodic in nature, meaning they repeat their values in regular intervals. This periodicity makes them ideal for applications in physics, engineering, and signal processing.

• Sine and Cosine: The cosine function is utilized in this exercise. Cosine provides the horizontal or x-coordinate of a point on the unit circle as it rotates around the origin. This cyclic nature allows for modeling phenomena such as sound waves, alternating current electricity, and more.


• Properties: The cosine function has an amplitude of 1, a period of \(2\pi\), and its range is from -1 to 1.

For our problem, we calculated \( \cos(3t^2 - 4) \), which modifies the phase of the cosine curve. By altering the input to the cosine function, we achieve a time-shifted or phase-shifted version of the standard cosine curve, making it adaptable for varied practical applications.

Exponential Functions

Exponential functions are crucial in mathematics for their capacity to model growth and decay processes. Defined typically as functions of the form \(f(x) = a \cdot e^{kx}\), where \( e \) is Euler's number (approximately 2.718), exponential functions have unique characteristics.

• Growth and Decay: These functions exhibit rapid increases or decreases. If the parameter \( k \) is positive, the function grows exponentially as \( x \) increases. Conversely, if \( k \) is negative, it depicts exponential decay.


• Continuous Compounding: Exponential functions are used to model processes like continuous compounding in finance, population growth in biology, and radioactive decay in physics.

In our problem, \( e^{3u} \) is part of the vector function, and through substitution, it becomes \( e^{9t^2 - 12} \). This substitution demonstrates how exponential growth factors can be used to stretch or shrink over different parameters, enabling calculations of variable growth dynamics in mathematical modeling.