Problem 61 - Essential Calculus Early Transcendentals

Question

When \(a \ne 0\), there are two solutions to \\(ax^2 + bx + c = 0\\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ $$ \int_{0}^{\pi / 2}\left(3 \sin ^{2} t \cos t \mathbf{i}+3 \sin t \cos ^{2} t \mathbf{j}+2 \sin t \cos t \mathbf{k}\right) d t $$

Step-by-Step Solution

Answer

The solution to the integral is the vector \(3\mathbf{i} - 3\mathbf{j} + \mathbf{k}\).

Step 1: Identify the Components of the Problem

We are given a vector function inside an integral. The components are \(3 \sin^2 t \cos t\) for the \(\mathbf{i}\)-component, \(3 \sin t \cos^2 t\) for the \(\mathbf{j}\)-component, and \(2 \sin t \cos t\) for the \(\mathbf{k}\)-component.

Step 2: Set Up the Integral for Each Component

We will evaluate the integral of each vector component separately over the interval \([0, \pi/2]\). The integrals are:1. \(\int_{0}^{\pi/2} 3 \sin^2 t \cos t \, dt\) for \(\mathbf{i}\)2. \(\int_{0}^{\pi/2} 3 \sin t \cos^2 t \, dt\) for \(\mathbf{j}\)3. \(\int_{0}^{\pi/2} 2 \sin t \cos t \, dt\) for \(\mathbf{k}\).

Step 3: Solve the Integral for the \(\mathbf{i}\)-component

The \(\mathbf{i}\)-component integral is \(\int_{0}^{\pi/2} 3 \sin^2 t \cos t \, dt\). Use the substitution \(u = \sin t\), \(du = \cos t \, dt\). The limits change from \(t = 0\) to \(u = 0\) and from \(t = \pi/2\) to \(u = 1\). The integral becomes \(\int_{0}^{1} 3u^2 \, du\), which evaluates to \(u^3\) from 0 to 1, giving \(1\). Multiply by 3 to obtain \(3\mathbf{i}\).

Step 4: Solve the Integral for the \(\mathbf{j}\)-component

The \(\mathbf{j}\)-component integral is \(\int_{0}^{\pi/2} 3 \sin t \cos^2 t \, dt\). Using the substitution \(u = \cos t\), \(du = -\sin t \, dt\), change the limits: \(t = 0\) to \(u = 1\) and \(t = \pi/2\) to \(u = 0\). The integral becomes \(-\int_{1}^{0} 3u^2 \, du\), or \(\int_{0}^{1} 3u^2 \, du\). Evaluate and get \(1\). Multiply by -3 (due to substitution) to obtain \(-3\mathbf{j}\).

Step 5: Solve the Integral for the \(\mathbf{k}\)-component

The \(\mathbf{k}\)-component integral is \(\int_{0}^{\pi/2} 2 \sin t \cos t \, dt\) which is \(\sin(2t)\) using the double-angle identity. The integral becomes \(\int_{0}^{\pi/2} \sin(2t) \, dt\). With the substitution \(u = 2t\), \(du = 2 \, dt\), and adjusting limits, this becomes \(\frac{1}{2}\int_{0}^{\pi} \sin(u) \, du\). Evaluating gives \(-\cos(u)\) from 0 to \(\pi\) or \(\cos(0) - \cos(\pi)\), which is 2. The result is \(\mathbf{k}\).

Step 6: Combine the Results

Combining all vector components, the result is \(3\mathbf{i} - 3\mathbf{j} + \mathbf{k}\). This is the solution to the original vector integral problem.

Key Concepts

Quadratic Formula

When dealing with quadratic equations of the form \(ax^2 + bx + c = 0\), the quadratic formula provides a straightforward method to find the roots. The formula is given by \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). This formula is derived using a method called completing the square and is applicable as long as \(a eq 0\).

• The term \(b^2-4ac\) is known as the discriminant.
• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution.
• A negative discriminant indicates two complex solutions.

This formula is extremely useful because it provides a quick way to find solutions without having to factor the quadratic equation, which can sometimes be difficult or impossible.

Integral Calculus

Understanding integral calculus is key to solving problems involving areas under curves and accumulation functions. In this exercise, we evaluated the integral of a vector function over a specific interval. When solving such integrals, it's useful to:

• Identify the components of the function that need to be integrated.
• Set up the integral for each component separately.

The fundamental theorem of calculus links the concept of differentiation with integration, providing a powerful tool for calculating areas and solving problems in physics and engineering.
Specifically, in vector calculus, we also need to consider each dimension separately and then combine the results to arrive at the vector-based solution. The integration process often involves techniques like substitution, which simplifies the integration by changing variables.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals that involve square roots or other complex expressions using trigonometric identities. This method is particularly useful when dealing with functions involving \(\sin\), \(\cos\), and \(\tan\).To use trigonometric substitution:

• Choose an appropriate substitution based on the function's form (e.g., \(x = \sin(\theta)\) for \(\sqrt{1-x^2}\)).
• Using the substitution, rewrite the integral in terms of the trigonometric function.
• Integrate using known trigonometric identities and formulas.
• Finally, revert back to the original variable to solve the integral.

In the step-by-step solution provided, trigonometric substitution was used to simplify vector components before integration. It helps in transforming a complex integral into a simpler form, making it easier to evaluate.