Problem 6 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-10 .\) \(\int_{0}^{1}\left[\frac{2}{\sqrt{1-t^{2}}} \mathbf{i}+\frac{\sqrt{3}}{1+t^{2}} \mathbf{k}\right] d t\)

Step-by-Step Solution

Answer

\( \pi \mathbf{i} + \frac{\sqrt{3}\pi}{4} \mathbf{k} \)

Step 1: Recognize the Integral Components

The integral is given as \( \int_{0}^{1} \left[ \frac{2}{\sqrt{1-t^{2}}} \mathbf{i} + \frac{\sqrt{3}}{1+t^{2}} \mathbf{k} \right] dt \). This is a vector integral, split into two main components: \( \frac{2}{\sqrt{1-t^2}} \mathbf{i} \) and \( \frac{\sqrt{3}}{1+t^2} \mathbf{k} \). We will evaluate each component separately.

Step 2: Integrate the \(\mathbf{i}\) Component

For the first integral, evaluate \( \int_{0}^{1} \frac{2}{\sqrt{1-t^{2}}} dt \). This integral represents the inverse sine function: \( 2 \sin^{-1}(t) \). Evaluate it from \( t = 0 \) to \( t = 1 \): \[ 2 \sin^{-1}(1) - 2 \sin^{-1}(0) = 2 \times \frac{\pi}{2} - 2 \times 0 = \pi. \]

Step 3: Integrate the \(\mathbf{k}\) Component

For the second integral, evaluate \( \int_{0}^{1} \frac{\sqrt{3}}{1+t^2} dt \). This is related to the inverse tangent function, \( \sqrt{3} \tan^{-1}(t) \). Evaluate the integral from \( t = 0 \) to \( t = 1 \): \[ \sqrt{3} \tan^{-1}(1) - \sqrt{3} \tan^{-1}(0) = \sqrt{3} \times \frac{\pi}{4} - 0 = \frac{\sqrt{3}\pi}{4}. \]

Step 4: Combine the Results

Combine the results from both components to express the final vector result. Thus, the evaluated integral is\[ \pi \mathbf{i} + \frac{\sqrt{3}\pi}{4} \mathbf{k}. \]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are crucial in calculus and appear frequently in integrals, like the ones evaluated in the given exercise. The inverse sine function, denoted as \( \sin^{-1}(x) \), is the angle whose sine is \( x \). This function is vital for integrating expressions like \( \int \frac{1}{\sqrt{1-x^2}} \, dx \). Here, it specifically is used for the term \( \frac{2}{\sqrt{1-t^2}} \), representing \( 2 \sin^{-1}(t) \) when integrated.
Inverse trigonometric functions include:

• \( \sin^{-1}(x) \) for arcsine, which results from integrals of the form \( \int \frac{1}{\sqrt{1-x^2}} \, dx \)
• \( \tan^{-1}(x) \) for arctangent, applicable to \( \int \frac{1}{1+x^2} \, dx \)
• Similar functions like \( \cos^{-1}(x) \) for arccosine and \( \cot^{-1}(x) \) for arccotangent

Inverse functions convert a trigonometric ratio back into an angle, useful in many real-world scenarios and physics problems. They help bridge the gap between algebraic expressions and geometric understanding.

Definite Integration

Definite integration is a fundamental concept in calculus that calculates the accumulated value of a function over a specific interval, providing critical insights into areas under curves and other applications.The symbol \( \int_a^b f(x) \, dx \) denotes the definite integral of \( f(x) \) from \( a \) to \( b \).
Key aspects include:

• **Upper and Lower Bounds**: \( a \) and \( b \) are the limits that define the interval.
• **Net Area Calculation**: Definite integrals can represent areas, including those that sit below the \( x \)-axis by considering signed area.
• For vector functions, each component is integrated separately. Solutions combine the results to form a vector result.

The exercise involves definite integrals of vector components, where each integral was evaluated individually and combined.This process confirms the integral of vector functions by evaluating each part over a specific interval.

Multivariable Calculus

Multivariable calculus extends fundamental calculus concepts to functions involving multiple variables, essential for understanding vector fields and complex systems.Essential aspects include:

• **Vectors**: Combinations of functions such as \( f(t) \mathbf{i} + g(t) \mathbf{k} \) in the exercise represent vector functions.
• **Components Integration**: Each component is integrated separately, then combined to determine the overall behavior of the vector function.
• **Applications**: Useful in physics for describing forces, velocity fields, and gradients across multi-dimensional spaces.

The exercise you solved illustrates key ideas of multivariable calculus by using vector components \( \mathbf{i} \) and \( \mathbf{k} \), each integrated separately over the given interval. This method provides a way to understand vector fields and phenomena involving more than one independent variable.