Problem 6 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \( \pi \mathbf{i} + \frac{\sqrt{3} \pi}{4} \mathbf{k} \).

Step 1: Recognize Integral Components

The integral is a definite integral of a vector function with respect to the variable \( t \), with limits from 0 to 1. The integrand is composed of two separate functions multiplied by constant unit vectors \( \mathbf{i} \) and \( \mathbf{k} \). Thus, we need to evaluate two integrals separately: 1) \( \int_{0}^{1} \frac{2}{\sqrt{1-t^2}} \, dt \) for the \( \mathbf{i} \) component and 2) \( \int_{0}^{1} \frac{\sqrt{3}}{1+t^2} \, dt \) for the \( \mathbf{k} \) component.

Step 2: Evaluate the First Integral

The integral \( \int_{0}^{1} \frac{2}{\sqrt{1-t^2}} \, dt \) can be recognized as a simple transformation of the standard integral form for the arcsine function. The antiderivative of \( \frac{1}{\sqrt{1-t^2}} \) is \( \sin^{-1}(t) \). Therefore, the integral evaluates to \( 2 \sin^{-1}(t) \) evaluated from 0 to 1. When you calculate: \[ 2 \left[ \sin^{-1}(1) - \sin^{-1}(0) \right] = 2 \left[ \frac{\pi}{2} - 0 \right] = \pi \].

Step 3: Evaluate the Second Integral

The integral for the \( \mathbf{k} \) component is \( \int_{0}^{1} \frac{\sqrt{3}}{1+t^2} \, dt \), which is a transformation of the standard integral form for the arctangent function. The antiderivative of \( \frac{1}{1+t^2} \) is \( \tan^{-1}(t) \). Therefore, the integral evaluates to \( \sqrt{3} \tan^{-1}(t) \) from 0 to 1. Calculate: \[ \sqrt{3} \left[ \tan^{-1}(1) - \tan^{-1}(0) \right] = \sqrt{3} \left[ \frac{\pi}{4} - 0 \right] = \frac{\sqrt{3} \pi}{4} \].

Step 4: Combine the Results into a Vector

Now, combine the results from the previous steps into a vector. The result for the \( \mathbf{i} \) component is \( \pi \), and the result for the \( \mathbf{k} \) component is \( \frac{\sqrt{3} \pi}{4} \). Therefore, the value of the integral is \[ \pi \mathbf{i} + \frac{\sqrt{3} \pi}{4} \mathbf{k} \].

Key Concepts

Definite Integrals

Definite integrals are crucial in vector calculus as they help calculate the accumulation of quantities over an interval. They differ from indefinite integrals in that they compute a specific value because of the limits of integration. When addressing a definite integral, such as \( \int_{0}^{1} \left[ \frac{2}{\sqrt{1-t^{2}}} \mathbf{i} + \frac{\sqrt{3}}{1+t^{2}} \mathbf{k} \right] dt \), it involves evaluating the integral of a function over the interval from 0 to 1.

• The integral bounds, from 0 to 1, signify the limits within which the function is considered.
• The goal is to find the net area under the curve represented by the function, incorporating both positive and negative parts.
• In vector functions, each component of the vector can be treated like a separate function being integrated over the given interval.

In the solution, this concept is evident as the components \( \mathbf{i} \) and \( \mathbf{k} \) are integrated separately with respect to \( t \), before combining them to form the final vector value of the integral.

Arc Functions

Arc functions, like arcsine and arctangent, are inverse trigonometric functions that play a significant role in evaluating definite integrals, particularly in problems involving symmetry and circular functions. For instance, in integration, we often use substituting strategies that transform complex expressions into simpler forms recognizable by their derivatives.

• The integral \( \int \frac{1}{\sqrt{1-t^2}} \, dt \) involves the arcsine function, whose antiderivative is \( \sin^{-1}(t) \).
• The integral \( \int \frac{1}{1+t^2} \, dt \) involves the arctangent function, having antiderivative \( \tan^{-1}(t) \).
• Using these functions allows us to unfold hidden angles in trigonometric expressions and derive their actual magnitude.

In the exercise, arcsine and arctangent functions simplified the evaluation, turning the integrals into expressions that are easier to handle and compute over the defined limits from 0 to 1.

Vector Functions

Vector functions extend the concept of definite integrals to multi-dimensional fields, allowing the calculation of accumulated effects in different directions. Unlike scalar functions, vector functions have components that can interact in various ways, and each component is considered separately when integrating.

• Each vector function is essentially a collection of individual functions corresponding to each direction represented by a unit vector, such as \( \mathbf{i} \) and \( \mathbf{k} \).
• In calculus, integrating vector functions necessitates separately handling each component function and then combining the results to produce the complete vector.
• This process facilitates comprehension of phenomena that have directionality, such as force fields and velocities.

In the given problem, the vector function comprised components \( \frac{2}{\sqrt{1-t^2}} \mathbf{i} \) and \( \frac{\sqrt{3}}{1+t^2} \mathbf{k} \), which were integrated separately and then combined to give \( \pi \mathbf{i} + \frac{\sqrt{3} \pi}{4} \mathbf{k} \). This result represents an accumulation of changes in the \( \mathbf{i} \) and \( \mathbf{k} \) directions over the interval [0,1].