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Chapter 12: Problem 6

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

Short Answer

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

Step by step solution

01

Identify the Vectors

The integral separates into two distinct components: \(\int_{0}^{1} \frac{2}{\sqrt{1-t^{2}}} \mathbf{i} \, dt\) and \(\int_{0}^{1} \frac{\sqrt{3}}{1+t^{2}} \mathbf{k} \, dt\). We will evaluate each integral separately, focusing first on the component along the \(\mathbf{i}\)-direction.
02

Evaluate the Integral for \(\mathbf{i}\) Component

To integrate \(\int_{0}^{1} \frac{2}{\sqrt{1-t^{2}}} \, dt\), note that this is the integral of the derivative of the arcsine function. Thus, this integral becomes: \[\int_{0}^{1} \frac{2}{\sqrt{1-t^{2}}} \, dt = 2\sin^{-1}(t) \Big|_{0}^{1}\]Calculating at the bounds, we find:\[2\left(\frac{\pi}{2} - 0\right) = \pi.\]
03

Evaluate the Integral for \(\mathbf{k}\) Component

Next, consider \(\int_{0}^{1} \frac{\sqrt{3}}{1+t^{2}} \, dt\). Recognize this as the integral of the derivative of the arctangent function:\[\int_{0}^{1} \frac{\sqrt{3}}{1+t^{2}} \, dt = \sqrt{3}\tan^{-1}(t) \Big|_{0}^{1}\]Evaluating the function at the limits gives:\[\sqrt{3}\left(\frac{\pi}{4} - 0\right) = \frac{\pi \sqrt{3}}{4}.\]
04

Combine the Results

Combine the results for the two components:- The \(\mathbf{i}\)-direction contributes \(\pi \mathbf{i}\).- The \(\mathbf{k}\)-direction contributes \(\frac{\pi \sqrt{3}}{4} \mathbf{k}\).Thus, the evaluated integral is:\[ \pi \mathbf{i} + \frac{\pi \sqrt{3}}{4} \mathbf{k}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integrals
Definite integrals are a fundamental concept in calculus, used to find the area under a curve between two specified limits. In our example, we work with the definite integral from 0 to 1.
  • The core idea is to compute the integral of a function over a specific interval.
  • Integrals can represent different quantities, such as areas, volumes, and other cumulative measures.
In this specific exercise, we evaluated the definite integrals of two separate functions. One function represented a vector component in the i-direction, and the other in the k-direction. Each was evaluated individually over the interval [0, 1], resulting in a clear numerical solution.
Arcsine Function
The arcsine function is the inverse of the sine function, represented as \( \sin^{-1}(x) \). It calculates the angle whose sine is x.
  • The derivative of arcsine, \( \frac{d}{dt}\left(\sin^{-1}(t)\right) = \frac{1}{\sqrt{1-t^{2}}} \), is useful in integral problems.
  • For the integral \( \int \frac{1}{\sqrt{1-t^{2}}} \), it translates to \( \sin^{-1}(t) \).
In our exercise, we had an integral involving \( \frac{2}{\sqrt{1-t^{2}}} \). Here, the arcsine function helped simplify it, given by multiplying by 2 and evaluating the definite integral from 0 to 1. Substituting the bounds, arcsine evolves to yield the \( \pi \) component of the result.
Arctangent Function
The arctangent function is the inverse of the tangent function, denoted as \( \tan^{-1}(x) \), and provides the angle whose tangent is x.
  • The derivative of arctangent, \( \frac{d}{dt}\left(\tan^{-1}(t)\right) = \frac{1}{1+t^{2}} \), assists in solving integrals.
  • When dealing with \( \int \frac{1}{1+t^{2}} \), it naturally integrates to \( \tan^{-1}(t) \).
In this problem, the function \( \frac{\sqrt{3}}{1+t^{2}} \) was integrated using the arctangent identity. The presence of \( \sqrt{3} \) scales the resulting integral value, leading to the calculation of \( \frac{\pi \sqrt{3}}{4} \), which contributes to the end solution.
Integral Evaluation
Integral evaluation is the process of computing the value of an integral, often with specific boundaries or limits.
  • Evaluating an integral can involve recognizing standard forms, such as arcsine or arctangent, to simplify the computation.
  • The final result may involve combining the computed results of multiple integrals.
In this exercise, evaluating the two integrals required identifying their forms as derivatives of inverse trigonometric functions, solving them individually, and then combining them. After integrating each part and combining them, the result was the vector solution \( \pi \mathbf{i} + \frac{\pi \sqrt{3}}{4} \mathbf{k} \). This outcome melds the contributions from both of the vector components evaluated.

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Most popular questions from this chapter

In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$\mathbf{r}(t)=t \mathbf{i}+(2 / 3) t^{3 / 2} \mathbf{k}, \quad 0 \leq t \leq 8$$

Show that a planet in a circular orbit moves with a constant speed. (Hint: This is a consequence of one of Kepler's laws.)

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves in Exercises. $$\mathbf{r}(t)=t \mathbf{i}+(a \cosh (t / a)) \mathbf{j}, \quad a > 0$$

Show that the ellipse \(x=a \cos t, y=b \sin t, a > b > 0,\) has its largest curvature on its major axis and its smallest curvature on its minor axis. (As in Exercise 17, the same is true for any ellipse.)

Rounding the answers to four decimal places, use a CAS to find \(\mathbf{v}, \mathbf{a}\) speed, \(\mathbf{T}, \mathbf{N}, \mathbf{B}, \kappa,\) and the tangential and normal components of acceleration for the curves at the given values of \(t\) $$\mathbf{r}(t)=\left(3 t-t^{2}\right) \mathbf{i}+\left(3 t^{2}\right) \mathbf{j}+\left(3 t+t^{3}\right) \mathbf{k}, \quad t=1$$
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