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Chapter 13: Problem 5

Evaluate the integrals in Exercises \(1-10\) $$ \int_{0}^{\pi / 3}[(\sec t \tan t) \mathbf{i}+(\tan t) \mathbf{j}+(2 \sin t \cos t) \mathbf{k}] d t $$

Short Answer

Expert verified
The evaluated integral is \( \mathbf{i} + \ln(2) \mathbf{j} + \frac{3}{4} \mathbf{k} \).

Step by step solution

01

Break Down the Integral by Vector Components

The given vector integral \( \int_{0}^{\pi / 3}[(\sec t \tan t) \mathbf{i}+(\tan t) \mathbf{j}+(2 \sin t \cos t) \mathbf{k}] \, dt \) can be separated into individual integrals for each component. We will solve each component separately:- For \( \mathbf{i} \): \( \int_{0}^{\pi / 3} (\sec t \tan t) \, dt \)- For \( \mathbf{j} \): \( \int_{0}^{\pi / 3} (\tan t) \, dt \)- For \( \mathbf{k} \): \( \int_{0}^{\pi / 3} (2 \sin t \cos t) \, dt \).
02

Integrate the \( \mathbf{i} \) Component

The integral for the \( \mathbf{i} \) component is \( \int_{0}^{\pi / 3} (\sec t \tan t) \, dt \). The antiderivative of \( \sec t \tan t \) is \( \sec t \). Thus: \[ \int (\sec t \tan t) \, dt = \sec t + C \]Evaluating from 0 to \( \pi/3 \):\( \sec(\pi/3) - \sec(0) \)which is equal to \( 2 - 1 = 1 \).
03

Integrate the \( \mathbf{j} \) Component

For the \( \mathbf{j} \) component, we have \( \int_{0}^{\pi/3} \tan t \, dt \). This integral can be rewritten using the identity \( \tan t = \frac{\sin t}{\cos t} \), leading to a logarithm after integration:\[ \int \tan t \, dt = -\ln|\cos t| + C \].Evaluating from 0 to \( \pi/3 \):\[ -\ln|\cos(\pi/3)| + \ln|\cos(0)| \]This simplifies to \(-\ln(1/2) + 0 = \ln(2) \).
04

Integrate the \( \mathbf{k} \) Component

The integral for the \( \mathbf{k} \) component is \( \int_{0}^{\pi/3} (2 \sin t \cos t) \, dt \). Using the double angle identity, \( 2 \sin t \cos t = \sin(2t) \), this integral becomes:\[ \int \sin(2t) \, dt = -\frac{1}{2}\cos(2t) + C \].Evaluating from 0 to \( \pi/3 \):\[ -\frac{1}{2}\cos(2(\pi/3)) - \left(-\frac{1}{2}\cos(0)\right) = -\frac{1}{2}(-\frac{1}{2}) + \frac{1}{2}(1) \].This results in \( \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \).
05

Combine the Results

Now that we have evaluated each component, we combine the results into a single vector:\[ \mathbf{i} \] component is 1,\[ \mathbf{j} \] component is \( \ln(2) \),\[ \mathbf{k} \] component is \( \frac{3}{4} \).Therefore, the evaluated integral is:\[ \mathbf{i} + \ln(2) \mathbf{j} + \frac{3}{4} \mathbf{k} \].

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Key Concepts

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

Vector Integration
Vector integration involves integrating vector-valued functions, where each component of the vector function is integrated separately. In this exercise, the vector function is composed of three components, represented by the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \). Each function is integrated within the same limits from 0 to \( \pi/3 \).
The process is simplified by breaking the integral into its component parts:
  • For the \( \mathbf{i} \) component, \( \sec t \tan t \) is integrated.
  • For the \( \mathbf{j} \) component, \( \tan t \) is integrated.
  • For the \( \mathbf{k} \) component, \( 2 \sin t \cos t \) is integrated.
After calculating each integral separately, the results are combined to form the result of the vector integral.
Antiderivatives
Antiderivatives, also known as indefinite integrals, involve finding a function whose derivative is the given function. In this problem, we are tasked with integrating each component to find its antiderivative.
  • The antiderivative of \( \sec t \tan t \) is \( \sec t \), which, when applied to the limits, simplifies the \( \mathbf{i} \) component.
  • The antiderivative of \( \tan t \) is \( -\ln |\cos t| \). The bounds transform this into a logarithmic difference for the \( \mathbf{j} \) component.
  • For \( 2 \sin t \cos t \), a trigonometric identity simplifies the integration process. The double angle identity helps in transforming and integrating this into \( -\frac{1}{2}\cos(2t) \).
These antiderivatives play a key role in calculating definite integrals by providing a means to evaluate the function at both the upper and lower bounds.
Trigonometric Identities
Trigonometric identities are fundamental in simplifying complex trigonometric expressions within integration and differentiation. They convert challenging integrals into more manageable forms.
In this exercise:
  • We used the identity \( \tan t = \frac{\sin t}{\cos t} \) to change the integral of \( \tan t \) into a logarithmic form.
  • The double angle formula \( 2 \sin t \cos t = \sin(2t) \) simplifies the integration of the \( \mathbf{k} \) component. This conversion makes the integral easier to handle, changing it into a standard sine integral.
Familiarity with these identities allows us to manipulate the integral components effectively, facilitating the computation of their antiderivatives.
Definite Integrals
Definite integrals evaluate the area under a curve within prescribed limits. Unlike indefinite integrals, they produce a numerical result.
In this scenario, we evaluate each vector component from 0 to \( \pi/3 \):
  • The \( \mathbf{i} \) component results in \( 2 - 1 = 1 \).
  • For the \( \mathbf{j} \) component, a logarithmic evaluation gives \( \ln(2) \).
  • The simplified function for the \( \mathbf{k} \) component yields \( \frac{3}{4} \).
By calculating these values at the specific bounds and subtracting the results, one achieves the final vector result. This process highlights the use of the Fundamental Theorem of Calculus in linking antiderivatives with definite integrals.

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Most popular questions from this chapter

In Exercises \(27-30\) , find the value(s) of \(t\) so that the tangent line to the given curve contains the given point. $$ \mathbf{r}(t)=-t \mathbf{i}+t^{2} \mathbf{j}+(\ln t) \mathbf{k} ; \quad(2,-5,-3) $$

A formula for the curvature of a parametrized plane curve $$ \begin{array}{c}{\text { a. Show that the curvature of a smooth curve } \mathbf{r}(t)=f(t) \mathbf{i}+} \\ {g(t) \mathbf{j} \text { defined by twice- differentiable functions } x=f(t) \text { and }} \\ {y=g(t) \text { is given by the formula }} \\ {\kappa=\frac{|\ddot{x} \ddot{y}-\dot{y} \ddot{x}|}{\left(\dot{x}^{2}+\dot{y}^{2}\right)^{3 / 2}}}\end{array} $$The dots in the formula denote differentiation with respect to \(t,\) one derivative for each dot. Apply the formula to find the curvatures of the following curves. $$ \begin{array}{l}{\text { b. } \mathbf{r}(t)=t \mathbf{i}+(\ln \sin t) \mathbf{j}, \quad 0 < t<\pi} \\ {\text { c. } \mathbf{r}(t)=\left[\tan ^{-1}(\sinh t)\right] \mathbf{i}+(\ln \cosh t) \mathbf{j}}\end{array} $$

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

Use Simpson's Rule with \(n=10\) to approximate the length of arc of \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k}\) from the origin to the point \((2,4,8) .\)

In Exercises \(31-36, \mathbf{r}(t)\) is the position of a particle in space at time \(t .\) Match each position function with one of the graphs A-F. $$ \mathbf{r}(t)=(t \sin t) \mathbf{i}+(t \cos t) \mathbf{j}+\left(\frac{t}{t^{2}+1}\right) \mathbf{k} $$
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