/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Evaluate the integrals. $$ \... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integrals. $$ \int_{0}^{\pi / 4}\left[\sec t \mathbf{i}+\tan ^{2} t \mathbf{j}-t \sin t \mathbf{k}\right] d t $$

Short Answer

Expert verified
\( \ln(\sqrt{2} + 1) \mathbf{i} + (1 - \pi/4) \mathbf{j} + (-\pi\sqrt{2}/8 + 1) \mathbf{k} \)

Step by step solution

01

Break Down the Integral

The given integral is a vector integral that consists of functions in the directions of the unit vectors \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \). This can be written as **\( \int_{0}^{\pi/4} \sec t \ dt \ \mathbf{i} + \int_{0}^{\pi/4} \tan^2 t \ dt \ \mathbf{j} - \int_{0}^{\pi/4} t \sin t \ dt \ \mathbf{k} \).** Each component of the vector is integrated separately.
02

Evaluate the Integral in the i-direction

Evaluate \( \int_{0}^{\pi/4} \sec t \ dt \). The antiderivative of \( \sec t \) is \( \ln |\sec t + \tan t| \).\[\int_{0}^{\pi/4} \sec t \ dt = \left[ \ln | \sec t + \tan t | \right]_{0}^{\pi/4}.\]At \( t = \pi/4 \), \( \sec(\pi/4) = \sqrt{2} \) and \( \tan(\pi/4) = 1 \), so the expression is \( \ln(\sqrt{2} + 1) \). At \( t = 0 \), both \( \sec(0) = 1 \) and \( \tan(0) = 0 \), so the expression is \( \ln 1 = 0 \). Thus, \[\text{Result for } \mathbf{i}\text{-component: } \ln(\sqrt{2} + 1).\]
03

Evaluate the Integral in the j-direction

Evaluate \( \int_{0}^{\pi/4} \tan^2 t \ dt \). We know \( \tan^2 t = \sec^2 t - 1 \). Thus, \[\int \tan^2 t \ dt = \int \sec^2 t \ dt - \int 1 \ dt = \tan t - t.\]Evaluating from 0 to \( \pi/4 \):\[\left[ \tan t - t \right]_{0}^{\pi/4} = (1 - \pi/4) - (0 - 0) = 1 - \pi/4.\]\( \text{Result for } \mathbf{j}\text{-component: } 1 - \pi/4 \).
04

Evaluate the Integral in the k-direction

Evaluate \( \int_{0}^{\pi/4} t \sin t \ dt \) using integration by parts.Let \( u = t \) and \( dv = \sin t \ dt \). Then \( du = dt \) and \( v = -\cos t \).Applying the integration by parts formula:\[\int u \ dv = uv - \int v \ du,\]\[(t (-\cos t)) \Bigg|_{0}^{\pi/4} - \int_{0}^{\pi/4} (-\cos t) \ dt = -t \cos t + \sin t \Bigg|_{0}^{\pi/4}.\]Evaluating, \(-\frac{\pi}{4} \cdot \frac{\sqrt{2}}{2} + %1 - \left(-0\cdot1 + 0\right)\), which gives: \(-\frac{\pi\sqrt{2}}{8} + 1\).\( \text{Result for } \mathbf{k}\text{-component: } -\frac{\pi\sqrt{2}}{8} + 1 \).
05

Combine the Results

Combine the results from each component:\[\left(\ln(\sqrt{2} + 1)\right) \mathbf{i} + \left(1 - \frac{\pi}{4}\right) \mathbf{j} + \left(-\frac{\pi\sqrt{2}}{8} + 1\right) \mathbf{k}.\]This is the final evaluated vector integral.

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Key Concepts

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

Integral Evaluation
Integral evaluation is the process of finding the antiderivative of a given function. An antiderivative, or an integral, is a function whose derivative is the original function that we are evaluating. In vector calculus, we sometimes deal with vector-valued functions that span multiple dimensions using unit vectors, such as
  • \( \mathbf{i} \) – represents the x-direction in a 3D space,
  • \( \mathbf{j} \) – represents the y-direction, and
  • \( \mathbf{k} \) – represents the z-direction.

Hence, when we have an integral involving vectors, like in our example, we treat each component separately, doing the integral for each individual function along its direction. By doing this, we are essentially decomposing the vector into its components, evaluating them independently, then summing them up for the final result.
This approach is foundational in vector calculus and provides a methodical way to handle and compute complex vector expressions efficiently.
Integration by Parts
Integration by parts is a technique derived from the product rule for derivatives. It is used to evaluate integrals where the standard methods are not easily applicable. The integration by parts formula is expressed as:
  • \( \int u \ dv = uv - \int v \ du \).
This method works best with integrals that are products of two functions. In our example, it was used for the \( \mathbf{k} \)-component, where one function was \( t \) and the other was \( \sin t \).
To apply the method:
  • Choose one function to be \( u \) (usually the one that becomes simpler when differentiated),
  • The other function becomes \( dv \).

Then, differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \). Finally, substitute these into the integration by parts formula. The process often simplifies otherwise difficult integrals by reducing the problem into more manageable components.
It helps us break down and solve integrals by turning the problem into a subtraction of two parts, relating closely to the choice of \( u \) and \( dv \).
Vector Integrals
Vector integrals extend the idea of integrating scalar functions to vectors, which are functions that have multiple components. These are crucial in physics and engineering for developing concepts such as work, flux, and circulation. In a vector integral, like the one from our example, each component a function is treated separately:
  • You integrate each directional component function using the methods suitable for scalars,
  • Combine the integrated results guided by the vector's original structure.

This is essential as it allows the computation of quantities that vary in multiple dimensions.
  • For instance, if the vector describes a field like velocity or force, the integral might represent total displacement or work done by a force.
This is often done using fundamental theorems and properties of vector calculus to greatly simplify complex physical systems. Vector integrals, by handling each component individually, provide an analytical approach to understanding and working with multi-dimensional systems.

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Most popular questions from this chapter

Solve the initial value problemsfor \(\mathbf{r}\) as a vector function of \(t .\) $$\text {Differential equation:}\quad \frac{d^{2} \mathbf{r}}{d t^{2}}=-(\mathbf{i}+\mathbf{j}+\mathbf{k})$$ $$\text{Initial conditions:}\begin{array}{l}{\mathbf{r}(0)=10 \mathbf{i}+10 \mathbf{j}+10 \mathbf{k} \text { and }} \\ {\left.\frac{d \mathbf{r}}{d t}\right|_{t=0}=\mathbf{0}}\end{array}$$

Show that the parabola \(y=a x^{2}, a \neq 0,\) has its largest curvature at its vertex and has no minimum curvature. (Note: Since the curvature of a curve remains the same if the curve is translated or rotated, this result is true for any parabola.)

Maximizing the curvature of a helix In Example \(5,\) we found the curvature of the helix \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}+b t \mathbf{k}\) \((a, b \geq 0)\) to be \(\kappa=a /\left(a^{2}+b^{2}\right) .\) What is the largest value \(\kappa\) can have for a given value of \(b ?\) Give reasons for your answer.

Where trajectories erest For a projectile fired from the ground at launch angle \(\alpha\) with initial speed \(v_{0},\) consider \(\alpha\) as a variable and \(v_{0}\) as a fixed constant. For each \(\alpha, 0<\alpha<\pi / 2,\) we obtable a parabolic trajectory as shown in the accompanying figure. Show that the points in the plane that give the maximum heights of these parabolic trajectories all lie on the ellipse $$x^{2}+4\left(y-\frac{v_{0}^{2}}{4 g}\right)^{2}=\frac{v_{0}^{4}}{4 g^{2}}$$ where \(x \geqq 0\)

In Exercises \(11-14,\) find the arc length parameter along the curve from the point where \(t=0\) by evaluating the integral $$ s=\int_{0}^{t}|\mathbf{v}(\tau)| d \tau $$ from Equation \((3) .\) Then find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=(1+2 t) \mathbf{i}+(1+3 t) \mathbf{j}+(6-6 t) \mathbf{k}, \quad-1 \leq t \leq 0 $$

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