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Chapter 11: Problem 66

Evaluate the following definite integrals. $$\int_{0}^{\pi / 4}\left(\sec ^{2} t \mathbf{i}-2 \cos t \mathbf{j}-\mathbf{k}\right) d t$$

Short Answer

Expert verified
Short Answer: The definite integral of the given vector function is given by the expression \((1\mathbf{i}-\sqrt{2}\mathbf{j}-\frac{\pi}{4}\mathbf{k})\).

Step by step solution

01

Integrate \(\sec^2 t\) with respect to \(\mathbf{i}\)

To perform this integral, recall that the indefinite integral of \(\sec^2 t\) is simply \(\tan t\). We have: $$\int_0^{\pi/4} \sec^2 t \ dt = \left[\tan t\right]_0^{\pi/4} = \tan\frac{\pi}{4} - \tan 0 = 1 - 0 = 1$$
02

Integrate \(-2\cos t\) with respect to \(\mathbf{j}\)

To perform this integral, recall that the indefinite integral of \(\cos t\) is \(\sin t\). We have: $$\int_0^{\pi/4} -2\cos t \ dt = -2\left[\sin t\right]_0^{\pi/4} = -2\left(\sin\frac{\pi}{4} - \sin 0\right) = -2\left(\frac{\sqrt{2}}{2} - 0\right) = -\sqrt{2}$$
03

Integrate \(-1\) with respect to \(\mathbf{k}\)

To perform this integral, recall that the indefinite integral of a constant is the constant multiplied by the variable. We have: $$\int_0^{\pi/4} -1 \ dt = -\left[t\right]_0^{\pi/4} = -\frac{\pi}{4} + 0 = -\frac{\pi}{4}$$
04

Combine the results

Now we combine the results from Steps 1-3 to get the final result: $$\int_{0}^{\pi / 4}\left(\sec ^{2} t \mathbf{i}-2 \cos t \mathbf{j}-\mathbf{k}\right) d t=(1\mathbf{i}-\sqrt{2}\mathbf{j}-\frac{\pi}{4}\mathbf{k})$$

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Most popular questions from this chapter

Suppose the vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is smooth on an interval containing the point \(t_{0} .\) The line tangent to \(\mathbf{r}(t)\) at \(t=t_{0}\) is the line parallel to the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) that passes through \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right) .\) For each of the following functions, find an equation of the line tangent to the curve at \(t=t_{0} .\) Choose an orientation for the line that is the same as the direction of \(\mathbf{r}^{\prime}\). $$\mathbf{r}(t)=\langle\sqrt{2 t+1}, \sin \pi t, 4\rangle ; t_{0}=4$$

Orthogonal lines Recall that two lines \(y=m x+b\) and \(y=n x+c\) are orthogonal provided \(m n=-1\) (the slopes are negative reciprocals of each other). Prove that the condition \(m n=-1\) is equivalent to the orthogonality condition \(\mathbf{u} \cdot \mathbf{v}=0,\) where \(\mathbf{u}\) points in the direction of one line and \(\mathbf{v}\) points in the direction of the other line..

Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are vectors in the plane. a. Use the Triangle Rule for adding vectors to explain why \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}| .\) This result is known as the Triangle Inequality. b. Under what conditions is \(|\mathbf{u}+\mathbf{v}|=|\mathbf{u}|+|\mathbf{v}| ?\)

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Verify that the Cauchy-Schwarz Inequality holds for \(\mathbf{u}=\langle 3,-5,6\rangle\) and \(\mathbf{v}=\langle-8,3,1\rangle\)

Relationship between \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) Consider the circle \(\mathbf{r}(t)=\langle a \cos t, a \sin t\rangle,\) for \(0 \leq t \leq 2 \pi\) where \(a\) is a positive real number. Compute \(\mathbf{r}^{\prime}\) and show that it is orthogonal to \(\mathbf{r}\) for all \(t\)
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