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Evaluate \( {\int }_{C}\left( {{x}^{2} + {y}^{2}}\right) {dx} + {2xydy} \), where:\n\n(a) \( C : x = t,\;y = {2t},\;0 \leq t \leq 1 \)\n\n(b) \( C : x = t,\;y = 2{t}^{2},\;0 \leq t \leq 1 \) | (a) Since \( {x}^{\prime }\left( t\right) = 1 \) and \( {y}^{\prime }\left( t\right) = 2 \), then\n\n\[{\int }_{C}\left( {{x}^{2} + {y}^{2}}\right) {dx} + {2xydy} = {\int }_{0}^{1}\left( {\left( {x{\left( t\right) }^{2} + y{\left( t\right) }^{2}}\right) {x}^{\prime }\left( t\right) + {2x}\left( t\right) y\left( t\right... | Yes |
Evaluate \( {\int }_{C}\left( {{x}^{2} + {y}^{2}}\right) {dx} + {2xydy} \), where \( C \) is the polygonal path from \( \left( {0,0}\right) \) to \( \left( {0,2}\right) \) to \( \left( {1,2}\right) \). | Solution: Write \( C = {C}_{1} \cup {C}_{2} \), where \( {C}_{1} \) is the curve given by \( x = 0, y = t \), \( 0 \leq t \leq 2 \) and \( {C}_{2} \) is the curve given by \( x = t, y = 2,0 \leq t \leq 1 \) (see Figure 4.1.5). Then\n\n\[ \n{\int }_{C}\left( {{x}^{2} + {y}^{2}}\right) {dx} + {2xydy} = {\int }_{{C}_{1}}\... | Yes |
Theorem 4.2. Let \( \mathbf{f}\left( {x, y}\right) = P\left( {x, y}\right) \mathbf{i} + Q\left( {x, y}\right) \mathbf{j} \) be a vector field, and let \( C \) be a smooth curve parametrized by \( x = x\left( t\right), y = y\left( t\right), a \leq t \leq b \) . Suppose that \( t = \alpha \left( u\right) \) for \( c \leq... | Proof: Since \( \alpha \left( u\right) \) is strictly increasing and maps \( \left\lbrack {c, d}\right\rbrack \) onto \( \left\lbrack {a, b}\right\rbrack \), then we know that \( t = \) \( \alpha \left( u\right) \) has an inverse function \( u = {\alpha }^{-1}\left( t\right) \) defined on \( \left\lbrack {a, b}\right\r... | Yes |
Evaluate the line integral \( {\int }_{C}\left( {{x}^{2} + {y}^{2}}\right) {dx} + {2xydy} \) from Example 4.2, Section 4.1, along the curve \( C : x = t, y = 2{t}^{2},0 \leq t \leq 1 \), where \( t = \sin u \) for \( 0 \leq u \leq \pi /2 \) . | Solution: First, we notice that \( 0 = \sin 0,1 = \sin \left( {\pi /2}\right) \), and \( \frac{dt}{du} = \cos u > 0 \) on \( \left( {0,\pi /2}\right) \) . So by Theorem 4.2 we know that if \( C \) is parametrized by\n\n\[ x = \sin u,\;y = 2{\sin }^{2}u,\;0 \leq u \leq \pi /2 \]\n\nthen \( {\int }_{C}\left( {{x}^{2} + {... | Yes |
In a region \( R \), the line integral \( {\int }_{C}\mathbf{f} \cdot d\mathbf{r} \) is independent of the path between any two points in \( R \) if and only if \( {\oint }_{C}\mathbf{f} \cdot d\mathbf{r} = 0 \) for every closed curve \( C \) which is contained in R. | Proof: Suppose that \( {\oint }_{C}\mathbf{f} \cdot d\mathbf{r} = 0 \) for every closed curve \( C \) which is contained in \( R \) . Let \( {P}_{1} \) and \( {P}_{2} \) be two distinct points in \( R \) . Let \( {C}_{1} \) be a curve in \( R \) going from \( {P}_{1} \) to \( {P}_{2} \), and let \( {C}_{2} \) be anothe... | Yes |
Theorem 4.4. (Chain Rule) If \( z = f\left( {x, y}\right) \) is a continuously differentiable function of \( x \) and \( y \), and both \( x = x\left( t\right) \) and \( y = y\left( t\right) \) are differentiable functions of \( t \), then \( z \) is a differentiable function of \( t \), and\n\n\[ \frac{dz}{dt} = \frac... | The proof is virtually identical to the proof of Theorem 2.2 from Section 2.4 (which uses the Mean Value Theorem), so we omit it. | No |
Theorem 4.5. Let \( \mathbf{f}\left( {x, y}\right) = P\left( {x, y}\right) \mathbf{i} + Q\left( {x, y}\right) \mathbf{j} \) be a vector field in some region \( R \), with \( P \) and \( Q \) continuously differentiable functions on \( R \) . Let \( C \) be a smooth curve in \( R \) parametrized by \( x = x\left( t\righ... | Proof: By definition of \( {\int }_{C}\mathbf{f} \cdot d\mathbf{r} \), we have\n\n\[{\int }_{C}\mathbf{f} \cdot d\mathbf{r} = {\int }_{a}^{b}\left( {P\left( {x\left( t\right), y\left( t\right) }\right) {x}^{\prime }\left( t\right) + Q\left( {x\left( t\right), y\left( t\right) }\right) {y}^{\prime }\left( t\right) }\rig... | Yes |
Recall from Examples 4.2 and 4.3 in Section 4.1 that the line integral \( {\int }_{C}\left( {{x}^{2} + }\right. \) \( \left. {y}^{2}\right) {dx} + {2xydy} \) was found to have the value \( \frac{13}{3} \) for three different curves \( C \) going from the point \( \left( {0,0}\right) \) to the point \( \left( {1,2}\righ... | Solution: We need to find a real-valued function \( F\left( {x, y}\right) \) such that\n\n\[ \frac{\partial F}{\partial x} = {x}^{2} + {y}^{2}\text{ and }\frac{\partial F}{\partial y} = {2xy}. \]\n\nSuppose that \( \frac{\partial F}{\partial x} = {x}^{2} + {y}^{2} \), Then we must have \( F\left( {x, y}\right) = \frac{... | Yes |
Example 4.6. Evaluate \( {\oint }_{C}{xdx} + {ydy} \) for \( C : x = 2\cos t, y = 3\sin t,0 \leq t \leq {2\pi } \) . | Solution: The vector field \( \mathbf{f}\left( {x, y}\right) = x\mathbf{i} + y\mathbf{j} \) has a potential \( F\left( {x, y}\right) \) :\n\n\[ \n\frac{\partial F}{\partial x} = x \Rightarrow F\left( {x, y}\right) = \frac{1}{2}{x}^{2} + g\left( y\right) ,\text{ so } \n\] \n\n\[ \n\frac{\partial F}{\partial y} = y \Righ... | Yes |
Evaluate \( {\oint }_{C}\left( {{x}^{2} + {y}^{2}}\right) {dx} + {2xydy} \), where \( C \) is the boundary (traversed counterclockwise) of the region \( R = \left\{ {\left( {x, y}\right) : 0 \leq x \leq 1,2{x}^{2} \leq y \leq {2x}}\right\} \) . | Solution: \( R \) is the shaded region in Figure 4.3.2. By Green’s Theorem, for\n\n\n\nFigure 4.3.2\n\n\( P\left( {x, y}\right) = {x}^{2} + {y}^{2} \) and \( Q\left( {x, y}\right) = {2xy} \), we have\n\n\[ \n{\oint }... | Yes |
Example 4.8. Let \( \mathbf{f}\left( {x, y}\right) = P\left( {x, y}\right) \mathbf{i} + Q\left( {x, y}\right) \mathbf{j} \), where\n\n\[ P\left( {x, y}\right) = \frac{-y}{{x}^{2} + {y}^{2}}\text{ and }Q\left( {x, y}\right) = \frac{x}{{x}^{2} + {y}^{2}}, \]\n\nand let \( R = \{ \left( {x, y}\right) : 0 < {x}^{2} + {y}^{... | This would seem to contradict Green’s Theorem. However, note that \( R \) is not the entire region enclosed by \( C \), since the point \( \left( {0,0}\right) \) is not contained in \( R \) . That is, \( R \) has a \ | Yes |
A torus \( T \) is a surface obtained by revolving a circle of radius \( a \) in the \( {yz} \) -plane around the \( z \) -axis, where the circle’s center is at a distance \( b \) from the z-axis \( \left( {0 < a < b}\right) \), as in Figure 4.4.3. Find the surface area of \( T \) . | Solution: For any point on the circle, the line segment from the center of the circle to that point makes an angle \( u \) with the \( y \) -axis in the positive \( y \) direction (see Figure 4.4.3(a)). And as the circle revolves around the \( z \) -axis, the line segment from the origin to the center of that circle sw... | Yes |
Theorem 4.8. (Divergence Theorem) Let \( \sum \) be a closed surface in \( {\mathbb{R}}^{3} \) which bounds a solid \( S \), and let \( \mathbf{f}\left( {x, y, z}\right) = {f}_{1}\left( {x, y, z}\right) \mathbf{i} + {f}_{2}\left( {x, y, z}\right) \mathbf{j} + {f}_{3}\left( {x, y, z}\right) \mathbf{k} \) be a vector fie... | The proof of the Divergence Theorem is very similar to the proof of Green’s Theorem, i.e. it is first proved for the simple case when the solid \( S \) is bounded above by one surface, bounded below by another surface, and bounded laterally by one or more surfaces. The proof can then be extended to more general solids.... | No |
Evaluate \( {\iint }_{\sum }\mathbf{f} \cdot d\mathbf{\sigma } \), where \( \mathbf{f}\left( {x, y, z}\right) = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \) and \( \sum \) is the unit sphere \( {x}^{2} + {y}^{2} + {z}^{2} = 1 \) | Solution: We see that div \( \mathbf{f} = 1 + 1 + 1 = 3 \), so\n\n\[ \n{\iint }_{\sum }\mathbf{f} \cdot d\mathbf{\sigma } = {\iiint }_{S}\operatorname{div}\mathbf{f}{dV} = {\iiint }_{S}{3dV} \]\n\n\[ \n= 3{\iiint }_{S}{1dV} = 3\operatorname{vol}\left( S\right) = 3 \cdot \frac{{4\pi }{\left( 1\right) }^{3}}{3} = {4\pi }... | Yes |
Theorem 4.9. If the flux of a vector field \( \mathbf{f} \) is zero through every closed surface containing a given point, then div \( \mathbf{f} = 0 \) at that point. | Proof: By formula (4.33), at the given point \( \left( {x, y, z}\right) \) we have\n\n\( \operatorname{div}\mathbf{f}\left( {x, y, z}\right) = \mathop{\lim }\limits_{{V \rightarrow 0}}\frac{1}{V}{\iint }_{\sum }\mathbf{f} \cdot d\mathbf{\sigma } \) for closed surfaces \( \sum \) containing \( \left( {x, y, z}\right) \)... | Yes |
Theorem 4.11. (Chain Rule) If \( w = f\\left( {x, y, z}\\right) \) is a continuously differentiable function of \( x, y \), and \( z \), and \( x = x\\left( t\\right), y = y\\left( t\\right) \) and \( z = z\\left( t\\right) \) are differentiable functions of \( t \), then \( w \) is a differentiable function of \( t \)... | \[ \frac{dw}{dt} = \frac{\\partial w}{\\partial x}\\frac{dx}{dt} + \frac{\\partial w}{\\partial y}\\frac{dy}{dt} + \frac{\\partial w}{\\partial z}\\frac{dz}{dt}. \] | Yes |
Evaluate \( {\int }_{C}f\left( {x, y, z}\right) {ds} \) . (Note: \( C \) is called a conical helix. See Figure 4.5.1). | Solution: Since \( {x}^{\prime }\left( t\right) = \sin t + t\cos t,{y}^{\prime }\left( t\right) = \cos t - t\sin t \), and \( {z}^{\prime }\left( t\right) = 1 \), we have\n\n\[ {x}^{\prime }{\left( t\right) }^{2} + {y}^{\prime }{\left( t\right) }^{2} + {z}^{\prime }{\left( t\right) }^{2}\; = \;\left( {{\sin }^{2}t + {2... | Yes |
Let \( \mathbf{f}\left( {x, y, z}\right) = x\mathbf{i} + y\mathbf{j} + {2z}\mathbf{k} \) be a vector field in \( {\mathbb{R}}^{3} \) . Using the same curve \( C \) from Example 4.12, evaluate \( {\int }_{C}\mathbf{f} \cdot d\mathbf{r} \) . | Solution: It is easy to see that \( F\left( {x, y, z}\right) = \frac{{x}^{2}}{2} + \frac{{y}^{2}}{2} + {z}^{2} \) is a potential for \( \mathbf{f}\left( {x, y, z}\right) \) (i.e. \( \nabla F = \mathbf{f} \) ). So by Theorem 4.12 we know that\n\n\[{\int }_{C}\mathbf{f} \cdot d\mathbf{r} = F\left( B\right) - F\left( A\ri... | Yes |
Theorem 4.14. (Stokes’ Theorem) Let \( \sum \) be an orientable surface in \( {\mathbb{R}}^{3} \) whose boundary is a simple closed curve \( C \), and let \( \mathbf{f}\left( {x, y, z}\right) = P\left( {x, y, z}\right) \mathbf{i} + Q\left( {x, y, z}\right) \mathbf{j} + R\left( {x, y, z}\right) \mathbf{k} \) be a smooth... | Proof: As the general case is beyond the scope of this text, we will prove the theorem only for the special case where \( \sum \) is the graph of \( z = z\left( {x, y}\right) \) for some smooth real-valued function \( z\left( {x, y}\right) \), with \( \left( {x, y}\right) \) varying over a region \( D \) in \( {\mathbb... | Yes |
Let \( \sum \) be the elliptic paraboloid \( z = \frac{{x}^{2}}{4} + \frac{{y}^{2}}{9} \) for \( z \leq 1 \), and let \( C \) be its boundary curve. Calculate \( {\oint }_{C}\mathbf{f} \cdot d\mathbf{r} \) for \( \mathbf{f}\left( {x, y, z}\right) = \left( {{9xz} + {2y}}\right) \mathbf{i} + \left( {{2x} + {y}^{2}}\right... | The surface is similar to the one in Example 4.14, except now the boundary curve \( C \) is the ellipse \( \frac{{x}^{2}}{4} + \frac{{y}^{2}}{9} = 1 \) laying in the plane \( z = 1 \) . In this case, using Stokes’ Theorem is easier than computing the line integral directly. As in Example 4.14, at each point \( \left( {... | Yes |
Determine if the vector field \( \mathbf{f}\left( {x, y, z}\right) = {xyz}\mathbf{i} + {xz}\mathbf{j} + {xy}\mathbf{k} \) has a potential in \( {\mathbb{R}}^{3} \). | Solution: Since \( {\mathbb{R}}^{3} \) is simply connected, we just need to check whether curl \( \mathbf{f} = \mathbf{0} \) throughout \( {\mathbb{R}}^{3} \), that is,\n\n\[ \frac{\partial R}{\partial y} = \frac{\partial Q}{\partial z},\;\frac{\partial P}{\partial z} = \frac{\partial R}{\partial x},\;\text{ and }\;\fr... | Yes |
Find (a) the gradient of \( \parallel \mathbf{r}{\parallel }^{2} \) (b) the divergence of \( \mathbf{r} \) (c) the curl of \( \mathbf{r} \) (d) the Laplacian of \( \parallel \mathbf{r}{\parallel }^{2} \) | Solution: (a) \( \nabla \parallel \mathbf{r}{\parallel }^{2} = {2x}\mathbf{i} + {2y}\mathbf{j} + {2z}\mathbf{k} = 2\mathbf{r} \)\n\n(b) \( \nabla \cdot \mathbf{r} = \frac{\partial }{\partial x}\left( x\right) + \frac{\partial }{\partial y}\left( y\right) + \frac{\partial }{\partial z}\left( z\right) = 1 + 1 + 1 = 3 \)\... | Yes |
Theorem 4.15. For any smooth real-valued function \( f\left( {x, y, z}\right) ,\nabla \times \left( {\nabla f}\right) = \mathbf{0} \) . | Proof: We see by the smoothness of \( f \) that\n\n\[ \nabla \times \left( {\nabla f}\right) = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac... | Yes |
Theorem 4.17. For any smooth vector field \( \mathbf{f}\left( {x, y, z}\right) ,\nabla \cdot \left( {\nabla \times \mathbf{f}}\right) = 0 \) . | The proof is straightforward and left as an exercise for the reader. | No |
Corollary 4.18. The flux of the curl of a smooth vector field \( \mathbf{f}\left( {x, y, z}\right) \) through any closed surface is zero. | Proof: Let \( \sum \) be a closed surface which bounds a solid \( S \) . The flux of \( \nabla \times \mathbf{f} \) through \( \sum \) is\n\n\[{\iint }_{\sum }\left( {\nabla \times \mathbf{f}}\right) \cdot d\mathbf{\sigma } = {\iiint }_{S}\nabla \cdot \left( {\nabla \times \mathbf{f}}\right) {dV}\;\text{ (by the Diverg... | Yes |
Show that \( \nabla \cdot \mathbf{E} = {4\pi \rho } \) . This is one of Maxwell’s Equations. \( {}^{10} \) In Gaussian (or CGS) units. | Solution: By the Divergence Theorem, we have\n\n\[ \n{\iiint }_{S}\nabla \cdot \mathbf{E}{dV} = {\iint }_{\sum }\mathbf{E} \cdot d\mathbf{\sigma }\n\]\n\n\[ \n= {4\pi }{\iiint }_{S}{\rho dV}\;\text{by Gauss’ Law, so combining the integrals gives}\n\]\n\n\[ \n{\iiint }_{S}\left( {\nabla \cdot \mathbf{E} - {4\pi \rho }}\... | Yes |
Example 4.19. In Example 4.17 we showed that \( \nabla \parallel \mathbf{r}{\parallel }^{2} = 2\mathbf{r} \) and \( \Delta \parallel \mathbf{r}{\parallel }^{2} = 6 \), where \( \mathbf{r}\left( {x, y, z}\right) = \) \( x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \) in Cartesian coordinates. Verify that we get the same answ... | Solution: Since \( \parallel \mathbf{r}{\parallel }^{2} = {x}^{2} + {y}^{2} + {z}^{2} = {\rho }^{2} \) in spherical coordinates, let \( F\left( {\rho ,\theta ,\phi }\right) = {\rho }^{2} \) (so that \( F\left( {\rho ,\theta ,\phi }\right) = \parallel \mathbf{r}{\parallel }^{2} \) ). The gradient of \( F \) in spherical... | Yes |
Consider the system of linear equations represented by the augmented matrix\n\n\[ B = \left\lbrack \begin{matrix} 1 & 2 & 3 & 4 \\ 0 & - 2 & - 4 & 6 \\ 1 & - 1 & 0 & 0 \end{matrix}\right\rbrack \]\n\nUse row operations to put \( B \) into row echelon form, then solve by backward substitution. Compare to the row-reduced... | Solution. The first operation is to replace row 3 with -1 times row 1, added to row 3.\n\n\( > > \% \) new row \( 3 = - 1 * \) row \( 1 + \) row 3\n\n\( > > \mathrm{B}\left( {3, : }\right) = \left( {-1}\right) * \mathrm{\;B}\left( {1, : }\right) + \mathrm{B}\left( {3, : }\right) \)\n\nans \( = \)\n\n\[ \begin{array}{ll... | Yes |
Example 2.1.2. Use left division to solve the system of equations with augmented matrix \( B \) .\n\n\[ B = \left\lbrack \begin{matrix} 1 & 2 & 3 & 4 \\ 0 & - 2 & - 4 & 6 \\ 1 & - 1 & 0 & 0 \end{matrix}\right\rbrack \] | Solution. To use left division, we need to extract the coefficient matrix and vector of right-side constants. Let’s call the coefficient matrix \( A \) and the right-side constants \( \mathbf{b} \) . (You have probably already noticed that Octave is case-sensitive.)\n\n\( > > \mathrm{B} = \left\lbrack \begin{array}{lll... | Yes |
Find an \( {LU} \) decomposition for\n\n\[ A = \left\lbrack \begin{matrix} 1 & 2 & 3 \\ 0 & - 2 & - 4 \\ 1 & - 1 & 0 \end{matrix}\right\rbrack \] | Solution. This is the same coefficient matrix we row-reduced in Example 2.1.1. We proceed the same way, carefully noting the multiplier used to obtain each 0 . The lower triangular \( L \) starts as an identity matrix, then the negative of each multiplier used in the elimination process is placed into the corresponding... | Yes |
Solve \( A\mathbf{x} = \mathbf{b} \), where \( A = \left\lbrack \begin{matrix} 1 & 2 & 3 \\ 0 & - 2 & - 4 \\ 1 & - 1 & 0 \end{matrix}\right\rbrack \) and \( \mathbf{b} = \left\lbrack \begin{array}{l} 4 \\ 6 \\ 0 \end{array}\right\rbrack \), using \( {LU} \) decomposition. | Solution. We already have the \( {LU} \) decomposition. Since \( L = \left\lbrack \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & {1.5} & 1 \end{matrix}\right\rbrack \), the first step\n\nis to solve:\n\[ \left\lbrack \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & {1.5} & 1 \end{matrix}\right\rbrack \cdot \left\lbrack \begi... | Yes |
Find an \( {LU} \) decomposition (with permutation) for\n\n\[ A = \left\lbrack \begin{array}{rrrr} - 7 & - 2 & 9 & 4 \\ - 4 & - 9 & 3 & 0 \\ - 3 & 4 & 6 & - 2 \\ 6 & 7 & - 4 & - 8 \end{array}\right\rbrack \] | Solution. We will use Octave for this.\n\n\[ \begin{array}{l} > > \mathrm{A} = \left\lbrack \begin{array}{llllllllllllllll} - 7 & - 2 & 9 & 4; & - 4 & - 9 & 3 & 0; & - 3 & 4 & 6 & - 2; & 6 & 7 & - 4 & - 8 \end{array}\right\rbrack \\ \mathrm{A} = \end{array} \]\n\n\[ \begin{array}{llll} - 7 & - 2 & 9 & 4 \end{array} \]\... | Yes |
Example 2.2.2. Use polyfit to find the least-squares parabola for the following data: | Solution. This is the same data as in Example 2.2.1. Re-enter the data values if necessary.\n\nWe use polyfit to determine the equation, then the polyval function to evaluate the polynomial at the given \( x \) -values.\n\n\( > > \mathrm{P} = \) polyfit \( \left( {\mathrm{{xdata}},\mathrm{{ydata}},2}\right) \% \) degre... | Yes |
Rotate the house graph through \( {90}^{ \circ } \) and \( {225}^{ \circ } \). | Solution. Note that each \( \theta \) must be converted to radians. Here we go:\n\n\( > > D = \left\lbrack \begin{array}{llllllllllllll} 1 & 1 & 3 & 3 & 2 & 1 & 3; & 2 & 0 & 0 & 2 & 3 & 2 & 2 \end{array}\right\rbrack \) ;\n\n\( > > \mathrm{x} = \mathrm{D}\left( {1, : }\right) \) ;\n\n\( > > \mathrm{y} = \mathrm{D}\left... | Yes |
Example 2.3.2. Reflect the house graph in the line \( y = x \) . | Solution. With the data matrix \( D \) and the original \( x \) and \( y \) vectors already defined, and using \( R \) as determined above, we have:\n\n\[ \mathrm{R} = \left\lbrack \begin{array}{llll} 0 & 1; & 1 & 0 \end{array}\right\rbrack \]\n\n\( \mathrm{R} = \)\n\n\( \begin{array}{ll} 0 & 1 \end{array} \)\n\n\( \be... | Yes |
Example 2.3.3. Expand the house graph by a factor of 2. | Solution. To scale by a factor of 2, we only need to multiply \( D \) by the matrix \( \left\lbrack \begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right\rbrack \) . | Yes |
Example 3.1.1. Let \( \mathop{\sum }\limits_{{n = 2}}^{\infty }{a}_{n} \) be the series whose \( n \) th term is \( {a}_{n} = \frac{1}{n\left( {n + 2}\right) } \) . Find the first ten terms, the first ten partial sums, and plot the sequence and partial sums. | Solution. To do this, we will define an index vector \( n \) from 2 to 11, then calculate the terms.\n\n\( \gg \mathrm{n} = {\left\lbrack \begin{array}{lll} 2 & : & {11} \end{array}\right\rbrack }^{\prime };\;\% \) index\n\n\( \gg \mathrm{a} = 1./\left( {\mathrm{n} \cdot * \left( {\mathrm{n} + 2}\right) }\right) \% \) ... | Yes |
Find the sum of the first 1000 terms of the harmonic series. | Solution. We only need to generate the terms as a vector, then take its sum.\n\n\( > > \mathrm{n} = \left\lbrack \begin{array}{lll} 1 & : & {1000} \end{array}\right\rbrack \)\n\n\( > > \mathrm{a} = 1./\mathrm{n} \) ;\n\n\( > > \operatorname{sum}\left( \mathrm{a}\right) \)\n\nans \( = {7.4855} \) | Yes |
Estimate \( {\int }_{0}^{\pi /2}{e}^{{x}^{2}}\cos \left( x\right) {dx} \) using Octave’s quad algorithm. | Solution. The correct syntax is quad(’ \( \mathrm{f} \) ’, a, b). We need to first define the function.\n\n\( > > \) function \( \mathrm{y} = \mathrm{f}\left( \mathrm{x}\right) \)\n\n\( \mathrm{y} = \exp \left( {\mathrm{x} \cdot {}^{ \land }2}\right) \cdot * \cos \left( \mathrm{x}\right) ; \)\n\nend\n\n\( > > \operator... | Yes |
Write an Octave script to calculate a midpoint rule approximation of\n\n\\[ \n{\\int }_{0}^{\\pi /2}{e}^{{x}^{2}}\\cos \\left( x\\right) {dx} \n\\]\n\nusing \\( n = {100} \\) . | Solution. The basic strategy is to use a for loop that adds an additional function value to a running total with each iteration. Then the final answer is found by multiplying the sum by \\( {\\Delta x} \\) .\n\nThe following code can be used. Switch to the editor tab and enter the code in a plain text file. Save it as ... | No |
Write a vectorized Octave script to calculate a midpoint rule approximation of\n\n\\[ \n{\\int }_{0}^{\\pi /2}{e}^{{x}^{2}}\\cos \\left( x\\right) {dx} \n\\]\n\nusing \\( n = {100} \\) . | Solution. Now our strategy is to create a vector of the \\( x \\) -coordinates of the midpoints. Then we evaluate \\( f \\) over this midpoint vector to obtain a vector of function values. The midpoint approximation is the sum of the components of the vector, multiplied by \\( {\\Delta x} \\) .\n\n---\n\nOctave Script ... | Yes |
Example 3.3.1. Graph three periods of a radius 2 cycloid. | Solution. The functions have period \( {2\pi } \), so we need \( 0 \leq t \leq {6\pi } \) to see three full cycles. We need to define the parameter \( t \) as a vector over this range, then we calculate \( x \) and \( y \), and plot \( x \) vs. \( y \) .\n\n\( > > \mathrm{t} = \) linspace \( \left( {0,6 * \mathrm{{pi}}... | Yes |
Plot the limaçon \( r = 1 - 2\sin \left( \theta \right) \) . | The needed commands are shown below and the graph is shown in Figure 3.3.\n\n\( > > \) theta \( = \) linspace \( \left( {0,2 * \mathrm{{pi}},{100}}\right) \) ;\n\n\( > > \mathrm{r} = 1 - 2 * \sin \left( \text{theta}\right) \) ;\n\n\( > > \mathrm{x} = \mathrm{r} \cdot * \cos \left( \text{theta}\right) \) ;\n\n\( > > \ma... | Yes |
Plot the curve defined by the equation\n\n\[ - {x}^{2} - {xy} + x + {y}^{2} - y = 1 \] | Solution. To define the function as \( f\left( {x, y}\right) = 0 \), we subtract 1 from both sides of the equation.\n\n\( > > \mathrm{f} = @\left( {\mathrm{x},\mathrm{y}}\right) - \mathrm{x} \cdot {}^{\hat{} }2 - \mathrm{x} \cdot * \mathrm{y} + \mathrm{x} + \mathrm{y} \cdot {}^{\hat{} }2 - \mathrm{y} - 1 \)\n\n\( \math... | No |
Find the equation of the line tangent to the graph of the circle \( {\left( x - 2\right) }^{2} + {y}^{2} = {25} \) , at the point \( \left( {-1,4}\right) \) . Plot a graph of the circle and the tangent line on the same axes. | Solution. To plot the circle, we’ll first define it as a function of the form \( f\left( {x, y}\right) = 0 \) .\n\n\[ > > \mathrm{f} = @\left( {\mathrm{x},\mathrm{y}}\right) \left( {\mathrm{x} - 2}\right) \cdot {}^{ \land }2 + \mathrm{y} \cdot {}^{ \land }2 - {25}\text{;} \]\n\nThe center of the circle is at \( \left( ... | Yes |
Example 3.4.1. Let \( f\left( x\right) = {x}^{3} + 3{x}^{2} - {10x} \) .\n\n(a) Evaluate \( f\left( \frac{1}{2}\right) \) . | Solution. The first step is to declare \( x \) as a symbolic variable with the command syms.\n\nsyms \( \mathrm{x} \) \( \% \) declare symbolic variable \( \mathrm{x} \)\n\nNow we define the expression. Notice that we do not need to worry about using elementwise operations here.\n\n\( \gg \mathrm{f} = {\mathrm{x}}^{ * ... | Yes |
Let \( f\left( x\right) = {x}^{2}\sin x \) . Find each of the following:\n\n(a) \( {f}^{\prime }\left( x\right) \)\n\n(b) \( \int f\left( x\right) {dx} \)\n\n(c) \( {\int }_{0}^{\pi /4}f\left( x\right) {dx} \) | Solution. First we define the expression.\n\n\( > > \mathrm{f} = \mathrm{x}{}^{ \land }2 * \sin \left( \mathrm{x}\right) \)\n\n\( \mathrm{f} = \left( \mathrm{{sym}}\right) \)\n\n\[ \begin{matrix} 2 \\ \mathrm{x} * \sin \left( \mathrm{x}\right) \end{matrix} \]\n\nNow, calculate the derivative using the diff command.\n\n... | Yes |
Show that\n\n\[ \n{\int }_{-r}^{r}2\sqrt{{r}^{2} - {x}^{2}}{dx} = \pi {r}^{2} \n\] | Solution. The integral, of course, represents the area of circle of radius \( r \) .\n\n\( > > \) syms \( \mathrm{x}\mathrm{r} \)\n\n\( > > \mathrm{f} = 2 * \operatorname{sqrt}\left( {\mathrm{r}\widehat{}2 - \mathrm{x}\widehat{}2}\right) \)\n\n\( \mathrm{f} = \left( \mathrm{{sym}}\right) \) ![c6a99c62-f1ae-42a9-87a0-43... | Yes |
Example 3.4.4. Let \( f\left( x\right) = {x}^{3} + 3{x}^{2} - {10x} \) . Graph \( f,{f}^{\prime } \), and \( {f}^{\prime \prime } \) on the same axes. | Solution. Once we’ve defined \( f \), plotting the function is as simple as typing explot \( \left( \mathrm{f}\right) \) .\n\n\( > > \) syms \( \mathrm{x} \)\n\n\( > > \mathrm{f} = {\mathrm{x}}^{ \land }3 + 3 * {\mathrm{x}}^{ \land }2 - {10} * \mathrm{x} \)\n\n\( \mathrm{f} = \left( \mathrm{{sym}}\right) \)\n\n\[ 3 + 3... | Yes |
Let \( {z}_{1} = 1 + {2i} \) and \( {z}_{2} = 2 - {3i} \) . Find each of the following:\n\n\[ \n{z}_{1} + {z}_{2},{z}_{2} - {z}_{1},{z}_{1} \cdot {z}_{2}\\text{, and}{z}_{1}/{z}_{2} \n\] | Solution. Octave has no difficulty dealing with complex arithmetic. The variables \( i \) and \( j \) \n\n(not case-sensitive) are both by default recognized as the imaginary unit \( \\sqrt{-1} \) . \n\nFirst define the variables: \n\n\[ \n> > \\mathrm{z}1 = 1 + 2\\mathrm{i} \n\] \n\n\[ \n> > \\mathrm{z}2 = 2 - 3\\math... | Yes |
Let \( {z}_{1} = 1 + {2i} \) and \( {z}_{2} = 2 - {3i} \) . Plot \( {z}_{1},{z}_{2} \), and the sum \( {z}_{1} + {z}_{2} \) in the complex plane. | Solution. We will show both variables and their sum on one set of axes.\n\n\( > > \mathrm{z}1 = 1 + 2\mathrm{i} \)\n\n\( > > \mathrm{z}2 = 2 - 3\mathrm{i} \) ;\n\n\( > > \operatorname{compass}\left( {\mathrm{z}1,{}^{\prime }{\mathrm{b}}^{\prime }}\right) \)\n\n\( > > \) hold on\n\n\( > > \operatorname{compass}\left( {\... | Yes |
Example 4.2.1. Graph \( \Gamma \left( {x + 1}\right) \) together on the same set of axes with the factorial function \( x \) !, for corresponding nonnegative integer values of \( x \) . | Solution. Both the gamma function and factorial function grow quite large very quickly, so we need to take care in selecting the domain. The gamma function is defined for positive and negative real numbers, while the factorial function is of course defined only for nonnegative integers. We will try the graph for \( x \... | Yes |
Create a vector \( Z \) of 1000 elements from the standard normal distribution. Use the transformation \( X = {Z\sigma } + \mu \) to generate a vector \( X \) of elements from a normal distribution with mean 400 and standard deviation 50 . Compare the means and variances of \( X \) and \( Z \) . Plot histograms of \( Z... | Solution. Here are the commands we need:\n\n\( \gg \% \) sample 1000 elements from a standard normal distribution\n\n\( > > \mathrm{Z} = \operatorname{randn}\left( {{1000},1}\right) \) ;\n\n\( > > \% \) transform the mean and standard deviation\n\n\( > > \mathrm{{mu}} = {400};\operatorname{sigma} = {50} \) ;\n\n\( > > ... | Yes |
Example 4.3.2. Let \( x = \{ 5,9,{18},{25},{32},{40},{53}\} \) and \( y = \{ {32},{28},{23},{20},{19},{18},9\} \) . Create a scatter plot of the data and calculate the correlation coefficient. Find the equation of the regression line and add it to the scatter plot. | Solution. First we enter the data and create the scatter plot.\n\n\( > > \mathrm{x} = \left\lbrack \begin{array}{lllllll} 5 & 9 & {18} & {25} & {32} & {40} & {53} \end{array}\right\rbrack \) ;\n\n\( > > \mathrm{y} = \left\lbrack \begin{array}{lllllll} {32} & {28} & {23} & {20} & {19} & {18} & 9 \end{array}\right\rbrack... | Yes |
Example 4.3.3. Plot binomial distributions for \( n = {10},{25} \), and 50 trials with probability of success \( p = {0.8} \) . What happens to the shape of the distribution as \( n \) increases? | Solution. The function binopdf \( \left( {\mathrm{x},\mathrm{n},\mathrm{p}}\right) \) gives the probability of \( x \) successes in \( n \) trials of a binomial experiment with a probability of success \( p \) on each trial. Load the statistics package to access this function. The distributions can be plotted with the ... | No |
For our random walk example, find the probability vector after five steps for each of these initial probability vectors: | Solution. We first form an array that records the probability of moving between positions.\n\n<table><thead><tr><th colspan=\ | No |
Find an equilibrium vector for the Markov chain with transition matrix\n\n\[ T = \left\lbrack \begin{array}{lll} {0.48} & {0.51} & {0.14} \\ {0.29} & {0.04} & {0.52} \\ {0.23} & {0.45} & {0.34} \end{array}\right\rbrack \] | Solution.\n\n\[ \begin{array}{l} > > \mathrm{T} = \left\lbrack \begin{array}{lllllllll} {0.48} & {0.51} & {0.14}; & {0.29} & {0.04} & {0.52}; & {0.23} & {0.45} & {0.34} \end{array}\right\rbrack \\ \mathrm{T} = \end{array} \]\n\n\[ \begin{array}{lll} {0.480000} & {0.510000} & {0.140000} \end{array} \]\n\n\[ \text{0.2900... | Yes |
Theorem 5.3.2. If \( A \) is an \( n \times n \) diagonalizable matrix and \( A = {S\Lambda }{S}^{-1} \) and \( k \) is a positive integer, then | \[ {A}^{k} = S{\Lambda }^{k}{S}^{-1} = S\left\lbrack \begin{array}{llll} {\lambda }_{1}^{k} & & & \\ & {\lambda }_{2}^{k} & & \\ & & \ddots & \\ & & & {\lambda }_{n}^{k} \end{array}\right\rbrack {S}^{-1} \] | Yes |
Example 5.3.3. Let \( A = \left\lbrack \begin{array}{rr} 7 & 8 \\ - 4 & - 5 \end{array}\right\rbrack \) . Find \( {A}^{100} \) . | Solution. Octave can solve such a problem easily.\n\n\[ \n> > \mathrm{A} = \left\lbrack \begin{array}{lll} 7 & 8 & \\ 5 & - 4 & - 5 \end{array}\right\rbrack \n\] \n\n\( \mathrm{A} = \)\n\n7\n\n\( - 4\; - 5 \)\n\n>> A^100\n\nans \( = \)\n\n\[ \n{1.0308}\mathrm{e} + {048}\;{1.0308}\mathrm{e} + {048} \]\n\n\[ \n- {5.1538}... | Yes |
Find an orthogonal diagonalization for \( A = \left\lbrack \begin{array}{rr} 2 & - 1 \\ - 1 & 2 \end{array}\right\rbrack \) . | Solution. \( A \) has eigenvalues 3 and 1 . The eigenvectors are \( \left\lbrack \begin{array}{r} 1 \\ - 1 \end{array}\right\rbrack \) and \( \left\lbrack \begin{array}{l} 1 \\ 1 \end{array}\right\rbrack \) . Notice that these are orthogonal. They are normalized by dividing by their length (both have length \( \sqrt{2}... | Yes |
Use Octave to orthogonally diagonalize \( A = \left\lbrack \begin{array}{rr} 3 & 3 \\ 3 & - 1 \end{array}\right\rbrack \) . | Solution. If an orthogonal diagonalization is possible, Octave will return the output of the eig(A) command in that format. This explains why Octave chooses normalized vectors that form an orthogonal set, when possible.\n\n\[ \begin{array}{l} > > \mathrm{A} = \left\lbrack \begin{array}{llll} 3 & 3; & 3 & - 1 \end{array... | Yes |
Theorem 5.4.1. Let \( A \) be an \( m \times n \) matrix. The square roots of the nonzero eigenvalues of \( {A}^{T}A \) and \( A{A}^{T} \) (they are the same) are called the singular values of \( A \), denoted \( {\sigma }_{1},{\sigma }_{2},\ldots ,{\sigma }_{r} \). Then \( A \) can be factored as \[ A = {U\sum }{V}^{T... | - \( U \) is \( m \times m \) and orthogonal\n- \( V \) is \( n \times n \) and orthogonal\n- \( \sum \) is \( m \times n \) and diagonal of the special form \[ \sum = \left\lbrack \begin{matrix} {\sigma }_{1} & & & & \vdots \\ & {\sigma }_{2} & & & 0 \\ & & \ddots & & \vdots \\ & & & {\sigma }_{r} & \\ \cdots & 0 & \c... | Yes |
Let \( A = \left\lbrack \begin{array}{rr} 4 & 4 \\ - 3 & 3 \end{array}\right\rbrack \) . Find the SVD via the simplified procedure outlined above, then compare to the results obtained using the Octave function svd. | Solution. We can readily verify that \( \operatorname{rank}\left( A\right) = 2 \), so the matrix should have two singular values.\n\n\[ \begin{array}{l} > > \mathrm{A} = \left\lbrack \begin{array}{llll} 4 & 4; & - 3 & 3 \end{array}\right\rbrack \\ \mathrm{A} = \end{array} \]\n\n\( \begin{array}{ll} 4 & 4 \end{array} \)... | Yes |
Find a linear equation of the form \( y = {ax} + b \) to model this data. | Solution. The given points yield a system \( A\mathbf{p} = \mathbf{y} \), with\n\n\[ A = \left\lbrack \begin{array}{rr} 5 & 1 \\ {10} & 1 \\ {12} & 1 \\ {18} & 1 \\ {21} & 1 \end{array}\right\rbrack ,\mathbf{p} = \left\lbrack \begin{array}{l} a \\ b \end{array}\right\rbrack ,\text{ and }\mathbf{y} = \left\lbrack \begin... | Yes |
Theorem 5.5.1. The Gram-Schmidt Process\n\nLet \( \\left\\{ {{\\mathbf{u}}_{1},{\\mathbf{u}}_{2},\\ldots ,{\\mathbf{u}}_{n}}\\right\\} \) be a linearly independent set. Then the following procedure will produce an orthogonal set \( \\left\\{ {{\\mathbf{v}}_{1},{\\mathbf{v}}_{2},\\ldots ,{\\mathbf{v}}_{n}}\\right\\} \) ... | \[ \n{\\mathbf{v}}_{1} = {\\mathbf{u}}_{1} \n\] \n\n\[ \n{\\mathbf{v}}_{2} = {\\mathbf{u}}_{2} - {\\operatorname{proj}}_{{\\mathbf{v}}_{1}}\\left( {\\mathbf{u}}_{2}\\right) \n\] \n\n\[ \n{\\mathbf{v}}_{3} = {\\mathbf{u}}_{3} - {\\operatorname{proj}}_{{\\mathbf{v}}_{1}}\\left( {\\mathbf{u}}_{3}\\right) - {\\operatorname... | Yes |
Theorem 5.5.3. Let \( A \) be a nonsingular square matrix. Then there exists an orthogonal matrix \( Q \) and an upper triangular matrix \( R \) such that \( A = {QR} \) . | Here’s how to find \( Q \) and \( R \) .\n\n1. Apply the Gram-Schmidt process to the columns of \( A \) . Use the resulting orthonormal vectors as columns of \( Q \) .\n\n2. Let \( R = \left\lbrack \begin{matrix} {\mathbf{q}}_{1} \cdot {\mathbf{a}}_{1} & {\mathbf{q}}_{1} \cdot {\mathbf{a}}_{2} & {\mathbf{q}}_{1} \cdot ... | Yes |
Theorem 5.5.5. The \( {QR} \) ALGORITHM\n\nLet \( A \) be an \( n \times n \) matrix with \( n \) real eigenvalues.\n\nSet \( {A}_{1} = A \) .\n\nFor each \( k = 1,2,3,\ldots \) do the following:\n\n(i) Find the \( {QR} \) decomposition of \( {A}_{k},{A}_{k} = {Q}_{k}{R}_{k} \) .\n\n(ii) Set \( {A}_{k + 1} = {R}_{k}{Q}... | As \( k \) increases, the matrices \( {A}_{k} \) approach an upper triangular form with the eigenvalues of \( A \) on the diagonal. | Yes |
Example 5.5.6. Apply three iterations of the \( {QR} \) algorithm to the matrix \( A = \left\lbrack \begin{array}{rrr} 5 & 7 & 0 \\ {10} & 8 & 0 \\ 5 & 6 & - 5 \end{array}\right\rbrack \) . | Solution. We will use the built-in \( {QR} \) -decomposition function, \( \left\lbrack {\mathrm{Q}\mathrm{R}}\right\rbrack = \mathrm{{qr}}\left( \mathrm{A}\right) \) .\n\n\( > > \mathrm{A}1 = \mathrm{A} \)\n\n\( \mathrm{A}1 = \)\n\n\[ \begin{array}{rrr} 5 & 7 & 0 \\ {10} & 8 & 0 \\ 5 & 6 & - 5 \end{array} \]\n\n\( > > ... | Yes |
Graph the surface \( f\left( {x, y}\right) = \left( {1 + {xy}}\right) \left( {x + y}\right) \) . | Solution. We begin using the same basic procedure outlined above, this time choosing to plot the function over \( \left\lbrack {-5,5}\right\rbrack \times \left\lbrack {-5,5}\right\rbrack \) . \n\n\( > > \% \) define the domain\n\n\( > > \mathrm{x} = \operatorname{linspace}\left( {-5,5,{30}}\right) \) ;\n\n\( > > \mathr... | Yes |
Graph the surface \( f\left( {x, y}\right) = \sqrt{9 - {x}^{2} - {y}^{2}} \) . | Solution. The function corresponds to the upper half of a radius-3 sphere. If we naively attempt to plot the function over \( \left\lbrack {-3,3}\right\rbrack \times \left\lbrack {-3,3}\right\rbrack \), we will run into trouble:\n\n\( > > \mathrm{x} = \operatorname{linspace}\left( {-3,3,{30}}\right) \) ;\n\n\( > > \mat... | Yes |
Graph the function \( f\left( x\right) = \ln \left( {x + y - 1}\right) \) | Solution. This function is defined only on the half-plane \( x + y > 1 \) . To graph such a function in Octave we must make a suitable change of variables. In this case, the domain of the function suggests the substitution \( u = x + y \) . We first create a \( {ux} \) -meshgrid, where \( u > 1 \) . Then, using this ch... | Yes |
Example 6.2.4. The function\n\n\[ \rho = 1 + \frac{1}{4}\sin \left( {5\phi }\right) \cos \left( {6\theta }\right) ,0 \leq \theta \leq {2\pi },0 \leq \phi \leq \pi \]\n\nin spherical coordinates is known as a bumpy sphere. Graph this function. | Solution. We use a \( {\theta \phi } \) -meshgrid to calculate \( \rho \) . Then we can calculate \( x, y \), and \( z \) using the standard spherical to rectangular coordinate identities.\n\n\( > > \% \) define phi (P) and theta (T)\n\n\( > > \) theta \( = \) linspace \( \left( {0,2 * \mathrm{{pi}},{30}}\right) \) ;\n... | Yes |
The curve in Figure 6.2 lies on the surface of a torus, defined parametrically as\n\n\[ x = \\left( {5 + \\cos \\left( u\\right) }\\right) \\cos \\left( v\\right) \]\n\n\[ y = \\left( {5 + \\cos \\left( u\\right) }\\right) \\sin \\left( v\\right) \]\n\n\[ z = \\sin \\left( u\\right) \]\n\nwhere \( u, v \\in \\left\\lbr... | Solution. We begin by defining the parameters.\n\n\( > > \\mathrm{u} = \) linspace \( \\left( {0,2 * \\mathrm{{pi}},{25}}\\right) \) ;\n\n\( > > \\mathrm{v} = \\mathrm{u} \)\n\n\( > > \\left\\lbrack {UV}\\right\\rbrack = \\operatorname{meshgrid}\\left( {u, v}\\right) \) ;\n\nCalculate \( x, y \), and \( z \) :\n\n\( > ... | Yes |
Graph the solid obtained by rotating \( f\left( x\right) = {x}^{2} - {4x} + 5 \), for \( 1 \leq x \leq 4 \), about the \( x \) -axis. | Solution. These commands will graph the surface.\n\n\( \gg \mathrm{x} = \operatorname{linspace}\left( {1,4,{25}}\right) \) ; \( \% \) define the domain\n\n\( > > \mathrm{f} = @\left( \mathrm{x}\right) \mathrm{x} \cdot 2 - 4 * \mathrm{x} + 5 \) ; \( \% \) define the function\n\n\( > > \mathrm{t} = \operatorname{linspace... | Yes |
Evaluate\n\n\[ \n{\iint }_{R}\left( {{x}^{2}y + {y}^{2}x}\right) {dA} \n\]\n\nover the region \( R \) bounded by the graphs of \( y = {x}^{2} \) and \( y = \sqrt{x} \) . | Solution. An analysis of the region of integration (Figure 6.13) shows that we need to evaluate the following iterated integral:\n\n\[ \n{\int }_{0}^{1}{\int }_{{x}^{2}}^{\sqrt{x}}\left( {{x}^{2}y + {y}^{2}x}\right) {dydx} \n\]\n\nWe need to evaluate over only part of the rectangle \( \left\lbrack {0,1}\right\rbrack \t... | Yes |
Use the change of variable formulas in Equations 6.4-6.6 to evaluate\n\n\\[ \n{\\int }_{0}^{1}{\\int }_{{x}^{2}}^{\\sqrt{x}}\\left( {{x}^{2}y + {y}^{2}x}\\right) {dydx} \n\\] | Solution.\n\n\\( > > \\mathrm{f}1 = @\\left( {\\mathrm{x},\\mathrm{y}}\\right) \\mathrm{x} \\cdot \\uparrow 2 \\cdot * \\mathrm{y} + \\mathrm{y} \\cdot \\uparrow 2 \\cdot * \\mathrm{x} \\) ;\n\n\\( > > \\mathrm{y}1 = @\\left( \\mathrm{x}\\right) \\mathrm{x} \\) . ^ 2;\n\n\\( > > \\mathrm{y}2 = @\\left( \\mathrm{x}\\rig... | Yes |
Write an Octave function file that computes\n\n\\[ \n{\\iint }_{R}f\\left( {x, y}\\right) {dA} \n\\]\n\nover the region \\( R \\) bounded by the graphs of \\( y = {y}_{1}\\left( x\\right), y = {y}_{2}\\left( x\\right), x = a \\), and \\( x = b \\), using the change of variables in Equations 6.4-6.6. | Solution. A function file is similar to a script; it is a plain text .m-file containing a series of Octave commands. To be recognized as a function file, the first line of code (excluding comments and white space) must be function. With the file placed in the load path, it can then be run from the command line like any... | Yes |
Example 6.4.1. Graph the vector field \( \mathbf{F}\left( {x, y}\right) = \langle - x, y\rangle \) . | Solution.\n\n\( > > \mathrm{x} = \operatorname{linspace}\left( {-2,2,{10}}\right) \) ;\n\n\( > > \mathrm{y} = \mathrm{x} \)\n\n\( > > \left\lbrack {XY}\right\rbrack = \operatorname{meshgrid}\left( {x, y}\right) \) ;\n\n\( > > \) quiver \( \left( {\mathrm{X},\mathrm{Y}, - \mathrm{X},\mathrm{Y}}\right) \) ;\n\n\( > > \) ... | Yes |
Example 6.4.2. Plot the vector field \( \mathbf{F}\left( {x, y, z}\right) = \langle 1,1, z\rangle \) . | Solution.\n\n\( > > \mathrm{x} = \operatorname{linspace}\left( {-3,3,{10}}\right) \) ;\n\n\( > > \mathrm{y} = \mathrm{x} \)\n\n\( > > \mathrm{z} = \mathrm{x} \)\n\n\( > > \left\lbrack {XYZ}\right\rbrack = \operatorname{meshgrid}\left( {x, y, z}\right) \) ;\n\n\( \gg \) quiver \( 3\left( {\mathrm{X},\mathrm{Y},\mathrm{Z... | Yes |
Plot the slope field along with several solutions of the differential equation\n\n\\[ \n\\frac{dy}{dx} = x \n\\] | Solution. The solution is \\( y = \\frac{1}{2}{x}^{2} + C \\) . For differential equations that cannot be solved so easily, plotting the slope field can be used to get a sense of the solutions. In this example, since we know the solution, we can show how the solutions follow the slope field.\n\nWe need to define the in... | Yes |
Solve\n\n\[ \n{y}^{\prime } = {e}^{-{3x}} - {3y},\;y\left( 0\right) = 1 \n\]\n\non \( \left\lbrack {0,3}\right\rbrack \) using a step size of 1 . | Solution. We will generate a series of approximate \( y \) -values at \( x = 0,1,2,3 \) . The value \( {y}_{0} \) is given. We compute the remaining values using Equation 6.7. Here is the first step:\n\n\[ \n{y}_{1} = {y}_{0} + {hf}\left( {{x}_{0},{y}_{0}}\right) \n\]\n\n\[ \n= 1 + \left( 1\right) f\left( {0,1}\right) ... | Yes |
Example 6.5.3. Solve\n\n\[ \n{y}^{\prime } = {e}^{-{3x}} - {3y},\;y\left( 0\right) = 1 \n\] \non \( \left\lbrack {0,3}\right\rbrack \) using a step size of 0.1 . | Solution. We will write a fairly general Octave script that can be easily modified for different functions, intervals, and step sizes.\n\nOctave Script 6.4: Euler's method\n\n\( 1\% \) Euler’s method solution for\n\n\( \% \;\mathrm{{dy}}/\mathrm{{dx}} = {\mathrm{e}}^{ \times }\left( {-3\mathrm{x}}\right) - 3\mathrm{y},... | Yes |
Use lsode to solve the differential equation\n\n\\[ \n\\frac{dx}{dt} = x\\left( {{t}^{2} + 1}\\right) \n\\] \n\non \\( \\left\\lbrack {0,2}\\right\\rbrack \\), with initial condition \\( x\\left( 0\\right) = 1 \\) . | Solution. To solve using lsode, we define the function listing \\( x \\) first, then \\( t \\) .\n\n\\( \\gg \\% \\) define the function, input values, and initial condition\n\n\\( > > \\mathrm{f} = @\\left( {\\mathrm{x},\\mathrm{t}}\\right) \\mathrm{x} \\cdot * \\left( {\\mathrm{t} \\cdot {}^{ \\land }2 + 1}\\right) ;... | Yes |
Use ode45 to solve the differential equation\n\n\[ \n\\frac{dx}{dt} = x\\left( {{t}^{2} + 1}\\right) \n\]\n\non \( \\left\\lbrack {0,2}\\right\\rbrack \), with initial condition \( x\\left( 0\\right) = 1 \) . Compare to the lsode solution from Example 6.5.4. | Solution. We need to redefine the function. MATLAB convention requires giving the independent variable first, the opposite of what lsode required.\n\n\( > > \\% \) define the function, input values, and initial condition\n\n\( > > \\mathrm{f} = @\\left( {\\mathrm{t},\\mathrm{x}}\\right) \\cdot \\mathrm{x} \\cdot * \\le... | Yes |
Find the general solution for \( {y}^{\prime } = {e}^{-{3x}} - {3y} \) . Then, find the particular solution if \( y\left( 0\right) = 1 \) . | Solution. First, the symbolic package must be installed and loaded. Refer to Section 3.4 for details.\n\nTo set things up, we declare \( y \) as a symbolic function of \( x \) .\n\n\( > > \) pkg load symbolic\n\n\( > > \) syms \( y\left( x\right) \)\n\nNow, define the differential equation. We do this using the equalit... | No |
Example 1.1.1. A sphere of solid gold has a mass of 100 kg and the density of gold is \( {19.3}\mathrm{\;g}/{\mathrm{{cm}}}^{3} \) . What is the radius of the sphere? | Solution. This problem is more involved. To answer this, let \( r \) be the unknown radius of the sphere in units of cm. The volume of the sphere is \( V = \frac{4}{3}\pi {r}^{3} \) . Since the sphere is solid gold, the density of gold is the ratio\n\n\[ \text{ density of gold } = \frac{\text{ mass of sphere }}{\text{ ... | Yes |
Example 1.2.1. A water pipe mounted to the ceiling has a leak. It is dripping onto the floor below and creates a circular puddle of water. The surface area of this puddle is increasing at a constant rate of 4 cm²/hour. Find the surface area and dimensions of the puddle after 84 minutes. | Solution. The quantity changing is \ | No |
Example 2.2.1. As the marketing director of Turboweb software, you have been asked to deliver a brief message at the annual stockholders meeting on the performance of your product. Your staff has assembled this tabular collection of data; how can you convey the content of this table most clearly? | One idea is to simply flash an overhead slide of this data to the audience; this can be deadly! A better idea is to use a visual aid. Suppose we let the variable \( x \) represent the week and the variable \( y \) represent the gross sales (in thousands of dollars) in week \( x \) . We can then plot the points \( \left... | No |
Example 2.3.1. Return to the tossed ball scenario on page 1. How do we decide where to draw a coordinate system in the picture? | Figure 2.7 on page 16 shows four natural choices of \( {xy} \) -coordinate system. To choose a coordinate system we must specify the origin. The four logical choices for the origin are either the top of the cliff, the bottom of the cliff, the launch point of the ball or the landing point of the ball. So, which choice d... | Yes |
Example 2.3.2. Michael and Aaron are running toward each other, beginning at opposite ends of a \( {10},{000}\mathrm{{ft}} \) . airport runway, as pictured in Figure 2.8 on page 17. Where and when will these guys collide? | Solution. This problem requires that we find the \ | No |
Example 2.4.1. You are in an airplane flying from Denver to New York. How far will you fly? To what extent will you travel north? To what extent will you travel east? | Consider two points \( P = \left( {{x}_{1},{y}_{1}}\right) \) and \( Q = \left( {{x}_{2},{y}_{2}}\right) \) in the \( {xy} \) coordinate system, where we assume that the units on each axis are the same; for example, both in units of \ | No |
Two cars depart from a four way intersection at the same time, one heading East and the other heading North. Both cars are traveling at the constant speed of 30 ft/sec. Find the distance (in miles) between the two cars after 1 hour 12 minutes. In addition, determine when the two cars would be exactly 1 mile apart. | Solution. Begin with a picture of the situation. We have indicated the locations of the two vehicles after \( t \) seconds and the distance \( d \) between them at time \( t \) . By the distance formula, the distance between them is\n\n\[ d = \sqrt{{\left( a - 0\right) }^{2} + {\left( 0 - b\right) }^{2}} \]\n\n\[ = \sq... | Yes |
(a) If \( x \neq 1 \) ,\n\n\[ \n\frac{{x}^{2} - 1}{x + 1} = \frac{{x}^{2} + \left( {-1}\right) 1}{x + 1} \n\] | \[ \n= \frac{{x}^{2}}{x} + \frac{-1}{1} \n\]\n\n\[ \n= x - 1 \n\] | Yes |
Glo-Tek Industries has designed a new halogen street light fixture for the city of Seattle. According to the product literature, when placed on a 50 foot light pole, the resulting useful illuminated area is a circular disc 120 feet in diameter. Assume the light pole is located 20 feet east and 40 feet north of the inte... | The illuminated area is a circular disc whose diameter and center are both known. Consequently, we really need to study the intersection of this circle with the two streets. Begin by imposing the pictured coordinate system; we will use units of feet for each axis. The illuminated region will be a circular disc centered... | Yes |
Problem 3.8. (a) Solve for \( x \) : | \[ \frac{{x}^{2} - {2x} + 1}{x + 5} = x - 2 \] | No |
Example 4.3.3. Consider the line \( \ell \) , in Figure 4.6, through the two points \( \mathrm{P} = \left( {1,1}\right) \) and \( \mathrm{Q} = \left( {4,5}\right) \) . Then the slope of \( \ell \) is \( m = 4/3 \) and \( \ell \) consists of all pairs of points \( \left( {x, y}\right) \) such that the coordinates \( x \... | Letting \( x = 0,1,6 \) and -1, we conclude that the following four points lie on the line \( \ell : \left( {0,\frac{4}{3}\left( {0 - 1}\right) + 1}\right) = \) \( \left( {0,\frac{-1}{3}}\right) ,\left( {1,\frac{4}{3}\left( {1 - 1}\right) + 1}\right) = \left( {1,1}\right) ,\left( {6,\frac{4}{3}\left( {6 - 1}\right) + 1... | Yes |
The yearly resident tuition at the University of Washington was $1827 in 1989 and $2907 in 1995. Assume that the tuition growth at the UW follows a linear model. What will be the tuition in the year 2000? When will yearly tuition at the University of Washington be $10,000? | Solution. If we consider a coordinate system where the \( x \) -axis represents the year and the \( y \) -axis represents dollars, we are given two data points: \( P = \left( {{1989},1,{827}}\right) \) and \( Q = \left( {{1995},2,{907}}\right) \) . Using the two-point formula for the equation of line through \( P \) an... | Yes |
Let \( \ell \) be a line in the plane passing through the points \( \left( {1,1}\right) \) and \( \left( {6, - 1}\right) \). Find a linear equation whose graph is a line parallel to \( \ell \) passing through 5 on the y-axis. Find a linear equation whose graph is perpendicular to \( \ell \) and passes through \( \left(... | Solution. Letting \( P = \left( {1,1}\right) \) and \( Q = \left( {6, - 1}\right) \), apply the \ | No |
1. Find a linear equation whose graph is the line along which the crop duster travels. | Take \( Q = \left( {-1,0}\right) \) and \( S = \left( {{1.5}, - 2}\right) = \) duster spotting point. Construct a line through \( Q \) and \( S \) . The slope is \( - {0.8} = m \) and the line equation becomes:\n\n\[ y = - {0.8x} - {0.8} \] | Yes |
Example 4.11.2. Olga is running in the xy-plane, and the coordinate are given in meters (so, for example, the point \( \left( {1,0}\right) \) is one meter from the origin \( \left( {0,0}\right) ) \) . She runs in a straight line, starting at the point \( \left( {3,5}\right) \) and running along the line \( y = - \frac{... | Solution. This example differs in some respects from the last example. In particular, instead of knowing where the runner is at two points in time, we only know one point, and have other information given to us about the speed and path of the runner. One approach is to use this new information to find where the runner ... | Yes |
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