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The base of a solid is the region between \( f\left( x\right) = {x}^{2} - 1 \) and \( g\left( x\right) = - {x}^{2} + 1 \). Its cross-sections perpendicular to the \( x \) -axis are equilateral triangles. Find the volume of the solid. | Solution For any value of \( x \), a cross-section is a triangle with base \( 2\left( {1 - {x}^{2}}\right) \) and height \( \sqrt{3}\left( {1 - {x}^{2}}\right) \), so the area of the cross-section is\n\n\[ \frac{1}{2}\left( \text{ base }\right) \left( \text{ height }\right) = \left( {1 - {x}^{2}}\right) \sqrt{3}\left( ... | Yes |
Find the volume of a right circular cone with base radius 10 and height 20. | Solution We can view this cone as produced by the rotation of the line \( y = x/2 \) rotated about the \( x \) -axis, as indicated in Figure 15.5.\n\nAt a particular point on the \( x \) -axis, the radius of the resulting cone is the \( y \) -coordinate of the corresponding point on the line \( y = x/2 \) . The area of... | Yes |
Find the volume of the object generated when the area between \( f\left( x\right) = x \) and \( g\left( x\right) = {x}^{2} \) is rotated around the \( x \) -axis, see Figure 15.6. | This solid has a \ | No |
Example 15.1.5 Suppose the area under \( y = - {x}^{2} + 1 \) between \( x = 0 \) and \( x = 1 \) is rotated around the \( x \) -axis. | Solution We’ll just set up integrals for each method.\n\nDisk method: \( {\int }_{0}^{1}\pi {\left( 1 - {x}^{2}\right) }^{2}{dx} = \frac{8}{15}\pi \) .\n\nShell method: \( {\int }_{0}^{1}{2\pi y}\sqrt{1 - y}{dy} = \frac{8}{15}\pi \) . | Yes |
Example 15.2.1 Let \( f\left( x\right) = \sqrt{{r}^{2} - {x}^{2}} \), the upper half circle of radius \( r \) . The length of this curve is half the circumference, namely \( {\pi r} \) . Let’s compute this with the arc length formula. | The derivative \( {f}^{\prime } \) is \( - x/\sqrt{{r}^{2} - {x}^{2}} \) so the integral is\n\n\[ \n{\int }_{-r}^{r}\sqrt{1 + \frac{{x}^{2}}{{r}^{2} - {x}^{2}}}{dx} = {\int }_{-r}^{r}\sqrt{\frac{{r}^{2}}{{r}^{2} - {x}^{2}}}{dx} = r{\int }_{-r}^{r}\sqrt{\frac{1}{{r}^{2} - {x}^{2}}}{dx}. \n\]\n\nUsing a trigonometric sub... | Yes |
Example 1.2. Consider the vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{RS} \) in \( {\mathbb{R}}^{3} \), where \( P = \left( {2,1,5}\right), Q = \left( {3,5,7}\right), R = \left( {1, - 3, - 2}\right) \) and \( S = \left( {2,1,0}\right) \) . Does \( \overrightarrow{PQ} = \overrightarrow{RS} \) ? | Solution: The vector \( \overrightarrow{PQ} \) is equal to the vector \( \mathbf{v} \) with initial point \( \left( {0,0,0}\right) \) and terminal point \( Q - P = \left( {3,5,7}\right) - \left( {2,1,5}\right) = \left( {3 - 2,5 - 1,7 - 5}\right) = \left( {1,4,2}\right) . \n\nSimilarly, \( \overrightarrow{RS} \) is equa... | Yes |
Theorem 1.2. For a vector \( \mathbf{v} = \left( {a, b, c}\right) \) in \( {\mathbb{R}}^{3} \), the magnitude of \( \mathbf{v} \) is:\n\n\[ \parallel \mathbf{v}\parallel = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}} \] | Proof: There are four cases to consider:\n\nCase 1: \( a = b = c = 0 \) . Then \( \mathbf{v} = \mathbf{0} \), so \( \parallel \mathbf{v}\parallel = 0 = \sqrt{{0}^{2} + {0}^{2} + {0}^{2}} = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}} \) .\n\nCase 2: exactly two of \( a, b, c \) are 0 . Without loss of generality, we assume that ... | Yes |
Theorem 1.3. Let \( \mathbf{v} = \left( {{v}_{1},{v}_{2}}\right) ,\mathbf{w} = \left( {{w}_{1},{w}_{2}}\right) \) be vectors in \( {\mathbb{R}}^{2} \), and let \( k \) be a scalar. Then (a) \( k\mathbf{v} = \left( {k{v}_{1}, k{v}_{2}}\right) \) | Proof: (a) Without loss of generality, we assume that \( {v}_{1},{v}_{2} > 0 \) (the other possibilities are handled in a similar manner). If \( k = 0 \) then \( k\mathbf{v} = 0\mathbf{v} = \mathbf{0} = \left( {0,0}\right) = \left( {0{v}_{1},0{v}_{2}}\right) = \left( {k{v}_{1}, k{v}_{2}}\right) \), which is what we nee... | Yes |
Example 1.4. Let \( \\mathbf{v} = \\left( {2,1, - 1}\\right) \) and \( \\mathbf{w} = \\left( {3, - 4,2}\\right) \) in \( {\\mathbb{R}}^{3} \) .\n\n(a) Find \( \\mathbf{v} - \\mathbf{w} \) . | Solution: \( \\mathbf{v} - \\mathbf{w} = \\left( {2 - 3,1 - \\left( {-4}\\right) , - 1 - 2}\\right) = \\left( {-1,5, - 3}\\right) \) | Yes |
Theorem 1.6. Let \( \mathbf{v} \) , \( \mathbf{w} \) be nonzero vectors, and let \( \theta \) be the angle between them. Then\n\n\[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{\parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel } \] | Proof: We will prove the theorem for vectors in \( {\mathbb{R}}^{3} \) (the proof for \( {\mathbb{R}}^{2} \) is similar). Let \( \mathbf{v} = \) \( \left( {{v}_{1},{v}_{2},{v}_{3}}\right) \) and \( \mathbf{w} = \left( {{w}_{1},{w}_{2},{w}_{3}}\right) \) . By the Law of Cosines (see Figure 1.3.2), we have\n\n\[ {\begin{... | Yes |
Find the angle \( \theta \) between the vectors \( \mathbf{v} = \left( {2,1, - 1}\right) \) and \( \mathbf{w} = \left( {3, - 4,1}\right) \). | \( \textit{Solution: Since }\textbf{v} \cdot \textbf{w} = \left( 2\right) \left( 3\right) + \left( 1\right) \left( {-4}\right) + \left( {-1}\right) \left( 1\right) = 1,\;\| \textbf{v}\| = \sqrt{6},\; \) and \( \;\| \textbf{w}\| = \sqrt{26},\; \) then\n\n\[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{\parallel \mat... | Yes |
Corollary 1.8. If \( \theta \) is the angle between nonzero vectors \( \mathbf{v} \) and \( \mathbf{w} \), then\n\n\[ \mathbf{v} \cdot \mathbf{w}\text{ is }\left\{ \begin{array}{ll} > 0 & \text{ for }{0}^{ \circ } \leq \theta < {90}^{ \circ } \\ 0 & \text{ for }\theta = {90}^{ \circ } \\ < 0 & \text{ for }{90}^{ \circ ... | By Corollary 1.8, the dot product can be thought of as a way of telling if the angle between two vectors is acute, obtuse, or a right angle, depending on whether the dot product is positive, negative, or zero, respectively. See Figure 1.3.3. | No |
Example 1.6. Are the vectors \( \mathbf{v} = \left( {-1,5, - 2}\right) \) and \( \mathbf{w} = \left( {3,1,1}\right) \) perpendicular? | Solution: Yes, \( \mathbf{v} \bot \mathbf{w} \) since \( \mathbf{v} \cdot \mathbf{w} = \left( {-1}\right) \left( 3\right) + \left( 5\right) \left( 1\right) + \left( {-2}\right) \left( 1\right) = 0 \) . | Yes |
Theorem 1.9. For any vectors \( \mathbf{u},\mathbf{v},\mathbf{w} \), and scalar \( k \), we have\n\n(a) \( \mathbf{v} \cdot \mathbf{w} = \mathbf{w} \cdot \mathbf{v} \) Commutative Law\n\n(b) \( \left( {k\mathbf{v}}\right) \cdot \mathbf{w} = \mathbf{v} \cdot \left( {k\mathbf{w}}\right) = k\left( {\mathbf{v} \cdot \mathb... | Proof: The proofs of parts (a)-(e) are straightforward applications of the definition of the dot product, and are left to the reader as exercises. We will prove part (f).\n\n(f) If either \( \mathbf{v} = \mathbf{0} \) or \( \mathbf{w} = \mathbf{0} \), then \( \mathbf{v} \cdot \mathbf{w} = 0 \) by part (c), and so the i... | No |
Theorem 1.10. For any vectors \( \\mathbf{v},\\mathbf{w} \), we have\n\n(a) \( \\parallel \\mathbf{v}{\\parallel }^{2} = \\mathbf{v} \\cdot \\mathbf{v} \)\n\n(b) \( \\parallel \\mathbf{v} + \\mathbf{w}\\parallel \\leq \\parallel \\mathbf{v}\\parallel + \\parallel \\mathbf{w}\\parallel \\; \) Triangle Inequality\n\n(c) ... | (a) Left as an exercise for the reader.\n\n(b) By part (a) and Theorem 1.9, we have\n\n\\[ \n{\\left\\| \\mathbf{v} + \\mathbf{w}\\right\\| }^{2} = \\left( {\\mathbf{v} + \\mathbf{w}}\\right) \\cdot \\left( {\\mathbf{v} + \\mathbf{w}}\\right) = \\mathbf{v} \\cdot \\mathbf{v} + \\mathbf{v} \\cdot \\mathbf{w} + \\mathbf{... | No |
Find \( \mathbf{i} \times \mathbf{j} \) . | Solution: Since \( \mathbf{i} = \left( {1,0,0}\right) \) and \( \mathbf{j} = \left( {0,1,0}\right) \), then\n\n\[ \mathbf{i} \times \mathbf{j} = \left( {\left( 0\right) \left( 0\right) - \left( 0\right) \left( 1\right) ,\left( 0\right) \left( 0\right) - \left( 1\right) \left( 0\right) ,\left( 1\right) \left( 1\right) -... | Yes |
Theorem 1.11. If the cross product \( \mathbf{v} \times \mathbf{w} \) of two nonzero vectors \( \mathbf{v} \) and \( \mathbf{w} \) is also a nonzero vector, then it is perpendicular to both \( \mathbf{v} \) and \( \mathbf{w} \) . | Proof: We will show that \( \left( {\mathbf{v} \times \mathbf{w}}\right) \cdot \mathbf{v} = 0 \) :\n\n\[ \left( {\mathbf{v} \times \mathbf{w}}\right) \cdot \mathbf{v} = \left( {{v}_{2}{w}_{3} - {v}_{3}{w}_{2},{v}_{3}{w}_{1} - {v}_{1}{w}_{3},{v}_{1}{w}_{2} - {v}_{2}{w}_{1}}\right) \cdot \left( {{v}_{1},{v}_{2},{v}_{3}}\... | Yes |
Corollary 1.12. If the cross product \( \mathbf{v} \times \mathbf{w} \) of two nonzero vectors \( \mathbf{v} \) and \( \mathbf{w} \) is also a nonzero vector, then it is perpendicular to the span of \( \mathbf{v} \) and \( \mathbf{w} \) . | The span of any two nonzero, nonparallel vectors \( \mathbf{v},\mathbf{w} \) in \( {\mathbb{R}}^{3} \) is a plane \( P \), so the above corollary shows that \( \mathbf{v} \times \mathbf{w} \) is perpendicular to that plane. As shown in Figure 1.4.2, there are two possible directions for \( \mathbf{v} \times \mathbf{w} ... | Yes |
Let \( \bigtriangleup {PQR} \) and \( {PQRS} \) be a triangle and parallelogram, respectively, as shown in Figure 1.4.3. Think of the triangle as existing in \( {\mathbb{R}}^{3} \), and identify the sides \( {QR} \) and \( {QP} \) with vectors \( \mathbf{v} \) and \( \mathbf{w} \), respectively, in \( {\mathbb{R}}^{3} ... | \[ b = \parallel \mathbf{v}\parallel \;\text{ and }\;h = \parallel \mathbf{w}\parallel \sin \theta \] \[ {A}_{PQR} = \frac{1}{2}\parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel \sin \theta \] \[ = \frac{1}{2}\parallel \mathbf{v} \times \mathbf{w}\parallel \] So since the area \( {A}_{PQRS} \) of the parallel... | Yes |
Theorem 1.13. Area of triangles and parallelograms\n\n(a) The area \( A \) of a triangle with adjacent sides \( \mathbf{v},\mathbf{w} \) (as vectors in \( {\mathbb{R}}^{3} \) ) is:\n\n\[ A = \frac{1}{2}\parallel \mathbf{v} \times \mathbf{w}\parallel \]\n\n(b) The area \( A \) of a parallelogram with adjacent sides \( \... | It may seem at first glance that since the formulas derived in Example 1.8 were for the adjacent sides \( {QP} \) and \( {QR} \) only, then the more general statements in Theorem 1.13 that the formulas hold for any adjacent sides are not justified. We would get a different formula for the area if we had picked \( {PQ} ... | No |
Calculate the area of the triangle \( \bigtriangleup {PQR} \), where \( P = \left( {2,4, - 7}\right), Q = \left( {3,7,{18}}\right) \) , and \( R = \left( {-5,{12},8}\right) \) . | Solution: Let \( \mathbf{v} = \overrightarrow{PQ} \) and \( \mathbf{w} = \overrightarrow{PR} \), as in Figure 1.4.4. Then\n\n\n\nFigure 1.4.4\n\n\( \mathbf{v} = \left( {3,7,{18}}\right) - \left( {2,4, - 7}\right) = \le... | Yes |
Calculate the area of the parallelogram PQRS, where \( P = \left( {1,1}\right), Q = \left( {2,3}\right) \) , \( R = \left( {5,4}\right) \), and \( S = \left( {4,2}\right) \) . | Solution: Let \( \mathbf{v} = \overrightarrow{SP} \) and \( \mathbf{w} = \overrightarrow{SR} \), as in Figure 1.4.5. Then\n\n\n\nFigure 1.4.5\n\n\( \mathbf{v} = \left( {1,1}\right) - \left( {4,2}\right) = \left( {-3, -... | Yes |
Theorem 1.14. For any vectors \( \mathbf{u},\mathbf{v},\mathbf{w} \) in \( {\mathbb{R}}^{3} \), and scalar \( k \), we have\n\n(a) \( \mathbf{v} \times \mathbf{w} = - \mathbf{w} \times \mathbf{v} \) Anticommutative Law | (a) By the definition of the cross product and scalar multipli-\ncation, we have:\n\n\[ \mathbf{v} \times \mathbf{w} = \left( {{v}_{2}{w}_{3} - {v}_{3}{w}_{2},{v}_{3}{w}_{1} - {v}_{1}{w}_{3},{v}_{1}{w}_{2} - {v}_{2}{w}_{1}}\right) \]\n\n\[ = - \left( {{v}_{3}{w}_{2} - {v}_{2}{w}_{3},{v}_{1}{w}_{3} - {v}_{3}{w}_{1},{v}_... | Yes |
Volume of a parallelepiped: Let the vectors \( \mathbf{u},\mathbf{v} \) , \( \mathbf{w} \) in \( {\mathbb{R}}^{3} \) represent adjacent sides of a parallelepiped \( P \), with \( \mathbf{u},\mathbf{v} \) , \( \mathbf{w} \) forming a right-handed system, as in Figure 1.4.7. Show that the volume of \( P \) is the scalar ... | Solution: Recall that the volume \( \operatorname{vol}\left( P\right) \) of a parallelepiped \( P \) is the area \( A \) of the base parallelogram times the height \( h \) . By Theorem 1.13(b), the area \( A \) of the base parallelogram is \( \parallel \mathbf{v} \times \mathbf{w}\parallel \) . And we can see that sinc... | Yes |
Find \( \mathbf{u} \times \left( {\mathbf{v} \times \mathbf{w}}\right) \) for \( \mathbf{u} = \left( {1,2,4}\right) ,\mathbf{v} = \left( {2,2,0}\right) ,\mathbf{w} = \left( {1,3,0}\right) \). | Solution: Since \( \mathbf{u} \cdot \mathbf{v} = 6 \) and \( \mathbf{u} \cdot \mathbf{w} = 7 \), then\n\n\[ \mathbf{u} \times \left( {\mathbf{v} \times \mathbf{w}}\right) = \left( {\mathbf{u} \cdot \mathbf{w}}\right) \mathbf{v} - \left( {\mathbf{u} \cdot \mathbf{v}}\right) \mathbf{w} \]\n\n\[ = 7\left( {2,2,0}\right) -... | Yes |
\[ \left| \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right| = \left( 1\right) \left( 4\right) - \left( 2\right) \left( 3\right) = 4 - 6 = - 2 \] | \[ \left| \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right| = \left( 1\right) \left( 4\right) - \left( 2\right) \left( 3\right) = 4 - 6 = - 2 \] | Yes |
Let \( \mathbf{v} = 4\mathbf{i} - \mathbf{j} + 3\mathbf{k} \) and \( \mathbf{w} = \mathbf{i} + 2\mathbf{k} \) . Then | \[ \mathbf{v} \times \mathbf{w} = \left| \begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & - 1 & 3 \\ 1 & 0 & 2 \end{array}\right| = \left| \begin{array}{rr} - 1 & 3 \\ 0 & 2 \end{array}\right| \mathbf{i} - \left| \begin{array}{rr} 4 & 3 \\ 1 & 2 \end{array}\right| \mathbf{j} + \left| \begin{array}{rr} 4 &... | Yes |
Find the volume of the parallelepiped with adjacent sides \( \mathbf{u} = \left( {2,1,3}\right) ,\mathbf{v} = \) \( \left( {-1,3,2}\right) ,\mathbf{w} = \left( {1,1, - 2}\right) \) | Solution: By Theorem 1.15, the volume \( \operatorname{vol}\left( P\right) \) of the parallelepiped\n\n\n\nFigure 1.4.9 \( P \)\n\n\( P \) is the absolute value of the scalar triple product of the three adjacent sides ... | Yes |
Example 1.18. Prove: \( \left( {\mathbf{u} \times \mathbf{v}}\right) \cdot \left( {\mathbf{w} \times \mathbf{z}}\right) = \left| \begin{matrix} \mathbf{u} \cdot \mathbf{w} & \mathbf{u} \cdot \mathbf{z} \\ \mathbf{v} \cdot \mathbf{w} & \mathbf{v} \cdot \mathbf{z} \end{matrix}\right| \) for all vectors \( \;\mathbf{u},\;... | Solution: Let \( \mathbf{x} = \mathbf{u} \times \mathbf{v} \) . Then\n\n\[ \left( {\mathbf{u} \times \mathbf{v}}\right) \cdot \left( {\mathbf{w} \times \mathbf{z}}\right) = \mathbf{x} \cdot \left( {\mathbf{w} \times \mathbf{z}}\right) \]\n\n\[ = \mathbf{w} \cdot \left( {\mathbf{z} \times \mathbf{x}}\right) \;\text{(by ... | Yes |
Write the line \( L \) through the point \( P = \left( {2,3,5}\right) \) and parallel to the vector \( \mathbf{v} = \left( {4, - 1,6}\right) \), in the following forms: (a) vector,(b) parametric,(c) symmetric. Lastly: (d) find two points on \( L \) distinct from \( P \) . | Solution: (a) Let \( \mathbf{r} = \left( {2,3,5}\right) \) . Then by formula (1.16), \( L \) is given by:\n\n\[ \mathbf{r} + t\mathbf{v} = \left( {2,3,5}\right) + t\left( {4, - 1,6}\right) ,\text{ for } - \infty < t < \infty \]\n\n(b) \( L \) consists of the points \( \left( {x, y, z}\right) \) such that\n\n\[ x = 2 + ... | Yes |
Write the line \( L \) through the points \( {P}_{1} = \left( {-3,1, - 4}\right) \) and \( {P}_{2} = \left( {4,4, - 6}\right) \) in parametric form. | Solution: By formula (1.21), \( L \) consists of the points \( \left( {x, y, z}\right) \) such that\n\n\[ \nx = - 3 + {7t},\;y = 1 + {3t},\;z = - 4 - {2t},\text{ for } - \infty < t < \infty \n\] | Yes |
Find the distance \( d \) from the point \( P = \left( {1,1,1}\right) \) to the line \( L \) in Example 1.20. | Solution: From Example 1.20, we see that we can represent \( L \) in vector form as: \( \mathbf{r} + t\mathbf{v} \), for \( \mathbf{r} = \left( {-3,1, - 4}\right) \) and \( \mathbf{v} = \left( {7,3, - 2}\right) \) . Since the point \( Q = \left( {-3,1, - 4}\right) \) is on \( L \), then for \( \mathbf{w} = \overrightar... | Yes |
Find the point of intersection (if any) of the following lines:\n\n\\[ \n\\frac{x + 1}{3} = \\frac{y - 2}{2} = \\frac{z - 1}{-1}\\text{ and }x + 3 = \\frac{y - 8}{-3} = \\frac{z + 3}{2} \n\\] | Solution: First we write the lines in parametric form, with parameters \\( s \\) and \\( t \\) :\n\n\\[ \nx = - 1 + {3s},\\;y = 2 + {2s},\\;z = 1 - s\\;\\text{and}\\,x = - 3 + t,\\;y = 8 - {3t},\\;z = - 3 + {2t} \n\\]\n\nThe lines intersect when \\( \\left( {-1 + {3s},2 + {2s},1 - s}\\right) = \\left( {-3 + t,8 - {3t},... | Yes |
Find the equation of the plane \( P \) containing the point \( \left( {-3,1,3}\right) \) and perpendicular to the vector \( \mathbf{n} = \left( {2,4,8}\right) \) . | By formula (1.25), the plane \( P \) consists of all points \( \left( {x, y, z}\right) \) such that:\n\n\[ 2\left( {x + 3}\right) + 4\left( {y - 1}\right) + 8\left( {z - 3}\right) = 0 \]\n\nIf we multiply out the terms in formula (1.25) and combine the constant terms, we get an equation of the plane in normal form:\n\n... | Yes |
Find the equation of the plane \( P \) containing the points \( \\left( {2,1,3}\\right) ,\\left( {1, - 1,2}\\right) \) and \( \\left( {3,2,1}\\right) \) . | Let \( Q = \\left( {2,1,3}\\right), R = \\left( {1, - 1,2}\\right) \) and \( S = \\left( {3,2,1}\\right) \) . Then for the vectors \( \\overrightarrow{QR} = \\left( {-1, - 2, - 1}\\right) \) and \( \\overrightarrow{QS} = \\left( {1,1, - 2}\\right) \), the plane \( P \) has a normal vector\n\n\[ \n\\mathbf{n} = \\overri... | Yes |
Theorem 1.19. Let \( Q = \left( {{x}_{0},{y}_{0},{z}_{0}}\right) \) be a point in \( {\mathbb{R}}^{3} \), and let \( P \) be a plane with normal form \( {ax} + {by} + {cz} + d = 0 \) that does not contain \( Q \) . Then the distance \( D \) from \( Q \) to \( P \) is:\n\n\[ D = \frac{\left| a{x}_{0} + b{y}_{0} + c{z}_{... | Proof: Let \( R = \left( {x, y, z}\right) \) be any point in the plane \( P \) (so that \( {ax} + {by} + {cz} + d = 0 \) ) and let \( \mathbf{r} = \overrightarrow{RQ} = \left( {{x}_{0} - x,{y}_{0} - y,{z}_{0} - z}\right) \) . Then \( \mathbf{r} \neq \mathbf{0} \) since \( Q \) does not lie in \( P \) . From the normal ... | Yes |
Find the distance \( D \) from \( \left( {2,4, - 5}\right) \) to the plane from Example 1.24. | Solution: Recall that the plane is given by \( {5x} - {3y} + z - {10} = 0 \) . So\n\n\[ D = \frac{\left| 5\left( 2\right) - 3\left( 4\right) + 1\left( -5\right) - {10}\right| }{\sqrt{{5}^{2} + {\left( -3\right) }^{2} + {1}^{2}}} = \frac{\left| -{17}\right| }{\sqrt{35}} = \frac{17}{\sqrt{35}} \approx {2.87} \] | Yes |
Find the line of intersection \( L \) of the planes \( {5x} - {3y} + z - {10} = 0 \) and \( {2x} + {4y} - z + 3 = 0 \) . | The plane \( {5x} - {3y} + z - {10} = 0 \) has normal vector \( {\mathbf{n}}_{1} = \left( {5, - 3,1}\right) \) and the plane \( {2x} + {4y} - z + 3 = 0 \) has normal vector \( {\mathbf{n}}_{2} = \left( {2,4, - 1}\right) \) . Since \( {\mathbf{n}}_{1} \) and \( {\mathbf{n}}_{2} \) are not scalar multiples, then the two ... | Yes |
Find the intersection of the sphere \( {x}^{2} + {y}^{2} + {z}^{2} = {169} \) with the plane \( z = {12} \) . | The sphere is centered at the origin and has radius \( {13} = \sqrt{169} \), so it does intersect the plane \( z = {12} \). Putting \( z = {12} \) into the equation of the sphere gives\n\n\[ {x}^{2} + {y}^{2} + {12}^{2} = {169} \]\n\n\[ {x}^{2} + {y}^{2} = {169} - {144} = {25} = {5}^{2} \]\n\nwhich is a circle of radiu... | Yes |
Is \( 2{x}^{2} + 2{y}^{2} + 2{z}^{2} - {8x} + {4y} - {16z} + {10} = 0 \) the equation of a sphere? | Solution: Dividing both sides of the equation by 2 gives\n\n\[ \n{x}^{2} + {y}^{2} + {z}^{2} - {4x} + {2y} - {8z} + 5 = 0 \n\]\n\n\[ \n\left( {{x}^{2} - {4x} + 4}\right) + \left( {{y}^{2} + {2y} + 1}\right) + \left( {{z}^{2} - {8z} + {16}}\right) + 5 - 4 - 1 - {16} = 0 \n\]\n\n\[ \n{\left( x - 2\right) }^{2} + {\left( ... | Yes |
Find the points(s) of intersection (if any) of the sphere from Example 1.28 and the line \( x = 3 + t, y = 1 + {2t}, z = 3 - t \) . | Solution: Put the equations of the line into the equation of the sphere, which was \( {\left( x - 2\right) }^{2} + \) \( {\left( y + 1\right) }^{2} + {\left( z - 4\right) }^{2} = {16} \), and solve for \( t \) :\n\n\[ \n{\left( 3 + t - 2\right) }^{2} + {\left( 1 + 2t + 1\right) }^{2} + {\left( 3 - t - 4\right) }^{2} = ... | Yes |
Find the intersection (if any) of the spheres \( {x}^{2} + {y}^{2} + {z}^{2} = {25} \) and \( {x}^{2} + {y}^{2} + (z - 2{)}^{2} = {16} \). | Solution: For any point \( \left( {x, y, z}\right) \) on both spheres, we see that\n\n\[ \n{x}^{2} + {y}^{2} + {z}^{2} = {25}\; \Rightarrow \;{x}^{2} + {y}^{2} = {25} - {z}^{2}\text{, and } \n\]\n\n\[ \n{x}^{2} + {y}^{2} + {\left( z - 2\right) }^{2} = {16}\; \Rightarrow \;{x}^{2} + {y}^{2} = {16} - {\left( z - 2\right)... | Yes |
Convert the point \( \left( {-2, - 2,1}\right) \) from Cartesian coordinates to (a) cylindrical and (b) spherical coordinates. | Solution: (a) \( r = \sqrt{{\left( -2\right) }^{2} + {\left( -2\right) }^{2}} = 2\sqrt{2},\;\theta = {\tan }^{-1}\left( \frac{-2}{-2}\right) = {\tan }^{-1}\left( 1\right) = \frac{5\pi }{4} \), since \( y = - 2 < 0. \)\n\n\( \therefore \left( {r,\theta, z}\right) = \left( {2\sqrt{2},\frac{5\pi }{4},1}\right) \)\n\n(b) \... | Yes |
Write the equation of the cylinder \( {x}^{2} + {y}^{2} = 4 \) in cylindrical coordinates. | Solution: Since \( r = \sqrt{{x}^{2} + {y}^{2}} \), then the equation in cylindrical coordinates is \( r = 2 \) . | Yes |
Write the equation \( {\left( x - 2\right) }^{2} + {\left( y - 1\right) }^{2} + {z}^{2} = 9 \) in spherical coordinates. | Solution: Multiplying the equation out gives\n\n\[ \n{x}^{2} + {y}^{2} + {z}^{2} - {4x} - {2y} + 5 = 9\\text{, so we get} \n\]\n\n\[ \n{\\rho }^{2} - {4\\rho }\\sin \\phi \\cos \\theta - {2\\rho }\\sin \\phi \\sin \\theta - 4 = 0\\text{, or } \n\]\n\n\[ \n{\\rho }^{2} - 2\\sin \\phi \\left( {2\\cos \\theta + \\sin \\th... | No |
Example 1.34. Describe the surface given by \( \theta = z \) in cylindrical coordinates. | Solution: This surface is called a helicoid. As the (vertical) \( z \) coordinate increases, so does the angle \( \theta \), while the radius \( r \) is unrestricted. So this sweeps out a (ruled!) surface shaped like a spiral staircase, where the spiral has an infinite radius. Figure 1.7.6 shows a section of this surfa... | Yes |
Define \( \mathbf{f} : \mathbb{R} \rightarrow {\mathbb{R}}^{3} \) by \( \mathbf{f}\left( t\right) = \left( {\cos t,\sin t, t}\right) \). | For each \( t \), the \( x \) - and \( y \) -coordinates of \( \mathbf{f}\left( t\right) \) are \( x = \cos t \) and \( y = \sin t \), so\n\n\[ \n{x}^{2} + {y}^{2} = {\cos }^{2}t + {\sin }^{2}t = 1.\n\]\n\nThus, the curve lies on the surface of the right circular cylinder \( {x}^{2} + {y}^{2} = 1 \). | Yes |
Let \( \mathbf{f}\left( t\right) = \left( {\cos t,\sin t, t}\right) \) . Then \( {\mathbf{f}}^{\prime }\left( t\right) = \left( {-\sin t,\cos t,1}\right) \) for all \( t \). | The tangent line \( L \) to the curve at \( \textbf{f}\left( {2\pi }\right) = \left( {1,0,{2\pi }}\right) \) is \( L = \textbf{f}\left( {2\pi }\right) + s\;{\textbf{f}}^{\prime }\left( {2\pi }\right) = \left( {1,0,{2\pi }}\right) + s\left( {0,1,1}\right) \), or in parametric form: \( x = 1, y = s, z = {2\pi } + s \) fo... | No |
Theorem 1.20. Let \( \mathbf{f}\left( t\right) \) and \( \mathbf{g}\left( t\right) \) be differentiable vector-valued functions, let \( u\left( t\right) \) be a differentiable scalar function, let \( k \) be a scalar, and let \( \mathbf{c} \) be a constant vector. Then\n\n(a) \( \frac{d}{dt}\left( \mathbf{c}\right) = \... | Proof: The proofs of parts (a)-(e) follow easily by differentiating the component functions and using the rules for derivatives from single-variable calculus. We will prove part (f), and leave the proof of part \( \left( \mathrm{g}\right) \) as an exercise for the reader.\n\n(f) Write \( \mathbf{f}\left( t\right) = \le... | No |
Suppose \( \mathbf{f}\left( t\right) \) is differentiable. Find the derivative of \( \parallel \mathbf{f}\left( t\right) \parallel \) . | Since \( \parallel \mathbf{f}\left( t\right) \parallel \) is a real-valued function of \( t \), then by the Chain Rule for real-valued functions, we know that \( \frac{d}{dt}\parallel \mathbf{f}\left( t\right) {\parallel }^{2} = 2\parallel \mathbf{f}\left( t\right) \parallel \frac{d}{dt}\parallel \mathbf{f}\left( t\rig... | Yes |
The spherical spiral \( \mathbf{f}\left( t\right) = \left( {\frac{\cos t}{\sqrt{1 + {a}^{2}{t}^{2}}},\frac{\sin t}{\sqrt{1 + {a}^{2}{t}^{2}}},\frac{-{at}}{\sqrt{1 + {a}^{2}{t}^{2}}}}\right) \) , for \( a \neq 0 \). | In the exercises, the reader will be asked to show that this curve lies on the sphere \( {x}^{2} + {y}^{2} + {z}^{2} = 1 \) and to verify directly that \( {\mathbf{f}}^{\prime }\left( t\right) \cdot \mathbf{f}\left( t\right) = 0 \) for all \( t \) . | No |
Example 1.39. Let \( \mathbf{r}\left( t\right) = \left( {5\cos t,3\sin t,4\sin t}\right) \) be the position vector of an object at time \( t \geq 0 \) . Find its (a) velocity and (b) acceleration vectors. | Solution: (a) \( \mathbf{v}\left( t\right) = \dot{\mathbf{r}}\left( t\right) = \left( {-5\sin t,3\cos t,4\cos t}\right) \)\n\n(b) \( \mathbf{a}\left( t\right) = \dot{\mathbf{v}}\left( t\right) = \left( {-5\cos t, - 3\sin t, - 4\sin t}\right) \) | Yes |
Example 1.40. Bézier curves are used in Computer Aided Design (CAD) to approximate the shape of a polygonal path in space (called the Bézier polygon or control polygon). For instance, given three points (or position vectors) \( {\mathbf{b}}_{0},{\mathbf{b}}_{1},{\mathbf{b}}_{2} \) in \( {\mathbb{R}}^{3} \), define\n\n\... | As an example, let \( {\mathbf{b}}_{0} = \left( {0,0,0}\right) ,{\mathbf{b}}_{1} = \left( {1,2,3}\right) \), and \( {\mathbf{b}}_{2} = \left( {4,5,2}\right) \) . Then the explicit formula for the Bézier curve is \( {\mathbf{b}}_{0}^{2}\left( t\right) = \left( {{2t} + 2{t}^{2},{4t} + {t}^{2},{6t} - 4{t}^{2}}\right) \), ... | Yes |
Find the length \( L \) of the helix \( \mathbf{f}\left( t\right) = \left( {\cos t,\sin t, t}\right) \) from \( t = 0 \) to \( t = {2\pi } \) . | Solution: By formula (1.41), we have\n\n\[ L = {\int }_{0}^{2\pi }\sqrt{{\left( -\sin t\right) }^{2} + {\left( \cos t\right) }^{2} + {1}^{2}}\;{dt} = {\int }_{0}^{2\pi }\sqrt{{\sin }^{2}t + {\cos }^{2}t + 1}\;{dt} = {\int }_{0}^{2\pi }\sqrt{2}\;{dt} \]\n\n\[ = \sqrt{2}\left( {{2\pi } - 0}\right) = 2\sqrt{2}\pi \] | Yes |
Example 1.42. The following are all equivalent parametrizations of the same curve: | To see that \( \mathbf{g}\left( s\right) \) is equivalent to \( \mathbf{f}\left( t\right) \), define \( \alpha : \left\lbrack {0,\pi }\right\rbrack \rightarrow \left\lbrack {0,{2\pi }}\right\rbrack \) by \( \alpha \left( s\right) = {2s} \) . Then \( \alpha \) is smooth, one-to-one, maps \( \left\lbrack {0,\pi }\right\r... | Yes |
Parametrize the helix \( \mathbf{f}\left( t\right) = \left( {\cos t,\sin t, t}\right) \), for \( t \) in \( \left\lbrack {0,{2\pi }}\right\rbrack \), by arc length. | Solution: By Example 1.41 and formula (1.43), we have\n\n\[ s = {\int }_{0}^{t}\begin{Vmatrix}{{\mathbf{f}}^{\prime }\left( u\right) }\end{Vmatrix}{du} = {\int }_{0}^{t}\sqrt{2}{du} = \sqrt{2}t\text{ for all }t\text{ in }\left\lbrack {0,{2\pi }}\right\rbrack .\n\]\n\nSo we can solve for \( t \) in terms of \( s : t = \... | Yes |
Theorem 1.22. Suppose that \( r = r\left( t\right) ,\theta = \theta \left( t\right) \) and \( z = z\left( t\right) \) are the cylindrical coordinates of a curve \( \mathbf{f}\left( t\right) \), for \( t \) in \( \left\lbrack {a, b}\right\rbrack \) . Then the arc length \( L \) of the curve over \( \left\lbrack {a, b}\r... | Proof: The Cartesian coordinates \( \left( {x\left( t\right), y\left( t\right), z\left( t\right) }\right) \) of a point on the curve are given by\n\n\[ x\left( t\right) = r\left( t\right) \cos \theta \left( t\right) ,\;y\left( t\right) = r\left( t\right) \sin \theta \left( t\right) ,\;z\left( t\right) = z\left( t\right... | Yes |
Find the arc length \( L \) of the curve whose cylindrical coordinates are \( r = {e}^{t} \) , \( \theta = t \) and \( z = {e}^{t} \), for \( t \) over the interval \( \left\lbrack {0,1}\right\rbrack \) . | Since \( {r}^{\prime }\left( t\right) = {e}^{t},{\theta }^{\prime }\left( t\right) = 1 \) and \( {z}^{\prime }\left( t\right) = {e}^{t} \), then\n\n\[ L = {\int }_{0}^{1}\sqrt{{r}^{\prime }{\left( t\right) }^{2} + r{\left( t\right) }^{2}{\theta }^{\prime }{\left( t\right) }^{2} + {z}^{\prime }{\left( t\right) }^{2}}{dt... | Yes |
The domain of the function\n\n\[ f\left( {x, y}\right) = {xy} \] | is all of \( {\mathbb{R}}^{2} \), and the range of \( f \) is all of \( \mathbb{R} \). | Yes |
The domain of the function\n\n\[ f\left( {x, y}\right) = \frac{1}{x - y} \] | is all of \( {\mathbb{R}}^{2} \) except the points \( \left( {x, y}\right) \) for which \( x = y \) . That is, the domain is the set \( D = \{ \left( {x, y}\right) : \( x \neq y\} \) . The range of \( f \) is all real numbers except 0 . | Yes |
The domain of the function\n\n\[ f\left( {x, y}\right) = \sqrt{1 - {x}^{2} - {y}^{2}} \]\n\nis the set \( D = \left\{ {\left( {x, y}\right) : {x}^{2} + {y}^{2} \leq 1}\right\} \) | since the quantity inside the square root is nonnegative if and only if \( 1 - \left( {{x}^{2} + {y}^{2}}\right) \geq 0 \) . We see that \( D \) consists of all points on and inside the unit circle in \( {\mathbb{R}}^{2} \) ( \( D \) is sometimes called the closed unit disk). The range of \( f \) is the interval \( \le... | Yes |
\[ \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {1,2}\right) }}\frac{xy}{{x}^{2} + {y}^{2}} = \frac{\left( 1\right) \left( 2\right) }{{1}^{2} + {2}^{2}} = \frac{2}{5} \] | since \( f\left( {x, y}\right) = \frac{xy}{{x}^{2} + {y}^{2}} \) is properly defined at the point \( \left( {1,2}\right) \) . | Yes |
[\\mathop{\\lim }\\limits_{{\\left( {x, y}\\right) \\rightarrow \\left( {0,0}\\right) }}\\frac{xy}{{x}^{2} + {y}^{2}}\\text{ does not exist }] | Note that we can not simply substitute \\( \\left( {x, y}\\right) = \\left( {0,0}\\right) \\) into the function, since doing so gives an indeterminate form 0/0 . To show that the limit does not exist, we will show that the function approaches different values as \\( \\left( {x, y}\\right) \\) approaches \\( \\left( {0,... | Yes |
Define a function \( f\left( {x, y}\right) \) on all of \( {\mathbb{R}}^{2} \) as follows:\n\n\[ f\left( {x, y}\right) = \left\{ \begin{array}{ll} 0 & \text{ if }\left( {x, y}\right) = \left( {0,0}\right) \\ \frac{{y}^{4}}{{x}^{2} + {y}^{2}} & \text{ if }\left( {x, y}\right) \neq \left( {0,0}\right) \end{array}\right. ... | then \( f\left( {x, y}\right) \) is continuous on all of \( {\mathbb{R}}^{2} \) . | Yes |
Find \( \frac{\partial f}{\partial x}\left( {x, y}\right) \) and \( \frac{\partial f}{\partial y}\left( {x, y}\right) \) for the function \( f\left( {x, y}\right) = {x}^{2}y + {y}^{3} \). | Solution: Treating \( y \) as a constant and differentiating \( f\left( {x, y}\right) \) with respect to \( x \) gives\n\n\[ \frac{\partial f}{\partial x}\left( {x, y}\right) = {2xy} \]\n\nand treating \( x \) as a constant and differentiating \( f\left( {x, y}\right) \) with respect to \( y \) gives\n\n\[ \frac{\parti... | Yes |
Find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) for the function \( f\left( {x, y}\right) = \frac{\sin \left( {x{y}^{2}}\right) }{{x}^{2} + 1} \). | Solution: Treating \( y \) as a constant and differentiating \( f\left( {x, y}\right) \) with respect to \( x \) gives\n\n\[ \frac{\partial f}{\partial x} = \frac{\left( {{x}^{2} + 1}\right) \left( {{y}^{2}\cos \left( {x{y}^{2}}\right) }\right) - \left( {2x}\right) \sin \left( {x{y}^{2}}\right) }{{\left( {x}^{2} + 1}\r... | Yes |
Find the partial derivatives \( \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{{\partial }^{2}f}{\partial {x}^{2}},\frac{{\partial }^{2}f}{\partial {y}^{2}},\frac{{\partial }^{2}f}{\partial y\partial x} \) and \( \frac{{\partial }^{2}f}{\partial x\partial y} \) for the function \( f\left( {x, y}\righ... | Solution: Proceeding as before, we have\n\n\[ \frac{\partial f}{\partial x} = {2xy}{e}^{{x}^{2}y} + {y}^{3}\;\frac{\partial f}{\partial y} = {x}^{2}{e}^{{x}^{2}y} + {3x}{y}^{2} \]\n\n\[ \frac{{\partial }^{2}f}{\partial {x}^{2}} = \frac{\partial }{\partial x}\left( {{2xy}{e}^{{x}^{2}y} + {y}^{3}}\right) \;\frac{{\partia... | Yes |
Find the equation of the tangent plane to the surface \( z = {x}^{2} + {y}^{2} \) at the point \( \left( {1,2,5}\right) \) . | For the function \( f\left( {x, y}\right) = {x}^{2} + {y}^{2} \), we have \( \frac{\partial f}{\partial x} = {2x} \) and \( \frac{\partial f}{\partial y} = {2y} \), so the equation of the tangent plane at the point \( \left( {1,2,5}\right) \) is\n\n\[ 2\left( 1\right) \left( {x - 1}\right) + 2\left( 2\right) \left( {y ... | Yes |
Find the equation of the tangent plane to the surface \( {x}^{2} + {y}^{2} + {z}^{2} = 9 \) at the point \( \left( {2,2, - 1}\right) \). | For the function \( F\left( {x, y, z}\right) = {x}^{2} + {y}^{2} + {z}^{2} - 9 \), we have \( \frac{\partial F}{\partial x} = {2x},\frac{\partial F}{\partial y} = {2y} \), and \( \frac{\partial F}{\partial z} = {2z} \) , so the equation of the tangent plane at \( \left( {2,2, - 1}\right) \) is\n\n\[ 2\left( 2\right) \l... | Yes |
Find the directional derivative of \( f\left( {x, y}\right) = x{y}^{2} + {x}^{3}y \) at the point \( \left( {1,2}\right) \) in the direction of \( \mathbf{v} = \left( {\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}}\right) \) . | Solution: We see that \( \nabla f = \left( {{y}^{2} + 3{x}^{2}y,{2xy} + {x}^{3}}\right) \), so\n\n\[ \n{D}_{\mathbf{v}}f\left( {1,2}\right) \; = \;\mathbf{v} \cdot \nabla f\left( {1,2}\right) \; = \;\left( {\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}}\right) \cdot \left( {{2}^{2} + 3{\left( 1\right) }^{2}\left( 2\right) ,2\l... | Yes |
In which direction does the function \( f\left( {x, y}\right) = x{y}^{2} + {x}^{3}y \) increase the fastest from the point \( \left( {1,2}\right) \) ? In which direction does it decrease the fastest? | Solution: Since \( \nabla f = \left( {{y}^{2} + 3{x}^{2}y,{2xy} + {x}^{3}}\right) \), then \( \nabla f\left( {1,2}\right) = \left( {{10},5}\right) \neq \mathbf{0} \) . A unit vector in that direction is \( \mathbf{v} = \frac{\nabla f}{\parallel \nabla f\parallel } = \left( {\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}}\right)... | Yes |
Example 2.17. The temperature \( T \) of a solid is given by the function \( T\left( {x, y, z}\right) = {e}^{-x} + {e}^{-{2y}} + \) \( {e}^{4z} \), where \( x, y, z \) are space coordinates relative to the center of the solid. In which direction from the point \( \left( {1,1,1}\right) \) will the temperature decrease t... | Solution: Since \( \nabla f = \left( {-{e}^{-x}, - 2{e}^{-{2y}},4{e}^{4z}}\right) \), then the temperature will decrease the fastest in the direction of \( - \nabla f\left( {1,1,1}\right) = \left( {{e}^{-1},2{e}^{-2}, - 4{e}^{4}}\right) \) . | Yes |
The function \( f\left( {x, y}\right) = {xy} \) has a critical point at \( \left( {0,0}\right) \) | But clearly \( f \) does not have a local maximum or minimum at \( \left( {0,0}\right) \) since any disk around \( \left( {0,0}\right) \) contains points \( \left( {x, y}\right) \) where the values of \( x \) and \( y \) have the same sign (so that \( f\left( {x, y}\right) = {xy} > 0 = f\left( {0,0}\right) \) ) and dif... | Yes |
Find all local maxima and minima of \( f\left( {x, y}\right) = {x}^{2} + {xy} + {y}^{2} - {3x} \) . | Solution: First find the critical points, i.e. where \( \nabla f = \mathbf{0} \) . Since\n\n\[ \frac{\partial f}{\partial x} = {2x} + y - 3\text{ and }\frac{\partial f}{\partial y} = x + {2y} \]\n\nthen the critical points \( \left( {x, y}\right) \) are the common solutions of the equations\n\n\[ {2x} + y - 3 = 0 \]\n\... | Yes |
Find all local maxima and minima of \( f\left( {x, y}\right) = {xy} - {x}^{3} - {y}^{2} \) . | Solution: First find the critical points, i.e. where \( \nabla f = \mathbf{0} \) . Since\n\n\[ \frac{\partial f}{\partial x} = y - 3{x}^{2}\;\text{ and }\;\frac{\partial f}{\partial y} = x - {2y} \]\nthen the critical points \( \left( {x, y}\right) \) are the common solutions of the equations\n\n\[ y - 3{x}^{2} = 0 \]\... | Yes |
Find all local maxima and minima of \( f\left( {x, y}\right) = {\left( x - 2\right) }^{4} + {\left( x - 2y\right) }^{2} \) . | Solution: First find the critical points, i.e. where \( \nabla f = \mathbf{0} \) . Since\n\n\[ \frac{\partial f}{\partial x} = 4{\left( x - 2\right) }^{3} + 2\left( {x - {2y}}\right) \text{ and }\frac{\partial f}{\partial y} = - 4\left( {x - {2y}}\right) \]\n\nthen the critical points \( \left( {x, y}\right) \) are the... | Yes |
Find all local maxima and minima of \( f\left( {x, y}\right) = \left( {{x}^{2} + {y}^{2}}\right) {e}^{-\left( {{x}^{2} + {y}^{2}}\right) } \) . | Solution: First find the critical points, i.e. where \( \nabla f = \mathbf{0} \) . Since\n\n\[ \frac{\partial f}{\partial x} = {2x}\left( {1 - \left( {{x}^{2} + {y}^{2}}\right) }\right) {e}^{-\left( {{x}^{2} + {y}^{2}}\right) } \]\n\n\[ \frac{\partial f}{\partial y} = {2y}\left( {1 - \left( {{x}^{2} + {y}^{2}}\right) }... | Yes |
Find all local maxima and minima of \( f\left( {x, y}\right) = {x}^{3} - {xy} - x + x{y}^{3} - {y}^{4} \). | Solution: First calculate the necessary partial derivatives:\n\n\[ \frac{\partial f}{\partial x} = 3{x}^{2} - y - 1 + {y}^{3},\;\frac{\partial f}{\partial y} = - x + {3x}{y}^{2} - 4{y}^{3} \]\n\n\[ \frac{{\partial }^{2}f}{\partial {x}^{2}} = {6x},\;\frac{{\partial }^{2}f}{\partial {y}^{2}} = {6xy} - {12}{y}^{2},\;\frac... | No |
For a rectangle whose perimeter is \( {20}\mathrm{\;m} \), find the dimensions that will maximize the area. | Solution: The area \( A \) of a rectangle with width \( x \) and height \( y \) is \( A = {xy} \) . The perimeter \( P \) of the rectangle is then given by the formula \( P = {2x} + {2y} \) . Since we are given that the perimeter \( P = {20} \), this problem can be stated as:\n\n\[ \n\text{Maximize :}f\left( {x, y}\rig... | Yes |
Theorem 2.7. Let \( f\left( {x, y}\right) \) and \( g\left( {x, y}\right) \) be smooth functions, and suppose that \( c \) is a scalar constant such that \( \nabla g\left( {x, y}\right) \neq \mathbf{0} \) for all \( \left( {x, y}\right) \) that satisfy the equation \( g\left( {x, y}\right) = c \) . Then to solve the co... | A rigorous proof of the above theorem requires use of the Implicit Function Theorem, which is beyond the scope of this text. \( {}^{11} \) Note that the theorem only gives a necessary condition for a point to be a constrained maximum or minimum. Whether a point \( \left( {x, y}\right) \) that satisfies \( \nabla f\left... | No |
For a rectangle whose perimeter is \( {20}\mathrm{\;m} \), use the Lagrange multiplier method to find the dimensions that will maximize the area. | Solution: As we saw in Example 2.24, with \( x \) and \( y \) representing the width and height, respectively, of the rectangle, this problem can be stated as:\n\n\[ \text{Maximize:}f\left( {x, y}\right) = {xy} \]\n\n\[ \text{given :}g\left( {x, y}\right) = {2x} + {2y} = {20} \]\n\nThen solving the equation \( \nabla f... | Yes |
Find the points on the circle \( {x}^{2} + {y}^{2} = {80} \) which are closest to and farthest from the point \( \left( {1,2}\right) \) . | The distance \( d \) from any point \( \left( {x, y}\right) \) to the point \( \left( {1,2}\right) \) is\n\n\[ d = \sqrt{{\left( x - 1\right) }^{2} + {\left( y - 2\right) }^{2}} \]\n\nand minimizing the distance is equivalent to minimizing the square of the distance. Thus the problem can be stated as:\n\n\[ \text{Maxim... | Yes |
Find the volume \( V \) under the plane \( z = {8x} + {6y} \) over the rectangle \( R = \left\lbrack {0,1}\right\rbrack \times \) \( \left\lbrack {0,2}\right\rbrack \) . | Solution: We see that \( f\left( {x, y}\right) = {8x} + {6y} \geq 0 \) for \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 2 \), so:\n\n\[ V = {\int }_{0}^{2}{\int }_{0}^{1}\left( {{8x} + {6y}}\right) {dxdy} \]\n\n\[ = {\int }_{0}^{2}\left( {4{x}^{2} + {\left. 6xy\right| }_{x = 0}^{x = 1}}\right) \;{dy} \]\n\n\[ = {\int }_{... | Yes |
Find the volume \( V \) under the surface \( z = {e}^{x + y} \) over the rectangle \( R = \left\lbrack {2,3}\right\rbrack \times \left\lbrack {1,2}\right\rbrack \) . | Solution: We know that \( f\left( {x, y}\right) = {e}^{x + y} > 0 \) for all \( \left( {x, y}\right) \), so\n\n\[ V = {\int }_{1}^{2}{\int }_{2}^{3}{e}^{x + y}{dxdy} \]\n\n\[ = {\int }_{1}^{2}\left( {\left. {e}^{x + y}\right| }_{x = 2}^{x = 3}\right) {dy} \]\n\n\[ = {\int }_{1}^{2}\left( {{e}^{y + 3} - {e}^{y + 2}}\rig... | Yes |
Evaluate \( {\int }_{0}^{2\pi }{\int }_{0}^{\pi }\sin \left( {x + y}\right) {dxdy} \). | Note that \( f\left( {x, y}\right) = \sin \left( {x + y}\right) \) is both positive and negative over the rectangle \( \left\lbrack {0,\pi }\right\rbrack \times \) \( \left\lbrack {0,{2\pi }}\right\rbrack \) . We can still evaluate the double integral:\n\n\[ \n{\int }_{0}^{2\pi }{\int }_{0}^{\pi }\sin \left( {x + y}\ri... | Yes |
Find the volume \( V \) under the plane \( z = {8x} + {6y} \) over the region \( R = \{ \left( {x, y}\right) : 0 \leq \) \( \left. {x \leq 1,0 \leq y \leq 2{x}^{2}}\right\} \) . | The region \( R \) is shown in Figure 3.2.2. Using vertical slices we get:\n\n\[ V = {\iint }_{R}\left( {{8x} + {6y}}\right) {dA} \]\n\n\[ = {\int }_{0}^{1}\left\lbrack {{\int }_{0}^{2{x}^{2}}\left( {{8x} + {6y}}\right) {dy}}\right\rbrack {dx} \]\n\n\[ = {\int }_{0}^{1}\left( {{8xy} + {\left. 3{y}^{2}\right| }_{y = 0}^... | Yes |
Find the volume \( V \) of the solid bounded by the three coordinate planes and the plane \( {2x} + y + {4z} = 4 \) . | The solid is shown in Figure 3.2.4(a) with a typical vertical slice. The volume \( V \) is given by \( \iint f\left( {x, y}\right) {dA} \), where \( f\left( {x, y}\right) = z = \frac{1}{4}\left( {4 - {2x} - y}\right) \) and the region \( R \), shown in Figure 3.2.4(b), is \( R = \{ \left( {x, y}\right) : 0 \leq x \leq ... | Yes |
Evaluate \( {\int }_{1}^{\infty }{\int }_{0}^{1/{x}^{2}}{2ydydx} \) . | \[ {\int }_{1}^{\infty }{\int }_{0}^{1/{x}^{2}}{2ydydx} = {\int }_{1}^{\infty }\left( {\left. {y}^{2}\right| }_{y = 0}^{y = 1/{x}^{2}}\right) {dx} \] \[ = {\int }_{1}^{\infty }{x}^{-4}{dx} = {\left. -\frac{1}{3}{x}^{-3}\right| }_{1}^{\infty } = 0 - \left( {-\frac{1}{3}}\right) = \frac{1}{3} \] | Yes |
Example 3.7. Evaluate \( {\int }_{0}^{3}{\int }_{0}^{2}{\int }_{0}^{1}\left( {{xy} + z}\right) {dxdydz} \) . | \[ \n{\int }_{0}^{3}{\int }_{0}^{2}{\int }_{0}^{1}\left( {{xy} + z}\right) \;{dx}\;{dy}\;{dz} = {\int }_{0}^{3}{\int }_{0}^{2}\left( {\left. \frac{1}{2}{x}^{2}y + xz\right| }_{x = 0}^{x = 1}\right) \;{dy}\;{dz} \]\n\[ = {\int }_{0}^{3}{\int }_{0}^{2}\left( {\frac{1}{2}y + z}\right) {dydz} \]\n\[ = {\int }_{0}^{3}\left(... | Yes |
Evaluate \( {\int }_{0}^{1}{\int }_{0}^{1 - x}{\int }_{0}^{2 - x - y}\left( {x + y + z}\right) {dzdydx} \) . | \[ {\int }_{0}^{1}{\int }_{0}^{1 - x}{\int }_{0}^{2 - x - y}\left( {x + y + z}\right) \;{dz}\;{dy}\;{dx}\; = \;{\int }_{0}^{1}{\int }_{0}^{1 - x}\left( {\left( {x + y}\right) z + {\left. \frac{1}{2}{z}^{2}\right| }_{z = 0}^{z = 2 - x - y}}\right) \;{dy}\;{dx} \]\n\[ = {\int }_{0}^{1}{\int }_{0}^{1 - x}\left( {\left( {x... | Yes |
Evaluate \( {\iint }_{R}{e}^{\frac{x - y}{x + y}}{dA} \), where \( R = \{ \left( {x, y}\right) : x \geq 0, y \geq 0, x + y \leq 1\} \) . | Solution: First, note that evaluating this double integral without using substitution is probably impossible, at least in a closed form. By looking at the numerator and denominator of the exponent of \( e \), we will try the substitution \( u = x - y \) and \( v = x + y \) . To use the change of variables formula (3.19... | Yes |
Find the volume \( V \) inside the paraboloid \( z = {x}^{2} + {y}^{2} \) for \( 0 \leq z \leq 1 \) . | \[ V = {\iint }_{R}\left( {1 - z}\right) {dA} = {\iint }_{R}\left( {1 - \left( {{x}^{2} + {y}^{2}}\right) }\right) {dA}, \] where \( R = \left\{ {\left( {x, y}\right) : {x}^{2} + {y}^{2} \leq 1}\right\} \) is the unit disk in \( {\mathbb{R}}^{2} \) (see Figure 3.5.2). In polar coordinates \( \left( {r,\theta }\right) \... | Yes |
Find the volume \( V \) inside the cone \( z = \sqrt{{x}^{2} + {y}^{2}} \) for \( 0 \leq z \leq 1 \) . | Solution: Using vertical slices, we see that\n\n\n\nFigure 3.5.3 \( z = \sqrt{{x}^{2} + {y}^{2}} \)\n\n\[ V = {\iint }_{R}\left( {1 - z}\right) {dA} = {\iint }_{R}\left( {1 - \sqrt{{x}^{2} + {y}^{2}}}\right) {dA}, \]... | Yes |
For \( a > 0 \), find the volume \( V \) inside the sphere \( S = {x}^{2} + {y}^{2} + {z}^{2} = {a}^{2} \) . | Solution: We see that \( S \) is the set \( \rho = a \) in spherical coordinates, so\n\n\[ \nV = {\iiint }_{S}{1dV} = {\int }_{0}^{2\pi }{\int }_{0}^{\pi }{\int }_{0}^{a}1{\rho }^{2}\sin {\phi d\rho d\phi d\theta } \n\]\n\n\[ \n= {\int }_{0}^{2\pi }{\int }_{0}^{\pi }\left( {\frac{{\rho }^{3}}{3}{\left. \right| }_{\rho ... | Yes |
Find the center of mass of the region \( R = \\left\\{ {\\left( {x, y}\\right) : 0 \\leq x \\leq 1,0 \\leq y \\leq 2{x}^{2}}\\right\\} \), if the density function at \( \\left( {x, y}\\right) \) is \( \\delta \\left( {x, y}\\right) = x + y \) . | Solution: The region \( R \) is shown in Figure 3.6.2. We have\n\n\n\nFigure 3.6.2\n\n\[ \nM = {\\iint }_{R}\\delta \\left( {x, y}\\right) {dA} \n\]\n\n\[ \n= {\\int }_{0}^{1}{\\int }_{0}^{2{x}^{2}}\\left( {x + y}\\r... | Yes |
Find the center of mass of the solid \( S = \left\{ {\left( {x, y, z}\right) : z \geq 0,{x}^{2} + {y}^{2} + {z}^{2} \leq {a}^{2}}\right\} \), if the density function at \( \left( {x, y, z}\right) \) is \( \delta \left( {x, y, z}\right) = 1 \) . | The solid \( S \) is just the upper hemisphere inside the sphere\n\n\n\nFigure 3.6.3\n\nof radius \( a \) centered at the origin (see Figure 3.6.3). So since the density function is a constant and \( S \) is symmetri... | Yes |
Let \( X \) represent a randomly selected real number in the interval \( \left( {0,1}\right) \) . We say that \( X \) has the uniform distribution on \( \left( {0,1}\right) \), with distribution function\n\n\[ F\left( x\right) = P\left( {X \leq x}\right) = \left\{ \begin{array}{ll} 1, & \text{ for }x \geq 1 \\ x, & \te... | and probability density function\n\n\[ f\left( x\right) = {F}^{\prime }\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ for }0 < x < 1 \\ 0, & \text{ elsewhere. } \end{array}\right. \] | Yes |
If you were to pick \( n > 2 \) random real numbers from the interval \( \\left( {0,1}\\right) \), what are the expected values for the smallest and largest of those numbers? | Let \( {U}_{1},\\ldots ,{U}_{n} \) be \( n \) continuous random variables, each representing a randomly selected real number from \( \\left( {0,1}\\right) \), i.e. each has the uniform distribution on \( \\left( {0,1}\\right) \). Define random variables \( X \) and \( Y \) by\n\n\[ \nX = \\min \\left( {{U}_{1},\\ldots ... | No |
Use a line integral to show that the lateral surface area \( A \) of a right circular cylinder of radius \( r \) and height \( h \) is \( {2\pi rh} \) . | Solution: We will use the right circular cylinder with base circle \( C \) given by \( {x}^{2} + {y}^{2} = {r}^{2} \) and with height \( h \) in the positive \( z \) direction (see Figure 4.1.3). Parametrize \( C \) as follows:\n\n\[ x = x\left( t\right) = r\cos t,\;y = y\left( t\right) = r\sin t,\;0 \leq t \leq {2\pi ... | Yes |
Theorem 4.1. For a vector field \( \mathbf{f}\left( {x, y}\right) = P\left( {x, y}\right) \mathbf{i} + Q\left( {x, y}\right) \mathbf{j} \) and a curve \( C \) with a smooth parametrization \( x = x\left( t\right), y = y\left( t\right), a \leq t \leq b \) and position vector \( \mathbf{r}\left( t\right) = x\left( t\righ... | If the vector field \( \mathbf{f}\left( {x, y}\right) \) represents the force moving an object along a curve \( C \), then the work \( W \) done by this force is\n\n\[W = {\int }_{C}\mathbf{f} \cdot \mathbf{T}{ds} = {\int }_{C}\mathbf{f} \cdot d\mathbf{r}.\n\n(4.16)\] | Yes |
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