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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.
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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\[ \frac{1}{2}\left( \text{ base }\right) \left( \text{ height }\right) = \left( {1 - {x}^{2}}\right) \sqrt{3}\left( {1 - {x}^{2}}\right) ,\]\n\nThus the total volume is\n\n\[ {\int }_{-1}^{1}\sqrt{3}{\left( 1 - {x}^{2}\right) }^{2}{dx} = \frac{16}{15}\sqrt{3} \]
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Yes
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Find the volume of a right circular cone with base radius 10 and height 20.
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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 the cross section is given by\n\n\[ \pi \cdot {\text{ radius }}^{2} = \pi {\left( \frac{x}{2}\right) }^{2} \]\n\nso the volume is given by\n\n\[ {\int }_{0}^{20}\pi \frac{{x}^{2}}{4}{dx} = \frac{\pi }{4}\frac{{20}^{3}}{3} = \frac{2000\pi }{3}. \]\n\nNote that we can instead do the calculation with a generic height and radius:\n\n\[ {\int }_{0}^{h}\pi \frac{{r}^{2}}{{h}^{2}}{x}^{2}{dx} = \frac{\pi {r}^{2}}{{h}^{2}}\frac{{h}^{3}}{3} = \frac{\pi {r}^{2}h}{3} \]
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Yes
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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.
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This solid has a \
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No
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Example 15.1.5 Suppose the area under \( y = - {x}^{2} + 1 \) between \( x = 0 \) and \( x = 1 \) is rotated around the \( x \) -axis.
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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: \( {\int }_{0}^{1}{2\pi y}\sqrt{1 - y}{dy} = \frac{8}{15}\pi \) .
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Yes
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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.
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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 substitution, we find the antiderivative, namely \( \arcsin \left( {x/r}\right) \) . Notice that the integral is improper at both endpoints, as the function \( \sqrt{1/\left( {{r}^{2} - {x}^{2}}\right) } \) is undefined when \( x = \pm r \) . So we need to compute\n\n\[ \n\mathop{\lim }\limits_{{D \rightarrow - {r}^{ + }}}{\int }_{D}^{0}\sqrt{\frac{1}{{r}^{2} - {x}^{2}}}{dx} + \mathop{\lim }\limits_{{D \rightarrow {r}^{ - }}}{\int }_{0}^{D}\sqrt{\frac{1}{{r}^{2} - {x}^{2}}}{dx}. \n\]\n\nThis is not difficult, and has value \( \pi \), so the original integral, with the extra \( r \) in front, has value \( {\pi r} \) as expected.
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Yes
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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 \( \overrightarrow{PQ} = \overrightarrow{RS} \) ?
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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 equal to the vector \( \mathbf{w} \) with initial point \( \left( {0,0,0}\right) \) and terminal point \( S - R = \left( {2,1,0}\right) - \left( {1, - 3, - 2}\right) = \left( {2 - 1,1 - \left( {-3}\right) ,0 - \left( {-2}\right) }\right) = \left( {1,4,2}\right) . \n\nSo \( \overrightarrow{PQ} = \mathbf{v} = \left( {1,4,2}\right) \) and \( \overrightarrow{RS} = \mathbf{w} = \left( {1,4,2}\right) . \n\n\( \therefore \overrightarrow{PQ} = \overrightarrow{RS} \)
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Yes
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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\[ \parallel \mathbf{v}\parallel = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}} \]
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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 \( a = b = 0 \) and \( c \neq 0 \) (the other two possibilities are handled in a similar manner). Then \( \mathbf{v} = \left( {0,0, c}\right) \), which is a vector of length \( \left| c\right| \) along the \( z \) -axis. So \( \parallel \mathbf{v}\parallel = \left| c\right| = \sqrt{{c}^{2}} = \sqrt{{0}^{2} + {0}^{2} + {c}^{2}} = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}} \) .\n\nCase 3: exactly one of \( a, b, c \) is 0 . Without loss of generality, we assume that \( a = 0, b \neq 0 \) and \( c \neq 0 \) (the other two possibilities are handled in a similar manner). Then \( \mathbf{v} = \left( {0, b, c}\right) \) , which is a vector in the \( {yz} \) -plane, so by the Pythagorean Theorem we have \( \parallel \mathbf{v}\parallel = \sqrt{{b}^{2} + {c}^{2}} = \) \( \sqrt{{0}^{2} + {b}^{2} + {c}^{2}} = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}}. \)\n\nCase 4: none of \( a, b, c \) are 0 . Without loss of generality, we can as-\n\nsume that \( a, b, c \) are all positive (the other seven possibilities are handled in a similar manner). Consider the points \( P = \left( {0,0,0}\right) \) , \( Q = \left( {a, b, c}\right), R = \left( {a, b,0}\right) \), and \( S = \left( {a,0,0}\right) \), as shown in Figure 1.1.8. Applying the Pythagorean Theorem to the right triangle \( \bigtriangleup {PSR} \) gives \( {\left| PR\right| }^{2} = {a}^{2} + {b}^{2} \) . A second application of the Pythagorean Theorem, this time to the right triangle \( \bigtriangleup {PQR} \),\n\ngives \( \parallel \mathbf{v}\parallel = \left| {PQ}\right| = \sqrt{{\left| PR\right| }^{2} + {\left| QR\right| }^{2}} = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}} \) .\n\nThis proves the theorem.
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Yes
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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) \( k\mathbf{v} = \left( {k{v}_{1}, k{v}_{2}}\right) \)
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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 needed to show. If \( k \neq 0 \), then \( \left( {k{v}_{1}, k{v}_{2}}\right) \) lies on a line with slope \( \frac{k{v}_{2}}{k{v}_{1}} = \frac{{v}_{2}}{{v}_{1}} \), which is the same as the slope of the line on which \( \mathbf{v} \) (and hence \( k\mathbf{v} \) ) lies, and \( \left( {k{v}_{1}, k{v}_{2}}\right) \) points in the same direction on that line as \( k\mathbf{v} \) . Also, by formula (1.3) the magnitude of \( \left( {k{v}_{1}, k{v}_{2}}\right) \) is \( \left| {\sqrt{{\left( k{v}_{1}\right) }^{2} + {\left( k{v}_{2}\right) }^{2}} = \sqrt{{k}^{2}{v}_{1}^{2} + {k}^{2}{v}_{2}^{2}} = \sqrt{{k}^{2}\left( {{v}_{1}^{2} + {v}_{2}^{2}}\right) } = }\right| k\left| {\;\sqrt{{v}_{1}^{2} + {v}_{2}^{2}} = }\right| k\left| {\;\parallel \mathbf{v}\parallel .\;\text{So}\;k\mathbf{v}\text{ and }\left( {k{v}_{1}, k{v}_{2}}\right) }\right| \) have the same magnitude and direction. This proves (a).
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Yes
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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 \( \\mathbf{v} - \\mathbf{w} \) .
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Solution: \( \\mathbf{v} - \\mathbf{w} = \\left( {2 - 3,1 - \\left( {-4}\\right) , - 1 - 2}\\right) = \\left( {-1,5, - 3}\\right) \)
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Yes
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Theorem 1.6. Let \( \mathbf{v} \) , \( \mathbf{w} \) be nonzero vectors, and let \( \theta \) be the angle between them. Then\n\n\[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{\parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel } \]
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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\[ {\begin{Vmatrix}\mathbf{v} - \mathbf{w}\end{Vmatrix}}^{2} = {\begin{Vmatrix}\mathbf{v}\end{Vmatrix}}^{2} + {\begin{Vmatrix}\mathbf{w}\end{Vmatrix}}^{2} - 2\begin{Vmatrix}\mathbf{v}\end{Vmatrix}\begin{Vmatrix}\mathbf{w}\end{Vmatrix}\cos \theta \]\n\n(note that equation (1.9) holds even for the \
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Yes
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Find the angle \( \theta \) between the vectors \( \mathbf{v} = \left( {2,1, - 1}\right) \) and \( \mathbf{w} = \left( {3, - 4,1}\right) \).
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\( \textit{Solution: Since }\textbf{v} \cdot \textbf{w} = \left( 2\right) \left( 3\right) + \left( 1\right) \left( {-4}\right) + \left( {-1}\right) \left( 1\right) = 1,\;\| \textbf{v}\| = \sqrt{6},\; \) and \( \;\| \textbf{w}\| = \sqrt{26},\; \) then\n\n\[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{\parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel } = \frac{1}{\sqrt{6}\sqrt{26}} = \frac{1}{2\sqrt{39}} \approx {0.08} \Rightarrow \theta = {85.41}^{ \circ } \]
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Yes
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Corollary 1.8. If \( \theta \) is the angle between nonzero vectors \( \mathbf{v} \) and \( \mathbf{w} \), then\n\n\[ \mathbf{v} \cdot \mathbf{w}\text{ is }\left\{ \begin{array}{ll} > 0 & \text{ for }{0}^{ \circ } \leq \theta < {90}^{ \circ } \\ 0 & \text{ for }\theta = {90}^{ \circ } \\ < 0 & \text{ for }{90}^{ \circ } < \theta \leq {180}^{ \circ } \end{array}\right. \]
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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.
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No
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Example 1.6. Are the vectors \( \mathbf{v} = \left( {-1,5, - 2}\right) \) and \( \mathbf{w} = \left( {3,1,1}\right) \) perpendicular?
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Solution: Yes, \( \mathbf{v} \bot \mathbf{w} \) since \( \mathbf{v} \cdot \mathbf{w} = \left( {-1}\right) \left( 3\right) + \left( 5\right) \left( 1\right) + \left( {-2}\right) \left( 1\right) = 0 \) .
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Yes
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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) \( \left( {k\mathbf{v}}\right) \cdot \mathbf{w} = \mathbf{v} \cdot \left( {k\mathbf{w}}\right) = k\left( {\mathbf{v} \cdot \mathbf{w}}\right) \) Associative Law\n\n(c) \( \mathbf{v} \cdot \mathbf{0} = 0 = \mathbf{0} \cdot \mathbf{v} \)\n\n(d) \( \mathbf{u} \cdot \left( {\mathbf{v} + \mathbf{w}}\right) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} \) Distributive Law\n\n(e) \( \left( {\mathbf{u} + \mathbf{v}}\right) \cdot \mathbf{w} = \mathbf{u} \cdot \mathbf{w} + \mathbf{v} \cdot \mathbf{w}\; \) Distributive Law\n\n(f) \( \left| {\mathbf{v} \cdot \mathbf{w}}\right| \leq \parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel \) Cauchy-Schwarz Inequality \( {}^{5} \)
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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 inequality holds trivially. So assume that \( \mathbf{v} \) and \( \mathbf{w} \) are nonzero vectors. Then by Theorem 1.6,\n\n\[ \mathbf{v} \cdot \mathbf{w} = \cos \theta \parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel \text{, so } \]\n\n\[ \left| {\mathbf{v} \cdot \mathbf{w}}\right| = \left| {\cos \theta }\right| \parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel \text{, so } \]\n\n\[ \left| {\mathbf{v} \cdot \mathbf{w}}\right| \leq \parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel \operatorname{since}\left| {\cos \theta }\right| \leq 1.\;\text{ QED } \]
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No
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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) \( \\parallel \\mathbf{v} - \\mathbf{w}\\parallel \\geq \\parallel \\mathbf{v}\\parallel - \\parallel \\mathbf{w}\\parallel \)
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(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{w} \\cdot \\mathbf{v} + \\mathbf{w} \\cdot \\mathbf{w} \n\\]\n\n\\( = \\parallel \\mathbf{v}{\\parallel }^{2} + 2\\left( {\\mathbf{v} \\cdot \\mathbf{w}}\\right) + \\parallel \\mathbf{w}{\\parallel }^{2} \\), so since \\( a \\leq \\left| a\\right| \\) for any real number \\( a \\), we have\n\n\\( \\leq \\parallel \\mathbf{v}{\\parallel }^{2} + 2\\left| {\\mathbf{v} \\cdot \\mathbf{w}}\\right| + \\parallel \\mathbf{w}{\\parallel }^{2} \\), so by Theorem 1.9(f) we have\n\n\\[ \n\\leq {\\begin{Vmatrix}\\mathbf{v}\\end{Vmatrix}}^{2} + 2\\begin{Vmatrix}\\mathbf{v}\\end{Vmatrix}\\begin{Vmatrix}\\mathbf{w}\\end{Vmatrix} + {\\begin{Vmatrix}\\mathbf{w}\\end{Vmatrix}}^{2} = {\\left( \\begin{Vmatrix}\\mathbf{v}\\end{Vmatrix} + \\begin{Vmatrix}\\mathbf{w}\\end{Vmatrix}\\right) }^{2}\\text{ and so } \n\\]\n\n\\( \\parallel \\mathbf{v} + \\mathbf{w}\\parallel \\leq \\parallel \\mathbf{v}\\parallel + \\parallel \\mathbf{w}\\parallel \\) after taking square roots of both sides, which proves (b).\n\n\\( \\left| {\\widehat{w}\\left( c\\right) \\text{ Since }\\mathbf{v} = \\mathbf{w} + \\left( {\\mathbf{v} - \\mathbf{w}}\\right) \\text{, then }\\parallel \\mathbf{v}\\parallel = \\parallel \\mathbf{w} + \\left( {\\mathbf{v} - \\mathbf{w}}\\right) \\parallel \\leq \\parallel \\mathbf{w}\\parallel + \\parallel \\mathbf{v} - \\mathbf{w}\\parallel \\text{ by the Triangle Inequality,}}\\right| \\) so subtracting \\( \\parallel \\mathbf{w}\\parallel \\) from both sides gives \\( \\parallel \\mathbf{v}\\parallel - \\parallel \\mathbf{w}\\parallel \\leq \\parallel \\mathbf{v} - \\mathbf{w}\\parallel . \\) QED
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No
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Find \( \mathbf{i} \times \mathbf{j} \) .
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Solution: Since \( \mathbf{i} = \left( {1,0,0}\right) \) and \( \mathbf{j} = \left( {0,1,0}\right) \), then\n\n\[ \mathbf{i} \times \mathbf{j} = \left( {\left( 0\right) \left( 0\right) - \left( 0\right) \left( 1\right) ,\left( 0\right) \left( 0\right) - \left( 1\right) \left( 0\right) ,\left( 1\right) \left( 1\right) - \left( 0\right) \left( 0\right) }\right) \]\n\n\[ = \left( {0,0,1}\right) \]\n\n\[ = \mathbf{k} \]
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Yes
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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 \( \mathbf{w} \) .
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Proof: We will show that \( \left( {\mathbf{v} \times \mathbf{w}}\right) \cdot \mathbf{v} = 0 \) :\n\n\[ \left( {\mathbf{v} \times \mathbf{w}}\right) \cdot \mathbf{v} = \left( {{v}_{2}{w}_{3} - {v}_{3}{w}_{2},{v}_{3}{w}_{1} - {v}_{1}{w}_{3},{v}_{1}{w}_{2} - {v}_{2}{w}_{1}}\right) \cdot \left( {{v}_{1},{v}_{2},{v}_{3}}\right) \]\n\n\[ = {v}_{2}{w}_{3}{v}_{1} - {v}_{3}{w}_{2}{v}_{1} + {v}_{3}{w}_{1}{v}_{2} - {v}_{1}{w}_{3}{v}_{2} + {v}_{1}{w}_{2}{v}_{3} - {v}_{2}{w}_{1}{v}_{3} \]\n\n\[ = {v}_{1}{v}_{2}{w}_{3} - {v}_{1}{v}_{2}{w}_{3} + {w}_{1}{v}_{2}{v}_{3} - {w}_{1}{v}_{2}{v}_{3} + {v}_{1}{w}_{2}{v}_{3} - {v}_{1}{w}_{2}{v}_{3} \]\n\n\( = 0 \), after rearranging the terms.\n\n\( \therefore \mathbf{v} \times \mathbf{w} \bot \mathbf{v} \) by Corollary 1.7.\n\nThe proof that \( \mathbf{v} \times \mathbf{w} \bot \mathbf{w} \) is similar. QED
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Yes
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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 \( \mathbf{w} \) .
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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 \( \mathbf{v} \times \mathbf{w} \), one the opposite of the other. It turns out (see Appendix B) that the direction of \( \mathbf{v} \times \mathbf{w} \) is given by the right-hand rule, that is, the vectors \( \mathbf{v},\mathbf{w},\mathbf{v} \times \mathbf{w} \) form a right-handed system. Recall from Section 1.1 that this means that you can point your thumb upwards in the direction of \( \mathbf{v} \times \mathbf{w} \) while rotating \( \mathbf{v} \) towards \( \mathbf{w} \) with the remaining four fingers.
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Yes
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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 \( {\mathbb{R}}^{3} \). Let \( \theta \) be the angle between \( \mathbf{v} \) and \( \mathbf{w} \). The area \( {A}_{PQR} \) of \( \bigtriangleup {PQR} \) is \( \frac{1}{2}{bh} \), where \( b \) is the base of the triangle and \( h \) is the height.
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\[ b = \parallel \mathbf{v}\parallel \;\text{ and }\;h = \parallel \mathbf{w}\parallel \sin \theta \] \[ {A}_{PQR} = \frac{1}{2}\parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel \sin \theta \] \[ = \frac{1}{2}\parallel \mathbf{v} \times \mathbf{w}\parallel \] So since the area \( {A}_{PQRS} \) of the parallelogram \( {PQRS} \) is twice the area of the triangle \( \bigtriangleup {PQR} \), then \[ {A}_{PQRS} = \parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel \sin \theta \]
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Yes
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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 \( \mathbf{v},\mathbf{w} \) (as vectors in \( {\mathbb{R}}^{3} \) ) is:\n\n\[ A = \parallel \mathbf{v} \times \mathbf{w}\parallel \]
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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 \( {PQ} \) and \( {PR} \) as the adjacent sides, but it can be shown (see Exercise 26) that the different formulas would yield the same value, so the choice of adjacent sides indeed does not matter, and Theorem 1.13 is valid.
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No
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Calculate the area of the triangle \( \bigtriangleup {PQR} \), where \( P = \left( {2,4, - 7}\right), Q = \left( {3,7,{18}}\right) \) , and \( R = \left( {-5,{12},8}\right) \) .
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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\( \mathbf{v} = \left( {3,7,{18}}\right) - \left( {2,4, - 7}\right) = \left( {1,3,{25}}\right) \) and \( \mathbf{w} = \left( {-5,{12},8}\right) - \left( {2,4, - 7}\right) = \) \( \left( {-7,8,{15}}\right) \), so the area \( A \) of the triangle \( \bigtriangleup {PQR} \) is\n\n\[ A = \frac{1}{2}\parallel \mathbf{v} \times \mathbf{w}\parallel = \frac{1}{2}\parallel \left( {1,3,{25}}\right) \times \left( {-7,8,{15}}\right) \parallel \]\n\n\[ = \frac{1}{2}\begin{Vmatrix}\left( {\left( 3\right) \left( {15}\right) - \left( {25}\right) \left( 8\right) ,\left( {25}\right) \left( {-7}\right) - \left( 1\right) \left( {15}\right) ,\left( 1\right) \left( 8\right) - \left( 3\right) \left( {-7}\right) }\right) \end{Vmatrix} \]\n\n\[ = \frac{1}{2}\parallel \left( {-{155}, - {190},{29}}\right) \parallel \]\n\n\[ = \frac{1}{2}\sqrt{{\left( -{155}\right) }^{2} + {\left( -{190}\right) }^{2} + {29}^{2}} = \frac{1}{2}\sqrt{60966} \]\n\n\[ A \approx {123.46} \]
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Yes
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Calculate the area of the parallelogram PQRS, where \( P = \left( {1,1}\right), Q = \left( {2,3}\right) \) , \( R = \left( {5,4}\right) \), and \( S = \left( {4,2}\right) \) .
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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\( \mathbf{v} = \left( {1,1}\right) - \left( {4,2}\right) = \left( {-3, - 1}\right) \) and \( \mathbf{w} = \left( {5,4}\right) - \left( {4,2}\right) = \left( {1,2}\right) \) . But these are vectors in \( {\mathbb{R}}^{2} \), and the cross product is only defined for vectors in \( {\mathbb{R}}^{3} \) . However, \( {\mathbb{R}}^{2} \) can be thought of as the subset of \( {\mathbb{R}}^{3} \) such that the \( z \) -coordinate is always 0 . So we can write \( \mathbf{v} = \left( {-3, - 1,0}\right) \) and \( \mathbf{w} = \left( {1,2,0}\right) \) . Then the area \( A \) of \( {PQRS} \) is\n\n\[ A = \parallel \mathbf{v} \times \mathbf{w}\parallel = \begin{Vmatrix}{\left( {-3, - 1,0}\right) \times \left( {1,2,0}\right) }\end{Vmatrix} \]\n\n\[ = \begin{Vmatrix}\left( {\left( {-1}\right) \left( 0\right) - \left( 0\right) \left( 2\right) ,\left( 0\right) \left( 1\right) - \left( {-3}\right) \left( 0\right) ,\left( {-3}\right) \left( 2\right) - \left( {-1}\right) \left( 1\right) }\right) \end{Vmatrix} \]\n\n\[ = \parallel \left( {0,0, - 5}\right) \parallel \]\n\n\[ A = 5 \]
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Yes
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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
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(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\[ = - \left( {{v}_{3}{w}_{2} - {v}_{2}{w}_{3},{v}_{1}{w}_{3} - {v}_{3}{w}_{1},{v}_{2}{w}_{1} - {v}_{1}{w}_{2}}\right) \]\n\n\[ = - \left( {{w}_{2}{v}_{3} - {w}_{3}{v}_{2},{w}_{3}{v}_{1} - {w}_{1}{v}_{3},{w}_{1}{v}_{2} - {w}_{2}{v}_{1}}\right) \]\n\n\[ = - \mathbf{w} \times \mathbf{v} \]\n\nNote that this says that \( \mathbf{v} \times \mathbf{w} \) and \( \mathbf{w} \times \mathbf{v} \) have the same magnitude but opposite direction (see Figure 1.4.6).
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Yes
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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 triple product \( \mathbf{u} \cdot \left( {\mathbf{v} \times \mathbf{w}}\right) \) .
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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 since \( \mathbf{v} \times \mathbf{w} \) is perpendicular to the base parallelogram determined by \( \mathbf{v} \) and \( \mathbf{w} \), then the height \( h \) is \( \parallel \mathbf{u}\parallel \cos \theta \), where \( \theta \) is the angle between \( \mathbf{u} \) and \( \mathbf{v} \times \mathbf{w} \) . By Theorem 1.6 we know that\n\n\[ \cos \theta = \frac{\mathbf{u} \cdot \left( {\mathbf{v} \times \mathbf{w}}\right) }{\parallel \mathbf{u}\parallel \parallel \mathbf{v} \times \mathbf{w}\parallel }\text{. Hence,}\]\n\n\[ \operatorname{vol}\left( P\right) = {Ah} \]\n\n\[ = \parallel \mathbf{v} \times \mathbf{w}\parallel \frac{\parallel \mathbf{u}\parallel \mathbf{u} \cdot \left( {\mathbf{v} \times \mathbf{w}}\right) }{\parallel \mathbf{u}\parallel \parallel \mathbf{v} \times \mathbf{w}\parallel } \]\n\n\[ = \mathbf{u} \cdot \left( {\mathbf{v} \times \mathbf{w}}\right) \]
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Yes
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Find \( \mathbf{u} \times \left( {\mathbf{v} \times \mathbf{w}}\right) \) for \( \mathbf{u} = \left( {1,2,4}\right) ,\mathbf{v} = \left( {2,2,0}\right) ,\mathbf{w} = \left( {1,3,0}\right) \).
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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\[ = 7\left( {2,2,0}\right) - 6\left( {1,3,0}\right) = \left( {{14},{14},0}\right) - \left( {6,{18},0}\right) \]\n\n\[ = \left( {8, - 4,0}\right) \]
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Yes
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\[ \left| \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right| = \left( 1\right) \left( 4\right) - \left( 2\right) \left( 3\right) = 4 - 6 = - 2 \]
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\[ \left| \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right| = \left( 1\right) \left( 4\right) - \left( 2\right) \left( 3\right) = 4 - 6 = - 2 \]
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Yes
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Let \( \mathbf{v} = 4\mathbf{i} - \mathbf{j} + 3\mathbf{k} \) and \( \mathbf{w} = \mathbf{i} + 2\mathbf{k} \) . Then
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\[ \mathbf{v} \times \mathbf{w} = \left| \begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & - 1 & 3 \\ 1 & 0 & 2 \end{array}\right| = \left| \begin{array}{rr} - 1 & 3 \\ 0 & 2 \end{array}\right| \mathbf{i} - \left| \begin{array}{rr} 4 & 3 \\ 1 & 2 \end{array}\right| \mathbf{j} + \left| \begin{array}{rr} 4 & - 1 \\ 1 & 0 \end{array}\right| \mathbf{k} = - 2\;\mathbf{i} - 5\;\mathbf{j} + \mathbf{k} \]
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Yes
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Find the volume of the parallelepiped with adjacent sides \( \mathbf{u} = \left( {2,1,3}\right) ,\mathbf{v} = \) \( \left( {-1,3,2}\right) ,\mathbf{w} = \left( {1,1, - 2}\right) \)
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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 (in any order). By Theorem 1.17,\n\n\[ \mathbf{u} \cdot \left( {\mathbf{v} \times \mathbf{w}}\right) = \left| \begin{array}{rrr} 2 & 1 & 3 \\ - 1 & 3 & 2 \\ 1 & 1 & - 2 \end{array}\right| \]\n\n\[ = 2\left| \begin{array}{rr} 3 & 2 \\ 1 & - 2 \end{array}\right| - 1\left| \begin{array}{rr} - 1 & 2 \\ 1 & - 2 \end{array}\right| + 3\left| \begin{array}{rr} - 1 & 3 \\ 1 & 1 \end{array}\right| \]\n\n\[ = 2\left( {-8}\right) - 1\left( 0\right) + 3\left( {-4}\right) = - {28}\text{, so } \]\n\n\[ \operatorname{vol}\left( P\right) = \left| {-{28}}\right| = {28}\text{.} \]
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Yes
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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 \( \;\mathbf{u},\;\mathbf{v},\;\mathbf{w},\;\mathbf{z}\; \) in \( \;{\mathbb{R}}^{3}. \)
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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\[ = \mathbf{w} \cdot \left( {\mathbf{z} \times \mathbf{x}}\right) \;\text{(by formula (1.12))} \]\n\n\[ = \mathbf{w} \cdot \left( {\mathbf{z} \times \left( {\mathbf{u} \times \mathbf{v}}\right) }\right) \]\n\n\[ = \mathbf{w} \cdot \left( {\left( {\mathbf{z} \cdot \mathbf{v}}\right) \mathbf{u} - \left( {\mathbf{z} \cdot \mathbf{u}}\right) \mathbf{v}}\right) \text{ (by Theorem 1.16) } \]\n\n\[ = \left( {\mathbf{z} \cdot \mathbf{v}}\right) \left( {\mathbf{w} \cdot \mathbf{u}}\right) - \left( {\mathbf{z} \cdot \mathbf{u}}\right) \left( {\mathbf{w} \cdot \mathbf{v}}\right) \]\n\n\[ = \left( {\mathbf{u} \cdot \mathbf{w}}\right) \left( {\mathbf{v} \cdot \mathbf{z}}\right) - \left( {\mathbf{u} \cdot \mathbf{z}}\right) \left( {\mathbf{v} \cdot \mathbf{w}}\right) \text{ (by commutativity of the dot product). } \]\n\n\[ = \left| \begin{array}{ll} \mathbf{u} \cdot \mathbf{w} & \mathbf{u} \cdot \mathbf{z} \\ \mathbf{v} \cdot \mathbf{w} & \mathbf{v} \cdot \mathbf{z} \end{array}\right| \]
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Yes
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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 \( P \) .
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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\[ x = 2 + {4t},\;y = 3 - t,\;z = 5 + {6t},\text{ for } - \infty < t < \infty \]\n\n(c) \( L \) consists of the points \( \left( {x, y, z}\right) \) such that\n\n\[ \frac{x - 2}{4} = \frac{y - 3}{-1} = \frac{z - 5}{6} \]\n\n(d) Letting \( t = 1 \) and \( t = 2 \) in part(b) yields the points \( \left( {6,2,{11}}\right) \) and \( \left( {{10},1,{17}}\right) \) on \( L \) .
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Yes
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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.
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Solution: By formula (1.21), \( L \) consists of the points \( \left( {x, y, z}\right) \) such that\n\n\[ \nx = - 3 + {7t},\;y = 1 + {3t},\;z = - 4 - {2t},\text{ for } - \infty < t < \infty \n\]
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Yes
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Find the distance \( d \) from the point \( P = \left( {1,1,1}\right) \) to the line \( L \) in Example 1.20.
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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 \( \mathbf{w} = \overrightarrow{QP} = \) \( \left( {1,1,1}\right) - \left( {-3,1, - 4}\right) = \left( {4,0,5}\right) \), we have:\n\n\[ \mathbf{v} \times \mathbf{w} = \left| \begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 7 & 3 & - 2 \\ 4 & 0 & 5 \end{array}\right| = \left| \begin{array}{rr} 3 & - 2 \\ 0 & 5 \end{array}\right| \mathbf{i} - \left| \begin{array}{rr} 7 & - 2 \\ 4 & 5 \end{array}\right| \mathbf{j} + \left| \begin{array}{ll} 7 & 3 \\ 4 & 0 \end{array}\right| \mathbf{k} = {15}\mathbf{i} - {43}\mathbf{j} - {12}\mathbf{k}\text{, so } \]\n\n\[ d = \frac{\parallel \mathbf{v} \times \mathbf{w}\parallel }{\parallel \mathbf{v}\parallel } = \frac{\parallel {15}\mathbf{i} - {43}\mathbf{j} - {12}\mathbf{k}\parallel }{\parallel \left( {7,3, - 2}\right) \parallel } = \frac{\sqrt{{15}^{2} + {\left( -{43}\right) }^{2} + {\left( -{12}\right) }^{2}}}{\sqrt{{7}^{2} + {3}^{2} + {\left( -2\right) }^{2}}} = \frac{\sqrt{2218}}{\sqrt{62}} = {5.98} \]
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Yes
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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\\]
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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}, - 3 + {2t}}\\right) \\) for some \\( s, t \\) :\n\n\\[ \n- 1 + {3s} = - 3 + t : \\Rightarrow t = 2 + {3s} \n\\]\n\n\\[ \n\\begin{matrix} 2 + {2s} = 8 \\ - {3t} : \\ \\Rightarrow 2 + {2s} = 8 \\ - 3\\left( {2 + {3s}}\\right) = 2 \\ - {9s} \\ \\Rightarrow {2s} = - {9s} \\ \\Rightarrow s = 0 \\ \\Rightarrow t = 2 + 3\\left( 0\\right) = 2 \\end{matrix} \n\\]\n\n\\[ \n1 - s = - 3 + {2t} : 1 - 0 = - 3 + 2\\left( 2\\right) \\Rightarrow 1 = 1\\checkmark \\text{(Note that we had to check this.)} \n\\]\n\nLetting \\( s = 0 \\) in the equations for the first line, or letting \\( t = 2 \\) in the equations for the second line, gives the point of intersection \\( \\left( {-1,2,1}\\right) \\) .
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Yes
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Find the equation of the plane \( P \) containing the point \( \left( {-3,1,3}\right) \) and perpendicular to the vector \( \mathbf{n} = \left( {2,4,8}\right) \) .
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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\[ {ax} + {by} + {cz} + d = 0 \]\n\n(1.26)\n\nFor example, the normal form of the plane in Example 1.23 is \( {2x} + {4y} + {8z} - {22} = 0 \) .
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Yes
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Find the equation of the plane \( P \) containing the points \( \\left( {2,1,3}\\right) ,\\left( {1, - 1,2}\\right) \) and \( \\left( {3,2,1}\\right) \) .
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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\[ \n\\mathbf{n} = \\overrightarrow{QR} \\times \\overrightarrow{QS} = \\left( {-1, - 2, - 1}\\right) \\times \\left( {1,1, - 2}\\right) = \\left( {5, - 3,1}\\right) \n\]\n\nSo using formula (1.25) with the point \( Q \) (we could also use \( R \) or \( S \) ), the plane \( P \) consists of all points \( \\left( {x, y, z}\\right) \) such that:\n\n\[ \n5\\left( {x - 2}\\right) - 3\\left( {y - 1}\\right) + \\left( {z - 3}\\right) = 0 \n\]\n\nor in normal form,\n\n\[ \n{5x} - {3y} + z - {10} = 0 \n\]
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Yes
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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\[ D = \frac{\left| a{x}_{0} + b{y}_{0} + c{z}_{0} + d\right| }{\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}} \]
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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 form equation for \( P \), we know that \( \mathbf{n} = \left( {a, b, c}\right) \) is a normal vector for \( P \) . Now, any plane divides \( {\mathbb{R}}^{3} \) into two disjoint parts. Assume that \( \mathbf{n} \) points toward the side of \( P \) where the point \( Q \) is located. Place \( \mathbf{n} \) so that its initial point is at \( R \), and let \( \theta \) be the angle between \( \mathbf{r} \) and In. Then \( {0}^{ \circ } < \theta < {90}^{ \circ } \), so \( \cos \theta > 0 \) . Thus, the distance \( D \) is \( \cos \theta \parallel \mathbf{r}\parallel = \left| {\cos \theta }\right| \parallel \mathbf{r}\parallel \) (see Figure 1.5.8).\n\nBy Theorem 1.6 in Section 1.3, we know that \( \cos \theta = \frac{\mathbf{n} \cdot \mathbf{r}}{\parallel \mathbf{n}\parallel \parallel \mathbf{r}\parallel } \), so\n\n\[ D = \left| {\cos \theta }\right| \parallel \mathbf{r}\parallel = \frac{\left| \mathbf{n} \cdot \mathbf{r}\right| }{\parallel \mathbf{n}\parallel \parallel \mathbf{r}\parallel }\parallel \mathbf{r}\parallel = \frac{\left| \mathbf{n} \cdot \mathbf{r}\right| }{\parallel \mathbf{n}\parallel } = \frac{\left| a\left( {x}_{0} - x\right) + b\left( {y}_{0} - y\right) + c\left( {z}_{0} - z\right) \right| }{\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}} \]\n\n\[ = \frac{\left| a{x}_{0} + b{y}_{0} + c{z}_{0} - \left( ax + by + cz\right) \right| }{\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}} = \frac{\left| a{x}_{0} + b{y}_{0} + c{z}_{0} - \left( -d\right) \right| }{\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}} = \frac{\left| a{x}_{0} + b{y}_{0} + c{z}_{0} + d\right| }{\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}} \]\n\nIf \( \mathbf{n} \) points away from the side of \( P \) where the point \( Q \) is located, then \( {90}^{ \circ } < \theta < {180}^{ \circ } \) and so \( \cos \theta < 0 \) . The distance \( D \) is then \( \left| {\cos \theta }\right| \parallel \mathbf{r}\parallel \), and thus repeating the same argument as above still gives the same result. QED
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Yes
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Find the distance \( D \) from \( \left( {2,4, - 5}\right) \) to the plane from Example 1.24.
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Solution: Recall that the plane is given by \( {5x} - {3y} + z - {10} = 0 \) . So\n\n\[ D = \frac{\left| 5\left( 2\right) - 3\left( 4\right) + 1\left( -5\right) - {10}\right| }{\sqrt{{5}^{2} + {\left( -3\right) }^{2} + {1}^{2}}} = \frac{\left| -{17}\right| }{\sqrt{35}} = \frac{17}{\sqrt{35}} \approx {2.87} \]
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Yes
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Find the line of intersection \( L \) of the planes \( {5x} - {3y} + z - {10} = 0 \) and \( {2x} + {4y} - z + 3 = 0 \) .
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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 planes are not parallel and hence will intersect. A point \( \left( {x, y, z}\right) \) on both planes will satisfy the following system of two equations in three unknowns:\n\n\[ {5x} - {3y} + z - {10} = 0 \]\n\n\[ {2x} + {4y} - z + 3 = 0 \]\n\nSet \( x = 0 \) (why is that a good choice?). Then the above equations are reduced to:\n\n\[ - {3y} + z - {10} = 0 \]\n\n\[ {4y} - z + 3 = 0 \]\n\nThe second equation gives \( z = {4y} + 3 \), substituting that into the first equation gives \( y = 7 \) . Then \( z = {31} \), and so the point \( \left( {0,7,{31}}\right) \) is on \( L \) . Since \( {\mathbf{n}}_{1} \times {\mathbf{n}}_{2} = \left( {-1,7,{26}}\right) \), then \( L \) is given by:\n\n\[ \mathbf{r} + t\left( {{\mathbf{n}}_{1} \times {\mathbf{n}}_{2}}\right) = \left( {0,7,{31}}\right) + t\left( {-1,7,{26}}\right) ,\text{ for } - \infty < t < \infty \]\n\nor in parametric form:\n\n\[ x = - t,\;y = 7 + {7t},\;z = {31} + {26t},\text{ for } - \infty < t < \infty \]
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Yes
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Find the intersection of the sphere \( {x}^{2} + {y}^{2} + {z}^{2} = {169} \) with the plane \( z = {12} \) .
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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 radius 5 centered at \( \left( {0,0,{12}}\right) \), parallel to the \( {xy} \)-plane (see Figure 1.6.2).
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Yes
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Is \( 2{x}^{2} + 2{y}^{2} + 2{z}^{2} - {8x} + {4y} - {16z} + {10} = 0 \) the equation of a sphere?
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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\[ \n{\left( x - 2\right) }^{2} + {\left( y + 1\right) }^{2} + {\left( z - 4\right) }^{2} = {16} \n\]\n\nwhich is a sphere of radius 4 centered at \( \left( {2, - 1,4}\right) \) .
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Yes
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Find the points(s) of intersection (if any) of the sphere from Example 1.28 and the line \( x = 3 + t, y = 1 + {2t}, z = 3 - t \) .
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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\[ \n{\left( 3 + t - 2\right) }^{2} + {\left( 1 + 2t + 1\right) }^{2} + {\left( 3 - t - 4\right) }^{2} = {16} \n\]\n\n\[ \n{\left( t + 1\right) }^{2} + {\left( 2t + 2\right) }^{2} + {\left( -t - 1\right) }^{2} = {16} \n\]\n\n\[ \n6{t}^{2} + {12t} - {10} = 0 \n\]\n\nThe quadratic formula gives the solutions \( t = - 1 \pm \frac{4}{\sqrt{6}} \) . Putting those two values into the equations of the line gives the following two points of intersection:\n\n\[ \n\left( {2 + \frac{4}{\sqrt{6}}, - 1 + \frac{8}{\sqrt{6}},4 - \frac{4}{\sqrt{6}}}\right) \text{ and }\left( {2 - \frac{4}{\sqrt{6}}, - 1 - \frac{8}{\sqrt{6}},4 + \frac{4}{\sqrt{6}}}\right) \n\]
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Yes
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Find the intersection (if any) of the spheres \( {x}^{2} + {y}^{2} + {z}^{2} = {25} \) and \( {x}^{2} + {y}^{2} + (z - 2{)}^{2} = {16} \).
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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\[ \n{x}^{2} + {y}^{2} + {\left( z - 2\right) }^{2} = {16}\; \Rightarrow \;{x}^{2} + {y}^{2} = {16} - {\left( z - 2\right) }^{2}\text{, so } \n\]\n\n\[ \n{16} - {\left( z - 2\right) }^{2} = {25} - {z}^{2}\; \Rightarrow \;{4z} - 4 = 9\; \Rightarrow \;z = {13}/4 \n\]\n\n\[ \n\Rightarrow \;{x}^{2} + {y}^{2} = {25} - {\left( {13}/4\right) }^{2} = {231}/{16} \n\]\n\n\( \therefore \) The intersection is the circle \( {x}^{2} + {y}^{2} = \frac{231}{16} \) of radius \( \frac{\sqrt{231}}{4} \approx {3.8} \) centered at \( \left( {0,0,\frac{13}{4}}\right) .\n
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Yes
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Convert the point \( \left( {-2, - 2,1}\right) \) from Cartesian coordinates to (a) cylindrical and (b) spherical coordinates.
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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) \( \rho = \sqrt{{\left( -2\right) }^{2} + {\left( -2\right) }^{2} + {1}^{2}} = \sqrt{9} = 3,\;\phi = {\cos }^{-1}\left( \frac{1}{3}\right) \approx {1.23} \) radians.\n\n\( \therefore \left( {\rho ,\theta ,\phi }\right) = \left( {3,\frac{5\pi }{4},{1.23}}\right) \)
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Yes
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Write the equation of the cylinder \( {x}^{2} + {y}^{2} = 4 \) in cylindrical coordinates.
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Solution: Since \( r = \sqrt{{x}^{2} + {y}^{2}} \), then the equation in cylindrical coordinates is \( r = 2 \) .
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Yes
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Write the equation \( {\left( x - 2\right) }^{2} + {\left( y - 1\right) }^{2} + {z}^{2} = 9 \) in spherical coordinates.
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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\[ \n{\\rho }^{2} - 2\\sin \\phi \\left( {2\\cos \\theta + \\sin \\theta }\\right) \\rho - 4 = 0 \n\]\n\nafter combining terms. Note that this actually makes it more difficult to figure out what the surface is, as opposed to the Cartesian equation where you could immediately identify the surface as a sphere of radius 3 centered at \( \\left( {2,1,0}\\right) \) .
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No
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Example 1.34. Describe the surface given by \( \theta = z \) in cylindrical coordinates.
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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 surface restricted to \( 0 \leq z \leq {4\pi } \) and \( 0 \leq r \leq 2 \) .
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Yes
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Define \( \mathbf{f} : \mathbb{R} \rightarrow {\mathbb{R}}^{3} \) by \( \mathbf{f}\left( t\right) = \left( {\cos t,\sin t, t}\right) \).
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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 \( {x}^{2} + {y}^{2} = 1 \).
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Yes
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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 \( t \).
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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: \( x = 1, y = s, z = {2\pi } + s \) for \( - \infty < s < \infty \) .
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No
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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) \( \frac{d}{dt}\left( \mathbf{c}\right) = \mathbf{0} \)\n\n(b) \( \frac{d}{dt}\left( {k\mathbf{f}}\right) = k\frac{d\mathbf{f}}{dt} \)\n\n(c) \( \frac{d}{dt}\left( {\mathbf{f} + \mathbf{g}}\right) = \frac{d\mathbf{f}}{dt} + \frac{d\mathbf{g}}{dt} \)\n\n(d) \( \frac{d}{dt}\left( {\mathbf{f} - \mathbf{g}}\right) = \frac{d\mathbf{f}}{dt} - \frac{d\mathbf{g}}{dt} \)\n\n(e) \( \frac{d}{dt}\left( {u\mathbf{f}}\right) = \frac{du}{dt}\mathbf{f} + u\frac{d\mathbf{f}}{dt} \)\n\n(f) \( \frac{d}{dt}\left( {\mathbf{f} \cdot \mathbf{g}}\right) = \frac{d\mathbf{f}}{dt} \cdot \mathbf{g} + \mathbf{f} \cdot \frac{d\mathbf{g}}{dt} \)\n\n(g) \( \frac{d}{dt}\left( {\mathbf{f} \times \mathbf{g}}\right) = \frac{d\mathbf{f}}{dt} \times \mathbf{g} + \mathbf{f} \times \frac{d\mathbf{g}}{dt} \)
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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 \( \mathbf{f}\left( t\right) = \left( {{f}_{1}\left( t\right) ,{f}_{2}\left( t\right) ,{f}_{3}\left( t\right) }\right) \) and \( \mathbf{g}\left( t\right) = \left( {{g}_{1}\left( t\right) ,{g}_{2}\left( t\right) ,{g}_{3}\left( t\right) }\right) \), where the component functions \( {f}_{1}\left( t\right) ,{f}_{2}\left( t\right) ,{f}_{3}\left( t\right) ,{g}_{1}\left( t\right) ,{g}_{2}\left( t\right) ,{g}_{3}\left( t\right) \) are all differentiable real-valued functions.\n\n\[ \frac{d}{dt}\left( {\mathbf{f}\left( t\right) \cdot \mathbf{g}\left( t\right) }\right) = \frac{d}{dt}\left( {{f}_{1}\left( t\right) {g}_{1}\left( t\right) + {f}_{2}\left( t\right) {g}_{2}\left( t\right) + {f}_{3}\left( t\right) {g}_{3}\left( t\right) }\right) \]\n\n\[ = \frac{d}{dt}\left( {{f}_{1}\left( t\right) {g}_{1}\left( t\right) }\right) + \frac{d}{dt}\left( {{f}_{2}\left( t\right) {g}_{2}\left( t\right) }\right) + \frac{d}{dt}\left( {{f}_{3}\left( t\right) {g}_{3}\left( t\right) }\right) \]\n\n\[ = \frac{d{f}_{1}}{dt}(t){g}_{1}(t) + {f}_{1}(t)\frac{d{g}_{1}}{dt}(t) + \frac{d{f}_{2}}{dt}(t){g}_{2}(t) + {f}_{2}(t)\frac{d{g}_{2}}{dt}(t) + \frac{d{f}_{3}}{dt}(t){g}_{3}(t) + {f}_{3}(t)\frac{d{g}_{3}}{dt}(t) \]\n\n\[ = \left( {\frac{d{f}_{1}}{dt}\left( t\right) ,\frac{d{f}_{2}}{dt}\left( t\right) ,\frac{d{f}_{3}}{dt}\left( t\right) }\right) \cdot \left( {{g}_{1}\left( t\right) ,{g}_{2}\left( t\right) ,{g}_{3}\left( t\right) }\right) \]\n\n\[ + \left( {{f}_{1}\left( t\right) ,{f}_{2}\left( t\right) ,{f}_{3}\left( t\right) }\right) \cdot \left( {\frac{d{g}_{1}}{dt}\left( t\right) ,\frac{d{g}_{2}}{dt}\left( t\right) ,\frac{d{g}_{3}}{dt}\left( t\right) }\right) \]\n\n\[ = \frac{d\mathbf{f}}{dt}\left( t\right) \cdot \mathbf{g}\left( t\right) + \mathbf{f}\left( t\right) \cdot \frac{d\mathbf{g}}{dt}\left( t\right) \text{ for all }t. \]
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No
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Suppose \( \mathbf{f}\left( t\right) \) is differentiable. Find the derivative of \( \parallel \mathbf{f}\left( t\right) \parallel \) .
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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 \( \frac{d}{dt}\parallel \mathbf{f}\left( t\right) {\parallel }^{2} = 2\parallel \mathbf{f}\left( t\right) \parallel \frac{d}{dt}\parallel \mathbf{f}\left( t\right) \parallel \). But \( \parallel \mathbf{f}\left( t\right) {\parallel }^{2} = \mathbf{f}\left( t\right) \cdot \mathbf{f}\left( t\right) \), so \( \frac{d}{dt}\parallel \mathbf{f}\left( t\right) {\parallel }^{2} = \frac{d}{dt}\left( {\mathbf{f}\left( t\right) \cdot \mathbf{f}\left( t\right) }\right) \). Hence, we have \[ 2\parallel \mathbf{f}\left( t\right) \parallel \frac{d}{dt}\parallel \mathbf{f}\left( t\right) \parallel = \frac{d}{dt}\left( {\mathbf{f}\left( t\right) \cdot \mathbf{f}\left( t\right) }\right) = {\mathbf{f}}^{\prime }\left( t\right) \cdot \mathbf{f}\left( t\right) + \mathbf{f}\left( t\right) \cdot {\mathbf{f}}^{\prime }\left( t\right) \text{by Theorem 1.20(f), so} \] \[ = 2{\mathbf{f}}^{\prime }\left( t\right) \cdot \mathbf{f}\left( t\right) \text{, so if}\parallel \mathbf{f}\left( t\right) \parallel \neq 0\text{then} \] \[ \frac{d}{dt}\parallel \mathbf{f}\left( t\right) \parallel = \frac{{\mathbf{f}}^{\prime }\left( t\right) \cdot \mathbf{f}\left( t\right) }{\parallel \mathbf{f}\left( t\right) \parallel }.\]
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Yes
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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 \( a \neq 0 \).
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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 \( t \) .
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No
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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.
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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) \( \mathbf{a}\left( t\right) = \dot{\mathbf{v}}\left( t\right) = \left( {-5\cos t, - 3\sin t, - 4\sin t}\right) \)
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Yes
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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\[ \n{\mathbf{b}}_{0}^{1}\left( t\right) = \left( {1 - t}\right) {\mathbf{b}}_{0} + t{\mathbf{b}}_{1} \]\n\n\[ \n{\mathbf{b}}_{1}^{1}\left( t\right) = \left( {1 - t}\right) {\mathbf{b}}_{1} + t{\mathbf{b}}_{2} \]\n\n\[ \n{\mathbf{b}}_{0}^{2}\left( t\right) = \left( {1 - t}\right) {\mathbf{b}}_{0}^{1}\left( t\right) + t{\mathbf{b}}_{1}^{1}\left( t\right) \]\n\n\[ \n= {\left( 1 - t\right) }^{2}{\mathbf{b}}_{0} + {2t}\left( {1 - t}\right) {\mathbf{b}}_{1} + {t}^{2}{\mathbf{b}}_{2} \]\n\nfor all real \( t \) . For \( t \) in the interval \( \left\lbrack {0,1}\right\rbrack \), we see that \( {\mathbf{b}}_{0}^{1}\left( t\right) \) is the line segment between \( {\mathbf{b}}_{0} \) and \( {\mathbf{b}}_{1} \), and \( {\mathbf{b}}_{1}^{1}\left( t\right) \) is the line segment between \( {\mathbf{b}}_{1} \) and \( {\mathbf{b}}_{2} \) . The function \( {\mathbf{b}}_{0}^{2}\left( t\right) \) is the Bézier curve for the points \( {\mathbf{b}}_{0},{\mathbf{b}}_{1},{\mathbf{b}}_{2} \) . Note from the last formula that the curve is a parabola that goes through \( {\mathbf{b}}_{0} \) (when \( t = 0 \) ) and \( {\mathbf{b}}_{2} \) (when \( t = 1 \) ).
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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 \( {\mathbf{b}}_{0}^{2}\left( t\right) = \left( {{2t} + 2{t}^{2},{4t} + {t}^{2},{6t} - 4{t}^{2}}\right) \), as shown in Figure 1.8.4, where the line segments are \( {\mathbf{b}}_{0}^{1}\left( t\right) \) and \( {\mathbf{b}}_{1}^{1}\left( t\right) \), and the curve is \( {\mathbf{b}}_{0}^{2}\left( t\right) \) .
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Yes
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Find the length \( L \) of the helix \( \mathbf{f}\left( t\right) = \left( {\cos t,\sin t, t}\right) \) from \( t = 0 \) to \( t = {2\pi } \) .
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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\[ = \sqrt{2}\left( {{2\pi } - 0}\right) = 2\sqrt{2}\pi \]
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Yes
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Example 1.42. The following are all equivalent parametrizations of the same curve:
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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 \( \left\lbrack {0,\pi }\right\rbrack \) onto \( \left\lbrack {0,{2\pi }}\right\rbrack \), and is strictly increasing (since \( {\alpha }^{\prime }\left( s\right) = 2 > 0 \) for all \( s \) ). Likewise, defining \( \alpha : \left\lbrack {0,1}\right\rbrack \rightarrow \left\lbrack {0,{2\pi }}\right\rbrack \) by \( \alpha \left( s\right) = {2\pi s} \) shows that \( \mathbf{h}\left( s\right) \) is equivalent to \( \mathbf{f}\left( t\right) \) .
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Yes
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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.
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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 \( s : t = \alpha \left( s\right) = \frac{s}{\sqrt{2}} \) .\n\n\( \therefore \;\mathbf{f}\left( s\right) = \left( {\cos \frac{s}{\sqrt{2}},\sin \frac{s}{\sqrt{2}},\frac{s}{\sqrt{2}}}\right) \) for all \( s \) in \( \left\lbrack {0,2\sqrt{2}\pi }\right\rbrack \) . Note that \( \begin{Vmatrix}{{\mathbf{f}}^{\prime }\left( s\right) }\end{Vmatrix} = 1 \) .
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Yes
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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 \( \left\lbrack {a, b}\right\rbrack \) is\n\n\[ L = {\int }_{a}^{b}\sqrt{{r}^{\prime }{\left( t\right) }^{2} + r{\left( t\right) }^{2}{\theta }^{\prime }{\left( t\right) }^{2} + {z}^{\prime }{\left( t\right) }^{2}}{dt} \]
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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\[ x\left( t\right) = r\left( t\right) \cos \theta \left( t\right) ,\;y\left( t\right) = r\left( t\right) \sin \theta \left( t\right) ,\;z\left( t\right) = z\left( t\right) \]\n\nso differentiating the above expressions for \( x\left( t\right) \) and \( y\left( t\right) \) with respect to \( t \) gives\n\n\[ \begin{array}{l} {x}^{\prime }\left( t\right) = {r}^{\prime }\left( t\right) \cos \theta \left( t\right) - r\left( t\right) {\theta }^{\prime }\left( t\right) \sin \theta \left( t\right) ,\;{y}^{\prime }\left( t\right) = {r}^{\prime }\left( t\right) \sin \theta \left( t\right) + r\left( t\right) {\theta }^{\prime }\left( t\right) \cos \theta \left( t\right) \end{array} \]\n\nand so\n\n\[ {x}^{\prime }{\left( t\right) }^{2} + {y}^{\prime }{\left( t\right) }^{2} = {\left( {r}^{\prime }\left( t\right) \cos \theta \left( t\right) - r\left( t\right) {\theta }^{\prime }\left( t\right) \sin \theta \left( t\right) \right) }^{2} + {\left( {r}^{\prime }\left( t\right) \sin \theta \left( t\right) + r\left( t\right) {\theta }^{\prime }\left( t\right) \cos \theta \left( t\right) \right) }^{2} \]\n\n\[ = {r}^{\prime }{\left( t\right) }^{2}\left( {{\cos }^{2}\theta + {\sin }^{2}\theta }\right) + r{\left( t\right) }^{2}{\theta }^{\prime }{\left( t\right) }^{2}\left( {{\cos }^{2}\theta + {\sin }^{2}\theta }\right) \]\n\n\[ \left. {-2{r}^{\prime }\left( t\right) r\left( t\right) {\theta }^{\prime }\left( t\right) \cos \theta \sin \theta + 2{r}^{\prime }\left( t\right) r\left( t\right) {\theta }^{\prime }\left( t\right) \cos \theta \sin \theta }\right. \]\n\n\[ = {r}^{\prime }{\left( t\right) }^{2} + r{\left( t\right) }^{2}{\theta }^{\prime }{\left( t\right) }^{2}\text{, and so } \]\n\n\[ L = {\int }_{a}^{b}\sqrt{{x}^{\prime }{\left( t\right) }^{2} + {y}^{\prime }{\left( t\right) }^{2} + {z}^{\prime }{\left( t\right) }^{2}}{dt} \]\n\n\[ = {\int }_{a}^{b}\sqrt{{r}^{\prime }{\left( t\right) }^{2} + r{\left( t\right) }^{2}{\theta }^{\prime }{\left( t\right) }^{2} + {z}^{\prime }{\left( t\right) }^{2}}{dt} \]\n\nQED
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Yes
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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 \( \left\lbrack {0,1}\right\rbrack \) .
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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\[ L = {\int }_{0}^{1}\sqrt{{r}^{\prime }{\left( t\right) }^{2} + r{\left( t\right) }^{2}{\theta }^{\prime }{\left( t\right) }^{2} + {z}^{\prime }{\left( t\right) }^{2}}{dt} \]\n\n\[ = {\int }_{0}^{1}\sqrt{{e}^{2t} + {e}^{2t}\left( 1\right) + {e}^{2t}}{dt} \]\n\n\[ = {\int }_{0}^{1}{e}^{t}\sqrt{3}{dt} = \sqrt{3}\left( {e - 1}\right) \]
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Yes
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The domain of the function\n\n\[ f\left( {x, y}\right) = {xy} \]
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is all of \( {\mathbb{R}}^{2} \), and the range of \( f \) is all of \( \mathbb{R} \).
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Yes
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The domain of the function\n\n\[ f\left( {x, y}\right) = \frac{1}{x - y} \]
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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 .
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Yes
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The domain of the function\n\n\[ f\left( {x, y}\right) = \sqrt{1 - {x}^{2} - {y}^{2}} \]\n\nis the set \( D = \left\{ {\left( {x, y}\right) : {x}^{2} + {y}^{2} \leq 1}\right\} \)
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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 \( \left\lbrack {0,1}\right\rbrack \) in \( \mathbb{R} \).
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Yes
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\[ \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {1,2}\right) }}\frac{xy}{{x}^{2} + {y}^{2}} = \frac{\left( 1\right) \left( 2\right) }{{1}^{2} + {2}^{2}} = \frac{2}{5} \]
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since \( f\left( {x, y}\right) = \frac{xy}{{x}^{2} + {y}^{2}} \) is properly defined at the point \( \left( {1,2}\right) \) .
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Yes
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[\\mathop{\\lim }\\limits_{{\\left( {x, y}\\right) \\rightarrow \\left( {0,0}\\right) }}\\frac{xy}{{x}^{2} + {y}^{2}}\\text{ does not exist }]
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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( {0,0}\\right) \\) along different paths in \\( {\\mathbb{R}}^{2} \\) . To see this, suppose that \\( \\left( {x, y}\\right) \\rightarrow \\left( {0,0}\\right) \\) along the positive \\( x \\) -axis, so that \\( y = 0 \\) along that path. Then\n\n\\[ f\\left( {x, y}\\right) = \\frac{xy}{{x}^{2} + {y}^{2}} = \\frac{x0}{{x}^{2} + {0}^{2}} = 0 \\]\n\nalong that path (since \\( x > 0 \\) in the denominator). But if \\( \\left( {x, y}\\right) \\rightarrow \\left( {0,0}\\right) \\) along the straight line \\( y = x \\) through the origin, for \\( x > 0 \\), then we see that\n\n\\[ f\\left( {x, y}\\right) = \\frac{xy}{{x}^{2} + {y}^{2}} = \\frac{{x}^{2}}{{x}^{2} + {x}^{2}} = \\frac{1}{2} \\]\n\nwhich means that \\( f\\left( {x, y}\\right) \\) approaches different values as \\( \\left( {x, y}\\right) \\rightarrow \\left( {0,0}\\right) \\) along different paths. Hence the limit does not exist.
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Yes
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Define a function \( f\left( {x, y}\right) \) on all of \( {\mathbb{R}}^{2} \) as follows:\n\n\[ f\left( {x, y}\right) = \left\{ \begin{array}{ll} 0 & \text{ if }\left( {x, y}\right) = \left( {0,0}\right) \\ \frac{{y}^{4}}{{x}^{2} + {y}^{2}} & \text{ if }\left( {x, y}\right) \neq \left( {0,0}\right) \end{array}\right. \]\n\nThen \( f\left( {x, y}\right) \) is well-defined for all \( \left( {x, y}\right) \) in \( {\mathbb{R}}^{2} \) (i.e. there are no indeterminate forms for any \( \left( {x, y}\right) ) \), and we see that\n\n\[ \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {a, b}\right) }}f\left( {x, y}\right) = \frac{{b}^{4}}{{a}^{2} + {b}^{2}} = f\left( {a, b}\right) \text{ for }\left( {a, b}\right) \neq \left( {0,0}\right) . \]\n\nSo since\n\n\[ \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {0,0}\right) }}f\left( {x, y}\right) = 0 = f\left( {0,0}\right) \text{ by Example }{2.8}, \]
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then \( f\left( {x, y}\right) \) is continuous on all of \( {\mathbb{R}}^{2} \) .
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Yes
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Find \( \frac{\partial f}{\partial x}\left( {x, y}\right) \) and \( \frac{\partial f}{\partial y}\left( {x, y}\right) \) for the function \( f\left( {x, y}\right) = {x}^{2}y + {y}^{3} \).
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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\[ \frac{\partial f}{\partial y}\left( {x, y}\right) = {x}^{2} + 3{y}^{2}. \]
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Yes
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Find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) for the function \( f\left( {x, y}\right) = \frac{\sin \left( {x{y}^{2}}\right) }{{x}^{2} + 1} \).
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Solution: Treating \( y \) as a constant and differentiating \( f\left( {x, y}\right) \) with respect to \( x \) gives\n\n\[ \frac{\partial f}{\partial x} = \frac{\left( {{x}^{2} + 1}\right) \left( {{y}^{2}\cos \left( {x{y}^{2}}\right) }\right) - \left( {2x}\right) \sin \left( {x{y}^{2}}\right) }{{\left( {x}^{2} + 1}\right) }^{2}} \]\n\nand treating \( x \) as a constant and differentiating \( f\left( {x, y}\right) \) with respect to \( y \) gives\n\n\[ \frac{\partial f}{\partial y} = \frac{{2xy}\cos \left( {x{y}^{2}}\right) }{{x}^{2} + 1}. \]
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Yes
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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 \( f\left( {x, y}\right) = {e}^{{x}^{2}y} + x{y}^{3} \) .
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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\[ \frac{{\partial }^{2}f}{\partial {x}^{2}} = \frac{\partial }{\partial x}\left( {{2xy}{e}^{{x}^{2}y} + {y}^{3}}\right) \;\frac{{\partial }^{2}f}{\partial {y}^{2}} = \frac{\partial }{\partial y}\left( {{x}^{2}{e}^{{x}^{2}y} + {3x}{y}^{2}}\right) \]\n\n\[ = {2y}{e}^{{x}^{2}y} + 4{x}^{2}{y}^{2}{e}^{{x}^{2}y} \]\n\n\[ = {x}^{4}{e}^{{x}^{2}y} + {6xy} \]\n\n\[ \frac{{\partial }^{2}f}{\partial x\partial y} = \frac{\partial }{\partial x}\left( {{x}^{2}{e}^{{x}^{2}y} + {3x}{y}^{2}}\right) \]\n\n\[ = {2x}{e}^{{x}^{2}y} + 2{x}^{3}y{e}^{{x}^{2}y} + 3{y}^{2} \]\n\n\[ = {2x}{e}^{{x}^{2}y} + 2{x}^{3}y{e}^{{x}^{2}y} + 3{y}^{2} \]
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Yes
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Find the equation of the tangent plane to the surface \( z = {x}^{2} + {y}^{2} \) at the point \( \left( {1,2,5}\right) \) .
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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\[ 2\left( 1\right) \left( {x - 1}\right) + 2\left( 2\right) \left( {y - 2}\right) - z + 5 = 0\text{, or} \]\n\n\[ {2x} + {4y} - z - 5 = 0. \]
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Yes
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Find the equation of the tangent plane to the surface \( {x}^{2} + {y}^{2} + {z}^{2} = 9 \) at the point \( \left( {2,2, - 1}\right) \).
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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\[ 2\left( 2\right) \left( {x - 2}\right) + 2\left( 2\right) \left( {y - 2}\right) + 2\left( {-1}\right) \left( {z + 1}\right) = 0\text{, or} \]\n\n\[ {2x} + {2y} - z - 9 = 0. \]
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Yes
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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 \( \mathbf{v} = \left( {\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}}\right) \) .
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Solution: We see that \( \nabla f = \left( {{y}^{2} + 3{x}^{2}y,{2xy} + {x}^{3}}\right) \), so\n\n\[ \n{D}_{\mathbf{v}}f\left( {1,2}\right) \; = \;\mathbf{v} \cdot \nabla f\left( {1,2}\right) \; = \;\left( {\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}}\right) \cdot \left( {{2}^{2} + 3{\left( 1\right) }^{2}\left( 2\right) ,2\left( 1\right) \left( 2\right) + {1}^{3}}\right) \; = \;\frac{15}{\sqrt{2}} \n\]
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Yes
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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?
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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 \( \mathbf{v} = \frac{\nabla f}{\parallel \nabla f\parallel } = \left( {\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}}\right) \) . Thus, \( f \) increases the fastest in the direction of \( \left( {\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}}\right) \) and decreases the fastest in the direction of \( \left( {\frac{-2}{\sqrt{5}},\frac{-1}{\sqrt{5}}}\right) \).
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Yes
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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 the fastest?
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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 \( - \nabla f\left( {1,1,1}\right) = \left( {{e}^{-1},2{e}^{-2}, - 4{e}^{4}}\right) \) .
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Yes
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The function \( f\left( {x, y}\right) = {xy} \) has a critical point at \( \left( {0,0}\right) \)
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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 different signs (so that \( f\left( {x, y}\right) = {xy} < 0 = f\left( {0,0}\right) ) \) . In fact, along the path \( y = x \) in \( {\mathbb{R}}^{2}, f\left( {x, y}\right) = {x}^{2} \), which has a local minimum at \( \left( {0,0}\right) \), while along the path \( y = - x \) we have \( f\left( {x, y}\right) = - {x}^{2} \), which has a local maximum at \( \left( {0,0}\right) \) . So \( \left( {0,0}\right) \) is an example of a saddle point, i.e. it is a local maximum in one direction and a local minimum in another direction.
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Yes
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Find all local maxima and minima of \( f\left( {x, y}\right) = {x}^{2} + {xy} + {y}^{2} - {3x} \) .
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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\n\[ x + {2y}\; = 0 \]\n\nwhich has the unique solution \( \left( {x, y}\right) = \left( {2, - 1}\right) \) . So \( \left( {2, - 1}\right) \) is the only critical point.\n\nTo use Theorem 2.6, we need the second-order partial derivatives:\n\n\[ \frac{{\partial }^{2}f}{\partial {x}^{2}} = 2,\;\frac{{\partial }^{2}f}{\partial {y}^{2}} = 2,\;\frac{{\partial }^{2}f}{\partial y\partial x} = 1 \]\n\nand so\n\n\[ D = \frac{{\partial }^{2}f}{\partial {x}^{2}}\left( {2, - 1}\right) \frac{{\partial }^{2}f}{\partial {y}^{2}}\left( {2, - 1}\right) - {\left( \frac{{\partial }^{2}f}{\partial y\partial x}\left( 2, - 1\right) \right) }^{2} = \left( 2\right) \left( 2\right) - {1}^{2} = 3 > 0 \]\n\nand \( \frac{{\partial }^{2}f}{\partial {x}^{2}}\left( {2, - 1}\right) = 2 > 0 \) . Thus, \( \left( {2, - 1}\right) \) is a local minimum.
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Yes
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Find all local maxima and minima of \( f\left( {x, y}\right) = {xy} - {x}^{3} - {y}^{2} \) .
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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\[ y - 3{x}^{2} = 0 \]\n\n\[ x - {2y} = 0 \]\n\nThe first equation yields \( y = 3{x}^{2} \), substituting that into the second equation yields \( x - 6{x}^{2} = 0 \) ,\n\nwhich has the solutions \( x = 0 \) and \( x = \frac{1}{6} \) . So \( x = 0 \Rightarrow y = 3\left( 0\right) = 0 \) and \( x = \frac{1}{6} \Rightarrow y = 3{\left( \frac{1}{6}\right) }^{2} = \frac{1}{12} \) .\n\nSo the critical points are \( \left( {x, y}\right) = \left( {0,0}\right) \) and \( \left( {x, y}\right) = \left( {\frac{1}{6},\frac{1}{12}}\right) \) .\n\nTo use Theorem 2.6, we need the second-order partial derivatives:\n\n\[ \frac{{\partial }^{2}f}{\partial {x}^{2}} = - {6x},\;\frac{{\partial }^{2}f}{\partial {y}^{2}} = - 2,\;\frac{{\partial }^{2}f}{\partial y\partial x} = 1 \]\n\nSo\n\n\[ D = \frac{{\partial }^{2}f}{\partial {x}^{2}}\left( {0,0}\right) \frac{{\partial }^{2}f}{\partial {y}^{2}}\left( {0,0}\right) - {\left( \frac{{\partial }^{2}f}{\partial y\partial x}\left( 0,0\right) \right) }^{2} = \left( {-6\left( 0\right) }\right) \left( {-2}\right) - {1}^{2} = - 1 < 0 \]\n\nand thus \( \left( {0,0}\right) \) is a saddle point. Also,\n\n\[ D = \frac{{\partial }^{2}f}{\partial {x}^{2}}\left( {\frac{1}{6},\frac{1}{12}}\right) \frac{{\partial }^{2}f}{\partial {y}^{2}}\left( {\frac{1}{6},\frac{1}{12}}\right) - {\left( \frac{{\partial }^{2}f}{\partial y\partial x}\left( \frac{1}{6},\frac{1}{12}\right) \right) }^{2} = \left( {-6\left( \frac{1}{6}\right) }\right) \left( {-2}\right) - {1}^{2} = 1 > 0 \]\n\nand \( \frac{{\partial }^{2}f}{\partial {x}^{2}}\left( {\frac{1}{6},\frac{1}{12}}\right) = - 1 < 0 \) . Thus, \( \left( {\frac{1}{6},\frac{1}{12}}\right) \) is a local maximum.
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Yes
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Find all local maxima and minima of \( f\left( {x, y}\right) = {\left( x - 2\right) }^{4} + {\left( x - 2y\right) }^{2} \) .
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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 common solutions of the equations\n\n\[ 4{\left( x - 2\right) }^{3} + 2\left( {x - {2y}}\right) = 0 \]\n\n\[ - 4\left( {x - {2y}}\right) = 0 \]\n\nThe second equation yields \( x = {2y} \), substituting that into the first equation yields \( 4{\left( 2y - 2\right) }^{3} = 0 \), which has the solution \( y = 1 \), and so \( x = 2\left( 1\right) = 2 \) . Thus, \( \left( {2,1}\right) \) is the only critical point.\n\nTo use Theorem 2.6, we need the second-order partial derivatives:\n\n\[ \frac{{\partial }^{2}f}{\partial {x}^{2}} = {12}{\left( x - 2\right) }^{2} + 2,\;\frac{{\partial }^{2}f}{\partial {y}^{2}} = 8,\;\frac{{\partial }^{2}f}{\partial y\partial x} = - 4 \]\n\nSo\n\n\[ D = \frac{{\partial }^{2}f}{\partial {x}^{2}}\left( {2,1}\right) \frac{{\partial }^{2}f}{\partial {y}^{2}}\left( {2,1}\right) - {\left( \frac{{\partial }^{2}f}{\partial y\partial x}\left( 2,1\right) \right) }^{2} = \left( 2\right) \left( 8\right) - {\left( -4\right) }^{2} = 0 \]\n\nand so the test fails. What can be done in this situation? Sometimes it is possible to examine the function to see directly the nature of a critical point. In our case, we see that \( f\left( {x, y}\right) \geq 0 \) for all \( \left( {x, y}\right) \), since \( f\left( {x, y}\right) \) is the sum of fourth and second powers of numbers and hence must be nonnegative. But we also see that \( f\left( {2,1}\right) = 0 \) . Thus \( f\left( {x, y}\right) \geq 0 = f\left( {2,1}\right) \) for all \( \left( {x, y}\right) \), and hence \( \left( {2,1}\right) \) is in fact a global minimum for \( f \) .
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Yes
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Find all local maxima and minima of \( f\left( {x, y}\right) = \left( {{x}^{2} + {y}^{2}}\right) {e}^{-\left( {{x}^{2} + {y}^{2}}\right) } \) .
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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\[ \frac{\partial f}{\partial y} = {2y}\left( {1 - \left( {{x}^{2} + {y}^{2}}\right) }\right) {e}^{-\left( {{x}^{2} + {y}^{2}}\right) } \]\n\nthen the critical points are \( \left( {0,0}\right) \) and all points \( \left( {x, y}\right) \) on the unit circle \( {x}^{2} + {y}^{2} = 1 \) .\n\nTo use Theorem 2.6, we need the second-order partial derivatives:\n\n\[ \frac{{\partial }^{2}f}{\partial {x}^{2}} = 2\left\lbrack {1 - \left( {{x}^{2} + {y}^{2}}\right) - 2{x}^{2} - 2{x}^{2}\left( {1 - \left( {{x}^{2} + {y}^{2}}\right) }\right) }\right\rbrack {e}^{-\left( {{x}^{2} + {y}^{2}}\right) } \]\n\n\[ \frac{{\partial }^{2}f}{\partial {y}^{2}} = 2\left\lbrack {1 - \left( {{x}^{2} + {y}^{2}}\right) - 2{y}^{2} - 2{y}^{2}\left( {1 - \left( {{x}^{2} + {y}^{2}}\right) }\right) }\right\rbrack {e}^{-\left( {{x}^{2} + {y}^{2}}\right) } \]\n\n\[ \frac{{\partial }^{2}f}{\partial y\partial x} = - {4xy}\left\lbrack {2 - \left( {{x}^{2} + {y}^{2}}\right) }\right\rbrack {e}^{-\left( {{x}^{2} + {y}^{2}}\right) } \]\n\nAt \( \left( {0,0}\right) \), we have \( D = 4 > 0 \) and \( \frac{{\partial }^{2}f}{\partial {x}^{2}}\left( {0,0}\right) = 2 > 0 \), so \( \left( {0,0}\right) \) is a local minimum. However, for points \( \left( {x, y}\right) \) on the unit circle \( {x}^{2} + {y}^{2} = 1 \), we have\n\n\[ D = \left( {-4{x}^{2}{e}^{-1}}\right) \left( {-4{y}^{2}{e}^{-1}}\right) - {\left( -4xy{e}^{-1}\right) }^{2} = 0 \]\n\nand so the test fails. If we look at the graph of \( f\left( {x, y}\right) \), as shown in Figure 2.5.2, it looks like we might have a local maximum for \( \left( {x, y}\right) \) on the unit circle \( {x}^{2} + {y}^{2} = 1 \) . If we switch to using polar coordinates \( \left( {r,\theta }\right) \) instead of \( \left( {x, y}\right) \) in \( {\mathbb{R}}^{2} \), where \( {r}^{2} = {x}^{2} + {y}^{2} \), then we see that we can write \( f\left( {x, y}\right) \) as a function \( g\left( r\right) \) of the variable \( r \) alone: \( g\left( r\right) = {r}^{2}{e}^{-{r}^{2}} \) . Then \( {g}^{\prime }\left( r\right) = {2r}\left( {1 - {r}^{2}}\right) {e}^{-{r}^{2}} \) , so it has a critical point at \( r = 1 \), and we can check that \( {g}^{\prime \prime }\left( 1\right) = - 4{e}^{-1} < 0 \), so the Second Derivative Test from single-variable calculus says that \( r = 1 \) is a local maximum. But \( r = 1 \) corresponds to the unit circle \( {x}^{2} + {y}^{2} = 1 \) . Thus, the points \( \left( {x, y}\right) \) on the unit circle \( {x}^{2} + {y}^{2} = 1 \) are local maximum points for \( f \) .
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Yes
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Find all local maxima and minima of \( f\left( {x, y}\right) = {x}^{3} - {xy} - x + x{y}^{3} - {y}^{4} \).
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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\[ \frac{{\partial }^{2}f}{\partial {x}^{2}} = {6x},\;\frac{{\partial }^{2}f}{\partial {y}^{2}} = {6xy} - {12}{y}^{2},\;\frac{{\partial }^{2}f}{\partial y\partial x} = - 1 + 3{y}^{2} \]\n\nNotice that solving \( \nabla f = \mathbf{0} \) would involve solving two third-degree polynomial equations in \( x \) and \( y \), which in this case can not be done easily.\n\nWe need to pick an initial point \( \left( {{x}_{0},{y}_{0}}\right) \) for our algorithm. Looking at the graph of \( z = f\left( {x, y}\right) \) over a large region may help (see Figure 2.6.1 below), though it may be hard to tell where the critical points are.\n\n\n\nFigure 2.6.1 \( f\left( {x, y}\right) = {x}^{3} - {xy} - x + x{y}^{3} - {y}^{4} \) for \( - {20} \leq x \leq {20} \) and \( - {20} \leq y \leq {20} \)\n\nNotice in the formulas (2.14) that we divide by \( D \) , so we should pick an initial point where \( D \) is not zero. And we can see that \( D\left( {0,0}\right) = \left( 0\right) \left( 0\right) - {\left( -1\right) }^{2} = - 1 \neq 0 \), so take \( \left( {0,0}\right) \) as our initial point. Since it may take a large number of iterations of Newton's algorithm to be sure that we are close enough to the actual critical point, and since the computations are quite tedious, we will let a computer do the computing. For this, we will write a simple program, using the Java programming language, which will take a given initial point as a parameter and then perform 100 iterations of Newton's algorithm. In each iteration the new point will be printed, so that we can see if there is convergence. The full code is shown in Listing 2.1.
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No
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For a rectangle whose perimeter is \( {20}\mathrm{\;m} \), find the dimensions that will maximize the area.
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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\[ \n\text{Maximize :}f\left( {x, y}\right) = {xy} \n\]\n\n\[ \n\text{given :}{2x} + {2y} = {20} \n\]\n\nThe reader is probably familiar with a simple method, using single-variable calculus, for solving this problem. Since we must have \( {2x} + {2y} = {20} \), then we can solve for, say, \( y \) in terms of \( x \) using that equation. This gives \( y = {10} - x \), which we then substitute into \( f \) to get \( f\left( {x, y}\right) = {xy} = x\left( {{10} - x}\right) = {10x} - {x}^{2} \) . This is now a function of \( x \) alone, so we now just have to maximize the function \( f\left( x\right) = {10x} - {x}^{2} \) on the interval \( \left\lbrack {0,{10}}\right\rbrack \) . Since \( {f}^{\prime }\left( x\right) = {10} - {2x} = 0 \Rightarrow x = 5 \) and \( {f}^{\prime \prime }\left( 5\right) = - 2 < 0 \), then the Second Derivative Test tells us that \( x = 5 \) is a local maximum for \( f \), and hence \( x = 5 \) must be the global maximum on the interval \( \left\lbrack {0,{10}}\right\rbrack \) (since \( f = 0 \) at the endpoints of the interval). So since \( y = {10} - x = 5 \), then the maximum area occurs for a rectangle whose width and height both are \( 5\mathrm{\;m} \) .
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Yes
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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 constrained optimization problem\n\n\[ \n\text{Maximize (or minimize):}f\left( {x, y}\right)\n\]\n\n\[ \n\text{given :}g\left( {x, y}\right) = c\text{,}\n\]\n\nfind the points \( \left( {x, y}\right) \) that solve the equation \( \nabla f\left( {x, y}\right) = \lambda \nabla g\left( {x, y}\right) \) for some constant \( \lambda \) (the number \( \lambda \) is called the Lagrange multiplier). If there is a constrained maximum or minimum, then it must be such a point.
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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 \( \nabla f\left( {x, y}\right) = \lambda \nabla g\left( {x, y}\right) \) for some \( \lambda \) actually is a constrained maximum or minimum can sometimes be determined by the nature of the problem itself.
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No
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For a rectangle whose perimeter is \( {20}\mathrm{\;m} \), use the Lagrange multiplier method to find the dimensions that will maximize the area.
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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 \( \nabla f\left( {x, y}\right) = \lambda \nabla g\left( {x, y}\right) \) for some \( \lambda \) means solving the equations\n\n\( \frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \) and \( \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y} \), namely:\n\n\[ y = {2\lambda } \]\n\n\[ x = {2\lambda } \]\n\nThe general idea is to solve for \( \lambda \) in both equations, then set those expressions equal (since they both equal \( \lambda \) ) to solve for \( x \) and \( y \) . Doing this we get\n\n\[ \frac{y}{2} = \lambda = \frac{x}{2} \Rightarrow x = y, \]\n\nso now substitute either of the expressions for \( x \) or \( y \) into the constraint equation to solve for \( x \) and \( y \) :\n\n\[ {20} = g\left( {x, y}\right) = {2x} + {2y} = {2x} + {2x} = {4x}\; \Rightarrow \;x = 5\; \Rightarrow \;y = 5 \]\n\nThere must be a maximum area, since the minimum area is 0 and \( f\left( {5,5}\right) = {25} > 0 \), so the point \( \left( {5,5}\right) \) that we found (called a constrained critical point) must be the constrained maximum.\n\n\( \therefore \) The maximum area occurs for a rectangle whose width and height both are \( 5\mathrm{\;m} \).
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Yes
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Find the points on the circle \( {x}^{2} + {y}^{2} = {80} \) which are closest to and farthest from the point \( \left( {1,2}\right) \) .
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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\[ \text{Maximize (and minimize):}f\left( {x, y}\right) = {\left( x - 1\right) }^{2} + {\left( y - 2\right) }^{2} \]\n\n\[ \text{given :}g\left( {x, y}\right) = {x}^{2} + {y}^{2} = {80} \]\n\nSolving \( \nabla f\left( {x, y}\right) = \lambda \nabla g\left( {x, y}\right) \) means solving the following equations:\n\n\[ 2\left( {x - 1}\right) = {2\lambda x} \]\n\n\[ 2\left( {y - 2}\right) = {2\lambda y} \]\n\nNote that \( x \neq 0 \) since otherwise we would get \( - 2 = 0 \) in the first equation. Similarly, \( y \neq 0 \) . So we can solve both equations for \( \lambda \) as follows:\n\n\[ \frac{x - 1}{x} = \lambda = \frac{y - 2}{y} \Rightarrow {xy} - y = {xy} - {2x} \Rightarrow y = {2x} \]\n\nSubstituting this into \( g\left( {x, y}\right) = {x}^{2} + {y}^{2} = {80} \) yields \( 5{x}^{2} = {80} \),\n\nso \( x = \pm 4 \) . So the two constrained critical points are \( \left( {4,8}\right) \) and \( \left( {-4, - 8}\right) \) . Since \( f\left( {4,8}\right) = {45} \) and \( f\left( {-4, - 8}\right) = {125} \), and since there must be points on the circle closest to and farthest from \( \left( {1,2}\right) \) , then it must be the case that \( \left( {4,8}\right) \) is the point on the circle closest to \( \left( {1,2}\right) \) and \( \left( {-4, - 8}\right) \) is the farthest from \( \left( {1,2}\right) \) (see Figure 2.7.1).
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Yes
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Find the volume \( V \) under the plane \( z = {8x} + {6y} \) over the rectangle \( R = \left\lbrack {0,1}\right\rbrack \times \) \( \left\lbrack {0,2}\right\rbrack \) .
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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\[ = {\int }_{0}^{2}\left( {4 + {6y}}\right) {dy} \]\n\n\[ = {\left. 4y + 3{y}^{2}\right| }_{0}^{2} \]\n\n\[ = {20} \]
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Yes
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Find the volume \( V \) under the surface \( z = {e}^{x + y} \) over the rectangle \( R = \left\lbrack {2,3}\right\rbrack \times \left\lbrack {1,2}\right\rbrack \) .
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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\[ = {\int }_{1}^{2}\left( {{e}^{y + 3} - {e}^{y + 2}}\right) {dy} \]\n\n\[ = {\left. {e}^{y + 3} - {e}^{y + 2}\right| }_{1}^{2} \]\n\n\[ = \;{e}^{5} - {e}^{4} - \left( {{e}^{4} - {e}^{3}}\right) \; = \;{e}^{5} - 2{e}^{4} + {e}^{3} \]
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Yes
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Evaluate \( {\int }_{0}^{2\pi }{\int }_{0}^{\pi }\sin \left( {x + y}\right) {dxdy} \).
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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\[ \n{\int }_{0}^{2\pi }{\int }_{0}^{\pi }\sin \left( {x + y}\right) {dxdy} = {\int }_{0}^{2\pi }\left( {-{\left. \cos \left( x + y\right) \right| }_{x = 0}^{x = \pi }}\right) {dy} \n\] \n\n\[ \n= {\int }_{0}^{2\pi }\left( {-\cos \left( {y + \pi }\right) + \cos y}\right) {dy} \n\] \n\n\[ \n= - {\left. \sin \left( y + \pi \right) + \sin y\right| }_{0}^{2\pi } = - \sin {3\pi } + \sin {2\pi } - \left( {-\sin \pi + \sin 0}\right) \n\] \n\n\[ \n= 0 \n\]
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Yes
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Find the volume \( V \) under the plane \( z = {8x} + {6y} \) over the region \( R = \{ \left( {x, y}\right) : 0 \leq \) \( \left. {x \leq 1,0 \leq y \leq 2{x}^{2}}\right\} \) .
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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\[ = {\int }_{0}^{1}\left( {{8xy} + {\left. 3{y}^{2}\right| }_{y = 0}^{y = 2{x}^{2}}}\right) {dx} \]\n\n\[ = {\int }_{0}^{1}\left( {{16}{x}^{3} + {12}{x}^{4}}\right) {dx} \]\n\n\[ = {\left. 4{x}^{4} + \frac{12}{5}{x}^{5}\right| }_{0}^{1} = 4 + \frac{12}{5} = \frac{32}{5} = {6.4} \]
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Yes
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Find the volume \( V \) of the solid bounded by the three coordinate planes and the plane \( {2x} + y + {4z} = 4 \) .
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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 \( R = \{ \left( {x, y}\right) : 0 \leq x \leq 2,0 \leq y \leq - {2x} + 4\} \) . Using vertical slices in \( R \) gives\n\n\[ V = {\iint }_{R}\frac{1}{4}\left( {4 - {2x} - y}\right) {dA} \]\n\n\[ = {\int }_{0}^{2}\left\lbrack {{\int }_{0}^{-{2x} + 4}\frac{1}{4}\left( {4 - {2x} - y}\right) {dy}}\right\rbrack {dx} \]\n\n\[ = {\int }_{0}^{2}\left( {-{\left. \frac{1}{8}{\left( 4 - 2x - y\right) }^{2}\right| }_{y = 0}^{y = - {2x} + 4}}\right) {dx} \]\n\n\[ = {\int }_{0}^{2}\frac{1}{8}{\left( 4 - 2x\right) }^{2}{dx} \]\n\n\[ = {\left. -\frac{1}{48}{\left( 4 - 2x\right) }^{3}\right| }_{0}^{2}\; = \;\frac{64}{48}\; = \;\frac{4}{3} \]
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Yes
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Evaluate \( {\int }_{1}^{\infty }{\int }_{0}^{1/{x}^{2}}{2ydydx} \) .
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\[ {\int }_{1}^{\infty }{\int }_{0}^{1/{x}^{2}}{2ydydx} = {\int }_{1}^{\infty }\left( {\left. {y}^{2}\right| }_{y = 0}^{y = 1/{x}^{2}}\right) {dx} \] \[ = {\int }_{1}^{\infty }{x}^{-4}{dx} = {\left. -\frac{1}{3}{x}^{-3}\right| }_{1}^{\infty } = 0 - \left( {-\frac{1}{3}}\right) = \frac{1}{3} \]
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Yes
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Example 3.7. Evaluate \( {\int }_{0}^{3}{\int }_{0}^{2}{\int }_{0}^{1}\left( {{xy} + z}\right) {dxdydz} \) .
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\[ \n{\int }_{0}^{3}{\int }_{0}^{2}{\int }_{0}^{1}\left( {{xy} + z}\right) \;{dx}\;{dy}\;{dz} = {\int }_{0}^{3}{\int }_{0}^{2}\left( {\left. \frac{1}{2}{x}^{2}y + xz\right| }_{x = 0}^{x = 1}\right) \;{dy}\;{dz} \]\n\[ = {\int }_{0}^{3}{\int }_{0}^{2}\left( {\frac{1}{2}y + z}\right) {dydz} \]\n\[ = {\int }_{0}^{3}\left( {\frac{1}{4}{y}^{2} + {\left. yz\right| }_{y = 0}^{y = 2}}\right) {dz} \]\n\[ = {\int }_{0}^{3}\left( {1 + {2z}}\right) {dz} \]\n\[ = {\left. z + {z}^{2}\right| }_{0}^{3} = {12} \]\n
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Yes
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Evaluate \( {\int }_{0}^{1}{\int }_{0}^{1 - x}{\int }_{0}^{2 - x - y}\left( {x + y + z}\right) {dzdydx} \) .
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\[ {\int }_{0}^{1}{\int }_{0}^{1 - x}{\int }_{0}^{2 - x - y}\left( {x + y + z}\right) \;{dz}\;{dy}\;{dx}\; = \;{\int }_{0}^{1}{\int }_{0}^{1 - x}\left( {\left( {x + y}\right) z + {\left. \frac{1}{2}{z}^{2}\right| }_{z = 0}^{z = 2 - x - y}}\right) \;{dy}\;{dx} \]\n\[ = {\int }_{0}^{1}{\int }_{0}^{1 - x}\left( {\left( {x + y}\right) \left( {2 - x - y}\right) + \frac{1}{2}{\left( 2 - x - y\right) }^{2}}\right) \;{dy}\;{dx} \]\n\[ = {\int }_{0}^{1}{\int }_{0}^{1 - x}\left( {2 - \frac{1}{2}{x}^{2} - {xy} - \frac{1}{2}{y}^{2}}\right) \;{dy}\;{dx} \]\n\[ = {\int }_{0}^{1}\left( {{2y} - \frac{1}{2}{x}^{2}y - {xy} - \frac{1}{2}x{y}^{2} - \frac{1}{6}{y}^{3}{\left. \right| }_{y = 0}^{y = 1 - x}}\right) {dx} \]\n\[ = {\int }_{0}^{1}\left( {\frac{11}{6} - {2x} + \frac{1}{6}{x}^{3}}\right) \;{dx} \]\n\[ = {\left. \frac{11}{6}x - {x}^{2} + \frac{1}{24}{x}^{4}\right| }_{0}^{1}\; = \;\frac{7}{8} \]
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Yes
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Evaluate \( {\iint }_{R}{e}^{\frac{x - y}{x + y}}{dA} \), where \( R = \{ \left( {x, y}\right) : x \geq 0, y \geq 0, x + y \leq 1\} \) .
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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), we need to write both \( x \) and \( y \) in terms of \( u \) and \( v \) . So solving for \( x \) and \( y \) gives \( x = \frac{1}{2}\left( {u + v}\right) \) and \( y = \frac{1}{2}\left( {v - u}\right) \) . In Figure 3.5.1 below, we see how the mapping \( x = x\left( {u, v}\right) = \frac{1}{2}\left( {u + v}\right), y = y\left( {u, v}\right) = \frac{1}{2}\left( {v - u}\right) \) maps the region \( {R}^{\prime } \) onto \( R \) in a one-to-one manner.\n\n\n\nFigure 3.5.1 The regions \( R \) and \( {R}^{\prime }\n\nNow we see that\n\n\[ J\left( {u, v}\right) = \left| \begin{array}{ll} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array}\right| = \left| \begin{array}{rr} \frac{1}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{1}{2} \end{array}\right| = \frac{1}{2} \Rightarrow \left| {J\left( {u, v}\right) }\right| = \left| \frac{1}{2}\right| = \frac{1}{2}, \]\n\nso using horizontal slices in \( {R}^{\prime } \), we have\n\n\[ {\iint }_{R}{e}^{\frac{x - y}{x + y}}{dA} = {\iint }_{{R}^{\prime }}f\left( {x\left( {u, v}\right), y\left( {u, v}\right) }\right) \left| {J\left( {u, v}\right) }\right| {dA} \]\n\n\[ = {\int }_{0}^{1}{\int }_{-v}^{v}{e}^{\frac{u}{v}}\frac{1}{2}{dudv} \]\n\n\[ = {\int }_{0}^{1}\left( {\left. \frac{v}{2}{e}^{\frac{u}{v}}\right| }_{u = - v}^{u = v}\right) {dv} \]\n\n\[ = {\int }_{0}^{1}\frac{v}{2}\left( {e - {e}^{-1}}\right) {dv} \]\n\n\[ = {\left. \frac{{v}^{2}}{4}\left( e - {e}^{-1}\right) \right| }_{0}^{1} = \frac{1}{4}\left( {e - \frac{1}{e}}\right) = \frac{{e}^{2} - 1}{4e} \]
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Yes
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Find the volume \( V \) inside the paraboloid \( z = {x}^{2} + {y}^{2} \) for \( 0 \leq z \leq 1 \) .
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\[ V = {\iint }_{R}\left( {1 - z}\right) {dA} = {\iint }_{R}\left( {1 - \left( {{x}^{2} + {y}^{2}}\right) }\right) {dA}, \] 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 \( \left( {r,\theta }\right) \) we know that \( {x}^{2} + {y}^{2} = {r}^{2} \) and that the unit disk \( R \) is the set \( {R}^{\prime } = \{ \left( {r,\theta }\right) \) : \( 0 \leq r \leq 1,0 \leq \theta \leq {2\pi }\} \) . Thus, \[ V = {\int }_{0}^{2\pi }{\int }_{0}^{1}\left( {1 - {r}^{2}}\right) {rdrd\theta } \] \[ = {\int }_{0}^{2\pi }{\int }_{0}^{1}\left( {r - {r}^{3}}\right) {drd\theta } \] \[ = {\int }_{0}^{2\pi }\left( {\frac{{r}^{2}}{2} - {\left. \frac{{r}^{4}}{4}\right| }_{r = 0}^{r = 1}}\right) \;{d\theta } \] \[ = {\int }_{0}^{2\pi }\frac{1}{4}{d\theta } \] \[ = \frac{\pi }{2} \]
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Yes
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Find the volume \( V \) inside the cone \( z = \sqrt{{x}^{2} + {y}^{2}} \) for \( 0 \leq z \leq 1 \) .
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Solution: Using vertical slices, we see that\n\n\n\nFigure 3.5.3 \( z = \sqrt{{x}^{2} + {y}^{2}} \)\n\n\[ V = {\iint }_{R}\left( {1 - z}\right) {dA} = {\iint }_{R}\left( {1 - \sqrt{{x}^{2} + {y}^{2}}}\right) {dA}, \]\n\nwhere \( R = \left\{ {\left( {x, y}\right) : {x}^{2} + {y}^{2} \leq 1}\right\} \) is the unit disk in \( {\mathbb{R}}^{2} \) (see Figure 3.5.3). In polar coordinates \( \left( {r,\theta }\right) \) we know that \( \sqrt{{x}^{2} + {y}^{2}} = r \) and that the unit disk \( R \) is the set \( {R}^{\prime } = \{ \left( {r,\theta }\right) : 0 \leq r \leq 1,0 \leq \theta \leq {2\pi }\} \) . Thus,\n\n\[ V = {\int }_{0}^{2\pi }{\int }_{0}^{1}\left( {1 - r}\right) {rdrd\theta } \]\n\n\[ = {\int }_{0}^{2\pi }{\int }_{0}^{1}\left( {r - {r}^{2}}\right) {drd\theta } \]\n\n\[ = {\int }_{0}^{2\pi }\left( {\frac{{r}^{2}}{2} - {\left. \frac{{r}^{3}}{3}\right| }_{r = 0}^{r = 1}}\right) \;{d\theta } \]\n\n\[ = {\int }_{0}^{2\pi }\frac{1}{6}{d\theta } \]\n\n\[ = \frac{\pi }{3} \]
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Yes
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For \( a > 0 \), find the volume \( V \) inside the sphere \( S = {x}^{2} + {y}^{2} + {z}^{2} = {a}^{2} \) .
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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\[ \n= {\int }_{0}^{2\pi }{\int }_{0}^{\pi }\left( {\frac{{\rho }^{3}}{3}{\left. \right| }_{\rho = 0}^{\rho = a}}\right) \sin {\phi d\phi d\theta } = {\int }_{0}^{2\pi }{\int }_{0}^{\pi }\frac{{a}^{3}}{3}\sin {\phi d\phi d\theta } \n\]\n\n\[ \n= {\int }_{0}^{2\pi }\left( {-{\left. \frac{{a}^{3}}{3}\cos \phi \right| }_{\phi = 0}^{\phi = \pi }}\right) {d\theta } = {\int }_{0}^{2\pi }\frac{2{a}^{3}}{3}{d\theta } = \frac{{4\pi }{a}^{3}}{3}. \n\]
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Yes
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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 \( \\delta \\left( {x, y}\\right) = x + y \) .
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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\[ \n= {\\int }_{0}^{1}{\\int }_{0}^{2{x}^{2}}\\left( {x + y}\\right) {dydx} \n\]\n\n\[ \n= {\\int }_{0}^{1}\\left( {{xy} + {\\left. \\frac{{y}^{2}}{2}\\right| }_{y = 0}^{y = 2{x}^{2}}}\\right) {dx} \n\]\n\n\[ \n= {\\int }_{0}^{1}\\left( {2{x}^{3} + 2{x}^{4}}\\right) {dx} \n\]\n\n\[ \n= {\\left. \\frac{{x}^{4}}{2} + \\frac{2{x}^{5}}{5}\\right| }_{0}^{1} = \\frac{9}{10} \n\]\nand\n\n\[ \n{M}_{x} = {\\iint }_{R}{y\\delta }\\left( {x, y}\\right) {dA}\\{M}_{y} = {\\iint }_{R}{x\\delta }\\left( {x, y}\\right) {dA} \n\]\n\n\[ \n= {\\int }_{0}^{1}{\\int }_{0}^{2{x}^{2}}y\\left( {x + y}\\right) {dydx}\\; = {\\int }_{0}^{1}{\\int }_{0}^{2{x}^{2}}x\\left( {x + y}\\right) {dydx} \n\]\n\n\[ \n= {\\int }_{0}^{1}\\left( {\\left. \\frac{x{y}^{2}}{2} + \\frac{{y}^{3}}{3}\\right| }_{y = 0}^{y = 2{x}^{2}}\\right) {dx}\\; = {\\int }_{0}^{1}\\left( {{x}^{2}y + {\\left. \\frac{x{y}^{2}}{2}\\right| }_{y = 0}^{y = 2{x}^{2}}}\\right) {dx} \n\]\n\n\[ \n= {\\int }_{0}^{1}\\left( {2{x}^{5} + \\frac{8{x}^{6}}{3}}\\right) {dx}\\; = {\\int }_{0}^{1}\\left( {2{x}^{4} + 2{x}^{5}}\\right) {dx} \n\]\n\n\[ \n= {\\left. \\frac{{x}^{6}}{3} + \\frac{8{x}^{7}}{21}\\right| }_{0}^{1} = \\frac{5}{7}\\; = {\\left. \\frac{2{x}^{5}}{5} + \\frac{{x}^{6}}{3}\\right| }_{0}^{1} = \\frac{11}{15}, \n\]\n\nso the center of mass \( \\left( {\\bar{x},\\bar{y}}\\right) \) is given by\n\n\[ \n\\bar{x} = \\frac{{M}_{y}}{M} = \\frac{{11}/{15}}{9/{10}} = \\frac{22}{27},\\;\\bar{y} = \\frac{{M}_{x}}{M} = \\frac{5/7}{9/{10}} = \\frac{50}{63}. \n\]
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Yes
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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 \( \delta \left( {x, y, z}\right) = 1 \) .
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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 symmetric about the \( z \) -axis, then it is clear that \( \bar{x} = 0 \) and \( \bar{y} = 0 \), so we need only find \( \bar{z} \) . We have\n\n\[ M = {\iiint }_{S}\delta \left( {x, y, z}\right) {dV} = {\iiint }_{S}{1dV} = \operatorname{Volume}\left( S\right) . \]\n\nBut since the volume of \( S \) is half the volume of the sphere of radius \( a \), which we know by Example 3.12 is \( \frac{{4\pi }{a}^{3}}{3} \), then \( M = \frac{{2\pi }{a}^{3}}{3} \) . And\n\n\[ {M}_{xy} = {\iiint }_{S}{z\delta }\left( {x, y, z}\right) {dV} \]\n\n\[ = {\iiint }_{S}{zdV}\text{, which in spherical coordinates is} \]\n\n\[ = {\int }_{0}^{2\pi }{\int }_{0}^{\pi /2}{\int }_{0}^{a}\left( {\rho \cos \phi }\right) {\rho }^{2}\sin {\phi d\rho d\phi d\theta } \]\n\n\[ = {\int }_{0}^{2\pi }{\int }_{0}^{\pi /2}\sin \phi \cos \phi \left( {{\int }_{0}^{a}{\rho }^{3}{d\rho }}\right) {d\phi d\theta } \]\n\n\[ = {\int }_{0}^{2\pi }{\int }_{0}^{\pi /2}\frac{{a}^{4}}{4}\sin \phi \cos {\phi d\phi d\theta } \]\n\n\[ {M}_{xy} = {\int }_{0}^{2\pi }{\int }_{0}^{\pi /2}\frac{{a}^{4}}{8}\sin {2\phi d\phi d\theta }\;\left( {\text{ since }\sin {2\phi } = 2\sin \phi \cos \phi }\right) \]\n\n\[ = {\int }_{0}^{2\pi }\left( {-{\left. \frac{{a}^{4}}{16}\cos 2\phi \right| }_{\phi = 0}^{\phi = \pi /2}}\right) {d\theta } \]\n\n\[ = {\int }_{0}^{2\pi }\frac{{a}^{4}}{8}{d\theta } \]\n\n\[ = \frac{\pi {a}^{4}}{4}, \]\n\nso\n\n\[ \bar{z} = \frac{{M}_{xy}}{M} = \frac{\frac{\pi {a}^{4}}{4}}{\frac{{2\pi }{a}^{3}}{3}} = \frac{3a}{8}. \]\n\nThus, the center of mass of \( S \) is \( \left( {\bar{x},\bar{y},\bar{z}}\right) = \left( {0,0,\frac{3a}{8}}\right) \) .
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Yes
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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\[ F\left( x\right) = P\left( {X \leq x}\right) = \left\{ \begin{array}{ll} 1, & \text{ for }x \geq 1 \\ x, & \text{ for }0 < x < 1 \\ 0, & \text{ for }x \leq 0, \end{array}\right. \]
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and probability density function\n\n\[ f\left( x\right) = {F}^{\prime }\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ for }0 < x < 1 \\ 0, & \text{ elsewhere. } \end{array}\right. \]
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Yes
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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?
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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\[ \nX = \\min \\left( {{U}_{1},\\ldots ,{U}_{n}}\\right) \\;\\text{ and }\\;Y = \\max \\left( {{U}_{1},\\ldots ,{U}_{n}}\\right) .\n\]\n\nThen it can be shown \( {}^{4} \) that the joint p.d.f. of \( X \) and \( Y \) is\n\n\[ \nf\\left( {x, y}\\right) = \\left\\{ \\begin{array}{ll} n\\left( {n - 1}\\right) {\\left( y - x\\right) }^{n - 2}, & \\text{ for }0 \\leq x \\leq y \\leq 1 \\\\ 0, & \\text{ elsewhere. } \\end{array}\\right.\n\]\n\n(3.56)\n\nThus, the expected value of \( X \) is\n\n\[ \n{EX} = {\\int }_{0}^{1}{\\int }_{x}^{1}n\\left( {n - 1}\\right) x{\\left( y - x\\right) }^{n - 2}{dydx}\n\]\n\n\[ \n= {\\int }_{0}^{1}\\left( {\\left. nx{\\left( y - x\\right) }^{n - 1}\\right| }_{y = x}^{y = 1}\\right) {dx}\n\]\n\n\[ \n= {\\int }_{0}^{1}{nx}{\\left( 1 - x\\right) }^{n - 1}{dx}\\text{, so integration by parts yields}\n\]\n\n\[ \n= - {\\left. x{\\left( 1 - x\\right) }^{n} - \\frac{1}{n + 1}{\\left( 1 - x\\right) }^{n + 1}\\right| }_{0}^{1}\n\]\n\n\[ \n{EX} = \\frac{1}{n + 1}\n\]\n\n\( {}^{4} \) See Ch. 6 in HOEL, PORT and STONE. and similarly (see Exercise 3) it can be shown that\n\n\[ \n{EY} = {\\int }_{0}^{1}{\\int }_{0}^{y}n\\left( {n - 1}\\right) y{\\left( y - x\\right) }^{n - 2}{dxdy} = \\frac{n}{n + 1}.\n\]
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No
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Use a line integral to show that the lateral surface area \( A \) of a right circular cylinder of radius \( r \) and height \( h \) is \( {2\pi rh} \) .
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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\[ x = x\left( t\right) = r\cos t,\;y = y\left( t\right) = r\sin t,\;0 \leq t \leq {2\pi } \]\n\nLet \( f\left( {x, y}\right) = h \) for all \( \left( {x, y}\right) \) . Then\n\n\[ A = {\int }_{C}f\left( {x, y}\right) {ds} = {\int }_{a}^{b}f\left( {x\left( t\right), y\left( t\right) }\right) \sqrt{{x}^{\prime }{\left( t\right) }^{2} + {y}^{\prime }{\left( t\right) }^{2}}{dt} \]\n\n\[ = {\int }_{0}^{2\pi }h\sqrt{{\left( -r\sin t\right) }^{2} + {\left( r\cos t\right) }^{2}}{dt} \]\n\n\[ = h{\int }_{0}^{2\pi }r\sqrt{{\sin }^{2}t + {\cos }^{2}t}{dt} \]\n\n\[ = {rh}{\int }_{0}^{2\pi }{1dt} = {2\pi rh} \]
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Yes
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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 \( \mathbf{r}\left( t\right) = x\left( t\right) \mathbf{i} + y\left( t\right) \mathbf{j} \) ,\n\n\[{\int }_{C}\mathbf{f} \cdot d\mathbf{r} = {\int }_{C}\mathbf{f} \cdot \mathbf{T}{ds}\]\n\n(4.15)\n\nwhere \( \mathbf{T}\left( t\right) = \frac{{\mathbf{r}}^{\prime }\left( t\right) }{\begin{Vmatrix}{\mathbf{r}}^{\prime }\left( t\right) \end{Vmatrix}} \) is the unit tangent vector to \( C \) at \( \left( {x\left( t\right), y\left( t\right) }\right) \) .
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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\[W = {\int }_{C}\mathbf{f} \cdot \mathbf{T}{ds} = {\int }_{C}\mathbf{f} \cdot d\mathbf{r}.\n\n(4.16)\]
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Yes
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