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Let \( {A}_{i} \) be the event the \( i \) th card is drawn on the first selection and let \( {B}_{i} \) be the event the card numbered \( i \) is drawn on the second selection (from the box). Determine \( P\left( {B}_{5}\right), P\left( {{A}_{1}{B}_{5}}\right) \), and \( P\left( {{A}_{1} \mid {B}_{5}}\right) \).
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From Example 3.4 (What is the conditioning event?), we have \( P\left( {A}_{i}\right) = 6/{10} \) and \( P\left( {{A}_{i}{A}_{j}}\right) = \) \( 3/{10} \) . This implies\n\n\[ P\left( {{A}_{i}{A}_{j}^{c}}\right) = P\left( {A}_{i}\right) - P\left( {{A}_{i}{A}_{j}}\right) = 3/{10} \]\n\n(3.22)\n\nNow we can draw card five on the second selection only if it is selected on the first drawing, so that \( {B}_{5} \subset {A}_{5} \) . Also \( {A}_{5} = {A}_{1}{A}_{5}\bigvee {A}_{1}^{c}{A}_{5} \) . We therefore have \( {B}_{5} = {B}_{5}{A}_{5} = {B}_{5}{A}_{1}{A}_{5}\bigvee {B}_{5}{A}_{1}^{c}{A}_{5} \) . By the law of total probability (CP2) (p. 66),\n\n\[ P\left( {B}_{5}\right) = P\left( {{B}_{5} \mid {A}_{1}{A}_{5}}\right) P\left( {{A}_{1}{A}_{5}}\right) + P\left( {{B}_{5} \mid {A}_{1}^{c}{A}_{5}}\right) P\left( {{A}_{1}^{c}{A}_{5}}\right) = \frac{1}{2} \cdot \frac{3}{10} + \frac{1}{3} \cdot \frac{3}{10} = \frac{1}{4} \]\n\n\( \left( {3.23}\right) \)\n\nAlso, since \( {A}_{1}{B}_{5} = {A}_{1}{A}_{5}{B}_{5} \),\n\n\[ P\left( {{A}_{1}{B}_{5}}\right) = P\left( {{A}_{1}{A}_{5}{B}_{5}}\right) = P\left( {{A}_{1}{A}_{5}}\right) P\left( {{B}_{5} \mid {A}_{1}{A}_{5}}\right) = \frac{3}{10} \cdot \frac{1}{2} = \frac{3}{20} \]\n\n(3.24)\n\nWe thus have\n\n\[ P\left( {{A}_{1} \mid {B}_{5}}\right) = \frac{3/{20}}{5/{20}} = \frac{6}{10} = P\left( {A}_{1}\right) \]\n\n\( \left( {3.25}\right) \)
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Yes
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\( \gg \mathrm{{pm}} = \operatorname{minprob}\left( {{0.1} * \left\lbrack \begin{array}{lll} 4 & 7 & 6 \end{array}\right\rbrack }\right) \)
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pm = 0.0720 0.1080 0.1680 0.2520 0.0480 0.0720 0.1120 0.1680
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No
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\( \gg \mathrm{{pm}} = {0.01} * \left\lbrack \begin{array}{llll} {15} & 5 & 2 & 1 \\ 8 & {25} & 5 & 1 \\ 8 & {12} & 1 & \end{array}\right\rbrack ;\;\% \) An arbitrary class \( \gg \operatorname{disp}\left( {\operatorname{imintest}\left( \mathrm{{pm}}\right) }\right) \)
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The class is NOT independent\nMinterms for which the product rule fails \( \begin{array}{llll} 1 & 1 & 1 & 0 \end{array} \) \( \begin{array}{llll} 1 & 1 & 1 & 0 \end{array} \)
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No
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Suppose the class \( \{ A, B, C\} \) is independent, with respective probabilities \( {0.4},{0.6},{0.8} \) . Determine \( P\left( {A \cup {BC}}\right) \) .
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The minterm expansion is\n\n\[ A \cup {BC} = M\left( {3,4,5,6,7}\right) \text{, so that}P\left( {A \cup {BC}}\right) = p\left( {3,4,5,6,7}\right) \]\n\n(4.10)\n\nIt is not difficult to use the product rule and the replacement theorem to calculate the needed minterm probabilities. Thus \( p\left( 3\right) = P\left( {A}^{c}\right) P\left( B\right) P\left( C\right) = {0.6} \cdot {0.6} \cdot {0.8} = {0.2280} \) . Similarly \( p\left( 4\right) = {0.0320}, p\left( 5\right) = {0.1280}, p\left( 6\right) = {0.0480}, p\left( 7\right) = {0.1920} \) . The desired probability is the sum of these, 0.6880 .\n\nAs an alternate approach, we write\n\n\[ A \cup {BC} = A\bigvee {A}^{c}{BC}\text{, so that}P\left( {A \cup {BC}}\right) = {0.4} + {0.6} \cdot {0.6} \cdot {0.8} = {0.6880} \]\n\n(4.11)
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Yes
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Consider the independent class \( \left\{ {{A}_{1},{A}_{2},{A}_{3},{B}_{1},{B}_{2},{B}_{3},{B}_{4}}\right\} \), with respective probabilities \( {0.4},{0.7},{0.3},{0.5},{0.8},{0.3},{0.6} \). Consider \( {M}_{3} \), minterm three for the class \( \left\{ {{A}_{1},{A}_{2},{A}_{3}}\right\} \), and \( {N}_{5} \), minterm five for the class \( \left\{ {{B}_{1},{B}_{2},{B}_{3},{B}_{4}}\right\} \). Then
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\[ P\left( {M}_{3}\right) = P\left( {{A}_{1}^{c}{A}_{2}{A}_{3}}\right) = {0.6} \cdot {0.7} \cdot {0.3} = {0.126}\text{and}P\left( {N}_{5}\right) = P\left( {{B}_{1}^{c}{B}_{2}{B}_{3}^{c}{B}_{4}}\right) = {0.5} \cdot {0.8} \cdot {0.7} \cdot {0.6} = {0.168} \] Also \[ P\left( {{M}_{3}{N}_{5}}\right) = P\left( {{A}_{1}^{c}{A}_{2}{A}_{3}{B}_{1}^{c}{B}_{2}{B}_{3}^{c}{B}_{4}}\right) = {0.6} \cdot {0.7} \cdot {0.3} \cdot {0.5} \cdot {0.8} \cdot {0.7} \cdot {0.6} = \left( {{0.6} \cdot {0.7} \cdot {0.3}}\right) \cdot \left( {{0.5} \cdot {0.8} \cdot {0.7} \cdot {0.6}}\right) = P\left( {M}_{3}\right) P\left( {N}_{5}\right) = {0.0212} \] The product rule shows the desired independence.
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Yes
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We wish to show that the pair \( \{ A, B\} \) is independent; i.e., the product rule \( P\left( {AB}\right) = P\left( A\right) P\left( B\right) \) holds.
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COMPUTATION\n\n\[ P\left( A\right) = P\left( {A}_{1}\right) + P\left( {A}_{2}\right) + P\left( {A}_{3}\right) = {0.3} + {0.4} + {0.1} = {0.8}\text{and}P\left( B\right) = P\left( {B}_{1}\right) + \]\n\n\[ P\left( {B}_{2}\right) = {0.2} + {0.5} = {0.7} \]\n\nNow\n\n\[ {AB} = \left( {{A}_{1}\bigvee {A}_{2}\bigvee {A}_{3}}\right) \left( {{B}_{1}\bigvee {B}_{2}}\right) = {A}_{1}{B}_{1}\bigvee {A}_{1}{B}_{2}\bigvee {A}_{2}{B}_{1}\bigvee {A}_{2}{B}_{2}\bigvee {A}_{3}{B}_{1}\bigvee {A}_{3}{B}_{2} \]\n\nBy additivity and pairwise independence, we have\n\n\[ P\left( {AB}\right) = P\left( {A}_{1}\right) P\left( {B}_{1}\right) + P\left( {A}_{1}\right) P\left( {B}_{2}\right) + P\left( {A}_{2}\right) P\left( {B}_{1}\right) + P\left( {A}_{2}\right) P\left( {B}_{2}\right) + \]\n\n\[ P\left( {A}_{3}\right) P\left( {B}_{1}\right) + P\left( {A}_{3}\right) P\left( {B}_{2}\right) = {0.3} \cdot {0.2} + {0.3} \cdot {0.5} + {0.4} \cdot {0.2} + {0.4} \cdot {0.5} + {0.1} \cdot \]\n\n\[ {0.2} + {0.1} \cdot {0.5} = {0.56} = P\left( A\right) P\left( B\right) \]\n\nThe product rule can also be established algebraically from the expression for \( P\left( {AB}\right) \), as follows:\n\n\[ P\left( {AB}\right) = P\left( {A}_{1}\right) \left\lbrack {P\left( {B}_{1}\right) + P\left( {B}_{2}\right) }\right\rbrack + P\left( {A}_{2}\right) \left\lbrack {P\left( {B}_{1}\right) + P\left( {B}_{2}\right) }\right\rbrack + P\left( {A}_{3}\right) \left\lbrack {P\left( {B}_{1}\right) + P\left( {B}_{2}\right) }\right\rbrack \]\n\n\[ = \left\lbrack {P\left( {A}_{1}\right) + P\left( {A}_{2}\right) + P\left( {A}_{3}\right) }\right\rbrack \left\lbrack {P\left( {B}_{1}\right) + P\left( {B}_{2}\right) }\right\rbrack = P\left( A\right) P\left( B\right) \]
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Yes
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Example 1.2.1 Suppose a human population is growing at \( 1\% \) per year and initially has 1,000,000 individuals. Let \( {P}_{t} \) denote the populations size \( t \) years after the initial population of \( {P}_{0} = 1,{000},{000} \) individuals. If one asks what the population will be in 50 years there are two options.
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Option 1. At \( 1\% \) per year growth, the dynamic equation would be\n\n\[ {P}_{t + 1} - {P}_{t} = {0.01}{P}_{t} \]\n\nand the corresponding iteration equation is\n\n\[ {P}_{t + 1} = {1.01}{P}_{t} \]\n\nWith \( {P}_{0} = 1,{000},{000},{P}_{1} = {1.01} \times 1,{000},{000} = 1,{010},{000},{P}_{2} = {1.011},{010},{000} = 1,{020},{100} \) and so on for 50 iterations.\n\nOption 2. Alternatively, one may write the solution\n\n\[ {P}_{t} = {1.01}^{t}\left( {1,{000},{000}}\right) \]\n\nso that\n\n\[ {P}_{50} = {1.01}^{50}\left( {1,{000},{000}}\right) = 1,{644},{631} \]\n\nThe algebraic form of the solution, \( {P}_{t} = {R}^{t}{P}_{0} \), with \( r > 0 \) and \( R > 1 \) is informative and gives rise to the common description of exponential growth attached to some populations. If \( r \) is negative and \( R = 1 + r < 1 \), the solution equation \( {P}_{t} = {R}^{t}{P}_{0} \) exhibits exponential decay.
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Yes
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Example 1.7.2 Simulation of chemical discharge into a lake. Begin with two one-liter beakers, a supply of distilled water and salt and a meter to measure conductivity in water. Place one liter of distilled water and \( {0.5}\mathrm{\;g} \) of salt in beaker \( \mathrm{F} \) (factory). Place one liter of distilled water in beaker L (lake). Repeatedly do\n\na. Measure and record the conductivity of the water beaker L.\n\nb. Remove \( {100}\mathrm{{ml}} \) of solution from beaker \( \mathrm{L} \) and discard.\n\nc. Transfer \( {100}\mathrm{{ml}} \) of salt water from beaker \( \mathrm{F} \) to beaker \( \mathrm{L} \) .\n\nThe conductivity of the salt water in beaker F should be about 1000 microsiemens \( \left( {\mu S}\right) \) . The conductivity of the water in beaker L should be initially 0 and increase as the concentration of salt in L increases. Data and a graph of the data are shown in Figure 1.20 and appears similar to the graph in Figure 1.19.
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In Exercise 1.7.4 you are asked to write and solve a mathematical model of this simulation and compare the solution with the data. ∎
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No
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Example 1.10.1 Jack A. Wolfe \( {}^{14} \) observed that leaves of trees growing in cold climates tend to be incised (have ragged edges) and leaves of trees growing in warm climates tend to have smooth edges (lacking lobes or teeth). He measured the percentages of species that have smooth margins among all species of the flora in many locations in eastern Asia. His data, as read from a graph in U. S. Geological Survey Professional Paper 1106, is presented in Figure 1.10.0.2.
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Figure for Example 1.10.0.2 Average temperature \( {\mathrm{C}}^{ \circ } \) vs percentage of tree species with smooth edge leaves in 33 forests in eastern Asia. The equation of the line is \( y = - {0.89} + {0.313x} \) .  The line, temp \( = {0.89} + {0.313}\% \) smooth is shown in Figure 1.10.0.2 and is close to the data. The line was used by Wolfe to estimate temperatures over the last 65 million years based on observed fossil leaf composition (Exercise 1.10.1).
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Yes
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On several nights during August and September in Ames, Iowa, some students listened to crickets chirping. They counted the number of chirps in a minute (chirp rate, \( R \) ) and also recorded the air (ambient) temperature \( \left( T\right) \) in \( {\mathrm{F}}^{ \circ } \) for the night. The data were collected between 9:30 and 10:00 pm each night, and are shown in the table and graph in Figure 1.24.
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These data also appear linear and the line through (65,100) and (75,145), \[ \frac{R - {100}}{T - {65}} = \frac{{145} - {100}}{{75} - {65}}, \; R = {4.5T} - {192.5} \] (1.32) lies close to the data. We can use the line to estimate temperature to be about \( {69.5}{\mathrm{\;F}}^{ \circ } \) if cricket chirp rate is 120 chirps/minute.
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Yes
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Example 2.2.1 Data for the percentage of U.S. population in 1955 that had antibodies to the polio virus as a function of age is shown in Table 2.2.1.1. The data show an interesting fact that a high percentage of the population in 1955 had been infected with polio. A much smaller percentage were crippled or killed by the disease.
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Although Table 2.2.1.1 is a function, it is only an approximation to a perhaps real underlying function. The order pair, \( \left( {{17.5},{72}}\right) \), signals that 72 percent of the people of age 17.5 years had antibodies to the polio virus. More accurately, \( \left( {{17.5},{72}}\right) \) signals that of a sample of people who had ages in the interval from age 15 to less than 20 , the percentage who tested positive to antibodies to the polio virus was greater than or equal to 71.5 and less than 72.5.
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Yes
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For the function, \( R \), defined by\n\n\[ R\left( x\right) = x + \frac{1}{x}\;x \neq 0 \]
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\[ R\left( {1 + 3}\right) = R\left( 4\right) = 4 + \frac{1}{4} = {4.25} \]\n\n\[ R\left( 1\right) = 1 + \frac{1}{1} = {2.0}\;\text{ and } \]\n\n\[ R\left( 3\right) = 3 + \frac{1}{3} = {3.3333}\cdots \]\n\n\[ R\left( 1\right) + R\left( 3\right) = 2 + {3.3333}\cdots = {5.3333}\cdots \neq {4.25} = R\left( 4\right) \]\n\nIn this case\n\n\[ R\left( {1 + 3}\right) \neq R\left( 1\right) + R\left( 3\right) \]
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Yes
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Show that polynomials are linear in their coefficients.
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Consider the following.\n\nLet \( \;P\left( x\right) = 7 - {3x} + 5{x}^{2},\; \) and \( \;Q\left( x\right) = - 2 + {4x} - {x}^{2} + 6{x}^{3} \) .\n\n\[ P\left( x\right) + Q\left( x\right) = \left( {7 - {3x} + 5{x}^{2}}\right) + \left( {-2 + {4x} - {x}^{2} + 6{x}^{3}}\right. \]\n\n\[ = \left( {7 - 2}\right) + \left( {-3 + 4}\right) x + \left( {5 - 1}\right) {x}^{2} + \left( {0 + 6}\right) {x}^{3} \]\n\n\[ = 5 + x + 4{x}^{2} + 6{x}^{3} \]\n\nThus \( P\left( x\right) + Q\left( x\right) \) is simply the polynomial obtained by adding corresponding coefficients in \( P\left( x\right) \) and \( Q\left( x\right) \) . Furthermore,\n\n\[ {13} \cdot P\left( x\right) = {13}\left( {7 - {3x} + 5{x}^{2}}\right) = {13} \cdot 7 - {13} \cdot {3x} + {13} \cdot 5{x}^{2} = {91} - {39x} + {65}{x}^{2}. \]\n\nThus \( {13} \cdot P\left( x\right) \) is simply the polynomial obtained by multiplying each coefficient of \( P\left( x\right) \) by 13 .
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Yes
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Example 2.5.1 If we use these equations to fit a line to the cricket data of Example 1.10.1 showing a relation between temperature and cricket chirp frequency, we get\n\n\[ y = {4.5008x} - {192.008}, \\text{ close to the line } y = {4.5x} - {192} \]\n\nthat we ’fit by eye’ using the two points, \( \\left( {{65},{100}}\\right) \) and \( \\left( {{75},{145}}\\right) \) .
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Explore 2.5.1 Technology. Your calculator or computer will hide all of the arithmetic of\n\nEquations 2.4 and give you the answer. The overall procedure is:\n\n1. Load the data. [Two lists, X and Y, say].\n\n2. Compute the coefficients of a first degree polynomial close to the data and store them in \( \\mathrm{P} \).\n\n3. Specify \( \\mathrm{X} \) coordinates and compute corresponding \( \\mathrm{Y} \) coordinates for the polynomial.\n\n4. Plot the original data and the computed polynomial.\n\nA MATLAB program to do this is:\n\nclose all;clc;clear\n\n\( X = \\left\\lbrack \\begin{array}{llllllllll} {67} & {73} & {78} & {61} & {66} & {66} & {67} & {77} & {74} & {76} \\end{array}\\right\\rbrack ; \)\n\n\( Y = \\left\\lbrack \\begin{array}{llllllllll} {109} & {136} & {160} & {87} & {103} & {102} & {108} & {154} & {144} & {150} \\end{array}\\right\\rbrack ; \)\n\n\( P = \\operatorname{polyfit}\\left( {X, Y,1}\\right) \)\n\n\( \\mathrm{{PX}} = \\left\\lbrack {{60} : {0.1} : {80}}\\right\\rbrack \) ;\n\n\( \\mathrm{{PY}} = \\operatorname{polyval}\\left( {\\mathrm{P},\\mathrm{{PX}}}\\right) \) ;\n\n\( \\operatorname{plot}\\left( {X, Y,\\prime + \\prime ,\\prime 1\\text{ in }{\\operatorname{ewidth}}^{\\prime },2}\\right) ;\\operatorname{hold}\\left( {\\prime \\text{ on’ }}\\right) ;\\;\\operatorname{plot}\\left( {{PX},{PY},\\prime 1\\text{ in }{\\operatorname{ewidth}}^{\\prime },2}\\right) \)
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Yes
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Exercise 2.5.1 Use Equations 2.3 to find the linear function that is the least squares fit to the data:
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\[ \left( {-2,5}\right) \;\left( {3,{12}}\right) \]
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No
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The graph of the inverse of \( F \) is the reflection of the graph of \( F \) about the diagonal line, \( y = x \) .
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The reflection of \( G \) with respect to the diagonal line, \( y = x \) consists of the points \( Q \) such that either \( Q \) is a point of \( G \) on the diagonal line, or there is a point \( P \) of \( G \) such that the diagonal line is the perpendicular bisector of the interval \( \overline{PQ} \) .
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Yes
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To compute the equation for the inverse of the function \( \mathrm{S} \) , \[ S\left( x\right) = {x}^{2}\;\text{ for }\;x \geq 0 \]
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let the equation for \( \mathrm{S} \) be written as \[ y = {x}^{2}\;\text{ for }\;x \geq 0 \] Then interchange \( y \) and \( x \) to obtain \[ x = {y}^{2}\;\text{ for }\;y \geq 0 \] and solve for \( \mathrm{y} \) . \[ x = {y}^{2}\;\text{ for }\;y \geq 0 \] \[ {y}^{2} = x\;\text{ for }\;y \geq 0 \] \[ {\left( {y}^{2}\right) }^{\frac{1}{2}} = {x}^{\frac{1}{2}} \] \[ {y}^{2\frac{1}{2}} = {x}^{\frac{1}{2}} \] \[ y = {x}^{\frac{1}{2}} \] Therefore \[ {S}^{-1}\left( x\right) = \sqrt{x}\;\text{ for }\;x \geq 0 \]
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Yes
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Example 2.6.3 The graph of function \( F\left( x\right) = 2{x}^{2} - {6x} + 3/2 \) shown in Figure 2.12 has two invertible portions, the left branch and the right branch. We compute the inverse of each of them.
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Let \( y = 2{x}^{2} - {6x} + 5/2 \), exchange symbols \( x = 2{y}^{2} - {6y} + 5/2 \), and solve for \( y \) . We use the steps of 'completing the square' that are used to obtain the quadratic formula.\n\n\[ x = 2{y}^{2} - {6y} + 5/2 \]\n\n\[ = 2\left( {{y}^{2} - {3x} + 9/4}\right) - 9/2 + 5/2 \]\n\n\[ = 2{\left( y - 3/2\right) }^{2} - 2 \]\n\n\[ {\left( y - 3/2\right) }^{2} = \frac{x + 2}{2} \]\n\n\[ y - 3/2 = \sqrt{\frac{x + 2}{2}}\;\text{ or }\; - \sqrt{\frac{x + 2}{2}} \]\n\n\[ y = \frac{3}{2} + \sqrt{\frac{x + 2}{2}} \] Right branch inverse.\n\n\[ y = \frac{3}{2} - \sqrt{\frac{x + 2}{2}} \] Left branch inverse.
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Yes
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Example 2.7.1 Two properties of the logarithm and exponential functions are\n\n\[ \n\\text{(a)}{\\log }_{b}{b}^{\\lambda } = \\lambda \\;\\text{and}\\;\\text{(b)}u = {b}^{{\\log }_{b}u} \n\]
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The logarithm function, \( L\\left( x\\right) = {\\log }_{b}\\left( x\\right) \) is the inverse of the exponential function, \( E\\left( x\\right) = {b}^{x} \), and the properties simply state that\n\n\[ \nL \\circ E = I\\;\\text{ and }E \\circ L = I \n\]
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Yes
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Is there an equation for such a signal? Yes, a very messy one!
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The graph of the following equation is shown in Figure 2.18(b). It is similar to the typical electrocardiogram in Figure 2.18(a).\n\n\[ H\left( t\right) = {25000}\frac{\left( {t + {0.05}}\right) t\left( {t - {0.07}}\right) }{\left( {1 + {\left( {20}t\right) }^{10}}\right) {2}^{\left( {2}^{\left( {40}t\right) }\right) }} + {0.15}{2}^{-{1600}{\left( t + {0.175}\right) }^{2}} + {0.25}{2}^{-{900}{\left( t - {0.2}\right) }^{2}} \]\n\n(2.12)\n\n\[ - {0.4} \leq t \leq {0.4} \]
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Yes
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Find the period, frequency, and amplitude of \[ H\left( t\right) = 3\sin \left( {{5t} + \pi /3}\right) \]
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Solution. Write \( H\left( t\right) = 3\sin \left( {{5t} + \pi /3}\right) \) as \[ H\left( t\right) = 3\sin \left( {\frac{2\pi }{{2\pi }/5}t + \pi /3}\right) . \] Then the amplitude of \( P \) is 3, and the period is \( {2\pi }/5 \) and the frequency is \( 5/\left( {2\pi }\right) \) .
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Yes
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At what rate was the Vibrio natriegens population of Section 1.1 growing at time \( T = {40} \) minutes?
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The average growth rate between times \( T = {32} \) and \( T = {48} \) is\n\n\[ \frac{{0.101} - {0.060}}{{48} - {32}} = {0.0026}\;\frac{\text{ Absorbance units }}{\text{ minute }}, \]\n\nand is a pretty good estimate of the growth rate at time \( T = {40} \), particularly because 40 is midway between 32 and 48.
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Yes
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Example 3.1.2 At what rate was the world human population increasing in 1980? Shown in Figure 3.3 are data for the twentieth century and a graph of an approximating function, \( F \) . A tangent to the graph of \( F \) at \( \left( {{1980}, F\left( {1980}\right) }\right) \) is drawn and has a slope of \( {0.0781} \cdot {10}^{9} = {78},{100},{000} \) .
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Now,\n\n\[ \text{ slope is }\;\frac{\text{ rise }}{\text{ run }}\; = \;\frac{\text{ change in population }}{\text{ change in years }}\; \approx \;\frac{\text{ people }}{\text{ year }}. \]\n\nThe units of slope, then, are people/year. Therefore,\n\n\[ \text{ slope } = {78},{100},{000}\;\frac{\text{ people }}{\text{ year }}. \]\n\nThe world population was increasing approximately 78,100,000 people per year in 1980.
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Yes
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Show that if \( a \) is a positive number, then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\sqrt{x} = \sqrt{a} \]
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Suppose \( \epsilon > 0 \) . Let \( \delta = \epsilon \sqrt{a} \) . Suppose \( x \) is in the domain of \( \sqrt{x} \) and \( 0 < \left| {x - a}\right| < \delta \) . Then\n\n\[ \left| {\sqrt{x} - \sqrt{a}}\right| \;\overset{\left( i\right) }{ = }\;\frac{\left| x - a\right| }{\sqrt{x} + \sqrt{a}}\;\overset{\left( ii\right) }{ \leq }\;\frac{\left| x - a\right| }{\sqrt{a}}\;\overset{\left( iii\right) }{ < }\;\frac{\delta }{\sqrt{a}}\;\overset{\left( iv\right) }{ = }\;\frac{\epsilon \sqrt{a}}{\sqrt{a}}\;\overset{\left( v\right) }{ = }\;\epsilon . \]
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Yes
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Theorem 3.2.1 : Limit is Unique Theorem. Suppose \( G \) is a function and\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}G\left( x\right) = {L}_{1}\;\text{ and }\;\mathop{\lim }\limits_{{x \rightarrow a}}G\left( x\right) = {L}_{2}. \]\n\nThen \( {L}_{1} = {L}_{2} \) .
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1. Suppose that \( {L}_{1} < {L}_{2} \).\n\n2. Let \( \epsilon = \left( {{L}_{2} - {L}_{1}}\right) /2 \).
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No
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At what rate is the light decreasing when \( d = 5\mathrm{\;{cm}} \) ?
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We find the derivative of \( L \) . For any value of \( d \) between 3 and \( {16}\mathrm{\;{cm}} \) ,\n\n\[ \n{L}^{\prime }\left( d\right) = \mathop{\lim }\limits_{{b \rightarrow d}}\frac{L\left( b\right) - L\left( d\right) \;\mathrm{{mW}}/{\mathrm{{cm}}}^{2}}{b - d\;\mathrm{\;{cm}}} \]\n\n\[ \n= \mathop{\lim }\limits_{{b \rightarrow d}}\frac{\frac{1.45}{b} - \frac{1.45}{d}}{b - d}\frac{\mathrm{{mW}}/{\mathrm{{cm}}}^{2}}{\mathrm{\;{cm}}} \]\n\n\[ \n= \mathop{\lim }\limits_{{d \rightarrow 5}}\left( {-{1.45}\frac{1}{d \times b}}\right) \;\frac{\mathrm{{mW}}/{\mathrm{{cm}}}^{2}}{\mathrm{\;{cm}}} \]\n\n\[ \n= \frac{-{1.45}}{{d}^{2}}\frac{\mathrm{{mW}}/{\mathrm{{cm}}}^{2}}{\mathrm{\;{cm}}} \]\n\n\[ \n\text{For}d = 5,\;{L}^{\prime }\left( 5\right) = {\left. \frac{-{1.45}}{{d}^{2}}\right| }_{d = 5} = \frac{-{1.45}}{{5}^{2}} = - {0.058}\;\frac{\mathrm{{mW}}/{\mathrm{{cm}}}^{2}}{\mathrm{\;{cm}}}\text{.} \]\n\nWe have just used a helpful notation: \( {\left. \right| }_{x = a} \)\n\n\[ \n\text{For any function,}G,{\left. \;G\left( x\right) \right| }_{x = a} = G\left( a\right) \text{.} \]\n\nIt is useful to compare the algebraic forms of \( L \) and \( {L}^{\prime } \),\n\n\[ \nL\left( d\right) = \frac{k}{d}\;\text{ and }\;{L}^{\prime }\left( d\right) = - k\frac{1}{{d}^{2}},\;\text{ and note that }\;{L}^{\prime }\left( d\right) = - \frac{1}{k}{L}^{2}\left( d\right) .\n\]
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Yes
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Example 3.3.2 Problem. The graphs of a function \( P \) and its derivative \( {P}^{\prime } \) are shown in Figure 3.16. Which graph is the graph of \( P \) ?
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Solution. We claim that graph 1 is not \( P \), because every tangent to graph 1 has negative slope and some \( y \) -coordinates of graph 2 are positive. Therefore graph 2 must be the graph of \( P \) .
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No
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Let \( P\left( t\right) = 3{t}^{4} - 5{t}^{2} + 2t + 7 \) . Then \( {P}^{\prime }\left( t\right) = {\left\lbrack P\left( t\right) \right\rbrack }^{\prime } \)
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\[ {P}^{\prime }\left( t\right) = {\left\lbrack 3{t}^{4} - 5{t}^{2} + 2t + 7\right\rbrack }^{\prime } \] Definition of \( P \) \[ = {\left\lbrack 3{t}^{4}\right\rbrack }^{\prime } - {\left\lbrack 5{t}^{2}\right\rbrack }^{\prime } + {\left\lbrack 2t\right\rbrack }^{\prime } + {\left\lbrack 7\right\rbrack }^{\prime } \] Sum Rule \[ = 3{\left\lbrack {t}^{4}\right\rbrack }^{\prime } - 5{\left\lbrack {t}^{2}\right\rbrack }^{\prime } + 2{\left\lbrack t\right\rbrack }^{\prime } + {\left\lbrack 7\right\rbrack }^{\prime } \] Constant Factor Rule \[ = 3 \times 4{t}^{3} - 5 \times 2{t}^{1} + 2{\left\lbrack t\right\rbrack }^{\prime } + {\left\lbrack 7\right\rbrack }^{\prime } \] \[ = 12{t}^{3} - 10t + 2 \times 1 + {\left\lbrack 7\right\rbrack }^{\prime } \] t Rule \[ = 12{t}^{3} - 10t + 2 + 0 \] Constant Rule
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Yes
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Example 3.5.2 It is useful to write some fractions in forms so that the Constant Factor Rule obviously applies. For example\n\n\[ P\left( t\right) = \frac{{t}^{3}}{7} = \frac{1}{7}{t}^{3} \]\n\nand to compute \( {P}^{\prime } \) you may write
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\n\[ {P}^{\prime }\left( t\right) = {\left\lbrack \frac{{t}^{3}}{7}\right\rbrack }^{\prime } = {\left\lbrack \frac{1}{7}{t}^{3}\right\rbrack }^{\prime } = \frac{1}{7}{\left\lbrack {t}^{3}\right\rbrack }^{\prime } = \frac{1}{7}3{t}^{2} = \frac{3}{7}{t}^{2} \]
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Yes
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Example 3.5.3 The derivative of a quadratic function is a linear function.
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Solution. Suppose \( P\left( t\right) = a{t}^{2} + {bt} + c \) where \( a, b \), and \( c \) are constants. Then\n\n\[ \n{P}^{\prime }\left( t\right) = {\left\lbrack a{t}^{2} + bt + c\right\rbrack }^{\prime } \n\]\n\n\[ \n= {\left\lbrack a{t}^{2}\right\rbrack }^{\prime } + {\left\lbrack bt\right\rbrack }^{\prime } + {\left\lbrack c\right\rbrack }^{\prime } \n\]\n\n\[ \n= a{\left\lbrack {t}^{2}\right\rbrack }^{\prime } + b{\left\lbrack t\right\rbrack }^{\prime } + {\left\lbrack c\right\rbrack }^{\prime } \n\]\n\n(3.32)\n\n\[ \n= {a2t} + b \times 1 + {\left\lbrack c\right\rbrack }^{\prime } \n\]\n\n\[ \n= {2at} + b + 0 \n\]\n\nWe see that \( {P}^{\prime }\left( t\right) = {2at} + b \) which is a linear function. -
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Yes
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In baseball, a 'pop fly' is hit and the ball leaves the bat traveling vertically at 30 meters per second. How high will the ball go and how much time does the catcher have to get in position to catch it?
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Using a formula from Section 3.6.1, the ball will be at a height \( s\left( t\right) = - {4.9}{t}^{2} + {30t} \) meters \( t \) seconds after it is released, where \( s\left( t\right) \) is the height above the point of impact with the bat. The velocity, \( v\left( t\right) = {s}^{\prime }\left( t\right) \) is\n\n\[ \n{\left\lbrack s\left( t\right) \right\rbrack }^{\prime } = {\left\lbrack -{4.9}{t}^{2} + {30}t\right\rbrack }^{\prime } = - {4.9} \cdot {2t} + {30} \cdot 1 = - {9.8t} + {30} \n\] \n\nThe ball will be at its highest position when the velocity \( v\left( t\right) = {s}^{\prime }\left( t\right) = 0 \) (which implies that the ball is not moving and identifies the time at which the ball is at its highest point, is not going up and is not going down).\n\n\[ \n{s}^{\prime }\left( t\right) = 0\;\text{ implies }\; - {9.8t} + {30} = 0,\;\text{ or }\;t = \frac{30}{9.8} \approx {3.1}\;\text{ seconds. } \n\] \n\nThe height of the ball at \( t = {3.1} \) seconds is \n\n\[ \ns\left( {3.1}\right) = - {4.9}{\left( {3.1}\right) }^{2} + {30} \times {3.1} \approx {45.9}\;\text{ meters } \n\] \n\nThus in about 3.1 seconds the ball reaches a height of about 45.9 meters. The catcher will have about 6 seconds to position to catch the ball. Furthermore, at time \( t = {6.2} \)\n\n\[ \n{s}^{\prime }\left( {6.2}\right) = - {9.8} \times {6.2} + {30} = - {30.76}\text{ meters/second. } \n\] \n\nThe velocity of the falling ball is \( \approx - {30.76}\mathrm{\;m}/\mathrm{s} \) when the catcher catches it. Its magnitude will be exactly \( {30}\mathrm{\;m}/\mathrm{s} \), the speed at which it left the bat. Why not \( {30.76} \) ?
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Yes
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A farmer's barn is 60 feet long on one side. He has 100 feet of fence and wishes to build a rectangular pen along that side of his barn. What should be the dimensions of the pen to maximize the area?
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Because there are 100 feet of fence, 2 * W + L = {100}. The area, A, is A = {LW}. Because 2W + L = {100}, L = {100} - {2W} and A = {LW} = \left( {{100} - {2W}}\right) W or A = {100W} - 2{W}^{2}. The graph of A vs W is a parabola with its highest point at the vertex. The tangent to the parabola at the vertex is horizontal, and we find a value of W for which {A}^{\prime }\left( W\right) = 0. {A}^{\prime }\left( W\right) = {\left\lbrack {100}W - 2{W}^{2}\right\rbrack }^{\prime } = {\left\lbrack {100}W\right\rbrack }^{\prime } - {\left\lbrack 2{W}^{2}\right\rbrack }^{\prime } = {100}{\left\lbrack W\right\rbrack }^{\prime } - 2{\left\lbrack {W}^{2}\right\rbrack }^{\prime } = {100} \times 1 - 2 \times {2W}. The optimum dimensions, W and L, are found by setting {A}^{\prime }\left( W\right) = 0, so that {A}^{\prime }\left( W\right) = {100} - {4W} = 0, W = {25}, L = {100} - {2W} = {50}. Thus the farmer should build a 25 by 50 foot pen.
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Yes
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Example 3.5.6 This problem is written on the assumption, to our knowledge untested, that spider webs have an optimum size. Seldom are they so small as \( 1\mathrm{\;{cm}} \) in diameter and seldom are they so large as \( 2\mathrm{\;m} \) in diameter. If they are one cm in diameter, there is a low probability of catching a flying insect; if they are \( 2\mathrm{\;m} \) in diameter they require extra strength to withstand wind and rain. We will examine circular webs, for convenience, and determine the optimum diameter for a web so that it will catch enough insects and not fall down.
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Solution. Assume a circular spider web of diameter, \( d \) . It is reasonable to assume that the amount of food gathered by the web is proportional to the area, \( A \), of the web. Because \( A = \pi {d}^{2}/4 \) , the amount of food gather is proportional to \( {d}^{2} \) . We also assume that the energy required to build and maintain a web of area \( A \) is proportional to \( {d}^{3} \) . (The basic assumption is that the work to build a square centimeter of web increases as the total web area increases because of the need to have stronger fibers. If, for example, the area of the fiber cross-section increases linearly with \( A \) and the mesh of the web is constant, the mass of the web increases as \( {d}^{3} \) .)\n\nWith these assumptions, the net energy, \( E \), available to the spider is of the form\n\n\[ E = \text{Energy from insects caught - Energy expended building the web} \]\n\n\[ = {k}_{1}{d}^{2} - {k}_{2}{d}^{3} \]\n\nwhere \( d \) is measured in centimeters and \( {k}_{1} \) and \( {k}_{2} \) are proportionality constants.\n\nFor illustration we will assume that \( {k}_{1} = {0.01} \) and \( {k}_{2} = {0.0001} \) . A graph of \( E\left( d\right) = {0.01}{d}^{2} - {0.0001}{d}^{3} \) is shown in Figure 3.23 where it can be seen that there are two points, \( A \) and \( B \), at which the graph has horizontal tangents. We find where the derivative is zero to locate \( A \) and \( B \) .\n\n\[ {E}^{\prime }\left( d\right) = {\left\lbrack {0.01}{d}^{2} - {0.0001}{d}^{3}\right\rbrack }^{\prime } \]\n\n\[ = {\left\lbrack {0.01}{d}^{2}\right\rbrack }^{\prime } - {\left\lbrack {0.0001}{d}^{3}\right\rbrack }^{\prime } \]\n\n\[ = {0.01}{\left\lbrack {d}^{2}\right\rbrack }^{\prime } - {0.0001}{\left\lbrack {d}^{3}\right\rbrack }^{\prime } \]\n\n(3.34)\n\n\[ = {0.01} \times {2d} - {0.0001} \times 3{d}^{2} \]\n\n\[ = {0.02d} - {0.0003}{d}^{2} \]\n\nThen \( {E}^{\prime }\left( d\right) = 0 \) yields\n\n\[ {0.02d} - {0.0003}{d}^{2} = 0 \]\n\n\[ d\left( {{0.02} - {0.0003d}}\right) = 0 \]\n\n\[ d = 0\;\text{ or }\;d = {0.02}/{0.0003} \doteq {66.7}\mathrm{\;{cm}} \]\n\nThe value \( d = 0 \) locates the local minimum at \( B \) and has an obvious interpretation: if there is no web there is no energy gain. At \( d = {66.7}\mathrm{\;{cm}}, E\left( d\right) = {14.8} \) (units unspecified) suggests that a positive net energy will accrue with a web of diameter \( {66.7}\mathrm{\;{cm}} \) and that weaving a web of \( {66.7}\mathrm{\;{cm}} \) diameter is the optimum strategy for the spider. Note that if our model and its parameters are correct we have determined in a rather short time what it took spiders many generations to work out. At the very least we could have moved the spiders ahead several generations with our model.
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Yes
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Example 3.6.2 Students measured height \( {vs} \) time of a falling bean bag using a Texas Instruments Calculator Based Laboratory Motion Detector, and the data are shown in Figure 3.26A. Average velocities were computed between data points and plotted against the midpoints of the data intervals in Figure 3.26B. Mid-time is \( \left( {{\operatorname{Time}}_{i + 1} + {\operatorname{Time}}_{i}}\right) /2 \) and Ave. Vel. is \( \left( \right. {\text{Height}}_{i + 1} \) - \( {\text{Height}}_{i})/\left( {{\text{Time}}_{i + 1} - {\text{Time}}_{i}}\right) \) .
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An equation of the line fit by least squares to the graph of Average Velocity \( {vs} \) Midpoint of time interval is\n\n\[ \n{v}_{\text{ave }} = - {849}{t}_{\text{mid }} + {126}\;\mathrm{\;{cm}}/\mathrm{s}. \n\]\n\nIf we assume a continuous model based on this data, we have\n\n\[ \n{s}^{\prime }\left( t\right) = - {849t} + {126},\;s\left( t\right) = \frac{-{849}}{2}{t}^{2} + {126t} + {s}_{0}\;\mathrm{{cm}} \n\]\n\nFrom Figure 3.26, the height of the first point is about 240 . We write\n\n\[ \ns\left( t\right) = \frac{-{849}}{2}{t}^{2} + {126t} + {240}\;\mathrm{\;{cm}} \n\]\n\nThe graph of \( s \) along with the original data is shown in Figure \( {3.26}\mathrm{C} \) . The match is good. Instead of \( g - {980}\mathrm{\;{cm}}/{\mathrm{s}}^{2} \) that applies to objects falling in a vacuum we have acceleration of the bean bag in air to be \( - {849}\mathrm{\;{cm}}/{\mathrm{{sec}}}^{2} \) .
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Yes
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Theorem 4.1.1 Locally Positive Theorem. If a function, \( f \), is continuous at a number a in its domain and \( f\left( a\right) \) is positive, then there is a positive number, \( \delta \), such that \( f\left( x\right) \) is positive for every number \( x \) in \( \left( {a - \delta, a + \delta }\right) \) and in the domain of \( f \) .
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1. Suppose the hypothesis of the Locally Positive Theorem.\n2. Let \( \epsilon = f\left( a\right) \) .\n3. Use the hypothesis that \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = f\left( a\right) \) .
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No
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Theorem 4.2.1 The Derivative Requires Continuity. If \( u \) is a function and \( {u}^{\prime }\left( t\right) \) exists at \( t = a \) then \( u \) is continuous at \( t = a \) .
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Proof. In Exercise 4.2.2 you are asked to give reasons for the following steps, \( \left( i\right) - \left( v\right) \) . Suppose the hypothesis of Theorem 4.2.1.\n\n\[ \left( {\mathop{\lim }\limits_{{b \rightarrow a}}u\left( b\right) }\right) - u\left( a\right) = \mathop{\lim }\limits_{{b \rightarrow a}}\left( {u\left( b\right) - u\left( a\right) }\right) \]\n\n(i)\n\n\[ = \mathop{\lim }\limits_{{b \rightarrow a}}\frac{u\left( b\right) - u\left( a\right) }{b - a} \times \left( {b - a}\right) \]\n\n\[ = \mathop{\lim }\limits_{{b \rightarrow a}}\frac{u\left( b\right) - u\left( a\right) }{b - a} \times \mathop{\lim }\limits_{{b \rightarrow a}}\left( {b - a}\right) \]\n\n(ii)\n\n(4.5)\n\n\[ = {u}^{\prime }\left( a\right) \mathop{\lim }\limits_{{b \rightarrow a}}\left( {b - a}\right) \]\n\n(iii)\n\n\[ = 0 \]\n\n(iv)\n\n\[ \left( {\mathop{\lim }\limits_{{b \rightarrow a}}u\left( b\right) }\right) = u\left( a\right) \]\n\n\( \left( v\right) \)\n\nEnd of proof.
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No
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Find the slope of the tangent to the circle, \( {x}^{2} + {y}^{2} = {13} \), at the point \( \left( {2,3}\right) \) .
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Solution First check that \( {2}^{2} + {3}^{2} = 4 + 9 = {13} \), so that \( \left( {2,3}\right) \) is indeed a point of the circle. Then solve for \( y \) in \( {x}^{2} + {y}^{2} = {13} \) to get\n\n\[ \n{y}_{1} = \sqrt{{13} - {x}^{2}} \n\] \n\nThen the Power Chain Rule with \( n = \frac{1}{2} \) yields\n\n\[ \n{y}_{1}^{\prime } = {\left\lbrack \sqrt{{13} - {x}^{2}}\right\rbrack }^{\prime } = {\left\lbrack {\left( {13} - {x}^{2}\right) }^{\frac{1}{2}}\right\rbrack }^{\prime } \n\] \n(i) Symbolic identity\n\n\[ \n= \frac{1}{2}{\left( {13} - {x}^{2}\right) }^{\frac{1}{2} - 1} \times {\left\lbrack {13} - {x}^{2}\right\rbrack }^{\prime } \n\] \n(ii) PCR, \( n = 1/{2u} = 1 - {x}^{2} \)\n\n\[ \n= \frac{1}{2}{\left( {13} - {x}^{2}\right) }^{-\frac{1}{2}} \times \left( {{\left\lbrack {13}\right\rbrack }^{\prime } - {\left\lbrack {x}^{2}\right\rbrack }^{\prime }}\right) \n\] \n(iii) Sum Rule\n\n\[ \n= \frac{1}{2}{\left( {13} - {x}^{2}\right) }^{-\frac{1}{2}} \times \left( {0 - {2x}}\right) \n\] \n(iv) Constant and Power Rules\n\n\[ \n= - x{\left( {13} - {x}^{2}\right) }^{-\frac{1}{2}} \n\] \n\nTo finish the computation, we compute \( {y}_{1}^{\prime } \) at \( x = 2 \) and get\n\n\[ \n{y}_{1}^{\prime }\left( 2\right) = \left( {-2}\right) {\left( {13} - {2}^{2}\right) }^{-\frac{1}{2}} = - \frac{2}{3} \n\] \n\nand the slope of the tangent to \( {x}^{2} + {y}^{2} = {13} \) at \( \left( {2,3}\right) \) is \( - 2/3 \) . An equation of the tangent is\n\n\[ \n\frac{y - 3}{x - 2} = - \frac{2}{3}\;\text{ or }\;y = - \frac{2}{3}x + 4\frac{1}{3} \n\]
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Yes
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A forester needs to get from point \( A \) on a road to point \( B \) in a forest (see diagram in Figure 4.4). She can travel \( 5\mathrm{\;{km}}/\mathrm{{hr}} \) on the road and \( 3\mathrm{\;{km}}/\mathrm{{hr}} \) in the forest. At what point, \( P \) , should she leave the road and enter the forest in order to minimize the time required to travel from \( A \) to \( B \) ?
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Solution. She might go directly from \( A \) to \( B \) through the forest; she might travel from \( A \) to \( C \) and then to \( B \) ; or she might, as illustrated by the dashed line, travel from \( A \) to a point, \( P \), along the road and then from \( P \) to \( B \) .\n\nAssume that the road is straight, the distance from \( B \) to the road is \( 5\mathrm{\;{km}} \) and the distance from \( A \) to the projection of \( B \) onto the road (point \( Q \) ) is \( 6\mathrm{\;{km}} \) . The point, \( P \), is where the forester leaves the road; let \( x \) be the distance from \( A \) to \( P \) . The basic relation between distance, speed, and time is that\n\nDistance \( \left( \mathrm{{km}}\right) = \) Speed \( \left( {\mathrm{{km}}/\mathrm{{hr}}}\right) \times \) Time \( \left( \mathrm{{hr}}\right) \)\n\nso that\n\n\[ \text{ Time } = \frac{\text{ Distance }}{\text{ Speed }} \]\n\nThe distance traveled and time required are\n\n<table><thead><tr><th></th><th>Along the road</th><th>In the forest</th></tr></thead><tr><td>Distance</td><td>\( x \)</td><td>\( \sqrt{{\left( 6 - x\right) }^{2} + {5}^{2}} \)</td></tr><tr><td>Time</td><td>\( \frac{x}{5} \)</td><td>\( \frac{\sqrt{{\left( 6 - x\right) }^{2} + {5}^{2}}}{3} \)</td></tr></table>\n\nThe total trip time, \( T \), is written as\n\n\[ T = \frac{x}{5} + \frac{\sqrt{{\left( 6 - x\right) }^{2} + {5}^{2}}}{3} \]\n\n(4.12)\n\nA graph of \( T \) vs \( x \) is shown in Figure 4.5. It appears that the lowest point on the curve occurs at about \( x = {2.5}\mathrm{\;{km}} \) and \( T = {2.5} \) hours. -
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Yes
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Find the slope of the graph of\n\n\\[ \sqrt{x} + \sqrt{5 - {y}^{2}} = 5\\;\\text{ at }\\left( {9,1}\\right) \\text{ and at }\\left( {4,2}\\right) \\]
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First we check to see that \\( \\left( {9,1}\\right) \\) satisfies the equation:\n\n\\[ \sqrt{9} + \sqrt{5 - {1}^{2}} = \sqrt{9} + \sqrt{4} = 3 + 2 = 5.\\;\\text{It checks.} \\]\n\nThen we assume there is a function \\( y\\left( x\\right) \\) such that\n\n\\[ \sqrt{x} + \sqrt{5 - {\\left( y\\left( x\\right) \\right) }^{2}} = 5\\;\\text{ and that }\\;y\\left( 9\\right) = 1. \\]\n\nWe convert the square root symbols to fractional exponents and differentiate using Leibnitz notation.\n\n\\[ {x}^{\\frac{1}{2}} + {\\left( 5 - {\\left( y\\left( x\\right) \\right) }^{2}\\right) }^{\\frac{1}{2}} = 5 \\]\n\n\\[ \\frac{d}{dx}\\left( {{x}^{\\frac{1}{2}} + {\\left( 5 - {\\left( y\\left( x\\right) \\right) }^{2}\\right) }^{\\frac{1}{2}}}\\right) = \\frac{d}{dx}5 \\]\n\n\\[ \\frac{d}{dx}{x}^{\\frac{1}{2}} + \\frac{d}{dx}{\\left( 5 - {\\left( y\\left( x\\right) \\right) }^{2}\\right) }^{\\frac{1}{2}} = 0 \\]\n\n\\[ \\frac{1}{2}{x}^{-\\frac{1}{2}} + \\frac{1}{2}{\\left( 5 - {y}^{2}\\right) }^{-\\frac{1}{2}}\\frac{d}{dx}\\left( {5 - {\\left( y\\left( x\\right) \\right) }^{2}}\\right) = 0\\;\\text{Pow} \\]\n\n\\[ \\frac{1}{2}{x}^{-\\frac{1}{2}} + \\frac{1}{2}{\\left( 5 - {y}^{2}\\right) }^{-\\frac{1}{2}}\\left( {\\frac{d}{dx}5 - \\frac{d}{dx}{\\left( y\\left( x\\right) \\right) }^{2}}\\right) \\; = \\;0\\;\\text{Sum Rul} \\]\n\n\\[ \\frac{1}{2}{x}^{-\\frac{1}{2}} + \\frac{1}{2}{\\left( 5 - {y}^{2}\\right) }^{-\\frac{1}{2}}\\left( {0 - {2y}\\frac{d}{dx}y\\left( x\\right) }\\right) = 0\\;\\text{Constant and Power Chain Rules} \\]\n\nNext we solve for \\( \\frac{d}{dx}y\\left( x\\right) \\) and get\n\n\\[ \\frac{d}{dx}y\\left( x\\right) = \\frac{\\sqrt{5 - {y}^{2}}}{{2y}{x}^{\\frac{1}{2}}}\\;\\text{ and evaluate at }\\left( {9,1}\\right) \\;{\\left. \\frac{\\sqrt{5 - {y}^{2}}}{{2y}{x}^{\\frac{1}{2}}}\\right| }_{\\left( {x, y}\\right) = \\left( {9,1}\\right) } = \\frac{1}{3} \\]\n\nSo the slope of the graph at \\( \\left( {9,1}\\right) \\) is \\( 1/3 \\) .
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No
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Theorem 5.1.1 If \( E\left( t\right) = {B}^{t} \) where \( B > 0 \), then\n\n\[ \n{E}^{\prime }\left( t\right) = {E}^{\prime }\left( 0\right) E\left( t\right) \n\]
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Proof: For \( E\left( t\right) = {B}^{t}, B > 0 \) ,\n\n\[ \n{E}^{\prime }\left( t\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{B}^{t + h} - {B}^{t}}{h} = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{B}^{t}{B}^{h} - {B}^{t}}{h} = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{B}^{t}\left( {{B}^{h} - 1}\right) }{h} \n\]\n\n\[ \n= {B}^{t}\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{B}^{h} - {B}^{0}}{h} = E\left( t\right) {E}^{\prime }\left( 0\right) \n\]\n\nEnd of proof.
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Yes
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Theorem 5.2.1 If \( {s}_{1} \leq {s}_{2} \leq {s}_{3} \leq \cdots \) is a bounded nondecreasing sequence of numbers there is a number \( s \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = s \) .
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To prove Theorem 5.2.1 we need a clear definition of limit of a sequence.\n\nDefinition 5.2.2 A number \( s \) is the limit of a number sequence \( {s}_{1},{s}_{2},{s}_{3}\cdots \) means that if \( \left( {u, v}\right) \) is an open interval containing \( s \) there is a positive integer \( N \) such that if \( n \) is an integer greater than \( N,{s}_{n} \) is in \( \left( {u, v}\right) \) . It is sometimes said that \( {s}_{1},{s}_{2},{s}_{3}\cdots \) converges to \( s \) or approaches \( s \) .
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Yes
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The limit of the sequence \( \left\{ {1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots }\right\} \) is zero.
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If \( \left( {u, v}\right) \) contains \( 0, v \) is greater than zero and there is \( {}^{2} \) a positive integer \( N \) that is greater than \( 1/v \) . Therefore, if \( n > N, n > 1/v \) so that \( 1/n < v \), and \( u < 0 < 1/n < v \), so \( \frac{1}{n} \) is in \( \left( {u, v}\right) \) ).
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No
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We can use Theorem 5.2.1 to show a useful result, that if \( a \) is a number and \( 0 < a < 1 \), then the sequence \( {x}_{n} = {a}^{n} \) converges to zero.
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Proof. We assume the alternate version of Theorem 5.2.1 that If \( {s}_{1} \geq {s}_{2} \geq {s}_{3} \geq \cdots \) is a ounded nonincreasing sequence of numbers there is a number \( s \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = s \) .\n\nBecause \( 0 < a < 1 \) and \( {x}_{n} = {a}^{n},{x}_{n + 1} = a \cdot {x}_{n} \) and \( {x}_{n + 1} < {x}_{n} \), and it follows that \( {x}_{n} \) is a nonincreasing (actually decreasing) sequence. Then \( \left\{ {x}_{n}\right\} \) approaches the greatest lower bound, \( s \) of \( \left\{ {x}_{n}\right\} \) .\n\nIf \( s < 0 \), then \( \left( {{2s},0}\right) \) is an open interval containing \( s \) and there is a number, \( {x}_{m} \) in \( \left\{ {x}_{n}\right\} \) in \( \left( {{2s},0}\right) \) . Then \( {x}_{m} = {a}^{m} \) is negative which is a contradiction.\n\nSuppose \( s > 0 \) . Because \( 0 < a < 1 \) ), the number \( s/a \) is greater than \( s \) and there is a number, \( {x}_{m} \) in \( \left\{ {x}_{n}\right\} \) such that \( {x}_{m} < {s}_{a} \) . Because \( s \) is a lower bound on \( \left\{ {x}_{n}\right\}, s \leq {x}_{m} \) . Then\n\n\[ s \leq {x}_{m} < s/a,\;s \leq {a}^{m} < s/a,\;s \cdot a < {a}^{m + 1} < s,\;{x}_{m + 1} < s. \]\n\nBut this contradicts the condition that \( s \) is a lower bound on \( \left\{ {x}_{n}\right\} \) .\n\nWe conclude that \( s \), the greatest lower bound on \( \left\{ {x}_{n}\right\} \), is zero and \( \left\{ {x}_{n}\right\} \) converges to 0 .
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Yes
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Example 5.4.1 We can also compute \( {E}^{\prime }\left( t\right) \) for \( E\left( t\right) = {2}^{t} \) .
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\[ {\left\lbrack {2}^{t}\right\rbrack }^{\prime } = {\left\lbrack {\left( {e}^{\ln 2}\right) }^{t}\right\rbrack }^{\prime } = {\left\lbrack {e}^{\left( {\ln 2}\right) t}\right\rbrack }^{\prime } = {e}^{\left( {\ln 2}\right) t}\ln 2 = {2}^{t}\ln 2 \]
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Yes
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Property 5.5.1 Proportional Growth or Decay. If \( P \) is a function defined by\n\n\[ P\left( t\right) = C{e}^{kt} \]\n\nwhere \( C \) and \( k \) are numbers, then\n\n\[ {P}^{\prime }\left( t\right) = {kP}\left( t\right) \]
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Proof that \( P\left( t\right) = C{e}^{kt} \) implies that \( {P}^{\prime }\left( t\right) = {kP}\left( t\right) \) :\n\n\[ {P}^{\prime }\left( t\right) = {\left\lbrack C{e}^{kt}\right\rbrack }^{\prime } = C{\left\lbrack {e}^{kt}\right\rbrack }^{\prime } = C{e}^{kt}k = {kC}{e}^{kt} = {kP}\left( t\right) \]
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Yes
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Property 5.5.2 Exponential Growth or Decay If \( P \) is a function and there is a number \( k \) for which\n\n\[ \n{P}^{\prime }\left( t\right) = {kP}\left( t\right) \;\text{ for all }t \geq 0 \n\]\n\nthen there is a number \( C \) for which\n\n\[ \nP\left( t\right) = C{e}^{kt} \n\]\n\nFurthermore,\n\n\[ \nC = P\left( 0\right) \;\text{ so that }\;P\left( t\right) = P\left( 0\right) {e}^{kt} \n\]
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In the preceding equations, \( k \) can be either positive or negative. When \( k \) is negative, it is more common to emphasize this and write \( - k \) and write \( P\left( t\right) = {e}^{-{kt}} \), where in this context it is understood that \( k \) is a positive number.
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Yes
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Example 5.5.1 A distinction between discrete and continuous models. Suppose in year 2000 a population is at 5 million people and the population growth rate (excess of births over deaths) is 6 percent per year. One interpretation of this is to let \( P\left( t\right) \) be the population size in millions of people at time \( t \) measured in years after 2000 and to write\n\n\[ P\left( 0\right) = 5\;{P}^{\prime }\left( t\right) = {0.06P}\left( t\right) \]\n\nThen, from the property of Exponential Growth or Decay 5.5.2, we may write\n\n\[ P\left( t\right) = P\left( 0\right) {e}^{0.06t} = 5{e}^{0.06t} \]\n\n\( P\left( t\right) = 5{e}^{0.06t} \) does not exactly match the hypothesis that ’population growth rate is 6 percent per year', however. By this equation, after one year,\n\n\[ P\left( 1\right) = 5{e}^{{0.06} \times 1} = 5{e}^{0.06} \doteq 5 \times {1.0618} \]\n\nThe consequence is that during the first year (and every year) there would be a 6.18 percent increase, a contradiction.
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The discrepancy lies with the model equation \( {P}^{\prime }\left( t\right) = {0.6P}\left( t\right) \) . Instead, we may write\n\n\[ P\left( 0\right) = 5\;{P}^{\prime }\left( t\right) = {kP}\left( t\right) \]\n\nwhere \( k \) is to be determined. Then from Exponential Growth or Decay 5.5.2 we may write\n\n\[ P\left( t\right) = 5{e}^{kt} \]\n\nNow impose that \( P\left( 1\right) = 5 \times {1.06} \), a 6 percent increase during the first year, and write\n\n\[ P\left( 1\right) = 5{e}^{k1} = 5 \times {1.06} \]\n\nThis leads to\n\n\[ {e}^{k \times 1} = {1.06} \]\n\nWe take the natural logarithm of both numbers and get\n\n\[ \ln \left( {e}^{k}\right) = \ln {1.06} \]\n\n\[ k = \ln {1.06} \]\n\n\[ \doteq {0.05827} \]\n\nThen\n\n\[ P\left( t\right) = 5{e}^{0.05827t} \]\n\ngives a description of the population \( t \) years after 2000 . Each annual population is 6 percent greater than that of the preceding year. The continuous model of growth is actually\n\n\[ P\left( 0\right) = 5\;{P}^{\prime }\left( t\right) = \left( {\ln {1.06}}\right) P\left( t\right) \;{P}^{\prime }\left( t\right) = {0.05827P}\left( t\right) \]
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Yes
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Mathematical Model 5.5.3 Potassium-40 decomposition. The rate of disintegration of \( {}^{40}\mathrm{K} \) is proportional to the amount of \( {}^{40}\mathrm{\;K} \) present.
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If we let \( K\left( t\right) \) be the amount of \( {}^{40}\mathrm{\;K} \) present \( t \) years after deposition of rock of volcanic origin and \( {K}_{0} \) the initial amount of \( {}^{40}\mathrm{\;K} \) present, then\n\n\[ K\left( 0\right) = {K}_{0},\;{K}^{\prime }\left( t\right) = - {rK}\left( t\right) \]\n\nwhere \( r \) is a positive constant. The minus sign reflects the disintegration of \( {}^{40}\mathrm{\;K} \) . From the equation we may write\n\n\[ K\left( t\right) = {K}_{0}{e}^{-{rt}} \]\n\nThe half-life of \( {}^{40}\mathrm{\;K} \) is \( {1.28} \times {10}^{9} \) years, meaning that \( {1.28} \times {10}^{9} \) years after deposition of the volcanic rock, the amount of \( {}^{40}\mathrm{\;K} \) in the rock will be \( \frac{1}{2}{K}_{0} \) . We use this information to evaluate \( r \) .\n\n\[ \frac{1}{2}{K}_{0} = {K}_{0}{e}^{-{r1280000000}} \]\n\n\[ \frac{1}{2} = {e}^{-{r1280000000}} \]\n\n\[ \ln \frac{1}{2} = - {r1280000000} \]\n\n\[ r = \frac{\ln 2}{1280000000} \]\n\n\[ K\left( t\right) = {K}_{0}{e}^{-\frac{\ln 2}{1280000000}t} \]
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Yes
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Suppose a runner is exhaling at the rate of 2 liters per second. Then the amount of air in her lungs is decreasing at the rate of two liters per second. If, furthermore, the \( {\mathrm{{CO}}}_{2} \) partial pressure in the exhaled air is \( {50}\mathrm{\;{mm}}\mathrm{{Hg}} \) (approx \( {0.114}\mathrm{\;g}{\mathrm{{CO}}}_{2}/ \) liter of air at body temperature of \( {310}\mathrm{\;K}{)}^{8} \), then she is exhaling \( {\mathrm{{CO}}}_{2} \) at the rate of \( {0.114}\mathrm{\;g}/ \) liter \( \times 2 \) liters \( /\mathrm{{sec}} = {0.228}\mathrm{\;g}/\mathrm{{sec}} \) .
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\( \blacksquare \)
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No
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Example 5.5.4 Classical Washout Curve. A barrel contains 100 liters of water and 300 grams of salt. You start a stream of pure water flowing into the barrel at 5 liters per minute, and a compensating stream of salt water flows from the barrel at 5 liters per minute. The solution in the barrel is 'well stirred' so that the salt concentration is uniform throughout the barrel at all times. Let \( S\left( t\right) \) be the amount of salt (grams) in the barrel \( t \) minutes after you start the flow of pure water into the barrel.
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Solution. First let us analyze \( S \) . We use Primitive Concept 2. The concentration of salt in the water flowing into the barrel is 0 . The concentration of salt in the water flowing out of the barrel is the same as the concentration \( C\left( t\right) \) of salt in the barrel which is\n\n\[ C\left( t\right) = \frac{S\left( t\right) }{100}\mathrm{\;g}/\mathrm{L} \]\n\nTherefore\n\nRate of change of salt \( = \) Rate salt enters \( - \) Rate salt leaves\n\nin the barrel the barrel the barrel\n\n\[ {S}^{\prime }\left( t\right) \; = \;{C}_{1}{R}_{1}\; - \;{C}_{2}{R}_{2} \]\n\n\[ {S}^{\prime }\left( t\right) \; = \;0 \times 5\; - \;\frac{S\left( t\right) }{100}\frac{\mathrm{{gr}}}{\mathrm{L}} \times 5\frac{\mathrm{L}}{\min } \]\n\nFurthermore, \( S\left( 0\right) = {300} \) . Thus\n\n\[ S\left( 0\right) = {300} \]\n\n\[ {S}^{\prime }\left( t\right) = - {0.05S}\left( t\right) . \]\n\nFrom the Exponential Growth and Decay property 5.5.2,\n\n\[ S\left( t\right) = {300}{e}^{-{0.05t}} \]
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Yes
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Problem. Suppose a 100 liter barrel is full of pure water and at time \( t = 0 \) minutes a stream of water flowing at 5 liters per minute and carrying \( 3\mathrm{\;g}/ \) liter of salt starts flowing into the barrel. Assume the salt is well mixed in the barrel and water overflows at the rate of 5 liters per minute. Let \( S\left( t\right) \) be the amount of salt in the barrel at time \( t \) minutes after the salt water starts flowing in.
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Solution: We analyze \( S \) ; again we use Primitive Concept 2. The concentration of salt in the inflow is \( 3\mathrm{\;g}/ \) liter. The concentration \( C\left( t\right) \) of salt in the tank at time \( t \) minutes is\n\n\[ C\left( t\right) = \frac{S\left( t\right) }{100} \]\n\nThe salt concentration in the outflow will also be \( C\left( t\right) \) . Therefore\n\nRate of change of salt \( = \) Rate salt enters \( - \) Rate salt leaves\n\nin the barrel the barrel the barrel\n\n\[ {S}^{\prime }\left( t\right) \; = \;{C}_{1}{R}_{1}\; - \;{C}_{2}{R}_{2} \]\n\n\( {S}^{\prime }\left( t\right) \; = \;3 \times 5\; - \;\frac{S\left( t\right) }{100}\;5 \)\n\nInitially the barrel is full of pure water, so\n\n\[ S\left( 0\right) = 0 \]\n\nWe now have\n\n\[ S\left( 0\right) = 0 \]\n\n(5.22)\n\n\[ {S}^{\prime }\left( t\right) = {15} - {0.05S}\left( t\right) \]\n\nThis equation is not in the form of \( {P}^{\prime }\left( t\right) = {kP}\left( t\right) \) because of the 15 . Proceed as follows.\n\nEquilibrium. Ask,’At what value, \( E \), of \( S\left( t\right) \) would \( {S}^{\prime }\left( t\right) = 0 \) ?’ That would require\n\n\[ 0 = {15} - {0.05E} = 0,\text{ or }E = {300}\mathrm{\;g}. \]\n\n\( E = {300}\mathrm{\;g} \) is the equilibrium level of salt in the barrel. We focus attention on the difference, \( D\left( t\right) \) , between the equilibrium level and the current level of salt. Thus\n\n\[ D\left( t\right) = {300} - S\left( t\right) \;\text{ and }\;S\left( t\right) = {300} - D\left( t\right) \]\n\nNow,\n\n\[ D\left( 0\right) = {300} - S\left( 0\right) = {300} - 0 = {300} \]\n\nFurthermore,\n\n\[ {S}^{\prime }\left( t\right) = {\left\lbrack {300} - D\left( t\right) \right\rbrack }^{\prime } = - {D}^{\prime }\left( t\right) \]\n\nWe substitute into Equations 5.22\n\n\[ \begin{aligned} S\left( 0\right) & = 0 & D\left( 0\right) & = {300} \\ {S}^{\prime }\left( t\right) & = {15} - {0.05S}\left( t\right) & - {D}^{\prime }\left( t\right) & = {15} - {0.05}\left( {{300} - D\left( t\right) }\right) \end{aligned} \]\n\nThe equations for \( D \) become\n\n\[ D\left( 0\right) = {300} \]\n\n\[ {D}^{\prime }\left( t\right) = - {0.05D}\left( t\right) \]\n\nThis is in the form of the Exponential Growth and Decay Property 5.5.2, and we write\n\n\[ D\left( t\right) = {300}{e}^{-{0.05t}} \]\n\nReturning to \( S\left( t\right) = {300} - D\left( t\right) \) we write\n\n\[ S\left( t\right) = {300} - D\left( t\right) = {300} - {300}{e}^{-{0.05t}} \]\n\nThe graph of \( S\left( t\right) = {300} - {300}{e}^{-{0.05t}} \) is shown in Figure 5.3. Curiously, the graph of \( S\left( t\right) \) is also called an exponential decay curve. \( S\left( t\right) \) is not decaying at all; \( S\left( t\right) \) is increasing. What is decaying exponentially is \( D\left( t\right) \), the remaining salt capacity.
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Yes
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Assume that \( {1000}\mathrm{\;w}/{\mathrm{m}}^{2} \) of light is striking the surface of a lake and that \( {40}\% \) of that light is reflected back into the atmosphere. We first solve the initial value problem\n\n\[ I\left( 0\right) = {600} \]\n\n\[ {I}^{\prime }\left( x\right) = - {KI}\left( x\right) \]
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to get\n\n\[ I\left( x\right) = {600}{e}^{-{Kx}} \]\n\nIf we have additional information that, say, the light intensity at a depth of \( {10}\mathrm{{meters}} \) is \( {400}\mathrm{\;W}/{\mathrm{m}}^{2} \) we can find the value of \( K \) . It must be that\n\n\[ I\left( {10}\right) = {600}{e}^{-K \times {10}} = {400} \]\n\nThe only unknown in the last equation is \( K \), and we solve\n\n\[ {600}{e}^{-K \times {10}} = {400} \]\n\n\[ {e}^{-{10K}} = {400}/{600} = 5/6 \]\n\n\[ \ln \left( {e}^{-{10K}}\right) = \ln \left( {2/3}\right) \]\n\n\[ - {10K} = \ln \left( {2/3}\right) \]\n\n\[ K \doteq {0.040557} \]\n\nThus we would say that\n\n\[ I\left( x\right) = {600}{e}^{-{0.040557x}} \]
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Yes
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Suppose a patient is prescribed to take \( {80}\mathrm{{mg}} \) of Sotolol, a drug that regularizes heart beat, once per day. Sotolol has a half-life in the body of 12 hrs. Compute the daily fluctuations of sotolol.
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Let \( {s}_{t}^{ - } \) be the amount of sotolol in the body at time, \( t \), just before the sotolol pill is taken and \( {s}_{t}^{ + } \) be the amount of sotolol in the body at time \( t \) just after the sotolol pill is taken. Then\n\n\[ \n{s}_{0}^{ - } = 0,\;{s}_{0}^{ + } = {80},\;{s}_{1}^{ - }\; = \;\frac{1}{4}{s}_{0}^{ + }\; = \;\frac{1}{4}{80}\; = \;{20}. \n\]\n\n\[ \n{s}_{t}^{ + } = {s}_{t}^{ - } + {80}\;{s}_{t + 1}^{ - }\; = \frac{1}{4}{s}_{t}^{ + } = \;\frac{1}{4}\left( {{s}_{t}^{ - } + {80}}\right) = \frac{1}{4}{s}_{t}^{ - } + {20} \n\]\n\nAfter some days, \( {s}_{t}^{ - } \) will reach approximate equilibrium, \( E : {s}_{t}^{ - } \doteq E \) and \( {s}_{t + 1}^{ - } \doteq E \) .\n\n\[ \n{s}_{0}^{ - } = 0,\;{s}_{t + 1}^{ - } = \frac{1}{4}{s}_{t}^{ - } + {20},\;E = \frac{1}{4}E + {20},\;E = {26.67} \n\]\n\nThus, approximately, \( \;{s}_{t}^{ - } = E = {26.67},\;{s}_{t}^{ + } = {s}_{t}^{ - } + {80} = E + {80} = {106.67} \) .\n\nso the system oscillates between \( {26.67}\mathrm{{mg}} \) and \( {106.67}\mathrm{{mg}} \), a four to one ratio.
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Yes
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Example 5.5.8 In Section 1.3 we showed the results of an experiment measuring the light decay as a function of depth. The data and a semilog graph of the data are shown in Figure 5.6.
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As shown in the figure, \[ {\log }_{10}{I}_{d} = - {0.4} - {0.087d} \] is a good approximation to the data. Therefore \[ {I}_{d} \doteq {10}^{-{0.4} - {0.087d}} \] \[ = {0.4} \times {0.82}^{d} \] which is the same result obtained in Section 1.3. As shown in the previous subsection, the relation \[ {I}^{\prime }\left( d\right) = - {kI}\left( d\right) \] corresponds to a process underlying light depletion in water.
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Yes
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Find the derivatives of\n\n\[ P\left( x\right) = {200}{e}^{-{3x}}\;Q\left( x\right) = \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2} \]
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Solutions:\n\n\[ {P}^{\prime }\left( x\right) = {\left\lbrack {200}{e}^{-{3x}}\right\rbrack }^{\prime }\;{Q}^{\prime }\left( x\right) = {\left\lbrack \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2}\right\rbrack }^{\prime }\;\text{ Logical Identity } \]\n\n\[ = {200}{\left\lbrack {e}^{-{3x}}\right\rbrack }^{\prime }\; = \frac{1}{\sqrt{2\pi }}{\left\lbrack {e}^{-{x}^{2}/2}\right\rbrack }^{\prime }\;\text{Constant Factor Rule} \]\n\n\[ = {200}{e}^{-{3x}}{\left\lbrack -3x\right\rbrack }^{\prime }\; = \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2}{\left\lbrack -{x}^{2}/2\right\rbrack }^{\prime }\text{Exponential Chain Rule} \]\n\n\[ = {200}{e}^{-{3x}}\left( {-3}\right) \; = \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2}\left( {-x}\right) \;\text{Constant Factor, Power Rules} \]
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Yes
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Logarithm functions to other bases have derivatives, but not as neat as \( 1/t \) . We compute \( {L}^{\prime }\left( t\right) \) for \( L\left( t\right) = {\log }_{b}t \) where \( b > 0 \) and \( b \neq 1 \) .
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\[ {\left\lbrack {\log }_{b}t\right\rbrack }^{\prime } = {\left\lbrack \frac{\ln t}{\ln b}\right\rbrack }^{\prime } \] \[ \left( i\right) \] \[ = \frac{1}{\ln b}{\left\lbrack \ln t\right\rbrack }^{\prime } \] (ii) (5.39) \[ = \frac{1}{\ln b}\frac{1}{t} \] (iii) We summarize this as \[ {\left\lbrack {\log }_{b}t\right\rbrack }^{\prime } = \frac{1}{\ln b}\frac{1}{t} \] (5.40)
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Yes
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Find the derivatives of a. \( y = \ln \left( \sqrt{1 - {t}^{2}}\right) \) b. \( y = \ln \left( \frac{1 - t}{1 + t}\right) \)
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\[ {\left\lbrack \ln \left( \sqrt{1 - {t}^{2}}\right) \right\rbrack }^{\prime } = {\left\lbrack \frac{1}{2}\ln \left( 1 - {t}^{2}\right) \right\rbrack }^{\prime }\;\text{ Logarithm Property } \] \[ = \frac{1}{2}{\left\lbrack \ln \left( 1 - {t}^{2}\right) \right\rbrack }^{\prime }\;\text{Constant Factor} \] \[ = \frac{1}{2}\frac{1}{1 - {t}^{2}}{\left\lbrack \left( 1 - {t}^{2}\right) \right\rbrack }^{\prime }\;\text{Generalized Logarithm Rule} \] \[ = \frac{1}{2}\frac{1}{1 - {t}^{2}}\left( {-{2t}}\right) \;\text{Sum, Constant, Constant Factor, and Power Rules} \] \[ \text{b.}{\left\lbrack \ln \left( \frac{1 - t}{1 + t}\right) \right\rbrack }^{\prime } = {\left\lbrack \ln \left( 1 - t\right) - \ln \left( 1 + t\right) \right\rbrack }^{\prime } \] Logarithm Property \[ = {\left\lbrack \ln \left( 1 - t\right) \right\rbrack }^{\prime } - {\left\lbrack \ln \left( 1 + t\right) \right\rbrack }^{\prime } \] Sum Rule \[ = \frac{1}{1 - t}{\left\lbrack 1 - t\right\rbrack }^{\prime } - \frac{1}{1 + t}{\left\lbrack 1 + t\right\rbrack }^{\prime } \] Generalized Logarithm Rule \[ = \frac{1}{1 - t}\left( {-1}\right) - \frac{1}{1 + t}\left( 1\right) \] Sum, Constant, and Power Rules \[ = \frac{-2}{1 - {t}^{2}} \]
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Yes
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Suppose we are to differentiate\n\n\\[ y\\left( t\\right) = \\left( {t + 2}\\right) \\left( {t + 1}\\right) \\left( {t - 1}\\right) \\]
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Proceeding indirectly, we first compute the derivative of the natural logarithm of \\( y \\) .\n\n\\[ \\ln \\left( {y\\left( t\\right) }\\right) = \\ln \\left( {\\left( {t + 2}\\right) \\left( {t + 1}\\right) \\left( {t - 1}\\right) }\\right) \\]\nLogical Identity\n\n\\[ \\ln \\left( {y\\left( t\\right) }\\right) = \\ln \\left( {t + 2}\\right) + \\ln \\left( {t + 1}\\right) + \\ln \\left( {t - 1}\\right) \\]\nLogarithm Property\n\n\\[ {\\left\\lbrack \\ln \\left( y\\left( t\\right) \\right) \\right\\rbrack }^{\\prime } = {\\left\\lbrack \\ln \\left( t + 2\\right) + \\ln \\left( t + 1\\right) + \\ln \\left( t - 1\\right) \\right\\rbrack }^{\\prime } \\]\nLogical Identity.\n\n\\[ \\frac{1}{y\\left( t\\right) }{y}^{\\prime }\\left( t\\right) = \\frac{1}{t + 2}{\\left\\lbrack t + 2\\right\\rbrack }^{\\prime } + \\frac{1}{t + 1}{\\left\\lbrack t + 1\\right\\rbrack }^{\\prime } + \\frac{1}{t - 1}{\\left\\lbrack t - 1\\right\\rbrack }^{\\prime }\\;\\text{ Logarithm chain rule. } \\]\n\n\\[ {y}^{\\prime }\\left( t\\right) = y\\left( t\\right) \\left( {\\frac{1}{t + 2} + \\frac{1}{t + 1} + \\frac{1}{t - 1}}\\right) \\;\\left\\lbrack t + C\\right\\rbrack }^{\\prime } = 1. \\]\n\n\\[ {y}^{\\prime }\\left( t\\right) = \\left( {t + 2}\\right) \\left( {t + 1}\\right) \\left( {t - 1}\\right) \\left( {\\frac{1}{t + 2} + \\frac{1}{t + 1} + \\frac{1}{t - 1}}\\right) \\;\\text{ Definition of }y. \\]\n\n\\[ {y}^{\\prime }\\left( t\\right) = \\left( {t + 1}\\right) \\left( {t - 1}\\right) + \\left( {t + 2}\\right) \\left( {t - 1}\\right) + \\left( {t + 2}\\right) \\left( {t + 1}\\right) \\;\\text{ Algebra. }\\;\\]
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Yes
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Theorem 6.1.1 Suppose \( u \) and \( v \) are two functions. Then for every number a for which \( {u}^{\prime }\left( a\right) \) and \( {v}^{\prime }\left( a\right) \) exist,\n\n\[{\left\lbrack u\left( t\right) \times v\left( t\right) \right\rbrack }_{t = a}^{\prime } = {u}^{\prime }\left( a\right) \times v\left( a\right) + u\left( a\right) \times {v}^{\prime }\left( a\right)\]
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The proof uses Theorem 4.2.1, The Derivative Requires Continuity, which in symbols is:\n\n\[ \mathop{\lim }\limits_{{b \rightarrow a}}\frac{u\left( b\right) - u\left( a\right) }{b - a} = {u}^{\prime }\left( a\right) \;\text{ exists implies that }\;\mathop{\lim }\limits_{{b \rightarrow a}}u\left( b\right) = u\left( a\right) .\n\nProof of Theorem 6.1.1.\n\n\[{\left\lbrack u\left( t\right) \times v\left( t\right) \right\rbrack }_{t = a}^{\prime } = \mathop{\lim }\limits_{{b \rightarrow a}}\frac{u\left( b\right) \times v\left( b\right) - u\left( a\right) \times v\left( a\right) }{b - a}\]\n\n\( \left( i\right) \)\n\n\[= \mathop{\lim }\limits_{{b \rightarrow a}}\frac{u\left( b\right) \times v\left( b\right) - u\left( a\right) \times v\left( b\right) + u\left( a\right) \times v\left( b\right) - u\left( a\right) \times v\left( a\right) }{b - a}\]\n\n(ii)\n\n\[= \mathop{\lim }\limits_{{b \rightarrow a}}\left( {\frac{u\left( b\right) - u\left( a\right) }{b - a} \times v\left( b\right) + u\left( a\right) \times \frac{v\left( b\right) - v\left( a\right) }{b - a}}\right)\]\n\n(iii)\n\n(6.4)\n\n\[= \mathop{\lim }\limits_{{b \rightarrow a}}\left( {\frac{u\left( b\right) - u\left( a\right) }{b - a} \times v\left( b\right) }\right) + u\left( a\right) \times \mathop{\lim }\limits_{{b \rightarrow a}}\frac{v\left( b\right) - v\left( a\right) }{b - a}\]\n\n\( \left( {iv}\right) \)\n\n\[= \mathop{\lim }\limits_{{b \rightarrow a}}\frac{u\left( b\right) - u\left( a\right) }{b - a} \times v\left( a\right) + u\left( a\right) \times \mathop{\lim }\limits_{{b \rightarrow a}}\frac{v\left( b\right) - v\left( a\right) }{b - a}\]\n\n\( \left( v\right) \)\n\n\[= {u}^{\prime }\left( a\right) \times v\left( a\right) + u\left( a\right) \times {v}^{\prime }\left( a\right)\]\n\n\( \left( {vi}\right) \)\n\nEnd of proof.
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Yes
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Theorem 6.1.2 Suppose \( u \) and \( v \) are functions and \( {u}^{\prime }\left( a\right) \) and \( {v}^{\prime }\left( a\right) \) exist and \( v\left( a\right) \neq 0 \) . Then\n\n\[ \n{\left\lbrack \frac{u\left( t\right) }{v\left( t\right) }\right\rbrack }_{t = a}^{\prime } = \frac{{u}^{\prime }\left( a\right) \times v\left( a\right) - u\left( a\right) \times {v}^{\prime }\left( a\right) }{{v}^{2}\left( a\right) }\n\]
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Proof of Theorem 6.1.2.\n\n\[ \n{\left\lbrack \frac{u\left( t\right) }{v\left( t\right) }\right\rbrack }_{t = a}^{\prime } = \mathop{\lim }\limits_{{b \rightarrow a}}\frac{\frac{u\left( b\right) }{v\left( b\right) } - \frac{u\left( a\right) }{v\left( a\right) }}{b - a}\n\]\n\n(i)\n\n\[ \n= \mathop{\lim }\limits_{{b \rightarrow a}}\frac{\frac{u\left( b\right) v\left( a\right) - u\left( a\right) v\left( b\right) }{b - a}}{v\left( b\right) v\left( a\right) }\n\]\n\n(ii)\n\n\[ \n= \mathop{\lim }\limits_{{b \rightarrow a}}\frac{\frac{u\left( b\right) v\left( a\right) - u\left( a\right) v\left( a\right) + u\left( a\right) v\left( a\right) - u\left( a\right) v\left( b\right) }{b - a}}{v\left( b\right) v\left( a\right) }\n\]\n\n(iii)\n\n(6.6)\n\n\[ \n= \mathop{\lim }\limits_{{b \rightarrow a}}\frac{\frac{u\left( b\right) - u\left( a\right) }{b - a}v\left( a\right) - u\left( a\right) \frac{v\left( b\right) - v\left( a\right) }{b - a}}{v\left( b\right) v\left( a\right) }\n\]\n\n\( \left( {iv}\right) \)\n\n\[ \n= \frac{\left( {\mathop{\lim }\limits_{{b \rightarrow a}}\frac{u\left( b\right) - u\left( a\right) }{b - a}}\right) v\left( a\right) - u\left( a\right) \mathop{\lim }\limits_{{b \rightarrow a}}\frac{v\left( b\right) - v\left( a\right) }{b - a}}{\left( {\mathop{\lim }\limits_{{b \rightarrow a}}v\left( b\right) }\right) v\left( a\right) }\n\]\n\n\( \left( v\right) \)\n\n\[ \n= \frac{{u}^{\prime }\left( a\right) \times v\left( a\right) - u\left( a\right) \times {v}^{\prime }\left( a\right) }{v\left( a\right) v\left( a\right) }\n\]\n\n( \( {vi} \) )\n\nEnd of proof.
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Yes
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The logistic function and its derivative. The logistic function\n\n\\[ \nP\\left( t\\right) = \\frac{{P}_{0}M{e}^{rt}}{M - {P}_{0} + {P}_{0}{e}^{rt}} \n\\]\n\ndescribes the size of a population of initial size \\( {P}_{0} \\) and low density relative growth rate \\( r \\) growing in an environment with limited carrying capacity \\( M \\).
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\[ \n{P}^{\\prime }\\left( t\\right) = {\\left\\lbrack \\frac{{P}_{0}M{e}^{rt}}{M - {P}_{0} + {P}_{0}{e}^{rt}}\\right\\rbrack }^{\\prime } \n\]\n\n\[ \n= {P}_{0}M{\\left\\lbrack \\frac{{e}^{rt}}{M - {P}_{0} + {P}_{0}{e}^{rt}}\\right\\rbrack }^{\\prime } \n\]\n\n\[ \n- {P}_{0}M\\frac{\\left( {M - {P}_{0} + {P}_{0}{e}^{rt}}\\right) {\\left\\lbrack {e}^{rt}\\right\\rbrack }^{\\prime } - {e}^{rt}{\\left\\lbrack M - {P}_{0} + {P}_{0}{e}^{rt}\\right\\rbrack }^{\\prime }}{{\\left( M - {P}_{0} + {P}_{0}{e}^{rt}\\right) }^{2}} \n\]\n\n\[ \n= {P}_{0}M\\frac{\\left( {M - {P}_{0} + {P}_{0}{e}^{rt}}\\right) {e}^{rt} \\times r - {e}^{rt}\\left( {0 + {P}_{0}{e}^{rt} \\times r}\\right) }{{\\left( M - {P}_{0} + {P}_{0}{e}^{rt}\\right) }^{2}} \n\]\n\n\[ \n= {P}_{0}M\\frac{\\left( {M - {P}_{0}}\\right) {e}^{rt}r}{{\\left( M - {P}_{0} + {P}_{0}{e}^{rt}\\right) }^{2}} \n\]\n\n\[ \n= r\\frac{{P}_{0}M{e}^{rt}}{M - {P}_{0} + {P}_{0}{e}^{rt}}\\frac{\\left( M - {P}_{0}\\right) }{M - {P}_{0} + {P}_{0}{e}^{rt}} \n\]\n\n\[ \n= {rP}\\left( t\\right) \\left( {1 - \\frac{P\\left( t\\right) }{M}}\\right) \n\]
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Yes
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\[ \text{a.}P\left( t\right) = {e}^{2t}\ln t\;{P}^{\prime }\left( t\right) = {\left\lbrack {e}^{2t}\ln t\right\rbrack }^{\prime } \]
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\[ = {\left\lbrack {e}^{2t}\right\rbrack }^{\prime }\ln t + {e}^{2t}{\left\lbrack \ln t\right\rbrack }^{\prime } \] \[ = {e}^{2t}2\ln t + {e}^{2t}\frac{1}{t} \]
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Yes
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Compute \( {F}^{\prime }\left( t\right) \) for \( F\left( t\right) = {\left( 1 - {t}^{2}\right) }^{3} \) .
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Let\n\n\[ G\left( z\right) = {z}^{3}\;\text{ and }\;u\left( t\right) = 1 - {t}^{2}.\;\text{ Then }\;F\left( t\right) = G\left( {u\left( t\right) }\right) .\n\n\]\n\n\[ {G}^{\prime }\left( z\right) = 3{z}^{2}\;\text{ and }\;{\left\lbrack u\left( t\right) \right\rbrack }^{\prime } = - {2t} \]\n\n\[ {G}^{\prime }\left( {u\left( t\right) }\right) = 3{\left( u\left( t\right) \right) }^{2} = 3{\left( 1 - {t}^{2}\right) }^{2}, \]\n\nand\n\n\[ {F}^{\prime }\left( t\right) = {G}^{\prime }\left( {u\left( t\right) }\right) {\left\lbrack u\left( t\right) \right\rbrack }^{\prime } = 3{\left( 1 - {t}^{2}\right) }^{2}\left( {-{2t}}\right) \]
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Yes
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Theorem 6.2.1 Chain Rule. Suppose \( G \) and \( u \) are functions that have derivatives and \( G\left( {u\left( t\right) }\right) \) is defined for all numbers \( t \) . Then \( G\left( {u\left( t\right) }\right) \) has a derivative for all \( t \) and\n\n\[{\left\lbrack G\left( u\left( t\right) \right) \right\rbrack }^{\prime } = {G}^{\prime }\left( {u\left( t\right) }\right) \times {\left\lbrack u\left( t\right) \right\rbrack }^{\prime }\]
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Proof: The argument is similar to that for the exponential chain rule. The difference is that we now have a general function \( G\left( u\right) \) rather than the specific functions \( {e}^{u} \) . We argue only for \( u \) an increasing function, and we need Theorem 4.2.1, The Derivative Requires Continuity.\n\nLet \( F = G \circ u\left( {F\left( t\right) = G\left( {u\left( t\right) }\right) \text{for all}t}\right) \) .\n\n\[{F}^{\prime }\left( a\right) = \mathop{\lim }\limits_{{b \rightarrow a}}\frac{F\left( b\right) - F\left( a\right) }{b - a}\]\n\n\[= \mathop{\lim }\limits_{{b \rightarrow a}}\frac{G\left( {u\left( b\right) }\right) - G\left( {u\left( a\right) }\right) }{b - a}\]\n\n\[= \mathop{\lim }\limits_{{b \rightarrow a}}\frac{G\left( {u\left( b\right) }\right) - G\left( {u\left( a\right) }\right) }{u\left( b\right) - u\left( a\right) }\mathop{\lim }\limits_{{b \rightarrow a}}\frac{u\left( b\right) - u\left( a\right) }{b - a}\;u\left( b\right) - u\left( a\right) \neq 0\]\n\n\[= {G}^{\prime }\left( {u\left( a\right) }\right) {u}^{\prime }\left( a\right)\]\n\nThe conclusion that\n\n\[\mathop{\lim }\limits_{{b \rightarrow a}}\frac{G\left( {u\left( b\right) }\right) - G\left( {u\left( a\right) }\right) }{u\left( b\right) - u\left( a\right) } = {G}^{\prime }\left( {u\left( a\right) }\right)\]\n\nrequires some support.\n\nIn Figure 6.5, the slope of the secant is \( \;\frac{G\left( {u\left( b\right) }\right) - G\left( {u\left( a\right) }\right) }{u\left( b\right) - u\left( a\right) } \) .\n\nBecause \( {u}^{\prime }\left( a\right) \) exists, \( u\left( b\right) \rightarrow u\left( a\right) \) as \( b \rightarrow a \) . The slope of the secant approaches the slope of the tangent as \( u\left( b\right) \rightarrow u\left( a\right) \), and\n\n\[\mathop{\lim }\limits_{{b \rightarrow a}}\frac{G\left( {u\left( b\right) }\right) - G\left( {u\left( a\right) }\right) }{u\left( b\right) - u\left( a\right) } = {G}^{\prime }\left( {u\left( a\right) }\right) .\]\n\nEnd of proof.
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Yes
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Compute the derivative of\n\n\\[ F\\left( t\\right) = {e}^{\\sqrt{\\ln t}} \\; t > 1 \\; \\text{ so that } \\; \\ln t > 0.\\]\n
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We peel the layers off from the outside. \\( F\\left( t\\right) \\) can be thought of as\n\n\\[ F\\left( t\\right) = G\\left( {H\\left( {K\\left( t\\right) }\\right) }\\right) ,\\; \\text{where} \\; G\\left( z\\right) = {e}^{z}, \\; H\\left( x\\right) = \\sqrt{x}, \\; \\text{and} \\; K\\left( t\\right) = \\ln t \\]\n\n\\[\n{\\left\\lbrack {e}^{\\sqrt{\\ln t}}\\right\\rbrack }^{\\prime } = {e}^{\\sqrt{\\ln t}}{\\left\\lbrack \\sqrt{\\ln t}\\right\\rbrack }^{\\prime } \\; G\\left( z\\right) = {e}^{z}, \\; {G}^{\\prime }\\left( z\\right) = {e}^{z}\n\\]\n\n\\[\n= {e}^{\\sqrt{\\ln t}}\\frac{1}{2\\sqrt{\\ln t}}{\\left\\lbrack \\ln t\\right\\rbrack }^{\\prime } \\; H\\left( x\\right) = \\sqrt{x}, \\; {H}^{\\prime }\\left( x\\right) = \\frac{1}{2\\sqrt{x}}\n\\]\n\n\\[\n= {e}^{\\sqrt{\\ln t}}\\frac{1}{2\\sqrt{\\ln t}}\\frac{1}{t} \\; K\\left( t\\right) = \\ln t, \\; {K}^{\\prime }\\left( t\\right) = \\frac{1}{t}\n\\]\n
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Yes
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Example 6.2.3 Find \( \frac{dy}{dt} \) for \( y\left( t\right) = {\left( 1 + {t}^{4}\right) }^{7}.y\left( t\right) \) is the composition of \( G\left( u\right) = {u}^{7} \) and \( u\left( t\right) = 1 + {t}^{4} \) .
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\[ \frac{dG}{du} = \frac{d}{du}{u}^{7}\; = 7{u}^{6} \] \[ \frac{du}{dt} = \frac{d}{dt}\left( {1 + {t}^{4}}\right) = 4{t}^{3} \] \[ \frac{dG}{dt} = \frac{dG}{du}\frac{du}{dt}\; = 7{u}^{6} \times 4{t}^{3} = 7{\left( 1 + {t}^{4}\right) }^{6}4{t}^{3} \]
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Yes
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The derivative of the inverse of a function. If \( g \) is an invertible function that has a nonzero derivative and \( h \) is its inverse, then for every number, \( t \), in the domain of \( g \) ,
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\[ {g}^{\prime }\left( {h\left( t\right) }\right) = \frac{1}{{h}^{\prime }\left( t\right) }\;\text{ and }\;{g}^{\prime }\left( t\right) = \frac{1}{{h}^{\prime }\left( {g\left( t\right) }\right) }.\]\n\nIf \( g \) is an invertible function and \( h \) is its inverse, then for every number, \( t \), in the domain of \( g \) ,\n\n\[ g\left( {h\left( t\right) }\right) = t \]\n\nWe differentiate both sides of this equation.\n\n\[ {\left\lbrack g\left( h\left( t\right) \right) \right\rbrack }^{\prime } = {\left\lbrack t\right\rbrack }^{\prime } \]\n\n\( {g}^{\prime }\left( {h\left( t\right) }\right) {h}^{\prime }\left( t\right) = 1\; \) Uses the Chain Rule\n\n\[ {h}^{\prime }\left( t\right) = \frac{1}{{g}^{\prime }\left( {h\left( t\right) }\right) }\;\text{ Assumes }{g}^{\prime }\left( {h\left( t\right) }\right) \neq 0 \]
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Yes
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Example 6.3.3 The function, \( h\left( t\right) = \sqrt{t}, t > 0 \) is the inverse of the function, \( g\left( x\right) = {x}^{2}, x > 0 \) .
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\[ {g}^{\prime }\left( x\right) = {2x} \] \[ {h}^{\prime }\left( t\right) = \frac{1}{{g}^{\prime }\left( {h\left( t\right) }\right) } = \frac{1}{{2h}\left( t\right) } = \frac{1}{2\sqrt{t}} \] a result that we obtained directly from the definition of derivative.
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Yes
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We illustrate the use of the derivative formulas for sine, cosine, and tangent by computing the derivatives of\n\n\[ \text{a.}y = 3\sin t\cos t\;\text{b.}y = {\sin }^{4}t\;\text{c.}y = \ln \left( {\tan t}\right) \;\text{d.}y = {e}^{\sin t} \]
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\[ \text{a.}\;{\left\lbrack 3\sin t\cos t\right\rbrack }^{\prime } = 3\left( {\sin t{\left\lbrack \cos t\right\rbrack }^{\prime } + {\left\lbrack \sin t\right\rbrack }^{\prime }\cos t}\right) \]\n\n\[ = 3\left( {\sin t\left( {-\sin t}\right) + \cos t\cos t}\right) \]\n\n\[ = - 3{\sin }^{2}t + 3{\cos }^{2}t \]
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No
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Example 7.2.2 The function \( F\left( t\right) = {e}^{-t/{10}}\sin t \) is an example of ’damped oscillation,’ an important type of vibration. Its graph is shown in Figure 7.4. The peaks and valleys of the oscillation are marked by values of \( t \) for which \( {F}^{\prime }\left( t\right) = 0 \) . We find them by
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\[ {\left\lbrack {e}^{-t/{10}}\sin t\right\rbrack }^{\prime } = {e}^{-t/{10}}{\left\lbrack \sin t\right\rbrack }^{\prime } + {\left\lbrack {e}^{-t/{10}}\right\rbrack }^{\prime }\sin t \] \[ = {e}^{-t/{10}}\cos t - \frac{1}{10}{e}^{-t/{10}}\sin t = {e}^{-t/{10}}\left( {\cos t - \frac{1}{10}\sin t}\right) \] Now \( {e}^{-t/{10}} > 0 \) for all \( t;{F}^{\prime }\left( t\right) = 0 \) implies that \[ \cos t - \frac{1}{10}\sin t = 0,\;\tan t = {10},\;t = \left( {\arctan {10}}\right) + {n\pi }\;\text{ for }n\text{ an integer. } \]
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Yes
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For \( u\left( t\right) = {kt} \) where \( k \) is a constant,
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\[ {\left\lbrack \sin \left( kt\right) \right\rbrack }^{\prime } = \cos \left( {kt}\right) \times {\left\lbrack kt\right\rbrack }^{\prime } = \cos \left( {kt}\right) \times k \]
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Yes
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Example 7.3.2 With repeated use of the chain rule, derivatives of some rather difficult and exotic functions can be computed. For example, find \( {y}^{\prime } \) for\n\n\[ y\left( t\right) = \ln \left( {\sin \left( {e}^{\cos t}\right) }\right) \]
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\[ {y}^{\prime }\left( t\right) = {\left\lbrack \ln \left( \sin \left( {e}^{\cos t}\right) \right) \right\rbrack }^{\prime }\;\text{Outside layer} \]\n\n\[ = \frac{1}{\sin \left( {e}^{\cos t}\right) } \times {\left\lbrack \sin \left( {e}^{\cos t}\right) \right\rbrack }^{\prime }\;G\left( u\right) = \ln u,\;{G}^{\prime }\left( u\right) = \frac{1}{u} \]\n\n\[ = \frac{1}{\sin \left( {e}^{\cos t}\right) } \times \cos \left( {e}^{\cos t}\right) \times {\left\lbrack {e}^{\cos t}\right\rbrack }^{\prime }\;G\left( u\right) = \sin u,\;{G}^{\prime }\left( u\right) = \cos u \]\n\n\[ = \frac{1}{\sin \left( {e}^{\cos t}\right) } \times \cos \left( {e}^{\cos t}\right) \times {e}^{\cos t} \times {\left\lbrack \cos t\right\rbrack }^{\prime }\;G\left( u\right) = {e}^{u},\;{G}^{\prime }\left( u\right) = {e}^{u} \]\n\n\[ = \frac{1}{\sin \left( {e}^{\cos t}\right) } \times \cos \left( {e}^{\cos t}\right) \times {e}^{\cos t} \times \left( {-\sin t}\right) \;G\left( u\right) = \cos u,\;{G}^{\prime }\left( u\right) = - \sin u \]\n\n\[ = - \sin t \times {e}^{\cos t} \times \cot \left( {e}^{\cos t}\right) \]
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Yes
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We show that \( y\left( t\right) = {e}^{-t}\cos {6t} \) solves \( {y}^{\prime \prime }\left( t\right) + 2{y}^{\prime }\left( t\right) + {37y}\left( t\right) = 0 \) .
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\[ {y}^{\prime }\left( t\right) = {\left\lbrack {e}^{-t}\cos t\right\rbrack }^{\prime } \] \[ = {\left\lbrack {e}^{-t}\right\rbrack }^{\prime }\cos {6t} + {e}^{-t}{\left\lbrack \cos 6t\right\rbrack }^{\prime } \] \( \left( i\right) \) \[ = \left( {-{e}^{-t}}\right) \cos {6t} + {e}^{-t}\left( {-\sin {6t}}\right) \times 6 \] (ii) \[ {y}^{\prime \prime }\left( t\right) = {\left\lbrack -{e}^{-t}\cos 6t - 6{e}^{-t}\sin 6t\right\rbrack }^{\prime } \] \[ = {\left\lbrack -{e}^{-t}\cos 6t\right\rbrack }^{\prime } - 6{\left\lbrack {e}^{-t}\sin 6t\right\rbrack }^{\prime } \] (iii) \[ = - {\left\lbrack {e}^{-t}\right\rbrack }^{\prime }\cos {6t} - {e}^{-t}{\left\lbrack \cos 6t\right\rbrack }^{\prime } \] (7.22) \[ - 6\left( {{\left\lbrack {e}^{-t}\right\rbrack }^{\prime }\sin {6t} + {e}^{-t}{\left\lbrack \sin 6t\right\rbrack }^{\prime }}\right) \] (iv) \[ = {e}^{-t}\cos {6t} + 6{e}^{-t}\sin {6t} + 6{e}^{-t}\sin {6t} - {36}{e}^{-t}\cos {6t} \] \( \left( v\right) \) \[ = {12}{e}^{-t}\sin {6t} - {35}{e}^{-t}\cos {6t} \] We next substitute \( y\left( t\right) = {e}^{-t}\cos {6t} \), and the computed values for \( {y}^{\prime }\left( t\right) \) and \( {y}^{\prime \prime }\left( t\right) \) into \( {y}^{\prime \prime } + 2{y}^{\prime } + {37y} \) and confirm the solution. \[ {y}^{\prime \prime } + \;2{y}^{\prime } + \;{37y} \] \[ {12}{e}^{-t}\sin {6t} - {35}{e}^{-t}\cos {6t} + 2\left( {-{e}^{-t}\cos {6t} - 6{e}^{-t}\sin {6t}}\right) + {37}{e}^{-t}\cos {6t} \] \[ = \left( {{12} - 2 \times 6}\right) {e}^{-t}\sin {6t}\; + \;\left( {-{35} - 2 + {37}}\right) {e}^{-t}\cos {6t}\; = 0 \]
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Yes
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Example 7.3.4 A searchlight is \( {400}\mathrm{\;m} \) from a straight beach and rotates at a constant rate once in two minutes. How fast is the beam moving along the beach when the beam is \( {600}\mathrm{\;m} \) from the nearest point, A, of the beach to the searchlight.
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Solution. See Figure 7.6. Let \( \theta \) measure the rotation of the light with \( \theta = 0 \) when the light is pointing toward A. Let \( x \) be the distance from A to the point where the beam strikes the beach.\n\nWe have\n\[ \tan \left( \theta \right) = \frac{x}{400} \]\n\nBoth \( \theta \) and \( x \) are functions of time and we write.\n\n\[ \tan \theta \left( t\right) = \frac{x\left( t\right) }{400} \]\n\nand differentiate both sides of the equation with respect to \( t \) . We get\n\n\[ {\left\lbrack \tan \left( \theta \left( t\right) \right) \right\rbrack }^{\prime } = {\left\lbrack \frac{x\left( t\right) }{400}\right\rbrack }^{\prime } \]\n\n\[ {\sec }^{2}\left( {\theta \left( t\right) }\right) {\left\lbrack \theta \left( t\right) \right\rbrack }^{\prime } = \frac{1}{400}{\left\lbrack x\left( t\right) \right\rbrack }^{\prime }\;\text{ Tangent Chain, Constant Factor } \]\n\n\[ {\sec }^{2}\left( {\theta \left( t\right) }\right) {\theta }^{\prime }\left( t\right) = \frac{1}{400}{x}^{\prime }\left( t\right) \]\n\nThe problem is to find \( {x}^{\prime }\left( t\right) \) when \( x\left( t\right) = {600} \) . When \( x\left( t\right) = {600},\tan \left( {\theta \left( t\right) }\right) = \frac{600}{400} = \frac{3}{2} \), and \( {\sec }^{2}\left( {\theta \left( t\right) }\right) = 1 + {\left( \frac{3}{2}\right) }^{2} \) . The light rotates once every two minutes, so \( {\theta }^{\prime }\left( t\right) = \frac{{2\pi }\text{ radians }}{2\text{ minutes }} = \pi \) radians/minute. Therefore when \( x\left( t\right) = {600} \), \n\n\[ {x}^{\prime }\left( t\right) = {400}\frac{13}{4}\pi \;\text{ meters }/\text{ minute } \]\n\nNote: Radian is a dimensionless measurement.
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Yes
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Suppose a body of mass \( m \) is suspended from a spring with spring constant \( k \). If \( m = {20}\mathrm{{gm}} = {0.020}\mathrm{{Kg}} \) and \( k = {0.125} \) Newtons/meter and the initial displacement, \( {y}_{0} = 5 \) \( \mathrm{{cm}} = {0.05}\mathrm{\;m} \), then what is the angular frequency, the equation of motion, the period of oscillation, and the frequency of oscillation?
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\[
\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{{0.125}\mathrm{{Kg}}}{{0.020}\mathrm{{Kg}} - \mathrm{m}/{\mathrm{s}}^{2}/\mathrm{m}}} = {2.5}/\mathrm{s}
\]
and
\[
y\left( t\right) = {0.05}\cos \left( {2.5t}\right)
\]
The period of oscillation is
\[
\frac{2\pi }{\omega } = \frac{2\pi }{{2.51}/\mathrm{s}} = {2.51}\mathrm{\;s}.
\]
and the frequency of oscillation is approximately
\[
{60}/{2.51} = {23.9}\;\text{oscillations per minute.}
\]
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Yes
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Suppose \( m = {20}\mathrm{{gm}} = {0.020}\mathrm{\;{kg}},\mathrm{r} = {0.06} \) Newtons/(meter/sec) and \( k = {0.125} \) Newtons/meter. Then the equation of damped motion is\n\n\[ {0.02}{y}^{\prime \prime }\left( t\right) + {0.06}{y}^{\prime }\left( t\right) + {0.125y}\left( t\right) = 0 \]
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We show that a solution to this equation is\n\n\[ y\left( t\right) = {e}^{-{1.5t}}\cos \left( {2t}\right) \]\n\n\[ {y}^{\prime }\left( t\right) = {\left\lbrack {e}^{-{1.5t}}\cos \left( 2t\right) \right\rbrack }^{\prime } \]\n\n\[ = {\left\lbrack {e}^{-{1.5t}}\right\rbrack }^{\prime }\cos \left( {2t}\right) + {e}^{-{1.5t}}{\left\lbrack \cos \left( 2t\right) \right\rbrack }^{\prime } \]\n\n\[ = {e}^{-{1.5t}}\left( {-{1.5}}\right) \cos {2t} + {e}^{-{1.5t}}\left( {-\sin {2t}}\right) 2 \]\n\n\[ = - {1.5}{e}^{-{1.5t}}\cos {2t} - 2{e}^{-{1.5t}}\left( {\sin {2t}}\right) \]\n\n\[ {y}^{\prime \prime }\left( t\right) = {\left\lbrack -{1.5}{e}^{-{1.5t}}\cos 2t - 2{e}^{-{1.5t}}\left( \sin 2t\right) \right\rbrack }^{\prime } \]\n\n\[ = - {1.5}\left( {{\left\lbrack {e}^{-{1.5t}}\right\rbrack }^{\prime }\cos {2t} + {e}^{-{1.5t}}{\left\lbrack \cos 2t\right\rbrack }^{\prime }}\right) - 2\left( {{\left\lbrack {e}^{-{1.5t}}\right\rbrack }^{\prime }\sin {2t} + {e}^{-{1.5t}}{\left\lbrack \sin 2t\right\rbrack }^{\prime }}\right) \]\n\n\[ = {\left( -{1.5}\right) }^{2}{e}^{-{1.5t}}\cos {2t} + 2\left( {-{1.5}}\right) \left( {-2}\right) {e}^{-{1.5t}}\sin {2t} - {2}^{2}{e}^{-{1.5}}\cos {2t} \]\n\n\[ = - {1.75}{e}^{-{1.5t}}\cos {2t} + 6{e}^{-{1.5t}}\sin {2t} \]\n\nNow we set up a table of coefficients and terms of Equation 7.29.\n\n\[ {0.02}{y}^{\prime \prime }\left( t\right) \; - {1.75}{e}^{-{1.5t}}\cos {2t} + 6{e}^{-{1.5t}}\sin {2t} \]\n\n\[ {0.06}\;{y}^{\prime }\left( t\right) \; - {1.5}{e}^{-{1.5t}}\cos {2t}\; - \;2{e}^{-{1.5t}}\sin {2t} \]\n\n\[ {0.125y}\left( t\right) \;{e}^{-{1.5t}}\cos {2t} \]\n\nAfter substitution into Equation 7.29 the coefficients of \( {e}^{-{1.5t}}\cos {2t} \) and \( {e}^{-{1.5t}}\sin {2t} \) are\n\n\[ {0.02} \times \left( {-{1.75}}\right) + {0.06} \times \left( {-{1.5}}\right) + {0.125} \times 1 = {0.0}\;\text{and}\;{0.02} \times 6 + {0.06} \times \left( {-2}\right) = {0.0} \]\n\nso\n\n\[ y\left( t\right) = {e}^{-{1.5t}}\cos \left( {2t}\right) \;\text{ solves }\;{0.02}{y}^{\prime \prime }\left( t\right) + {0.06}{y}^{\prime }\left( t\right) + {0.125y}\left( t\right) = 0. \]
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Yes
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Example 8.1.1 A graph of the cubic function, \( P\left( t\right) = {t}^{3} \) is shown in Figure 8.1.1.1 (and in Explore Figure 8.1.1), together with the tangent to the graph at the point \( \left( {0,0}\right) .P\left( t\right) = {t}^{3} \) is an increasing function: If \( a < b \) then \( {a}^{3} < {b}^{3} \) .
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However, the tangent to the graph of \( P \) at \( \left( {0,0}\right) \) is horizontal. \( {\left. {P}^{\prime }\left( 0\right) = 3{t}^{2}\right| }_{t = 0} = {30}^{2} = 0 \) . So that \( {P}^{\prime } \) is not everywhere positive.
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No
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Example 8.1.2 The natural logarithm, ln, is an increasing function.
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By the previous theorem, ln is increasing if \( {\ln }^{\prime } \) is positive. \( \ln x \) is only defined for \( x > 0 \), and\n\n\[{\left\lbrack \ln x\right\rbrack }^{\prime } = \frac{1}{x} > 0\;\text{ for }\;x > 0\]
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Yes
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During what time interval is the concentration of penicillin in the blood increasing? What is the maximum concentration of penicillin?
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Compute\n\n\[ \n{C}^{\prime }\left( t\right) = {\left\lbrack 5{e}^{-{0.2t}} - 5{e}^{-{0.3t}}\right\rbrack }^{\prime } \n\]\n\n\[ \n= 5{e}^{-{0.2t}}\left( {-{0.2}}\right) - 5{e}^{-{0.3t}}\left( {-{0.3}}\right) \n\]\n\n\[ \n= 5{e}^{-{0.3t}}\left( {-{0.2}{e}^{0.1t} + {0.3}}\right) \n\]\n\nBecause \( \;5 > 0\; \) and \( \;{e}^{-{0.3t}} > 0\; \) for all \( t,{C}^{\prime }\left( t\right) \) is positive if \( - {0.2}{e}^{0.1t} + {0.3} \) is positive. The following inequalities are equivalent:\n\n\[ \n- {0.2}{e}^{0.1t} + {0.3} > 0 \n\]\n\n\[ \n{1.5} > {e}^{0.1t} \n\]\n\n\[ \n\ln {1.5} > \ln \left( {e}^{0.1t}\right) \n\]\n( \( \ln t \) is an increasing function.)\n\n\[ \n\ln {1.5} > {0.1t} \n\]\n\n\[ \n{10}\ln {1.5} > t \n\]\n\nWe can conclude that \( {C}^{\prime }\left( t\right) \) is positive if \( 0 \leq t < {10}\ln {1.5} \doteq {4.05} \) . Thus serum penicillin is increasing for about four hours after ingestion of the penicillin pill into the intestinal track.
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Yes
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Theorem 5.2.2. If \( {a}_{1},{a}_{2},\cdots ,{a}_{n} \) is a sequence of \( n \) positive numbers then\n\n\[ \frac{{a}_{1} + {a}_{2} + \cdots {a}_{n}}{n} \geq \sqrt[n]{{a}_{1}{a}_{2}\cdots {a}_{n}} \]\n\nwith equality only when \( {a}_{1} = {a}_{2} = \cdots = {a}_{n} \) .
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Proof of Theorem 5.2.2. We proceed by induction. First we prove that if \( {a}_{1} \) and \( {a}_{2} \) are two positive numbers then \( \left( {{a}_{1} + {a}_{2}}\right) /2 \geq \sqrt[2]{{a}_{1}{a}_{2}} \) .\n\n\[ {\left( {a}_{1} - {a}_{2}\right) }^{2} \geq 0 \]\n\n\[ {a}_{1}^{2} - 2{a}_{1}{a}_{2} + {a}_{2}^{2} \geq 0 \]\n\n\[ {a}_{1}^{2} + 2{a}_{1}{a}_{2} + {a}_{2}^{2} \geq 4{a}_{1}{a}_{2} \]\n\n\[ \frac{{\left( {a}_{1} + {a}_{2}\right) }^{2}}{4} \geq {a}_{1}{a}_{2} \]\n\n\[ \frac{{a}_{1} + {a}_{2}}{2} \geq \sqrt[2]{{a}_{1}{a}_{2}} \]\n\nFurthermore, \( {\left( {a}_{1} - {a}_{2}\right) }^{2} = 0 \) only when \( {a}_{1} = {a}_{2} \) and equality holds in each expression of the\n\nprevious array only when \( {a}_{1} = {a}_{2} \) . The statement in Theorem 5.2.2 is valid with \( n = 2 \) .\n\nNow suppose \( n \) is an integer, \( n \geq 3 \), and Equation 8.1 is valid for sequences of length \( n - 1 \) and \( {a}_{1},{a}_{2},\cdots {a}_{n} \) is a sequence of positive numbers of length \( n \) . We assume without loss of generality that \( {a}_{n} \) is the smallest number in \( {a}_{1},{a}_{2},\cdots {a}_{n} \), and consider the sequence \( {b}_{k} = {a}_{k}/{a}_{n}, k = 1, n \) . Then \( {b}_{k} \geq 1 \) and \( {b}_{n} = 1 \) . Consequently,\n\n\[ t = {b}_{1}{b}_{2}\cdots {b}_{n - 1}{b}_{n} = {b}_{1}{b}_{2}\cdots {b}_{n - 1} \geq 1 \]\n\nwith equality only for \( {b}_{1} = {b}_{2} = \cdots = {b}_{n - 1} = 1 \) . Observe that \( {b}_{1} = {b}_{2} = \cdots = {b}_{n - 1} = 1 \) only if \( {a}_{1} = {a}_{2} = \cdots = {a}_{n}. \)\n\nFrom Equations 8.2 and 8.3 we know that \( 1 + {\left( n - 1\right) }^{n}\sqrt[{-1}]{t} \geq n\sqrt[n]{t}\; \) for \( \;t \geq 1 \) with equality only for \( t = 1 \) . Therefore\n\n\[ 1 + \left( {n - 1}\right) \left( \sqrt[{n - 1}]{{b}_{1}{b}_{2}\cdots {b}_{n - 1}}\right) \geq n\sqrt[n]{{b}_{1}{b}_{2}\cdots {b}_{n}} \]\n\nBy the induction hypothesis\n\n\[ {b}_{1} + {b}_{2} + \cdots {b}_{n - 1} \geq \left( {n - 1}\right) \left( \sqrt[{n - 1}]{{b}_{1}{b}_{2}\cdots {b}_{n - 1}}\right) ,\]\n\nso that\n\n\[ 1 + {b}_{1} + {b}_{2} + \cdots {b}_{n - 1} \geq 1 + \left( {n - 1}\right) \left( \sqrt[{n - 1}]{{b}_{1}{b}_{2}\cdots {b}_{n - 1}}\right) \geq n\sqrt[n]{{b}_{1}{b}_{2}\cdots {b}_{n}} \]\n\n\[ 1 + {b}_{1} + {b}_{2} + \cdots {b}_{n - 1} \geq n\sqrt[n]{{b}_{1}{b}_{2}\cdots {b}_{n}} \]\n\n\[ {a}_{n} + {a}_{1}/{a}_{n} + {a}_{2}/{a}_{n} + \cdots + {a}_{n - 1}/{a}_{n} \geq n\sqrt[n]{{a}_{1}{a}_{2}\cdots {a}_{n}} \]\n\n\[ {a}_{1} + {a}_{2} + \cdots + {a}_{n} \geq n\sqrt[n]{{a}_{1}{a}_{2}\cdots {a}_{n}} \]\n\n\[ \frac{{a}_{1} + {a}_{2} + \cdots + {a}_{n}}{n} \geq \sqrt[n]{{a}_{1}{a}_{2}\cdots {a}_{n}} \]\n\nwith equality only when \( {a}_{1} = {a}_{2} = \cdots = {a}_{n} \). - End of proof of Theorem 5.2.2.
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Yes
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Theorem 8.2.1 If \( \left( {c, f\left( c\right) }\right) \) is an interior local maximum for a function, \( f \) , and \[ \text{if}{f}^{\prime }\left( c\right) \text{exists, then}\;{f}^{\prime }\left( c\right) = 0\text{.} \] (Equivalently, if the graph of \( f \) has a tangent at an interior local maximum \( \left( {c, f\left( c\right) }\right) \), then that tangent is horizontal.)
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Proof. Suppose \( \left( {c, f\left( c\right) }\right) \) is an interior local maximum for \( f \), and \( \left( {p, q}\right) \) is an interval in D containing \( c \) for which \( f\left( x\right) \leq f\left( c\right) \) for all \( x \) in \( \left( {p, q}\right) \) . We wish to show that \( {f}^{\prime }\left( c\right) = 0 \) . See Figure 8.5. Suppose \( p < b < c \) . Then \( b - c < 0 \), and because \( c \) is a local maximum \[ f\left( b\right) \leq f\left( c\right) \;\text{ and }\;f\left( b\right) - f\left( c\right) \leq 0\;\text{ and }\;\frac{f\left( b\right) - f\left( c\right) }{b - c} \geq 0 \] It follows that \[ {f}^{\prime - }\left( c\right) = \mathop{\lim }\limits_{{b \rightarrow {c}^{ - }}}\frac{f\left( b\right) - f\left( c\right) }{b - c} \geq 0. \] Similar analysis shows that \[ {f}^{\prime + }\left( c\right) = \mathop{\lim }\limits_{{b \rightarrow {c}^{ + }}}\frac{f\left( b\right) - f\left( c\right) }{b - c} \leq 0. \] If \( {f}^{\prime }\left( c\right) \) exists, \( {f}^{\prime }\left( c\right) = {f}^{\prime - }\left( c\right) \) and \( {f}^{\prime }\left( c\right) = {f}^{\prime + }\left( c\right) \) . Therefore, \( {f}^{\prime }\left( c\right) = 0 \) . End of Proof.
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Yes
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Suppose you are going to make a rectangular box with open top from a 3 meter by 4 meter sheet of tin. One procedure for doing so would be to cut squares of side \( x \) from each corner as shown in Figure 8.6A, and to fold the ’tabs’ up. Four pieces of area \( {x}^{2} \) would be discarded. What value of \( x \) will maximize the volume of the box you construct in this way? What will be the volume of the box of largest volume?
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After the corners are cut, the 'core' of the tin that will make the bottom of the box will be of length \( 4 - {2x} \) and width \( 3 - {2x} \) . The height of the box will be \( x \) and the volume of the box formed will be\n\n\[ V = \left( {4 - {2x}}\right) \left( {3 - {2x}}\right) x\;0 \leq x \leq \frac{3}{2} \]\n\n\( 0 \leq x \leq \frac{3}{2} \) insures that no side of the box is a negative number. A graph of \( V \) appears in Figure 8.6B. The dashed line shows an extension of the graph of \( y = \left( {4 - {2x}}\right) \left( {3 - {2x}}\right) x \) that is not part of the graph of \( V \) . Compute \( {V}^{\prime } \) :\n\n\[ {V}^{\prime } = {\left\lbrack \left( 4 - 2x\right) \left( 3 - 2x\right) x\right\rbrack }^{\prime } \]\n\n\[ = {\left\lbrack \left( 4 - 2x\right) \right\rbrack }^{\prime }\left( {3 - {2x}}\right) x + \left( {4 - {2x}}\right) {\left\lbrack \left( 3 - 2x\right) x\right\rbrack }^{\prime } \]\n\n\[ = {\left\lbrack \left( 4 - 2x\right) \right\rbrack }^{\prime }\left( {3 - {2x}}\right) x + \left( {4 - {2x}}\right) {\left\lbrack \left( 3 - 2x\right) \right\rbrack }^{\prime }x + \left( {4 - {2x}}\right) \left( {3 - {2x}}\right) {\left\lbrack x\right\rbrack }^{\prime } \]\n\n\[ {V}^{\prime } = 8{x}^{2} - {28x} + {12} \]\n\nFind the critical points:\n\n- \( {\mathbf{V}}^{\prime } = \mathbf{0} \)\n\n\[ {V}^{\prime } = 0\;\text{ implies }\;8{x}^{2} - {28x} + {12} = 0\;\text{ implies }\;x = 3\;\text{ or }\;x = \frac{1}{2} \]\n\n\( x = \frac{1}{2} \) is in the domain of \( V \) but \( x = 3 \) is not.\n\n- \( {\mathbf{V}}^{\prime } \) does not exist. No such points. \( V \) is a cubic function and has derivatives at every point.\n\n- End points. The end points are \( x = 0 \) and \( x = \frac{3}{2} \)\n\nThe three critical values of \( x \) are \( 0,\frac{1}{2} \) and \( \frac{3}{2} \)\n\nFind the maximum \( V \) :\n\n\[ V\left( 0\right) = \left( {4 - 2 \times 0}\right) \left( {3 - 2 \times 0}\right) \times 0 = 0 \]\n\n\[ V\left( \frac{1}{2}\right) = \left( {4 - 2\frac{1}{2}}\right) \left( {3 - 2\frac{1}{2}}\right) \frac{1}{2} = 3 \]\n\n\[ V\left( \frac{3}{2}\right) = \left( {4 - 2\frac{3}{2}}\right) \left( {3 - 2\frac{3}{2}}\right) \frac{3}{2} = 0 \]\n\nThe maximum volume is 3 and occurs with \( x = \frac{1}{2} \) .
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Yes
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Suppose a box with a square base and closed top and bottom is to have a volume of 8 cubic meters. What dimensions of the box will minimize the surface area of the box?
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Solution. Let \( x \) be the length of one side of the square base and \( y \) be the height of the box. Then the volume and surface area of the box are\n\n\[ V = {x}^{2}y\;S = 2{x}^{2} + {4xy} \]\n\nBecause \( V \) is specified to be 8 cubic meters\n\n\[ 8 = {x}^{2}y\;\text{ so that }\;y = \frac{8}{{x}^{2}} \]\n\nWe substitute \( \frac{8}{{x}^{2}} \) for \( y \) in the expression for \( S \) and get\n\n\[ S = 2{x}^{2} + {4x}\frac{8}{{x}^{2}} = 2{x}^{2} + \frac{32}{x} \]\n\nThe domain of \( S \) is \( x > 0 \) (there are no endpoints, \( x = 0 \) is not allowed by the \( x \) in the denominator and there is no upper limit on \( x \) ).\n\nThen\n\n\[ {S}^{\prime }\left( x\right) = {\left\lbrack 2{x}^{2} + \frac{32}{x}\right\rbrack }^{\prime } \]\n\n\[ = {\left\lbrack 2{x}^{2}\right\rbrack }^{\prime } + {\left\lbrack {32}{x}^{-1}\right\rbrack }^{\prime } \]\n\n\[ = {4x} - {32}{x}^{-2} \]\n\n\( {S}^{\prime }\left( x\right) \) exists for all \( x > 0 \) and \( {S}^{\prime }\left( x\right) = 0 \) yields\n\n\[ {4x} - {32}{x}^{-2} = 0,\;4{x}^{3} - {32} = 0,\;x = 2 \]\n\nThus we conclude that the base of the box should be 2 by 2 and because \( {x}^{2}y = 8 \) the height \( y \) of the box should also be 2 . Examination of the graph of \( S\left( x\right) = 2{x}^{2} + \frac{32}{x} \) in Example Figure 8.2.2.2B suggests it is a minimum (and not, for example, a maximum!).
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Yes
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Of all cans of volume equal to 1 , which has the smallest surface area?
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Assume that the 'can' (cylinder) in Example Figure 8.2.3.3 has volume equal to 1. The area of the top end of the can is \( \pi {r}^{2} \) and the circumference of the top lid is \( {2\pi r} \) . The volume, \( V \), of the can is\n\n\[ V = \pi {r}^{2}h \]\n\nand the surface area, \( S \), of the can is\n\n\[ S = {2\pi }{r}^{2} + {2\pi rh \]\n\nThe requirement that the volume be 1 yields\n\n\[ 1 = \pi {r}^{2}h\;\text{ and solving for }h\text{ yields }\;h = \frac{1}{\pi {r}^{2}} \]\n\nWe substitute this value into the expression for \( S \) and obtain\n\n\[ S = {2\pi }{r}^{2} + {2\pi r}\frac{1}{\pi {r}^{2}}\;S = {2\pi }{r}^{2} + \frac{2}{r} \]\n\nThe domain for \( S \) is \( r > 0 \) (there are no endpoints).\n\nFrom the graph of \( S \) in Figure 8.2.3.3B it appears that there is a single minimum at about \( r = {0.5} \) . We find the critical points.\n\n\[ {S}^{\prime }\left( r\right) = {\left\lbrack 2\pi {r}^{2} + \frac{2}{r}\right\rbrack }^{\prime } = {\left\lbrack 2\pi {r}^{2}\right\rbrack }^{\prime } + {\left\lbrack 2{r}^{-1}\right\rbrack }^{\prime } = {4\pi r} - 2{r}^{-2} \]\n\nThe requirement \( {S}^{\prime }\left( r\right) = 0 \) yields\n\n\[ {4\pi r} - 2{r}^{-2} = 0,\;{4\pi }{r}^{3} - 2 = 0,\;r = \frac{1}{\sqrt[3]{2\pi }} \doteq {0.542} \]\n\nRecall that \( h = \frac{1}{\pi {r}^{2}} \) so that the ratio of \( h \) to \( r \) (height to radius ratio) that gives minimum surface\n\narea is\n\n\[ \frac{h}{r} = {\left. \frac{1/\pi {r}^{2}}{r}\right| }_{r = 1/\sqrt[3]{2\pi }} = {\left. \frac{1}{\pi {r}^{3}}\right| }_{r = 1/\sqrt[3]{2\pi }} = 2 \]\n\nThus the height should be twice the radius, or equal to the diameter.
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Yes
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Snell's Law When you see a fish in a lake it typically is below where it appears to be. A spear, arrow, bullet, rock or other projectile launched toward the image of the fish that you see will pass above the fish. The different speeds of light in air and in water cause the light beam traveling from the fish to your eye to bend at the surface of the lake. The apparent location of the fish is marked as a dotted fish in Figure 8.2.4.4.
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Pierre Fermat asserted in 1662 that the path of the light beam will be that path that minimizes the total time of travel in the two media. The speed of light in water, \( {v}_{2} \), is about 0.75 times the speed of light in air, \( {v}_{1} \) . Suppose \( d \) is the horizontal distance between your eye and the fish, \( {h}_{1} \) is the height of your eye above the water and \( {h}_{2} \) is the depth of the fish below the surface of the lake. Finally let \( x \) be the horizontal distance between your eye and the point at which the beam passes through the surface of the lake.\n\nThe distances the light ray travels in air and water are\n\n\[ \text{Air distance:}\;\sqrt{{h}_{1}^{2} + {x}^{2}}\;\text{Water distance:}\;\sqrt{{h}_{2}^{2} + {\left( d - x\right) }^{2}} \]\n\nThe times that the light spends traversing air and traversing water are (distance/velocity)\n\n\[ \text{Air time:}\;\frac{\sqrt{{h}_{1}^{2} + {x}^{2}}}{{v}_{1}}\;\text{Water time:}\;\frac{\sqrt{{h}_{2}^{2} + {\left( d - x\right) }^{2}}}{{v}_{2}} \]\n\nThe total time, \( T \), for the ray to travel from the fish to your eye is\n\n\[ T = \frac{\sqrt{{h}_{1}^{2} + {x}^{2}}}{{v}_{1}} + \frac{\sqrt{{h}_{2}^{2} + {\left( d - x\right) }^{2}}}{{v}_{2}}\;0 \leq x \leq d \]\n\nOur task is to find the value of \( x \) that minimizes \( T \) . A graph of \( T \) is shown for values \( {v}_{1} = 1 \) , \( {v}_{2} = {0.75}, d = {10} \) meters, \( {h}_{1} = 2 \) meters, \( {h}_{2} = 3 \) meters. As tempting as it may be, it is not the time to get clever and square each term in the previous equation, for it is not generally true that \( {\left( a + b\right) }^{2} = {a}^{2} + {b}^{2} \) . Instead we compute \( {T}^{\prime } \) directly and find that\n\n\[ {T}^{\prime } = \frac{\frac{x}{\sqrt{{h}_{1}^{2} + {x}^{2}}}}{{v}_{1}} - \frac{\frac{d - x}{\sqrt{{h}_{2}^{2} + {\left( d - x\right) }^{2}}}}{{v}_{2}} \]\n\nSetting \( {T}^{\prime } = 0 \) and solving for \( x \) is not advised. We take a qualitative approach instead, and return to the geometry and identify two angles, \( {\theta }_{1} \) and \( {\theta }_{2} \), the angles the light ray makes with a vertical line through the point of intersection with the surface. They are marked in both figures 8.2.4.4 and 8.2.4.4. It may be seen that\n\n\[ \frac{x}{\sqrt{{h}_{1}^{2} + {x}^{2}}} = \sin {\theta }_{1}\;\text{ and }\;\frac{d - x}{\sqrt{{h}_{2}^{2} + {\left( d - x\right) }^{2}}} = \sin {\theta }_{2} \]\n\nand\n\n\[ {T}^{\prime } = \frac{\sin {\theta }_{1}}{{v}_{1}} - \frac{\sin {\theta }_{2}}{{v}_{2}} \]\n\nObserve from the geometry that as \( x \) moves from 0 to \( d,{\theta }_{1} \) increases from 0 to a positive number and \( {\theta }_{2} \) decreases from a positi
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Yes
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A 10 meter ladder leans against a wall. The foot of the ladder slips horizontally at the rate of 1 meter per minute. At what rate does the top of the ladder descend when the top is 6 meters from the ground?
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Draw a picture. See Figure 8.4.1.1. Let \( x \) be the distance from the wall to the foot of the ladder and let \( y \) be the distance from the ground to the top of the ladder. We are asked to find \( {y}^{\prime } \)\n\nat a certain instant. Because the top of the ladder is descending, we expect our answer \( \left( {y}^{\prime }\right) \) to be negative. \( \mathrm{x} \) and \( \mathrm{y} \) are intrinsically related.\n\n\[ \n{x}^{2} + {y}^{2} = {10}^{2} \n\]\n\nFigure for Example 8.4.1.1 Ladder leaning against a wall and sliding down the wall.\n\n\n\n\( x \) and \( y \) are changing with time, and\n\n\[ \n{\left( x\left( t\right) \right) }^{2} + {\left( y\left( t\right) \right) }^{2} = {10}^{2} \n\]\n\nDifferentiate with respect to \( t \) (use the chain rule, twice):\n\n\[ \n{2x}\left( t\right) {x}^{\prime }\left( t\right) + {2y}\left( t\right) {y}^{\prime }\left( t\right) = 0 \n\]\n\n’When’ The instant, \( {t}_{0} \), specified in the problem is defined by \( y = 6 \) . At that instant \( {x}^{2} + {6}^{2} = {10}^{2} \) , so that \( x = 8 \) . The problem also specifies \( {x}^{\prime }\left( t\right) = 1 \) for all \( t \) . The actual value of \( {t}_{0} \) is not required; we know that \( x\left( {t}_{0}\right) = 8, y\left( {t}_{0}\right) = 6 \), and \( {x}^{\prime }\left( {t}_{0}\right) = 1 \) . Therefore\n\n\[ \n\left( {2 \times 8 \times 1}\right) + \left( {2 \times 6 \times {y}^{\prime }}\right) = 0\;\text{ and }\;{y}^{\prime } = - \frac{4}{3}\;\text{ meter }/\mathrm{{min}} \n\]
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Yes
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In an aqueous solution the concentrations of \( {\mathrm{H}}^{ + } \) and \( {\mathrm{{OH}}}^{ - } \) ions satisfy\n\n\[ \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack = {10}^{-{14}} \]\n\nIf in a certain lake the \( \mathrm{{pH}} \) is \( 6\left( {\left\lbrack {\mathrm{H}}^{ + }\right\rbrack = {10}^{-6}}\right) \) and is decreasing at the rate of \( {0.1}\mathrm{{pH}} \) units per year, at what rate is the hydroxyl concentration, \( \left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack \), increasing?
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Solution: It is useful to take the \( {\log }_{10} \) of the two sides of the previous equation:\n\n\[ {\log }_{10}\left( {\left\lbrack {\mathrm{H}}^{ + }\right\rbrack \left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack }\right) = {\log }_{10}{10}^{-{14}} \]\n\n\[ {\log }_{10}\left\lbrack {\mathrm{H}}^{ + }\right\rbrack + {\log }_{10}\left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack = - {14} \]\n\n\[ \mathrm{{pH}} + {\log }_{10}\left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack = - {14} \]\n\n\[ \mathrm{{pH}} + \frac{\ln \left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack }{\ln {10}} = - {14} \]\n\n\nNow we take derivatives of each term:\n\n\[ {\left\lbrack \mathrm{{pH}}\right\rbrack }^{\prime } + {\left\lbrack \frac{\ln \left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack }{\ln {10}}\right\rbrack }^{\prime } = {\left\lbrack -{14}\right\rbrack }^{\prime } \]\n\n\[ {\left\lbrack \mathrm{{pH}}\right\rbrack }^{\prime } + \frac{1}{\ln {10}}\frac{1}{\left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack }{\left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack }^{\prime } = 0 \]\n\nAt the given instant, \( \left( {\left\lbrack {\mathrm{H}}^{ + }\right\rbrack = {10}^{-6}}\right) \) so that\n\n\[ \left\lbrack {10}^{-6}\right\rbrack \left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack = {10}^{-{14}}\;\text{ and }\;\left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack = {10}^{-8} \]\n\nAlso, by the hypothesis, \( {\left\lbrack \mathrm{{pH}}\right\rbrack }^{\prime } = - {0.1} \) . Therefore\n\n\[ {\left\lbrack \mathrm{{pH}}\right\rbrack }^{\prime } + \frac{1}{\ln {10}}\frac{1}{\left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack }{\left\lbrack \left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack \right\rbrack }^{\prime } = 0 \]\n\n\[ = - {0.1} + \frac{1}{\ln {10}}\frac{1}{{10}^{-8}}{\left\lbrack \left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack \right\rbrack }^{\prime } = 0 \]\n\n\[ {\left\lbrack {\mathrm{{OH}}}^{ - }\right\rbrack }^{\prime } = {0.1}\left( {\ln {10}}\right) {10}^{-8} \doteq {0.23}{10}^{-8}\frac{\text{ ions }}{\text{ liter }}\text{ per second } \]
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Yes
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Solve \( x{e}^{-x} = a \) for \( a = {0.2} \) and \( a = 2 \) .
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Graphs help explore the problem. Shown in Figure 8.10A is a graph of \( y = x{e}^{-x} \) . It is apparent that the highest point of the graph has \( y \) -coordinate about \( y = {0.37} \) . We are relieved of solving \( x{e}^{-x} = 2 \) ; there is no solution. The dashed line at \( y = {0.2} \) does intersect the graph of \( y = x{e}^{-x} \) ; at two points so there are two solutions to \( x{e}^{-x} = {0.2},{r}_{1} \) at about \( x = {0.3} \) and \( {r}_{2} \) at about \( x = {2.5} \) .\n\nIteration. There is a simple scheme that sometimes works. \( x{e}^{-x} = {0.2} \) is equivalent to \( x = {0.2}{e}^{x} \) . We can guess \( {x}_{0} = {0.3} \) as an estimate of \( {r}_{1} \), and compute a new estimate \( {x}_{1} = {0.2}{e}^{{x}_{0}} = {0.2}{e}^{0.3} = {0.2700} \) . Then compute \( {x}_{2} = {0.2}{e}^{{x}_{1}} = {0.2}{e}^{0.2700} = {0.2620} \) . Continue and we find that\n\n\[ \n{x}_{3} = {0.2599},\;{x}_{4} = {0.2594},\;{x}_{5} = {0.2592},\;{x}_{6} = {0.2592}.\n\]\n\nCorrect to four decimal places, \( x = {0.2592} \) is a solution to \( x{e}^{-x} = {0.2} \) . This scheme fails for finding the root \( {r}_{2} \) near \( x = {2.5} \) and an alternate scheme is suggested in Exercise 8.5.1.
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Yes
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Example 8.7.1 Usually more than one derivative formula may be used in a single step toward computing a derivative. It is important to know which steps are being used, and a slow but sure way to insure proper use is to use but a single formula for each step. We illustrate with \( P\left( t\right) = {\left\lbrack {e}^{{t}^{2}} + {e}^{-t}\right\rbrack }^{4}. \)
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\[ {P}^{\prime }\left( t\right) = {\left\lbrack {\left\lbrack {e}^{{t}^{2}} + \sin t\right\rbrack }^{4}\right\rbrack }^{\prime } \] Notational identity.\n\n\[ = 4{\left\lbrack {e}^{{t}^{2}} + \sin t\right\rbrack }^{3}{\left\lbrack {e}^{{t}^{2}} + \sin t\right\rbrack }^{\prime } \] Equation 8.31\n\n\[ = 4{\left\lbrack {e}^{{t}^{2}} + \sin t\right\rbrack }^{3}\left( {{\left\lbrack {e}^{{t}^{2}}\right\rbrack }^{\prime } + {\left\lbrack \sin t\right\rbrack }^{\prime }}\right) \] Equation 8.27\n\n\[ = 4{\left\lbrack {e}^{{t}^{2}} + \sin t\right\rbrack }^{3}\left( {{e}^{{t}^{2}}{\left\lbrack {t}^{2}\right\rbrack }^{\prime } + {\left\lbrack \sin t\right\rbrack }^{\prime }}\right) \] Equation 8.32\n\n\[ = 4{\left\lbrack {e}^{{t}^{2}} + \sin t\right\rbrack }^{3}\left( {{e}^{{t}^{2}}\left( {2t}\right) + {\left\lbrack \sin t\right\rbrack }^{\prime }}\right) \] Equation 8.21\n\n\[ = 4{\left\lbrack {e}^{{t}^{2}} + \sin t\right\rbrack }^{3}\left( {{e}^{{t}^{2}}\left( {2t}\right) + \cos t}\right) \] Equation 8.24.
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Yes
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Find (approximately) the area of the region \( R \) in Figure 9.3B bounded by the graph of the normal distribution, \( f \), the \( \mathrm{X} \) -axis, the vertical line, \( x = - 1 \) and the vertical line, \( x = 1 \) .
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1. Because of the symmetry of the region \( R \) about the \( Y \) -axis, it is sufficient to find the area of that portion of \( R \) that is to the right of the \( Y \) -axis, and then to multiply by 2 . The portion \( R \) to the right of the \( Y \) -axis is shown in Figure 9.4A.\n\n2. Compute the sum of the areas of the upper rectangles in Figure 9.4A. The width of each rectangle is 0.2 and the heights of the rectangles are determined by the density function, f. The area of the leftmost rectangle, based on the \( X \) interval \( \left\lbrack {0,{0.2}}\right\rbrack \), is\n\n\[ f\left( 0\right) \times {0.2} = \frac{1}{\sqrt{2\pi }}{e}^{-\frac{{0}^{2}}{2}} \times {0.2} = {0.079788} \]\n\nThe area of the second rectangle from the left, based on the \( X \) interval \( \left\lbrack {{0.2},{0.4}}\right\rbrack \), is\n\n\[ f\left( {0.2}\right) \times {0.2} = \frac{1}{\sqrt{2\pi }}{e}^{-\frac{{0.2}^{2}}{2}} \times {0.2} = {0.078209} \]\n\nExplore 9.1.5 Find the areas of the remaining three rectangles in Figure 9.4A.\n\n3. The sum of the areas of the five rectangles is 0.356234 ; twice the sum is 0.712469 . This is an over estimate of the area of \( R \) .\n\n4. An under estimate of the area can be computed by summing the areas of the lower rectangles in Figure 9.4B. The leftmost lower rectangle, based on the \( X \) interval \( \left\lbrack {0,{0.2}}\right\rbrack \) has the same area as the second upper rectangle from the left, based on the \( X \) interval \( \left\lbrack {{0.2},{0.4}}\right\rbrack \), and we computed that area to be 0.078209 . The next three rectangles are similarly related to upper rectangles, and their areas are precisely the areas you computed in the preceding problem.\n\nThe area of the right most lower rectangle is\n\n\[ \frac{1}{\sqrt{2\pi }}{e}^{-\frac{{1}^{2}}{2}} \times {0.2} = {0.07358} \]\n\nand the sum of the five areas is 0.324840 ; twice the sum is 0.649680 .\n\n5. We now know that the area of \( R \), is greater than 0.649680 and less than 0.712469 . A good estimate of the area of \( R \) is the average of these two bounds,\n\n\[ \frac{{0.649680} + {0.712469}}{2} = {0.681075}. \]\n\nThe probability of being within one standard deviation of the mean in a normal distribution is 0.68268949 correct to eight digits.
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No
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Find the area of the region \( \mathbf{R} \) illustrated in Example Figure 9.2.1.1A that is bounded by the parabola \( \mathbf{y} = {\mathbf{t}}^{\mathbf{2}} \), the \( \mathbf{t} \) -axis, and the line \( \mathbf{t} = \mathbf{x} \) where \( x \) is a positive number.
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Solution. The sum of the areas of the rectangles in Figure 9.2.1.1B approximates the area of R. The interval \( \left\lbrack {0, x}\right\rbrack \) on the horizontal axis has been partitioned into \( n \) subintervals by the numbers\n\n\[ 0,\;\frac{x}{n},\;2\frac{x}{n}\;\cdots \;\left( {n - 1}\right) \frac{x}{n},\;n\frac{x}{n} \]\n\nThe width of each rectangle in the figure is \( \frac{x}{n} \) . The height of each rectangle is the height of the parabola \( y = {t}^{2} \) at the right end of the base of the rectangle. The heights of the rectangles, from left to right, are\n\n\[ {\left( \frac{x}{n}\right) }^{2},\;{\left( 2\frac{x}{n}\right) }^{2},\;\cdots \;{\left( \left( n - 1\right) \frac{x}{n}\right) }^{2},\;{\left( n\frac{x}{n}\right) }^{2} \]\n\nThe sum, \( {U}_{n} \), of the areas of the rectangles is\n\n\[ {U}_{n} = {\left( \frac{x}{n}\right) }^{2} \times \frac{x}{n} + {\left( 2\frac{x}{n}\right) }^{2} \times \frac{x}{n} + \cdots + {\left( \left( n - 1\right) \frac{x}{n}\right) }^{2} \times \frac{x}{n} + {\left( n\frac{x}{n}\right) }^{2} \times \frac{x}{n} \]\n\n\[ = \left( {{1}^{2} + {2}^{2} + \cdots + {\left( n - 1\right) }^{2} + {n}^{2}}\right) \frac{{x}^{3}}{{n}^{3}} \]\n\n\[ = \frac{n\left( {n + 1}\right) \left( {{2n} + 1}\right) }{6}\frac{{x}^{3}}{{n}^{3}} \]\n\n\[ = \frac{2{n}^{3} + 3{n}^{2} + n}{6{n}^{3}}{x}^{3} \]\n\n\[ = \left( {\frac{1}{3} + \frac{1}{2n} + \frac{1}{6{n}^{2}}}\right) {x}^{3} \]\n\nThis is powerful medicine. The region \( R \) covered with \( n = {50} \) rectangles is illustrated in Figure 9.2.1.1C. The sum of the areas of the fifty rectangles, \( {U}_{50} \), is close to the area of \( R \), and\n\n\[ {U}_{50} = \left( {\frac{1}{3} + \frac{1}{2 \times {50}} + \frac{1}{6 \times {50}^{2}}}\right) {x}^{3} = \left( {\frac{1}{3} + {0.01} + {0.000067}}\right) {x}^{3} \doteq \frac{1}{3}{x}^{3} \]\n\nFurthermore,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{U}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\frac{1}{3} + \frac{1}{2n} + \frac{1}{6{n}^{2}}}\right) {x}^{3} = \frac{{x}^{3}}{3}. \]\n\nIn Section 9.4 the area of \( R \) is defined to be this limit, so that:\n\nFor the region \( \mathrm{R} \) bounded by the parabola \( y = {t}^{2} \), the \( t \) -axis, and the line\n\n\[ t = x, \]\n\n\[ \text{the area of}R\text{is}\;\frac{{x}^{3}}{3}\text{.} \]\n\n(9.7)
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Yes
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Find approximately the area of the region \( \mathbf{R} \) illustrated in Example Figure 9.2.2.2A bounded by the cubic \( \mathbf{y} = {\mathbf{x}}^{\mathbf{2}}\left( {\mathbf{2} - \mathbf{x}}\right) \), and the \( \mathbf{X} \) -axis.
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The cubic \( y = {x}^{2}\left( {2 - x}\right) = 2{x}^{2} - {x}^{3} \) intersects the \( x \) -axis at \( x = 0 \) and \( x = 2 \) . The rectangles in Figure 9.2.2 partition \( \left\lbrack {0,2}\right\rbrack \) into ten intervals and the sum of the areas of the rectangles\n\napproximates the area of the region \( R \) . The rectangles are a mixture of lower and upper rectangles. The base of each rectangle is 0.2 and its height is the height of the left edge. For \( k = 0 \) to 9, the height of the \( k + 1 \) st rectangle, \( \left\lbrack {k \times {0.2},\left( {k + 1}\right) \times {0.2}}\right\rbrack \), is \( 2 \times {\left( k \times {0.2}\right) }^{3} - {\left( k \times {0.2}\right) }^{2} \) . The area of the \( k + 1 \) st rectangle is\n\n\[ \left\lbrack {2 \times {\left( k \times {0.2}\right) }^{2} - {\left( k \times {0.2}\right) }^{3}}\right\rbrack \times {0.2} \]\n\nThe sum of the areas of the rectangles is\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{9}\left\lbrack {2 \times {\left( k \times {0.2}\right) }^{2} - {\left( k \times {0.2}\right) }^{3}}\right\rbrack \times {0.2} = \]\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{9}\left\lbrack {2{\left( {0.2}\right) }^{2} \times {k}^{3} - {\left( {0.2}\right) }^{3} \times {k}^{2}}\right\rbrack \times {0.2} = \]\n\n\[ 2{\left( {0.2}\right) }^{3}\mathop{\sum }\limits_{{k = 0}}^{9}{k}^{3} - {\left( {0.2}\right) }^{4}\mathop{\sum }\limits_{{k = 0}}^{9}{k}^{2} = \]\n\n\[ 2{\left( {0.2}\right) }^{3}\frac{9\left( {9 + 1}\right) \left( {2 \times 9 + 1}\right) }{6} - {\left( {0.2}\right) }^{4}\frac{{9}^{2}{\left( 9 + 1\right) }^{2}}{4} = {1.32} \]
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Yes
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1. First, one may have computed the upper approximating sum to the area of the region bounded by the graphs of \( y = \sqrt{t}, y = 0 \), and \( t = 4 \) using 20 subintervals, as shown in Figure 9.3.1.1. The upper sum is \( {5.51557}\cdots \) . The error box for this computation is shown to the right of the region. Because each rectangle is of width \( \frac{4}{20} = {0.2} \), the error box is of width 0.2. The height of the error box is \( \sqrt{4} - \sqrt{0} = 2 \) so the error in the approximation,5.51557... is no larger than \( 2 * {0.2} = {0.4} \) .
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2. On the other hand, one may wish to compute the area of the region in Figure 9.3.1.1 correct to 0.01 , and need to know how many intervals are required to insure that accuracy. For any number, \( n \), of intervals, \[ \text{The error box is of height}\;\sqrt{4} - \sqrt{0} = 2\;\text{and width}\;\frac{4 - 0}{n} \] Thus, \[ \text{The size of the error box is}\;2 \times \frac{4 - 0}{n} = \frac{8}{n} \] Because the error is sure to be less than the size of the error box, our desired accuracy will be obtained if the size of the error box is less than 0.01 . Therefore we require \( \frac{8}{n} < {0.01} \) so that \( \frac{8}{0.01} < n \), or \( n = {800} \) . We will find that the actual error is only approximately one-half the size of the error box, and that \( n = {400} \) will almost give the required accuracy.
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Yes
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Problem. How much work was done?
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Solution. The force against the plunger on the \( k \) th step was\n\n\[ \n{F}_{k} \doteq \left( {{P}_{k} - {P}_{0}}\right) \;\mathrm{N}/{\mathrm{{cm}}}^{2} \times {5.5}\;{\mathrm{\;{cm}}}^{2}.\n\]\n\nThe total work done, \( W \), was approximately\n\n\[ \nW \doteq \mathop{\sum }\limits_{{k = 1}}^{9}\left( {{P}_{k} - {P}_{0}}\right) \;\mathrm{N}/{\mathrm{{cm}}}^{2} \times {5.5}\;{\mathrm{\;{cm}}}^{2} \times \left( {{x}_{k} - {x}_{k - 1}}\right)\n\]\n\n(9.11)\n\n\[ \n= \mathop{\sum }\limits_{{k = 1}}^{9}\left( {{P}_{k} - {P}_{0}}\right) \times {5.5} \times {0.9}\;\mathrm{\;N} - \mathrm{{cm}}\n\]\n\nUsing the data from Figure 9.12 we calculate\n\n\[ \nW \doteq {383}\mathrm{\;N} - \mathrm{{cm}} = {3.83}\mathrm{\;N} - \mathrm{m}\n\]\n\nThe number \( {5.5}{\mathrm{\;{cm}}}^{2} \times {0.9}\mathrm{\;{cm}} \) is, except for measurement error, equal to \( {4.95} \doteq 5{\mathrm{\;{cm}}}^{3} \), or the \( 5\mathrm{{ml}} \) marked on the syringe and in the data. Therefore the work done is\n\n\[ \nW \doteq \mathop{\sum }\limits_{{k = 1}}^{9}\left( {{P}_{k} - {P}_{0}}\right) \times \left( {{V}_{k - 1} - {V}_{k}}\right) = - \mathop{\sum }\limits_{{k = 1}}^{9}\left( {{P}_{k} - {P}_{0}}\right) \times \left( {{V}_{k} - {V}_{k - 1}}\right) \;\mathrm{N} - \mathrm{{cm}}\n\]\n\n(9.12)
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Yes
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We use Definition 9.4.1 to compute the area of the region \( R \) bounded by \( y = {x}^{2}\left( {2 - x}\right) = 2{x}^{2} - {x}^{3}, y = 0, x = 0 \), and \( x = 2 \) .
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\[ {\int }_{0}^{2}2{x}^{2} - {x}^{3}{dx} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{{k = 1}}^{n}\left( {2{\left( k\left( 2/n\right) \right) }^{2} - {\left( k\left( 2/n\right) \right) }^{3}}\right) \times 2/n = 2\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\mathop{\sum }\limits_{{k = 1}}^{n}{k}^{2}}\right) \frac{{2}^{3}}{{n}^{3}} - \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\mathop{\sum }\limits_{{k = 1}}^{n}{k}^{3}}\right) \frac{{2}^{4}}{{n}^{4}} = 2\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{n\left( {n + 1}\right) \left( {{2n} + 1}\right) }{6}\frac{{2}^{3}}{{n}^{3}} - \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{n}^{2}{\left( n + 1\right) }^{2}}{4}\frac{{2}^{4}}{{n}^{4}} = 2\mathop{\lim }\limits_{{n \rightarrow \infty }}{2}^{3}\left( {\frac{1}{3} + \frac{1}{2n} + \frac{1}{6{n}^{2}}}\right) - \mathop{\lim }\limits_{{n \rightarrow \infty }}{2}^{4}\left( {\frac{1}{4} + \frac{1}{2n} + \frac{1}{4{n}^{2}}}\right) = 2 \times \frac{{2}^{3}}{3} - \frac{{2}^{4}}{4} = \frac{4}{3} \]
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Yes
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We use Definition 9.4.1 to compute the area of the region bounded by \( y = {e}^{t} \) , \( y = 0, t = 0 \), and \( t = x \) . The region for which we wish to compute the area is shown in Figure 9.4.3.3 and we wish to compute\n\n\[ \n{\int }_{0}^{x}{e}^{t}{dt} \n\]
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Assume the interval \( \left\lbrack {0, x}\right\rbrack \) to be partitioned into \( n \) equal subintervals, each of length \( x/n \) and we let \( h = x/n \) . The \( {kth} \) such interval has endpoints on the \( t \) axis at \( \left( {k - 1}\right) \times h \) and \( k \times h \) . The area of the \( k \) th rectangle is\n\n\[ \n{e}^{\left( {k - 1}\right) \times \left( {x/n}\right) } \times \frac{x}{n} = {e}^{\left( {k - 1}\right) \times \left( h\right) } \times h, \n\]\n\nand the sum of the areas of the \( n \) rectangles is\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{n}{e}^{\left( {k - 1}\right) \times h} \times h \n\]\n\nOur job is to make sense of\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{n}{e}^{\left( {k - 1}\right) \times h} \times h\;\text{ because }\;{\int }_{0}^{x}{e}^{t}{dt}\; = \;\mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{{k = 1}}^{n}{e}^{\left( {k - 1}\right) \times h} \times h. \n\]\n\nObserve that\n\[ \n{e}^{\left( {k - 1}\right) \times h} \times h = {\left( {e}^{h}\right) }^{k - 1} \times h \n\]\n\nso that\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{n}{e}^{\left( {k - 1}\right) \times h} \times h = \left\lbrack {\mathop{\sum }\limits_{{k = 1}}^{n}{\left( {e}^{h}\right) }^{k - 1}}\right\rbrack \times h. \n\]\n\nWe write the part in [ ]'s in long format with a generic term as\n\n\[ \nk = 1\;k = 2\;k\;k = n \n\]\n\n\[ \n{\left( {e}^{h}\right) }^{0} + {\left( {e}^{h}\right) }^{1} + \cdots + {\left( {e}^{h}\right) }^{k - 1} + \cdots + {\left( {e}^{h}\right) }^{n - 1} \n\]\n\nWith \( a = {e}^{h} \) the preceding sum is the geometric series shown in Equation 9.15 and we conclude that\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{n}{\left( {e}^{h}\right) }^{k - 1} \times h = \frac{{\left( {e}^{h}\right) }^{n} - 1}{{e}^{h} - 1} \times h \n\]\n\n\[ \n= \left( {{\left( {e}^{x/n}\right) }^{n} - 1}\right) \frac{h}{{e}^{h} - 1} \n\]\n\n\[ \n= \left( {{e}^{x} - 1}\right) \times \frac{1}{\frac{{e}^{h} - 1}{h}} \n\]\n\nNow the problem is to evaluate\n\n\[ \n{\int }_{0}^{x}{e}^{t}{dt} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{{k = 1}}^{n}{e}^{\left( {k - 1}\right) \times \left( h\right) } \times h = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{e}^{x} - 1}\right) \times \frac{1}{\frac{{e}^{h} - 1}{h}} = \left( {{e}^{x} - 1}\right) \times \frac{1}{\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{e}^{h} - 1}{h}} \n\]\n\nBy its definition, Definition 5.2.1 on page 219, the number \( e \) has the property that\n\n\[ \n\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{e}^{h} - 1}{h} = 1 \n\]\n\nWe conclude that\n\[ \n{\int }_{0}^{x}{e}^{t}{dt} = {e}^{x} - 1 \n\]
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Yes
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Use the trigonometric identity Bolt out of the Blue! \( {}^{6} \)\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{n}\sin {k\theta } = \frac{-\cos \left( {{n\theta } + \frac{\theta }{2}}\right) + \cos \frac{\theta }{2}}{2\sin \frac{\theta }{2}}\;\text{ and }\;\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\sin \left( h\right) }{h} = 1 \]
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\n\n\( {}^{6} \) The Blue: Use \( 2\sin x\sin y = - \cos \left( {x + y}\right) + \cos \left( {x - y}\right) \) . Let \( S = \mathop{\sum }\limits_{{k = 1}}^{n}\sin {k\theta } \) . Then\n\n\[ {2S}\sin \frac{\theta }{2} = 2\mathop{\sum }\limits_{{k = 1}}^{n}\sin {k\theta }\sin \frac{\theta }{2} = \mathop{\sum }\limits_{{k = 1}}^{n}\left( {-\cos \left( {{k\theta } + \theta /2}\right) + \cos \left( {{k\theta } - \theta /2}\right) }\right) \]\n\n\[ \left( {-\cos \left( {\theta + \theta /2}\right) + \cos \left( {\theta - \theta /2}\right) }\right) + \left( {-\cos \left( {{2\theta } + \theta /2}\right) + \cos \left( {{2\theta } - \theta /2}\right) }\right) + \cdots + \left( {-\cos \left( {{n\theta } + \theta /2}\right) + \cos \left( {{n\theta } - \theta /2}\right) }\right) \]\n\n\[ \left( {-\cos \left( {{3\theta }/2}\right) + \cos \left( {\theta /2}\right) }\right) + \left( {-\cos \left( {{5\theta }/2}\right) + \cos \left( {{3\theta }/2}\right) }\right) + \cdots + \left( {-\cos \left( {{n\theta } + \theta /2}\right) + \cos \left( {{n\theta } - \theta /2}\right) }\right) \]\n\n\[ = - \cos \left( {{n\theta } + \theta /2}\right) + \cos \left( {\theta /2}\right) \]
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Yes
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Property 9.5.1 Linearity of the integral. Suppose \( f \) and \( g \) are integrable functions defined on an interval \( \left\lbrack {a, b}\right\rbrack \) and \( c \) is a number. Then\n\n\[ \text{A.}\;{\int }_{a}^{b}\left\lbrack {f\left( t\right) + g\left( t\right) }\right\rbrack {dt} = {\int }_{a}^{b}f\left( t\right) {dt} + {\int }_{a}^{b}g\left( t\right) {dt} \]\n\n(9.20)\n\n\[ \text{3.}\;{\int }_{a}^{b}c \times f\left( t\right) {dt} = c \times {\int }_{a}^{b}f\left( t\right) {dt} \]\n\n(9.21)
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The reasons the integral has the two linearity properties is that the approximating sum also has the two properties. The integral being the limit of the approximating sums inherits the linearity properties of the approximating sums. For example, if \( f \) and \( g \) are functions defined on an interval \( \left\lbrack {a, b}\right\rbrack \) and \( c \) is a number, and \( a = {t}_{0} < {t}_{1} < {t}_{2}\cdots < {t}_{n - 1} < {t}_{n} = b \) is a partition of \( \left\lbrack {a, b}\right\rbrack \) and\n\n\[ {t}_{0} \leq {\tau }_{1} \leq {t}_{1}\;{t}_{1} \leq {\tau }_{2} \leq {t}_{2}\;\cdots \;{t}_{k - 1} \leq {\tau }_{k} \leq {t}_{k}\;\cdots \;{t}_{n - 1} \leq {\tau }_{n} \leq {t}_{n} \]\n\nthen\n\nA. \( \mathop{\sum }\limits_{{k = 1}}^{n}\left\lbrack {f\left( {\tau }_{k}\right) + g\left( {\tau }_{k}\right) }\right\rbrack \times \left( {{t}_{k} - {t}_{k - 1}}\right) = \mathop{\sum }\limits_{{k = 1}}^{n}f\left( {\tau }_{k}\right) \times \left( {{t}_{k} - {t}_{k - 1}}\right) + \mathop{\sum }\limits_{{k = 1}}^{n}g\left( {\tau }_{k}\right) \times \left( {{t}_{k} - {t}_{k - 1}}\right) \), \n\nB. \( \;\mathop{\sum }\limits_{{k = 1}}^{n}c \times f\left( {\tau }_{k}\right) \times \left( {{t}_{k} - {t}_{k - 1}}\right) = c\mathop{\sum }\limits_{{k = 1}}^{n}f\left( {\tau }_{k}\right) \times \left( {{t}_{k} - {t}_{k - 1}}\right) \) .\n\nEach approximating sum for \( {\int }_{a}^{b}f\left( t\right) + g\left( t\right) {dt} \) is the sum of approximating sums for \( {\int }_{a}^{b}f\left( t\right) {dt} \) and \( {\int }_{a}^{b}g\left( t\right) {dt} \), and each approximating sum for \( {\int }_{a}^{b}c \times f\left( t\right) {dt} \) is \( c \) times an approximating sum for \( {\int }_{a}^{b}f\left( t\right) {dt} \)
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Yes
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\[ \text{A.}{\int }_{0}^{1}\left\lbrack {{t}^{2} + t}\right\rbrack {dt}\;\text{B.}{\int }_{0}^{2}2 \times {tdt} \]
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A. By Equation 9.17 in Exercise 9.4.1, \( {\int }_{0}^{1}{t}^{2}{dt} = \frac{1}{3} \), and \( {\int }_{0}^{1}{tdt} = \frac{1}{2} \). From Property 9.5.1 A, \[ {\int }_{0}^{1}\left\lbrack {{t}^{2} + t}\right\rbrack {dt} = {\int }_{0}^{1}{t}^{2}{dt} + {\int }_{0}^{1}{tdt} = \frac{1}{3} + \frac{1}{2} = \frac{5}{6} \] B. By Equation 9.17 in Exercise 9.4.1, \( {\int }_{0}^{2}{tdt} = 2 \). From Property 9.5.1 B, \[ {\int }_{0}^{2}{2tdt} = 2 \times {\int }_{0}^{2}{tdt} = 2 \times 2 = 4 \]
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Yes
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For example, a ball thrown vertically may have a positive velocity as it ascends, but will then have a negative velocity as it descends. If one throws the ball with a vertical velocity of \( {19.6}\mathrm{\;m}/\mathrm{s} \), then the velocity \( t \) seconds later will be \( {19.6} - {9.8}\mathrm{\;{tm}}/\mathrm{s} \) where the term \( {9.8t} \) is the change in velocity due to gravity. A graph of the velocity is shown in Figure 9.16.
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At \( t = 2 \) seconds the velocity is zero, the ball is at its maximum height, and that height is the area of the triangle marked ’ + ’ in Figure 9.16. Between \( t = 2 \) and \( t = 4 \) seconds the velocity is negative, the motion of the ball is downward and the displacement is the area of the triangle marked ’-’ in Figure 9.16. At \( t = 4 \) the ball has fallen to its original starting point. The net displacement is zero and that is \( {\int }_{0}^{4}v\left( t\right) {dt} \), the sum of the ’areas’ of the two triangles treating the area of the second triangle as negative (see Exercise 9.5.9). The distance traveled by the ball is the sum of the areas of the two triangles, both treated as positive.
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No
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