| [ | |
| { | |
| "problem_text": "How large should we take $n$ in order to guarantee that the Trapezoidal and Midpoint Rule approximations for $\\int_1^2(1 / x) d x$ are accurate to within 0.0001 ?", | |
| "answer_latex": " 41", | |
| "answer_number": "41", | |
| "unit": " ", | |
| "source": "calculus", | |
| "problemid": " 7.7.2", | |
| "comment": " with solution", | |
| "solution": "We saw in the preceding calculation that $\\left|f^{\\prime \\prime}(x)\\right| \\leqslant 2$ for $1 \\leqslant x \\leqslant 2$, so we can take $K=2$, $a=1$, and $b=2$ in 3 . Accuracy to within 0.0001 means that the size of the error should be less than 0.0001 . Therefore we choose $n$ so that\r\n$$\r\n\\frac{2(1)^3}{12 n^2}<0.0001\r\n$$\r\nSolving the inequality for $n$, we get\r\nor\r\n$$\r\n\\begin{aligned}\r\n& n^2>\\frac{2}{12(0.0001)} \\\\\r\n& n>\\frac{1}{\\sqrt{0.0006}} \\approx 40.8\r\n\\end{aligned}\r\n$$\r\nThus $n=41$ will ensure the desired accuracy." | |
| }, | |
| { | |
| "problem_text": "Find the length of the cardioid $r=1+\\sin \\theta$.", | |
| "answer_latex": " 8", | |
| "answer_number": "8", | |
| "unit": " ", | |
| "source": "calculus", | |
| "problemid": " 10.4.4", | |
| "comment": " with solution", | |
| "solution": "\nThe cardioid's full length is given by the parameter interval $0 \\leqslant \\theta \\leqslant 2 \\pi$, $$\r\n\\begin{aligned}\r\nL & =\\int_0^{2 \\pi} \\sqrt{r^2+\\left(\\frac{d r}{d \\theta}\\right)^2} d \\theta=\\int_0^{2 \\pi} \\sqrt{(1+\\sin \\theta)^2+\\cos ^2 \\theta} d \\theta \\\\\r\n& =\\int_0^{2 \\pi} \\sqrt{2+2 \\sin \\theta} d \\theta\r\n\\end{aligned}\r\n$$\r\nWe could evaluate this integral by multiplying and dividing the integrand by $\\sqrt{2-2 \\sin \\theta}$, or we could use a computer algebra system. In any event, we find that the length of the cardioid is $L=8$.\n" | |
| }, | |
| { | |
| "problem_text": "Estimate the volume of the solid that lies above the square $R=[0,2] \\times[0,2]$ and below the elliptic paraboloid $z=16-x^2-2 y^2$. Divide $R$ into four equal squares and choose the sample point to be the upper right corner of each square $R_{i j}$. ", | |
| "answer_latex": " 34", | |
| "answer_number": "34", | |
| "unit": " ", | |
| "source": "calculus", | |
| "problemid": "15.1.1 ", | |
| "comment": " with solution", | |
| "solution": "\nThe paraboloid is the graph of $f(x, y)=16-x^2-2 y^2$ and the area of each square is $\\Delta A=1$. Approximating the volume by the Riemann sum with $m=n=2$, we have\r\n$$\r\n\\begin{aligned}\r\nV & \\approx \\sum_{i=1}^2 \\sum_{j=1}^2 f\\left(x_i, y_j\\right) \\Delta A \\\\\r\n& =f(1,1) \\Delta A+f(1,2) \\Delta A+f(2,1) \\Delta A+f(2,2) \\Delta A \\\\\r\n& =13(1)+7(1)+10(1)+4(1)=34\r\n\\end{aligned}\r\n$$" | |
| }, | |
| { | |
| "problem_text": "Find the average value of the function $f(x)=1+x^2$ on the interval $[-1,2]$.", | |
| "answer_latex": " 2", | |
| "answer_number": "2", | |
| "unit": " ", | |
| "source": "calculus", | |
| "problemid": " 6.5.1", | |
| "comment": " with equation", | |
| "solution": "With $a=-1$ and $b=2$ we have\r\n$$\r\n\\begin{aligned}\r\nf_{\\text {ave }} & =\\frac{1}{b-a} \\int_a^b f(x) d x=\\frac{1}{2-(-1)} \\int_{-1}^2\\left(1+x^2\\right) d x \\\\\r\n& =\\frac{1}{3}\\left[x+\\frac{x^3}{3}\\right]_{-1}^2=2\r\n\\end{aligned}\r\n$$" | |
| }, | |
| { | |
| "problem_text": "Find the area of the region enclosed by the parabolas $y=x^2$ and $y=2 x-x^2$", | |
| "answer_latex": " $\\frac{1}{3}$", | |
| "answer_number": "0.333333333333333", | |
| "unit": " ", | |
| "source": "calculus", | |
| "problemid": " 6.1.2", | |
| "comment": "with solution ", | |
| "solution": "\nWe first find the points of intersection of the parabolas by solving their equations simultaneously. This gives $x^2=2 x-x^2$, or $2 x^2-2 x=0$. Thus $2 x(x-1)=0$, so $x=0$ or 1 . The points of intersection are $(0,0)$ and $(1,1)$. By analyzing these two equations, we can see that the top and bottom boundaries are\r\n$$\r\ny_T=2 x-x^2 \\quad \\text { and } \\quad y_B=x^2\r\n$$\r\nThe area of a typical rectangle is\r\n$$\r\n\\left(y_T-y_B\\right) \\Delta x=\\left(2 x-x^2-x^2\\right) \\Delta x\r\n$$\r\nand the region lies between $x=0$ and $x=1$. So the total area is\r\n$$\r\n\\begin{aligned}\r\nA & =\\int_0^1\\left(2 x-2 x^2\\right) d x=2 \\int_0^1\\left(x-x^2\\right) d x \\\\\r\n& =2\\left[\\frac{x^2}{2}-\\frac{x^3}{3}\\right]_0^1=2\\left(\\frac{1}{2}-\\frac{1}{3}\\right)=\\frac{1}{3}\r\n\\end{aligned}\r\n$$\n" | |
| }, | |
| { | |
| "problem_text": "The region $\\mathscr{R}$ enclosed by the curves $y=x$ and $y=x^2$ is rotated about the $x$-axis. Find the volume of the resulting solid.", | |
| "answer_latex": " $\\frac{2\\pi}{15}$", | |
| "answer_number": "0.41887902047", | |
| "unit": " ", | |
| "source": "calculus", | |
| "problemid": " 6.2.4", | |
| "comment": " with solution", | |
| "solution": "\nThe curves $y=x$ and $y=x^2$ intersect at the points $(0,0)$ and $(1,1)$. A cross-section in the plane $P_x$ has the shape of a washer (an annular ring) with inner radius $x^2$ and outer radius $x$, so we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle:\r\n$$\r\nA(x)=\\pi x^2-\\pi\\left(x^2\\right)^2=\\pi\\left(x^2-x^4\\right)\r\n$$\r\nTherefore we have\r\n$$\r\n\\begin{aligned}\r\nV & =\\int_0^1 A(x) d x=\\int_0^1 \\pi\\left(x^2-x^4\\right) d x \\\\\r\n& =\\pi\\left[\\frac{x^3}{3}-\\frac{x^5}{5}\\right]_0^1=\\frac{2 \\pi}{15}\r\n\\end{aligned}\r\n$$\n" | |
| }, | |
| { | |
| "problem_text": "Use Simpson's Rule with $n=10$ to approximate $\\int_1^2(1 / x) d x$.", | |
| "answer_latex": " 0.693150", | |
| "answer_number": "0.693150", | |
| "unit": " ", | |
| "source": "calculus", | |
| "problemid": " 7.7.4", | |
| "comment": " with solution", | |
| "solution": "Putting $f(x)=1 / x, n=10$, and $\\Delta x=0.1$ in Simpson's Rule, we obtain\r\n$$\r\n\\begin{aligned}\r\n\\int_1^2 \\frac{1}{x} d x & \\approx S_{10} \\\\\r\n& =\\frac{\\Delta x}{3}[f(1)+4 f(1.1)+2 f(1.2)+4 f(1.3)+\\cdots+2 f(1.8)+4 f(1.9)+f(2)] \\\\\r\n& =\\frac{0.1}{3}\\left(\\frac{1}{1}+\\frac{4}{1.1}+\\frac{2}{1.2}+\\frac{4}{1.3}+\\frac{2}{1.4}+\\frac{4}{1.5}+\\frac{2}{1.6}+\\frac{4}{1.7}+\\frac{2}{1.8}+\\frac{4}{1.9}+\\frac{1}{2}\\right) \\\\\r\n& \\approx 0.693150\r\n\\end{aligned}\r\n$$\r\n" | |
| }, | |
| { | |
| "problem_text": "Use the Midpoint Rule with $m=n=2$ to estimate the value of the integral $\\iint_R\\left(x-3 y^2\\right) d A$, where $R=\\{(x, y) \\mid 0 \\leqslant x \\leqslant 2,1 \\leqslant y \\leqslant 2\\}$.", | |
| "answer_latex": " -11.875", | |
| "answer_number": "-11.875", | |
| "unit": " ", | |
| "source": "calculus", | |
| "problemid": " 15.1.3", | |
| "comment": " ", | |
| "solution": "\nIn using the Midpoint Rule with $m=n=2$, we evaluate $f(x, y)=x-3 y^2$ at the centers of the four subrectangles. So $\\bar{X}_1=\\frac{1}{2}, \\bar{X}_2=\\frac{3}{2}, \\bar{y}_1=\\frac{5}{4}$, and $\\bar{y}_2=\\frac{7}{4}$. The area of each subrectangle is $\\Delta A=\\frac{1}{2}$. Thus\r\n$$\r\n\\begin{aligned}\r\n\\iint_R\\left(x-3 y^2\\right) d A & \\approx \\sum_{i=1}^2 \\sum_{j=1}^2 f\\left(\\bar{x}_i, \\bar{y}_j\\right) \\Delta A \\\\\r\n& =f\\left(\\bar{x}_1, \\bar{y}_1\\right) \\Delta A+f\\left(\\bar{x}_1, \\bar{y}_2\\right) \\Delta A+f\\left(\\bar{x}_2, \\bar{y}_1\\right) \\Delta A+f\\left(\\bar{x}_2, \\bar{y}_2\\right) \\Delta A \\\\\r\n& =f\\left(\\frac{1}{2}, \\frac{5}{4}\\right) \\Delta A+f\\left(\\frac{1}{2}, \\frac{7}{4}\\right) \\Delta A+f\\left(\\frac{3}{2}, \\frac{5}{4}\\right) \\Delta A+f\\left(\\frac{3}{2}, \\frac{7}{4}\\right) \\Delta A \\\\\r\n& =\\left(-\\frac{67}{16}\\right) \\frac{1}{2}+\\left(-\\frac{139}{16}\\right) \\frac{1}{2}+\\left(-\\frac{51}{16}\\right) \\frac{1}{2}+\\left(-\\frac{123}{16}\\right) \\frac{1}{2} \\\\\r\n& =-\\frac{95}{8}=-11.875\r\n\\end{aligned}\r\n$$\r\nThus we have\r\n$$\r\n\\iint_R\\left(x-3 y^2\\right) d A \\approx-11.875\r\n$$\n" | |
| }, | |
| { | |
| "problem_text": "The base radius and height of a right circular cone are measured as $10 \\mathrm{~cm}$ and $25 \\mathrm{~cm}$, respectively, with a possible error in measurement of as much as $0.1 \\mathrm{~cm}$ in each. Use differentials to estimate the maximum error in the calculated volume of the cone.", | |
| "answer_latex": " $20\\pi$", | |
| "answer_number": "62.8318530718", | |
| "unit": " $\\mathrm{~cm}^3$", | |
| "source": "calculus", | |
| "problemid": " 14.4.5", | |
| "comment": "with solution ", | |
| "solution": "The volume $V$ of a cone with base radius $r$ and height $h$ is $V=\\pi r^2 h / 3$. So the differential of $V$ is\r\n$$\r\nd V=\\frac{\\partial V}{\\partial r} d r+\\frac{\\partial V}{\\partial h} d h=\\frac{2 \\pi r h}{3} d r+\\frac{\\pi r^2}{3} d h\r\n$$\r\nSince each error is at most $0.1 \\mathrm{~cm}$, we have $|\\Delta r| \\leqslant 0.1,|\\Delta h| \\leqslant 0.1$. To estimate the largest error in the volume we take the largest error in the measurement of $r$ and of $h$. Therefore we take $d r=0.1$ and $d h=0.1$ along with $r=10, h=25$. This gives\r\n$$\r\nd V=\\frac{500 \\pi}{3}(0.1)+\\frac{100 \\pi}{3}(0.1)=20 \\pi\r\n$$\r\nThus the maximum error in the calculated volume is about $20 \\pi \\mathrm{cm}^3 \\approx 63 \\mathrm{~cm}^3$.\r\n" | |
| }, | |
| { | |
| "problem_text": "A force of $40 \\mathrm{~N}$ is required to hold a spring that has been stretched from its natural length of $10 \\mathrm{~cm}$ to a length of $15 \\mathrm{~cm}$. How much work is done in stretching the spring from $15 \\mathrm{~cm}$ to $18 \\mathrm{~cm}$ ?", | |
| "answer_latex": " 1.56", | |
| "answer_number": "1.56", | |
| "unit": " $\\mathrm{~J}$", | |
| "source": "calculus", | |
| "problemid": " 6.4.3", | |
| "comment": " with equation", | |
| "solution": "According to Hooke's Law, the force required to hold the spring stretched $x$ meters beyond its natural length is $f(x)=k x$. When the spring is stretched from $10 \\mathrm{~cm}$ to $15 \\mathrm{~cm}$, the amount stretched is $5 \\mathrm{~cm}=0.05 \\mathrm{~m}$. This means that $f(0.05)=40$, so\r\n$$\r\n0.05 k=40 \\quad k=\\frac{40}{0.05}=800\r\n$$\r\nThus $f(x)=800 x$ and the work done in stretching the spring from $15 \\mathrm{~cm}$ to $18 \\mathrm{~cm}$ is\r\n$$\r\n\\begin{aligned}\r\nW & \\left.=\\int_{0.05}^{0.08} 800 x d x=800 \\frac{x^2}{2}\\right]_{0.05}^{0.08} \\\\\r\n& =400\\left[(0.08)^2-(0.05)^2\\right]=1.56 \\mathrm{~J}\r\n\\end{aligned}\r\n$$" | |
| } | |
| ] |