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Math 3010 § 1. Treibergs Final Exam Name: Practice Problems March 30 Math 3010 x 1. Final Exam Name: Practice Problems Treibergs March 30, 2018 Here are some problems soluble by methods encountered in the course. I have tried to select problems ranging over the topics we've encountered. Admittedly, they were chosen because they're fascinating to me. As such, they may have solutions that are longer than the questions you might expect on an exam. But some of them are samples of homework problems. Here are a few of my references. References. • Carl Boyer, A History of the Calculus and its Conceptual Development, Dover Publ. Inc., New York, 1959; orig. publ. The Concepts of the Calculus, A Critical and Historical Discussion of the Derivative and the Integral, Hafner Publ. Co. Inc., 1949. • Carl Boyer, A History of Mathematics, Princeton University Press, Princeton 1985 • Lucas Bunt, Phillip Jones, Jack Bedient, The Historical Roots of Elementary mathematics, Dover, Mineola 1988; orig. publ. Prentice- Hall, Englewood Cliffs 1976 • David Burton, The History of Mathmatics, An Introduction, 7th ed., McGraw Hill, New York 2011 • Ronald Calinger, Classics of Mathematics, Prentice-Hall, Englewood Cliffs 1995; orig. publ. Moore Publ. Co. Inc., 1982 • Victor Katz, A History of Mathematics An introduction, 3rd. ed., Addison-Wesley, Boston 2009 • Morris Kline, Mathematical Thought from Ancient to Modern Times, Oxford University Press, New York 1972 • John Stillwell, Mathematics and its History, 3rd ed., Springer, New York, 2010. • Dirk Struik, A Concise History of Mathematics, Dover, New York 1967 • Harry N. Wright, First Course in Theory of Numbers, Dover, New York, 1971; orig. publ. John Wiley & Sons, Inc., New York, 1939. 1. Use Fermat's method of Adequality to find x > 0 where f(x) has a local maximum, where p x f(x) = : 1 + x2 Fermat assumes that x and x + E bracket the local max and have equal function values. Thus p p x x + E = 1 + x2 1 + (x + E)2 which implies by squaring 2 2 1 + (x + E)2 x = 1 + x2 (x + E) 2 2 1 + x2 + (2x + E)E x = 1 + x2 (x + E) 2 2 2 1 + x2 x + 2 1 + x2 (2x + E)Ex + (2x + E)2E2 = 1 + x2 x + 1 + x2 E 2 2 1 + x2 (2x + E)E + (2x + E)2E2x = 1 + x2 E 1 Dividing by E yields 2 1 + x2 Ex + (2x + E)2Ex = (1 + x2)(1 − 3x2) Neglecting the E terms which vanish (in the limit), 0 = 1 − 3x2 (1) 1 1 the roots are x = ±p and the max occurs when x = p . As a reality check, we find the 3 3 max using calculus. The derivative 1 − 1 p x 2 2x x f 0(x) = 2 − 1 + x2 (1 + x2)2 1 3 3 1 − 2 2 2 2 x + x − 2x = (1 + x2)2 1 − 3x2 = p 2 x (1 + x2)2 Thus again, f 0(x) = 0 when (1) holds so the maximizing point is the one computed already. 2. Use Fermat's method of Ad-equality to find the tangent line and slope of the function at x. f(x) = x2 − x3: We seek the tangent line at B = (x; f(x)). A nearby point on the curve is A = (x+e; f(x+e)) where e = CI. Let t = EO. This time A and B bracket the point of tangency. Thus Fermat ad-equated the slopes of the triangles 4BCE and 4AIE BC AI = CE IE yielding f(x) f(x + e) = : t + x t + x + e Substituting the function, x2 − x3 (x + e)2 − (x + e)3 x2 − x3 + (2x − 3x2)e + (1 − 3x)e2 − e3 = = : t + x t + x + e t + x + e 2 Cross multiplying yields x2 − x3 (t + x + e) = x2 − x3 + (2x − 3x2)e + (1 − 3x)e2 − e3 (t + x): Cancelling, x2 − x3 e = (2x − 3x2)e + (1 − 3x)e2 − e3 (t + x): Dividing by e and collecting terms yields x2 − x3 − (2x − 3x2)(t + x) = (1 − 3x)e − e2 (t + x): Neglecting e terms gives x2 − x3 − (2x − 3x2)(t + x) = 0: In other words, the slope of the tangent line is f(x) x2 − x3 = = 2x − 3x2 t + x t + x and x2 − x3 − (2x − 3x2)x 2x2 − x t = = : 2x − 3x2 2 − 3x Of course, this agrees with the modern computations f 0(x) = 2x − 3x2 and since the slope is the derivative, f(x) x2 − x3 2x2 − x t = − x = − x = : f 0(x) 2x − 3x2 2 − 3x p 3. Use Fermat's method of Ad-equality to find the slope of f(x) = x2 − x at x > 0. [University of Utah, Math 3010 homework problem, April 13, 2018.] The tangent line is approximated by a secant line through two infinitesimally close points at A = (x; f(x)) and B = (x + E; f(x + E)). Let T = (t; 0) be the intersection point of the tangent line with the x-axis and C = (x; 0) and D = (x + E; 0). Then the slopes of the two similar triangles 4ACT ∼ 4BDT are equal f(x) f(x + E) = : x − t x + E − t Cross multiplying p p (x2 − x)(x + E − t) = f(x)(x + E − t) = f(x + E)(x − t) = ((x + E)2 − x + E)(x − t): Isolate the square root p p (x2 − x)(x + E − t) − (x + E)2(x − t) = −(x − t) x + E: Square p 2 (x2 − x)(x + E − t) − (x + E)2(x − t) = (x − t)2(x + E) p p (x2 − x)2(x + E − t)2 − 2(x2 − x)(x − t)(x + E − t)(x + E)2 + (x + E)4(x − t)2 = (x − t)2(x + E) 3 Expand in E p (x2 − x)2(x − t)2 + 2(x − t)E + E2 p −2(x2 − x)(x − t)(x + E − t)x2 + 2xE + E2 +x4 + 4x3E + 6x2E2 + 4xE3 + E4(x − t)2 = (x − t)2(x + E): Write A, B, C, D, G, H for functions of x and t, p p (x2 − x)2(x − t)2 + 2(x2 − x)2(x − t)E + AE2 p p −2(x2 − x)(x − t)2x2 − 2(x2 − x)(x − t)x2 + 2x(x − t)E + BE2 + CE3 +x4(x − t)2 + 4x3(x − t)2E + DE2 + GE3 + HE4 = (x − t)2x + (x − t)2E We note that the E0 terms cancel p p (x2 − x)2(x − t)2 − 2(x2 − x)(x − t)2x2 + x4(x − t)2 p 2 = (x − t)2(x2 − x) − x2 = (x − t)2x: We divide out E and drop any remaining terms with a factor E. p p 2(x2 − x)2(x − t) − 2(x2 − x)(x − t)x2 + 2x(x − t) + 4x3(x − t)2 = (x − t)2: If x 6= t which happens except when x = 1 and f(1) = 0 we may divide by x − t p p p 2(x2 − x)2 − 2(x2 − x)x2 = 1 + 4x(x2 − x) − 4x3(x − t): which implies p p 2(x2 − x) x x − t = − p : 1 − 4x x Hence the slope is p f(x) 1 − 4x x 1 = − p = 2x − p : x − t 2 x 2 x 0 Of course this agrees with f (x) computed the modern way. This function is contnuousp at x > 0 so it continues to provide the derivative at x = 1. x − t blows up when 1 − 4x x = 0 at x = 2−4=3. Since f(2−4=3) 6= 0 this simply means that the slope is zero at this point. Indeed, it is a minimum point for f(x). Bo Zhu recommends the following procedure. Instead of isolating the square, rewrite the equation as p p p x2(x + E − t) − (x + E)2(x − t) + ( x + E − x)(x − t) − xE = 0 x2(x − t) + x2E − (x2 + 2xE + E2)(x − t)+ p p p p x + E + x p +( x + E − x)p p (x − t) − xE = 0 x + E + x E p x2E − (2xE + E2)(x − t) + p p (x − t) − xE = 0 x + E + x Divide by E and set the remaining E = 0. 1 p x2 − 2x(x − t) + p (x − t) − x = 0 2 x which implies p x2 − x 1 = 2x − p x − t 2 x as before. 4 4. Following Roberval, find the are under one half of one arch of the cycloid x(t) = a(t − sin t); y(t) = a(1 − cos t) generated by following a point on the edge of a circle of radius a as it rolls along the x-axis. First show using Cavalieri's principle that the area between the cycloid and its companion curve x(t) = at; y(t) = a (1 − cos t) equals half the area of its generating circle. Then show, again using Cavalieri's principle that the companion curve bisects the area of the rectangle containing the half cycloid arch. The cycloid AQC is generated by rolling the circle of radius a to the right. As the circle rolls through arc AF , the base of the wheel moves a distance AN. The point that was at A and rolled to Q has moved up to height F above line AB. The distance Q is behind the contact point N is PQ which equals the distance FR. Thus the horizontal slice QP of the area between the cycloid and the companion curve AP XC has the same length as the horizontal slice RF of the half disk ARY DF A. It follows that the area of the half disk equals the area between the cycloid and the companion curve. The companion curve is determined so that the height NP is the same as the height of F when the circle has rolled to base at N.
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