MATH 18.01, FALL PROBLEM SET # 8

MATH 18.01, FALL 01 - PROBLEM SET # 8 Professor: Jared Speck Due: by 1:45pm on Tuesday 11-7-1 (in the boxes outside of Room -55 during the day; stick it under the door if the room is locked; write your name, recitation instructor, and recitation meeting days/time on your homework) 18.01 Supplementary Notes (including Exercises and Solutions) are available on the course web page: http://math.mit.edu/classes/18.01/. This is where to find the exercises labeled 1A, 1B, etc. You will need these to do the homework. Part I consists of exercises given and solved in the Supplementary Notes. It will be graded quickly, checking that all is there and the solutions not copied. Part II consists of problems for which solutions are not given; it is worth more points. Some of these problems are longer multi-part exercises given here because they do not fit conveniently into an exam or short-answer format. See the guidelines below for what collaboration is acceptable, and follow them. To encourage you to keep up with the lectures, both Part I and Part II tell you for each problem on which day you will have the needed background for it. You are encouraged to use graphing calculators, software, etc. to check your answers and to explore calculus. However, (unless otherwise indicated) we strongly discourage you from using these tools to solve problems, perform computations, graph functions, etc. An extremely important aspect of learning calculus is developing these skills. And you will not be allowed to use any such tools on the exams. Part I (0 points) Notation: The problems come from three sources: the Supplementary Notes, the Simmons book, and problems that are described in full detail inside of this pset. I refer to the former two sources using abbreviations such as the following ones:.1 = Section.1 of the Simmons textbook; Notes G = Section G of the Supplementary Notes; Notes 1A: 1a, = Exercises 1a and in the Exercise Section 1A of the Supplementary Notes; Section.4: 13 = Problem 13 in Section.4 of Simmons, etc. Lecture 4. (Thurs., Nov. 15) Integrating rational functions, partial fractions. Read: Read 10.6, Notes F. Homework: Notes 5E:, 3, 5, 6, 10h (complete the square). Lecture 5. (Fri., Nov. 16) Integration by parts. Reduction formulas. Read: 10.7. Homework: Notes 5F: 1a, d, followed by b, 3. 1

MATH 18.01, FALL 01 - PROBLEM SET # 8 Lecture 6. (Tues., Nov. 0) Parametric equations; arc length. Surface area. Read: 17.1, 7.5, 7.6. Homework: Notes 4E:, 3, 8; Notes 4F: 1d, 4, 5, 8; Notes 4G:, 5. If a curve is given by x = x(t), y = y(t) then to find its arc length, use ds in the form ds = (dx ) ( ) dy (dx) + (dy) = + and integrate from start to finish: from t = t 0 to t = t 1.

MATH 18.01, FALL 01 - PROBLEM SET # 8 3 Part II (50 points) Directions and Rules: Collaboration on problem sets is encouraged, but: i) Attempt each part of each problem yourself. Read each portion of the problem before asking for help. If you don t understand what is being asked, ask for help interpreting the problem and then make an honest attempt to solve it. ii) Write up each problem independently. On both Part I and II exercises you are expected to write the answer in your own words. You must show your work; bare solutions will receive very little credit. iii) Write on your problem set whom you consulted and the sources you used. If you fail to do so, you may be charged with plagiarism and subject to serious penalties. iv) It is illegal to consult materials from previous semesters. 0. (not until due date; 3 points) Write the names of all the people you consulted or with whom you collaborated and the resources you used, or say none or no consultation. This includes visits outside recitation to your recitation instructor. If you don t know a name, you must nevertheless identify the person, as in, tutor in Room -106, or the student next to me in recitation. Optional: note which of these people or resources, if any, were particularly helpful to you. This Problem 0 will be assigned with every problem set. Its purpose is to make sure that you acknowledge (to yourself as well as others) what kind of help you require and to encourage you to pay attention to how you learn best (with a tutor, in a group, alone). It will help us by letting us know what resources you use. 1. (Thurs., Nov. 15; partial fractions; 3 + = 5 points) a) Derive a formula for sec x dx by first writing sec x = cos x (verify this), and then making a 1 sin x substitution for sin x and using partial fractions. (Your final answer must be expressed in terms of x.) Solution: We first use the identity cos x = 1 sin x to deduce that sec x = 1 cos x = cos x cos x = cos x 1 sin x. We then make the change of variable u = sin x, du = cos x dx to deduce that sec x dx = = cos x 1 sin x dx du 1 u. To evaluate the latter integral, we first decompose 1/(1 u ) using the method of partial fractions: Using this decomposition, we compute that 1 1 u = 1 ( 1 1 + u + 1 ). 1 u

4 MATH 18.01, FALL 01 - PROBLEM SET # 8 du 1 u = 1 ( 1 1 + u + 1 ) du 1 u = 1 (ln 1 + u ln 1 u ) = 1 ln 1 + u 1 u = 1 ln 1 + sin x 1 sin x. b) Convert the formula into the one we derived in class by multiplying the fraction in the answer on both top and bottom by 1 + sin x. (Note that (1/) ln v = ln v.) Solution: We first note that Taking square roots, we deduce that Hence, 1 + sin x 1 sin x = 1 + sin x 1 sin x 1 + sin x 1 + sin x = (1 + sin x) 1 sin x (1 + sin x) =. cos x 1 + sin x 1 sin x = 1 + sin x cos x = 1 cos x + sin x cos x = sec x + tan x. sec x dx = 1 ( ) ln 1 + sin x 1 + sin x 1 sin x + C = ln + C 1 sin x = ln sec x + tan x + C. The above formula is the one we derived in class.. (Fri., Nov. 16; integration by parts; 10 points) Let R be the region in the first quadrant of the (x, y) plane (x 0, y 0) under the first hump of the curve y = e x cos x and above the x axis. The region R is rotated about the y axis to generate a solid of revolution S. Use the shell method to find the volume of S. Solution: The first hump is the region under the graph of y = e x cos x with 0 x π/. Recall that the volume of a thin shell is dv = π shell radius shell height shell thickness = πxe x cos x dx. Hence, the total volume is V = dv = π/ 0 πxe x cos x dx It will take us several steps to evaluate the integral. To this end, we want to use integration by parts with u = x, dv = e x cos x, du = dx. But first, we have to compute v.

MATH 18.01, FALL 01 - PROBLEM SET # 8 5 In order to compute v, we have to use integration by parts yet again. We set w = e x, dw = e x dx, dz = cos x dx, z = sin x. Then using the integration by parts formula w dz = wz z dw, we compute that v = e x cos x dx = w dz = wz z dw = e x sin x e x sin x dx. To handle the last integral on the right-hand side, we again use integration by parts (the complete details are omitted) to compute that e x sin x dx = e x cos x + e x cos x dx = e x cos x + v. Inserting this formula for e x sin x dx back into the above identity for v, we deduce that We can therefore solve for v algebraically: v = v = e x sin x + e x cos x v. e x cos x dx = 1 (ex sin x + e x cos x). Later, we will need the following antiderivative, which can be computed using the same strategy that we used to compute v (the complete details are omitted): e x sin x dx = 1 (ex sin x e x cos x). We now return to the original integral. We use integration by parts in the form u dv = uv v du together with the above formulas for v = e x cos x dx and e x sin x dx to deduce that xe x cos x dx = uv v du = 1 (xex sin x + xe x cos x) 1 e x sin x + e x cos x dx = 1 (xex sin x + xe x cos x) 1 ex sin x + C. Finally, inserting the factor of π and the bounds of integration, we conclude that π/ V = πxe x cos x dx = π (xe x sin x + xe x cos x) πe x sin x x=π/ x=0 0 ( ) π = π e π/. 3. (Fri., Nov. 16; integration by parts; 5 points) Recall that d Consider the following sequence of integration by parts: dx d cosh x = sinh x and sinh x = cosh x. dx

6 MATH 18.01, FALL 01 - PROBLEM SET # 8 e x sinh x dx = e x cosh x e x cosh x dx = e x cosh x e x sinh x + e x sinh x dx. By canceling e x sinh x dx from each side, one would conclude that e x cosh x = e x sinh x and hence cosh x = sinh x. However, these last two equalities are false. Explain the mistake in reasoning. Solution: The mistake stems from the fact that the two antiderivatives e x sinh x dx on each side of the equation only have to agree with each other up to a constant C. There is no reason that C has to be equal to 0 in general. In fact, in this particular problem C = 1. Equivalently, e x cosh x e x sinh x = 1. To see this, we use the definitions cosh x = ex +e x and sinh x = ex e x to compute that ( ) ( ) e e x cosh x e x sinh x = e x x + e x e e x x e x = ex + 1 ex 1 = 1. 4. (Tues., Nov. 0; parametric equations, arc length, surface area; + 4 + 4 = 10 points) In this problem, you will investigate a parametric curve in the (x, y) plane called a cycloid. You already encountered essentially the same curve in problem 4 of pset 3b. The curve is defined by x = t sin t, y = 1 cos t. The curve describes the motion of a point P = (x, y) on the outer rim of a wheel of radius 1 as the wheel rolls to the right at a constant angular velocity (in problem 4 of pset 3b the wheel was rolling to the left). Note that the point P is initially at the bottom of wheel (the center of the wheel starts at (0, 1)). a) Express dy/dx in terms of t and find the values of t and (x, y) where dy/dx is infinite. Then sketch the cycloid (in the (x, y) plane) for 0 t 4π. Solution: Using the trigonometric identities we compute that ( ) t 1 cos t = sin, sin t = sin(t/) cos(t/),

MATH 18.01, FALL 01 - PROBLEM SET # 8 7 dx = 1 cos t dy = sin t, dy dx = dy dy dx = = sin t 1 cos t = cot(t/). dx = sin(t/) cos(t/) sin (t/) It follows that dy/dx is infinite when t = mπ, where m is an integer. At these values of t, y = 1 cos(mπ) = 0 and x = mπ sin(mπ) = mπ. The graph of the cycloid is given in the figure below. Figure 1. Graph of the inverted cycloid for 0 t 4π b) Compute the arc length of one full period of the cycloid (0 t π). This will tell you the total distance traveled by the point P in the plane after one full period. Solution: We first note the trigonometric identity

8 MATH 18.01, FALL 01 - PROBLEM SET # 8 sin ( t ) = 1 (1 cos(t)). Using this identity, we compute that dx = 1 cos t, dy = sin t, ds = (dx ) (dx) + (dy) = + = (1 cos t) + sin t = (1 cos t) ( = t ) sin. Using the change of variable u = t/, du = 1, we compute that ( ) dy Arc Length = = = 4 = 4 π t=0 π t=0 π u=0 π ds ( t ) sin u=0 sin u du sin u du = 4( cos u) π 0 = 4 ( cos π + cos 0) = 8. c) The region under one full period of the cycloid (0 t π) and above the x axis is revolved about the x axis to generate a solid of revolution S. Find the surface area of S. Solution: We divide the solid into many thin conical strips whose axes of symmetry are parallel to the x axis. The surface area of such a strip of radius y and slant edge length ds is da = shell circumference surface edge length = πyds. To find the total surface area A, we just add up the surface areas of the thin strips (i.e., we integrate):

MATH 18.01, FALL 01 - PROBLEM SET # 8 9 A = da = πy ds = t=π = 8π = 16π t=0 t=π t=0 π π (1 cos t) 3/ ds ( ) t sin 3 u=0 sin 3 u du. In our calculations above, we have used the trig identities and the u substitution from part b). To evaluate the integral π u=0 sin3 u du, we make the change of variables v = cos u, dv = sin u du. We then compute sin 3 u du = sin u sin u du = (1 cos u) sin u du = (1 v ) dv = v3 3 v = 1 3 cos3 u cos u. Inserting the bounds of integration, we deduce that π u=0 sin 3 u du = 1 3 cos3 u cos u u=π u=0 = 1 3 cos3 π cos π 1 3 cos3 0 + cos 0 = 4 3. Combining the above calculations, we deduce that π A = 16π sin 3 u du = 16π 4 u=0 3 = 64π 3. 5. (Tues., Nov. 0; parametric equations; 5 points) Consider the curve defined by the parametric equations x = y = t 1 t 1 (cos v)(ln v) dv, (sin v)(ln v) dv

10 MATH 18.01, FALL 01 - PROBLEM SET # 8 for t 1. Find the arc length of the portion of the curve that starts from the origin in the (x, y) plane and ends at the first non-origin point where the curve is vertical. Solution: Using FTC, we compute that dx = (cos t)(ln t), dy = (sin t)(ln t). The first non-origin point where the curve is vertical is the first non-origin point where dx value corresponding to this point is t = π/. We then compute that = 0. The t (dx ) ( ) dy ds = + = cos t ln t + sin t ln t = ln t = ln t = ln t (when t 1). The arc length of the portion of the curve of interest is therefore π/ Arc Length = ds = ln t = t ln t t π/ 1 t=1 = π ( π ) ln π + 1. In the above calculations, we used integration by parts with u = ln t, du = /t, dv =, v = t to derive ln t = u dv = uv v du = t ln t = t ln t t + C. 6. (Tues., Nov. 0; parametric equations; 1 + 7 + 4 = 1 points) A standard ellipse is a closed curve. That is, when you trace it out, it ends up where it started after one full orbit. In the simplest models of our solar system, a detailed analysis of Newton s laws of motion shows the following fact (which was first shown by Newton and which is now known as his solution to his classic two-body problem): a planet traces out a standard ellipse as it revolves around the sun. However, more accurate models take into account an effect called apsidal precession, which distorts the ellipses. Precession is caused by many factors including the gravitational force exerted by other planets, the fact that planets are not perfect point particles, and the fact that Newton s theory of gravity is not as accurate as Einstein s theory of gravity (general relativity). In part c), you will investigate a simple model of precession. In this problem, you will not investigate the cause of the precession (investigating the cause is something you would do in an advanced course on celestial mechanics); your goal will only be to understand the shape of the curve. In parts b) and c), you are strongly encouraged to use computer software such as Maple or Mathematica to assist you with plotting and computations.

MATH 18.01, FALL 01 - PROBLEM SET # 8 11 a) Consider the following ellipse in parametric form for 0 t π : x = cos t, y = sin t. This ellipse describes 1 a planet that is orbiting around the sun, which is located at the focal point F = ( 3, 0). Find an algebraic equation in x and y for the curve. Then plot the portion of the curve corresponding to the parameter range 0 t π. Is the curve traced out clockwise or counterclockwise as t increases? Solution: To find the algebraic equation for the curve, we note that ( x ) 1 = sin t + cos t = y x + = 4 + y. This is an ellipse centered at the origin with major x radius equal to, minor y radius equal to 1, and focal points (± 3, 0). The parameter range 0 t π corresponds to one full orbit. The ellipse is traced out counterclockwise as t increases. The graph of the ellipse is given in the figure below. b) Let θ be a constant. Given any point (x, y) in the plane, define R θ (x, y) to be the rotated point in the plane with coordinates R θ (x, y) = ( (x + 3) cos θ y sin θ 3, (x + ) 3) sin θ + y cos θ. In linear algebra, you will learn that R θ (x, y) is simply the rotation of the point (x, y) by θ radians counterclockwise 3 about the focus F = ( 3, 0). More precisely, both (x, y) and R θ (x, y) lie on the same circle of radius r = (x + 3) + y centered at F, but R θ (x, y) has been rotated counterclockwise by θ about F (notice that F itself doesn t move at all, that is, R θ ( 3, 0) = ( 3, 0) ). Using these facts and your insight from part a), plot the following rotated ellipse in parametric form for 0 t π when θ = π/4 : x = ( cos t + 3) cos θ sin t sin θ 3, y = ( cos t + 3) sin θ + sin t cos θ. Then find an algebraic relation in x and y for the curve. Hint: Given any point (x, y) on the rotated ellipse, define ( x, ỹ) = R θ (x, y). Then ( x, ỹ) is a point on the original non-rotated ellipse from part a). Then do the same exercises when θ = π/ and θ = 3π/4. 1 The actual orbits of the planets in our solar system are much more circular than the ellipse in this problem. A completely precise solution of Newton s two-body problem shows that both bodies revolve about their common center of mass. However, if one body (in this case the sun) is much more massive than the other, then the common center of mass approximately coincides with the location of the heavier body. That is, the heavier body approximately stays in place. 3 Some people in the UK call this widdershins.

1 MATH 18.01, FALL 01 - PROBLEM SET # 8 Figure. Graph of the ellipse Solution: The graphs of the three ellipses with θ = π/4, π/, 3π/4 are given in the figures below. To find the algebraic equation for the curve, given any point (x, y) on the rotated ellipse, we define ( x, ỹ) = R θ (x, y) ( = (x + 3) cos θ + y sin θ 3, (x + ) 3) sin θ + y cos θ. Then ( x, ỹ) is a point on the original non-rotated ellipse, and hence x 4 + ỹ = 1. Inserting the formula for ( x, ỹ) in terms of (x, y), we deuce that { x = (x + 3) cos θ + y sin θ 3}, { ỹ = (x + 3) sin θ + y cos θ}. Thus, the algebraic equation satisfied by the curve is

MATH 18.01, FALL 01 - PROBLEM SET # 8 13 Figure 3. Graph of the ellipse with θ = π/4 1 = x 4 + ỹ { } (x + 3) cos θ + y sin θ 3 = 4 + { (x + 3) sin θ + y cos θ}. Thus, in the cases θ = π/4, π/, 3π/4, we expand the above relation and collect like terms to deduce that the algebraic equations are respectively (the many tedious computations are omitted): 5 8 x + 5 8 y + 5 8 x + 5 8 y + ( 5 4 ( 5 4 ) 1 3 3 x 4 ) 1 3 + 3 x + 4 ( 3 4 ( 3 4 ) 1 3 + 3 y 3 4 4 xy 3 13 + 4 11 3y + x + 1 4 y + 3x 1 ) 1 3 3 y + 3 4 4 xy + 3 4 8 = 0, 4 = 0, 13 + 8 = 0. c) Now consider the case when the angle depends on t : θ(t) = t/8. You can think of this curve as an ellipse that is continuously being rotated about its focal point F as it is traced out, where the angular velocity of the rotation is constantly π/8. Some possible physical causes of such precession were

14 MATH 18.01, FALL 01 - PROBLEM SET # 8 Figure 4. Graph of the ellipse with θ = π/ mentioned above. Sketch the following curve in parametric form for 0 t π (with the help of a computer if you want): x = ( cos t + 3) cos(θ(t)) sin t sin(θ(t)) 3, y = ( cos t + 3) sin(θ(t)) + sin t cos(θ(t)). Carefully plot the starting and end points of the curve. Does the curve end up where it started as t varies from 0 to π? Finally, extend your plot to the range 0 t 16π. Solution: The curve does not end up where it started as t varies from 0 to π. The graphs of the precessing ellipse with θ(t) = t/8 are given in the two figures below.

MATH 18.01, FALL 01 - PROBLEM SET # 8 15 Figure 5. Graph of the ellipse with θ = 3π/4

16 MATH 18.01, FALL 01 - PROBLEM SET # 8 Figure 6. Graph of the precessing ellipse for 0 t π

MATH 18.01, FALL 01 - PROBLEM SET # 8 17 Figure 7. Graph of the precessing ellipse for 0 t 16π