MATH 18.01, FALL PROBLEM SET # 3A

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1 MATH 18.01, FALL PROBLEM SET # 3A Professor: Jared Speck Due: by Friday 1:45pm on (in the boxes outside of Room 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) Supplementary Notes (including Exercises and Solutions) are available on the course web page: 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 (15 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: 2.1 = Section 2.1 of the Simmons textbook; Notes G = Section G of the Supplementary Notes; Notes 1A: 1a, 2 = Exercises 1a and 2 in the Exercise Section 1A of the Supplementary Notes; Section 2.4: 13 = Problem 13 in Section 2.4 of Simmons, etc. Lecture 7. (Thurs., Sept. 20) Linear and quadratic approximations. Read: Notes A. Homework: Notes 2A: 2, 3, 7, 11, 12ade. Lecture 8. (Tues., Sept. 25) Curve-sketching. Read: 4.1, 4.2. Homework: Notes 2B: 1, 2aeh, 4, 6ab, 7ab. 1

2 2 MATH 18.01, FALL PROBLEM SET # 3A Part II (40 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 2-106, 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. (Sept. 26.; quadratic approximations; = 8 points) Your electric guitar speaker has a continuous dial that varies from 0 to 11. Assume that the volume output v of the speaker is a twice differentiable function of the dial number d. Using your decibel meter, you have measured the volume output of your speaker at the following three settings: d = 8 d = 9 d = 10 v (in decibels) You are very curious to know the volume output when d = 11, but since you live in a dorm, you don t dare turn the dial to 11. Luckily, you are enrolled in and are therefore able to estimate the output using the following procedure. a) Estimate v (9) and v (10) by using difference quotients of the form f(x) f(x x) x for x = 1. We compute that v (9) v (10) v(9) v(8) = , 9 8 v(10) v(9) = b) Use part a) and a similar procedure to estimate v (10). We compute that

3 MATH 18.01, FALL PROBLEM SET # 3A 3 v (10) v (10) v (9) c) Derive a quadratic approximation of v(d) near d = 10. In your quadratic approximation, you can use the approximate values for v and v from parts a) and b). Then use your quadratic approximation to estimate v(11). We use the quadratic approximation formula to deduce that when d is near d 0, we have v(d) v(d 0 ) + v (d 0 )(d d 0 ) v (d 0 )(d d 0 ) 2 Setting d 0 = 10 and d = 11, we therefore have that (d 10) (.426)(d 10)2. v(d) (11 10) + 1 (.426)(11 10)2 2 = (Sept. 26.; limits; continuity; quadratic approximations; = 10 points) In this problem, you will investigate what happens to the surface area of a sphere when you slightly squash it. A perfect unit sphere can be visualized in the following way: consider the set of points in the (x, y) plane that satisfy the equation x 2 + y 2 = 1. This curve describes a sphere of radius 1 centered at the origin. Now imagine that this circle is continuously rotated about the y axis until it has spun a full 360 degrees. The resulting 3 d object that is traced out is a sphere of radius 1 in three-dimensional space centered at the origin. Now instead of the circle, consider the curve ( y ) 2 x 2 + = 1, a where a is some positive constant with 0 < a < 1. This curve describes an ellipse in the (x, y) plane centered at the origin. The horizontal axis of the ellipse has length 2 and the vertical axis has length 2a. Now imagine that this ellipse is rotated about the y axis in the manner described above. The resulting 3 d object that gets traced out is called an oblate ellipsoid, which means that it is a sphere that has been squashed in one direction (in this case the y axis is the squashed direction). This geometric object comes up when one studies the shape of the earth: the earth s rotation gives it a slightly squashed character (in reality the earth is quite bumpy, and therefore the oblate ellipsoid model of the earth may not be accurate enough for all applications). There is an important quantity associated to the oblate ellipsoid called the eccentricity E. It is 0 for a perfect sphere, and for a oblate ellipsoid, it measures how squashed it is: E = 1 a 2.

4 4 MATH 18.01, FALL PROBLEM SET # 3A In vector calculus, you will develop the tools necessary to calculate the surface area of the oblate ellipsoid. Here, I will only provide you with the formula. The surface area S can be expressed in terms of E as follows: where S = 2π ( 1 + f(e) (1 E 2 ) ), f(e) = { tanh 1 (E) E if E 0, 1 if E = 0. In the above formula, tanh 1 (E) is the inverse of the hyperbolic tangent function. That is, y = tanh 1 (E) if and only if tanh(y) = E, where tanh(e) = sinh(e)/ cosh(e). a) Show that f(e) is continuous at E = 0. We first investigate lim y 0 tanh(y)/y. We use the linear approximations e y 1 + y and e y 1 y for y near 0. Therefore, sinh(y) = 1 2 (ey e y ) y, cosh(y) = 1 2 (ey + e y ) 1, and tanh(y) = sinh(y)/ cosh(y) = y + O(y 2 ). It follows that lim y 0 y + O(y 2 ) tanh(y)/y = lim tanh(y)/y = lim y 0 y 0 y = lim {1 + O(y)} = 1. y 0 We now use the above limit to investigate the limit of interest. We set y = tanh 1 (E), which implies that tanh(y) = E. We note that when E 0, it follows that y 0. Therefore, with the help of the above limit, we compute that Therefore, y lim E 0 tanh 1 (E)/E = lim y 0 tanh(y) = 1. Hence, f(e) is continuous at E = 0 by definition. b) Show that (d/de) tanh 1 (E) = 1/(1 E 2 ). lim f(e) = lim E 0 E 0 tanh 1 (E)/E = 1 = f(0). We set y = tanh 1 (E). Our goal is to compute dy/de. To this end, we consider the equation E = tanh(y). We take the derivative with respect to E of both sides and use the quotient rule, the chain rule, and the relation cosh 2 y sinh 2 (y) = 1 in order to deduce that

5 MATH 18.01, FALL PROBLEM SET # 3A 5 Therefore, 1 = d { dy d tanh(y) dy de = dy ( )} sinh(y) dy cosh(y) de = cosh(y) d d sinh(y) sinh(y) cosh(y) dy dy cosh 2 dy (y) de = cosh2 (y) sinh 2 (y) cosh 2 (y) 1 = cosh 2 (y) dy de. dy de dy de = cosh2 (y). We would like to express the right-hand side in terms of E. To this end, use first use the relation cosh 2 y sinh 2 (y) = 1 to deduce that that cosh 2 (y) = sinh2 (y). We can use the first and last = cosh2 y 1 tanh 2 (y) tanh 2 y equalities in this equation to solve for cosh 2 (y) in terms of tanh 2 (y) = E 2 : We conclude that as desired. cosh 2 (y) = 1 1 tanh 2 (y) = 1 1 E. 2 dy de = 1 1 E 2 c) Find the quadratic approximation to 1/(1 E 2 ) near E = 0. That is, find the constants A 0, A 1 and A 2 such that 1/(1 E 2 ) = A 0 + A 1 E + A 2 E 2 + O(E 3 ) near E = 0. In class, we saw that 1/(1 x) = (1 x) 1 = 1 + x + O(x 2 ) for x near 0. Applying this expansion with x = E 2, we deduce that 1/(1 E 2 ) = 1+E 2 +O(E 4 ). It follows that A 0 = 1, A 1 = 0, and A 2 = 1. d) Imagine that you have made a cubic approximation to tanh 1 (E) near E = 0. That is, assume that you have approximated tanh 1 (E) = B 0 + B 1 E + B 2 E 2 + B 3 E 3 + O(E 4 ) for constants B 0, B 1, B 2, B 3. Assume that B 0 +B 1 E +B 2 E 2 is the usual quadratic approximation to tanh 1 (E) near E = 0. Assume in addition that the derivative of your cubic approximation to tanh 1 (E) is precisely the quadratic approximation to 1/(1 E 2 ) = (d/de) tanh 1 (E) that you found in part b). Under these assumptions, find the constants B 0, B 1, B 2, B 3. We take the derivative of the cubic approximation and set it equal to the quadratic approximation, which was found in part c):

6 6 MATH 18.01, FALL PROBLEM SET # 3A d de (B 0 + B 1 E + B 2 E 2 + B 3 E 3 ) = B 1 + 2B 2 E + 3B 3 E 2 = 1 + E 2. We then equate powers of E to discover that B 1 = 1, B 2 = 0, B 3 = 1/3. The last thing to determine is B 0. By assumption, B 0 is the term one gets from the usual quadratic approximation of tanh 1 (E) near E = 0. That is, B 0 = tanh 1 (0) = 0. In total, the cubic approximation to tanh 1 (E) near E = 0 is tanh 1 (E) = E + (1/3)E 3 + O(E 4 ). e) Use part d) to find the quadratic approximation to f(e) near E = 0. We simply divide the cubic approximation from part d) by E to discover that f(e) = tanh 1 (E) E = E2 + O(E 3 ). f) Use part e) to find the quadratic approximation to S near E = 0. Remark: The quadratic approximation procedure outlined in this problem can be rigorously justified. It turns out that the quadratic approximations you are deriving are in some sense the best possible ones. These ideas fall under the umbrella of Taylor series, which we will discuss at the end of the course. Using part e), we see that Therefore, f(e) (1 E 2 ) = ( ) 3 E2 + O(E 3 ) ( 1 E 2 ) ) = E2 + O(E 3 ). S = 2π ( 1 + f(e)(1 E 2 ) ) ( = 2π ) 3 E2 + O(E 3 ) = 4π 4π 3 E2 + O(E 3 ). g) When E is positive but near 0, does your quadratic approximation predict that the surface area of the ellipsoid will be greater than or less than the surface area of the perfect sphere of radius 1 (the perfect sphere has E = 0)? When E is positive and near 0, the quadratic approximation part from f) predicts that S(E) < 4π because of the 4π 3 E2 term. 3. (Sept. 28.; = 10 points) Section 4.1: 18abc, 20ab

7 MATH 18.01, FALL PROBLEM SET # 3A 7 18) We recall that f(x) = x m (1 x) n, where m, n are positive integers. a) If m is even and x is near 0, then f(x) is non-negative because x m is a square and (1 x) n is near 1. Since f(0) = 0, x = 0 must be a minimum. b) Similarly, if n is even and x is near 1, then f(x) is non-negative because (1 x) n is a square and x m is near 1. Since f(1) = 0, x = 1 must be a minimum. c) Using the product and chain rules, we compute that f (x) = mx m 1 (1 x) n n(1 x) n 1. To find the critical points of f, we set f (x) = 0, which yields the equation mx m 1 (1 x) n nx m (1 x) n 1 = 0. Assuming that x 0 and x 1, we may divide both sides by mx m 1 (1 x) n 1, thus arriving at the following equation: f (x) mx m 1 (1 x) n 1 = (1 x) n m x = 0. The above equation has only the solution x = m, which means that aside from the possibility of m+n x = 0 or x = 1, x = m is the only critical point of f. Note that the expression (1 x) n x is strictly m+n m positive for 0 < x < m and strictly negative for 1 > x > m. Therefore, the first equality in the m+n m+n previous equation implies that the same statement is also true for f (x). Hence, f(x) is increasing for 0 < x < m and decreasing for 1 > x > m. We conclude that x = m is a maximum. m+n m+n m+n 20) a) See the figure below. b) See the figure below. 4. (Sept. 28.; = 9 points) Section 4.2: 12, 16, 18 12) Set y = f(x) = 12/x 2 12/x. Note that there is a discontinuity at x = 0. We compute that y = f (x) = 24x x 2, y = f (x) = 72x 4 24x 3. To find the inflection points, we set y = 0 and multiply the equation by x 4 (for convenience) to derive the equation 72 24x = 0. The above equation has only the solution x = 3, which is the inflection point. For x < 3, we have that f (x) is positive, while for x > 3, we have that f (x) is negative. The graph of f(x) is therefore concave up for x < 3 and concave down for x > 3. See the figure below for a sketch. 16) The question is: Is it possible for a function f(x) defined for all x to have the properties i) f(x) > 0, ii) f (x) < 0, and iii) f (x) < 0? The answer is no.

8 8 MATH 18.01, FALL PROBLEM SET # 3A Figure 1. Section 4.1: 20 a) Here is an explanation. By assumption, f (0) is equal to some negative number N. Since f (x) < 0, f (x) is a decreasing function for all x. Therefore, f (x) < N holds for all x > 0. Let us give a geometric interpretation of this. Let L be the tangent line to the graph of f at x = 0. Then L has a (negative) slope equal to N and is above the x axis at x = 0, so L must intersect the x axis at some point x 0 > 0. Since f (x) < N holds for all x > 0, the graph of f is more negatively sloped than L for all x > 0. It follows that the graph of f lies below the line L for all x > 0. Therefore, the graph of f(x) must intersect the x axis at some x value in between x = 0 and x 0. This contradicts property i) and therefore shows that the three properties are not mutually compatible. 18) Note that x 2 + y 2 = a 2 describes a circle of radius a centered at the origin. We differentiate each side of the equation with respect to x and use the chain rule relation (d/dx)(y 2 ) = 2yy to obtain 2x + 2yy = 0. We then differentiate the above equation with respect to x to obtain

9 MATH 18.01, FALL PROBLEM SET # 3A 9 Figure 2. Section 4.1: 20 b) We now use the above equation to solve for y : 2 + 2(y ) 2 + 2yy = 0. y = (y ) 2 1 y y. Note that the above formula agrees with the graph of the circle: above the x axis, y is positive and the circle is concave down. This agrees with the above formula, which shows that y < 0 when y > 0. On the other hand, below the x axis, y is negative and the circle is concave up. This also agrees with the above formula, which shows that y > 0 when y < 0.

10 10 MATH 18.01, FALL PROBLEM SET # 3A Figure 3. Section 4.2: 12

MATH 18.01, FALL PROBLEM SET # 2

MATH 18.01, FALL PROBLEM SET # 2 MATH 18.01, FALL 2012 - PROBLEM SET # 2 Professor: Jared Speck Due: by Thursday 4:00pm on 9-20-12 (in the boxes outside of Room 2-255 during the day; stick it under the door if the room is locked; write