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 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 (30 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 4. (Thurs., Sept. 13) Chain rule; higher derivatives. Read: 3.3, 3.6. Homework: Notes 1F: 1ab, 2, 6, 7bd. Homework: Notes 1J: 1abf. Homework: Notes 1G: 1b, 5ab. Lecture 5. (Fri., Sept. 14) Implicit differentiation; inverse functions. Read: 3.5; Notes G: Section 5; 9.5 (bottom pp. 313-315). Homework: Notes 1F: 3, 5, 8c. 1
2 MATH 18.01, FALL 2012 - PROBLEM SET # 2 Homework: Notes 1A: 5b. Homework: Notes 5A: 1abc (just sin, cos, sec), 3fh. Lecture 6. (Tues., Sept. 18) Exponentials and logs: definitions, algebra, applications; logarithmic differentiation; hyperbolic functions. Read: Notes: Section X (8.2 has some of this); 8.3 to middle p. 267; 8.4 to top p. 271. Read: 9.7 to p. 326. Homework: Notes 1H: 1, 2, 3a, 5b. Homework: Notes 1I: 1cdefm, 4a. Homework: Notes 5A: 5abc (in 5a, you don t have to find the critical points or inflection points; just provide a rough sketch).
MATH 18.01, FALL 2012 - PROBLEM SET # 2 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 dont 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. (now; slope, basic curve sketching; 1 + 2 = 3 points) a) Find an even function E(x) and an odd function O(x) such that the function f(x) = x 2 /(x + 1) can be decomposed as f(x) = E(x) + O(x). Solution: As we saw in the previous problem set, we have E(x) = 1 (f(x) + f( x)) and O(x) = 2 1 (f(x) f( x)). Therefore, we compute that E(x) = 1(f(x)+f( x)) = 1 2 2 2 (x2 /(x + 1) + x 2 /( x + 1)) = x 2 /(1 x 2 ) and O(x) = 1(f(x) f( x)) = 1 2 2 (x2 /(x + 1) x 2 /( x + 1)) = x 3 /(1 x 2 ). b) Graph E(x). In the same picture, also provide a rough sketch of the graph of E (x) without performing any computations (except to check your answer if you want). Then do the same thing for O(x) and O (x). Solution: See Figure 1 and Figure 2 below. 2. (Sept. 13; chain rule, higher derivatives; 4 points) A function f(x) is said to be concave up at x 0 if f (x 0 ) > 0 (we will study this notion in detail later in the course). Suppose that f(x) is a differentiable function with f(x 0 ) > 0 for all x 0. Suppose in addition that f(x) is concave up at all x 0. Show that the function g(x) = (f(x)) 2 is concave up at every x 0. Solution: We use the chain rule to compute that g (x) = 2f(x)f (x). We then differentiate this equation with respect to x and use the product rule to compute that g (x) = 2f (x) f (x) + 2f(x)f (x) = 2(f (x)) 2 + 2f(x)f (x). By assumption, f(x 0 ) > 0 and f (x 0 ) > 0 holds for all x 0. Therefore, g (x 0 ) 2f(x 0 )f (x 0 ) > 0 holds for all x 0. This is the desired conclusion.
4 MATH 18.01, FALL 2012 - PROBLEM SET # 2 Figure 1. Graphs of E(x) (black) and E (x) (blue) 3. (Sept. 14; chain rule and inverse trig differentiation; 3 + 3 = 6 points) Compute the following derivatives: a) (d/dx)[arctan(x 3 + 2x + 1)] 4. Solution: We set w = x 3 + 2x + 1, v = arctan w, and u = v 4. Then by the chain rule, we have that (d/dx)[arctan(x 3 + 2x + 1)] 4 = (d/dx)u = du dv dv dw dw dx 1 1 + w 2 (3x2 + 2) = 4v 5 = { 4[arctan(x 3 + 2x + 1)] 5} { 1 1 + (x 3 + 2x + 1) 2 } (3x 2 + 2).
MATH 18.01, FALL 2012 - PROBLEM SET # 2 5 Figure 2. Graphs of O(x) (black) and O (x) (blue) b) (d/dy) ( cos 2 y sin 2 y ). Do this is in two different ways: i) First do it directly. ii) Then use a trig identity to write cos 2 y sin 2 y = f(2y) for some trig function f (and then compute (d/dy)f(2y)). Show that your two answers agree. Solution: i) We use the chain rule to compute that d ( cos 2 y sin 2 y ) = d ( cos 2 y ) d ( sin 2 y ) dy dy dy = 2 cos y( sin y) 2 sin y(cos y) = 4 sin y cos y. ii) Our second solution uses both of the trig identities cos 2 y sin 2 y = cos(2y) and sin(2y) = 2 sin y cos y. We now use the chain rule and these two identities to compute that
6 MATH 18.01, FALL 2012 - PROBLEM SET # 2 which agrees with our first solution. d ( cos 2 y sin 2 y ) = d dy dy cos(2y) d = sin(2y) dy (2y) = 2 sin(2y) = 4 sin y cos y, 4. (Sept.14; implicit differentiation; slope; 5 + 7 = 12 points) Consider the following curve in the (x, y) plane: (1/3)y 3 + xy 2 + x 3 + y + x = 5/3. a) Find all of the points (if there are any...) P = (x 0, y 0 ) on the curve such that the tangent line to the curve at P is horizontal. Solution: We first differentiate both sides of the equation with respect to x and use implicit differentiation (we view y as a function of x and use the notation y = dy/dx) to deduce that We then solve for y : y 2 y + 2yy x + y 2 + 3x 2 + y + 1 = 0. y = 3x2 + y 2 + 1 y 2 + 2xy + 1. The points P of interest are exactly the points (x 0, y 0 ) on the curve such that y (x 0 ) = 0. However, because the numerator 3x 2 + y 2 + 1 in the formula for y is strictly positive, it is never possible that y = 0. So there are no points P. b) Find the one point Q on the curve such that the tangent line to the curve at Q is vertical. Remark: You only have to find the one point; you don t have to prove that there is only one Q. In order to prove that there is only one, you need some tools that we haven t yet developed. Solution: We can view x as an implicit function of y (i.e., x = g(y)). The points Q of interest where the tangent line is vertical are exactly the points (x 0, y 0 ) on the curve where dx dy y 0 = 0. Performing a similar calculation to the one from part a), we deduce that (0.0.1) dx dy = + 2xy + 1 y2 3x 2 + y 2 + 1. Thus, to find the points Q of interest, we set the numerator equal to 0 : y 2 +2xy+1 = 0. In addition, in order for such points Q to lie on the curve, they must satisfy the equation (1/3)y 3 +xy 2 +x 3 +y+x = 5/3. In other words, the points Q of interest simultaneously satisfy the equations y 2 + 2xy + 1 = 0, (1/3)y 3 + xy 2 + x 3 + y + x = 5/3.
MATH 18.01, FALL 2012 - PROBLEM SET # 2 7 To solve the above system, we use the first equation to substitute (2xy + 1) for y 2 in the second equation to derive the equation (2/3)xy 2 2x 2 y + x 3 + (2/3)y = 5/3. We again substitute (2xy + 1) for y 2 in the new equation to deduce that (2/3)x 2 y + (2/3)x + x 3 + (2/3)y = 5/3. We now note that the equation y 2 + 2xy + 1 = 0 can be solved via the quadratic formula to yield y = x ± x 2 1. In particular, this equation has real solutions if and only if x 2 1. We now substitute this expression for y into the previous equation to deduce that (5/3)x 3 ± (1 x) x 2 1 5/3 = 0. The above equation has the solution x 0 = 1. The equation y = x ± x 2 1 yields that the corresponding y value is y 0 = 1. In summary, Q = (x 0, y 0 ) = (1, 1) is a solution to the system. Later in the course you will have the tools needed to show that it is the only solution. 5. (Sept. 14; inverse trig functions and differentiation; 2 + 2 + 2 = 6 points) The function θ = cos 1 (x) is the inverse of cos θ for 0 θ π. Similarly, the function φ = sin 1 (x) is the inverse of sin φ for π/2 φ π/2. a) Use implicit differentiation to find a formula for (d/dx) cos 1 (x). Pay particular attention to the sign of the square root. Solution: If θ = cos 1 (x), then cos θ = x. We want to compute d/dx cos 1 (x), or equivalently, θ (x). We now differentiate the equation cos θ = x with respect to x and use implicit differentiation (where θ is viewed as a function of x) to deduce that (d/dx) cos θ = (sin θ)θ = (d/dx)x = 1. We then solve for θ to deduce that θ = 1/ sin θ. We need to express the right-hand side in terms of x. To this end, we note the identity sin 2 θ + cos 2 θ = 1. Since cos θ = x, we have sin 2 θ = 1 x 2. Note that when we are computing θ = cos 1 (x), we are considering only those θ values with 0 θ π. For such θ, we have sin θ 0. Therefore, we can use the previous equation to solve for sin θ in terms of x as follows: sin θ = 1 x 2. Plugging this into the formula θ = 1/ sin θ, we finally conclude that θ (x) = d/dx cos 1 (x) = 1/ 1 x 2. b) Simplify the expression sin 1 (x) + cos 1 (x) as much as possible. Solution: Because cos(θ) = sin(π/2 θ), if θ = cos 1 (x), then sin(π/2 θ) = cos(θ) = x. Since 0 θ π we have π/2 π/2 θ π/2, so sin 1 (x) = π/2 θ. Therefore, sin 1 (x)+cos 1 (x) = π/2. c) Explain how you can use parts a) and b) to quickly derive a formula for (d/dx) sin 1 (x) by doing very little additional work (you should be able to do the computation in your head). Solution: Since sin 1 (x)+cos 1 (x) is constantly π/2 by part b), we have that (d/dx) { sin 1 (x) + cos 1 (x) } = 0. Therefore, (d/dx) sin 1 (x) = (d/dx) cos 1 (x) = 1/ 1 x 2. 6. (Sept. 18; logarithmic differentiation; 2 + 2 + 2 = 6 points)
8 MATH 18.01, FALL 2012 - PROBLEM SET # 2 a) Given two functions u and v, set y = uv. Note the identity log y = log u + log v. Use this identity and logarithmic differentiation to give an alternative derivation (i.e., different than the derivation we discussed in class) of the product rule (uv) = u v + uv. Solution: We differentiate the relation log y = log u + log v with respect to x to deduce that Therefore, y y = d dx log y = d u (log u + log v) = dx u + v v. { } u y = y u + v v { } u = uv u + v v = u v + v u. b) Modify the approach from part a) in order to give an alternative derivation of the quotient rule ( u v ) = u v uv v 2. Solution: We set y = u/v and note the identity log y = log u log v. Therefore, we can repeat the calculations from part a) and change the + sign to a sign in the appropriate spots to deduce that { } u y = y u v v { } u = u/v u v v = u /v uv /(v 2 ) = u v uv v 2. c) Suppose now that you are given n functions u 1 (x), u 2 (x), u n (x). Modify the approach from part a) in order to derive a formula for the derivative of the product u 1 u 2 u n, i.e., a formula for (d/dx)(u 1 u 2 u n ). Solution: Set y = u 1 u 2 u n and note the identity log y = log u 1 + log u 2 + + log u n. We then differentiate this relation and use to the chain rule to deduce that Therefore, y y = d dx log y = d dx (log u 1 + log u 2 + + log u n ) = u 1 + u 2 + + u n. u 1 u 2 u n
MATH 18.01, FALL 2012 - PROBLEM SET # 2 9 which is the desired relation. { } u y = y 1 + u 2 + + u n u 1 u 2 u n { u = u 1 u 2 u 1 n + u 2 + + u n u 1 u 2 u n = u 1u 2 u n + u 1 u 2 u n + + u 1 u 2 u n 1 u n, 7. (Sept. 18; 2+2+2+2+2 = 10 points) Section 8.2: 8ac, 10, 11; Section 8.4: 18, 20 (only compute dy/dx). Solution (Section 8.2): 8a) Let M 1 = (2/3) log 10 (E 1 /E 0 ), M 2 = (2/3) log 10 (E 2 /E 0 ), where E 1 and E 2 are the energies released by the two earthquakes. Let us assume that E 2 > E 1. Using the information given in the problem, we have that 1 = M 2 M 1. We deduce that log 10 (E 2 /E 0 ) log 10 (E 1 /E 0 ) = 3/2. Using the properties of log, it follows that the left-hand side of the previous equation is equal to log 10 (E 2 /E 1 ). Therefore, we can raise 10 to both sides of the equation and use the identity 10 log 10 x = x to deduce that E 2 /E 1 = 10 log 10 (E 2/E 1 ) = 10 3/2 31.62. 8c) If M 2 = 6 and E 0 = 7 10 3, then we use straightforward algebra to deduce that E 2 = E 0 10 (3/2)M 2 = 7 10 3 10 9 = 7 10 6 kwh. For a city that uses 3 10 5 kwh of electric energy every day, the magnitude 6 earthquake would provide 7 10 6 /(3 10 5 ) = 23 + (1/3) days supply of energy. 10) Suppose that log 3 2 = p/q, where p, q are positive integers. Then q log 3 2 = p. The basic properties of log imply that log 3 2 q = p. We raise 3 to both sides of the equation to deduce that 2 q = 3 log 3 2q = 3 p. This is impossible since 2 q is even and 3 p is odd. Therefore, it is not possible to express log 3 2 = p/q, where p, q are positive integers. 11) Note that log(1/2) = log 2 < 0. Therefore, 1 log(1/2) > 2 log(1/2). Hence, the mistake in the book s proof lies in the very first line. Solution (Section 8.4): 18) Since y = {(x + 1)(x 2)(2x + 7)} 1/3, we use the basic properties of log to deduce that log y = 1 {log(x + 1) + log(x 2) + log(2x + 7)}. We now use logarithmic differentiation and the chain rule 3 to deduce that } Therefore, y y = d dx log y = 1 d {log(x + 1) + log(x 2) + log(2x + 7)} 3 dx = 1 { 1 3 x + 1 + 1 x 2 + 2 }. 2x + 7 y = y d dx log y = 1 { 1 3 {(x + 1)(x 2)(2x + 7)}1/3 x + 1 + 1 x 2 + 2 }. 2x + 7
10 MATH 18.01, FALL 2012 - PROBLEM SET # 2 20) Since y = x x, we use the basic properties of log to deduce that log y = x log x. We then use logarithmic differentiation, the chain rule, and the product rule to deduce that Therefore, y y = d dx log y = x d dx log x + ( d x) log x dx = x 1 + 1 log x = 1 + log x. x y = y d dx log y = (1 + log x)xx.