MATH 100, Section 110 Midterm 2 November 4, 2014 Page 1 of 7

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1 MATH 00, Section 0 Midterm 2 November 4, 204 Page of 7 Midterm 2 Duration: 45 minutes This test has 5 questions on 7 pages, for a total of 40 points. Read all the questions carefully before starting to work. Q and Q2 are short-answer questions; put your answer in the boes provided. All other questions are long-answer; you should give complete arguments and eplanations for all your calculations; answers without justifications will not be marked. Continue on the back of the previous page if you run out of space. Attempt to answer all questions for partial credit. This is a closed-book eamination. None of the following are allowed: documents, cheat sheets or electronic devices of any kind (including calculators, cell phones, etc.) First Name: Last Name: Student-No: Section: Signature: Question: Total Points: Score: Student Conduct during Eaminations. Each eamination candidate must be prepared to produce, upon the request of the invigilator or eaminer, his or her UBCcard for identification. 2. Eamination candidates are not permitted to ask questions of the eaminers or invigilators, ecept in cases of supposed errors or ambiguities in eamination questions, illegible or missing material, or the like. 3. No eamination candidate shall be permitted to enter the eamination room after the epiration of one-half hour from the scheduled starting time, or to leave during the first half hour of the eamination. Should the eamination run forty-five (45) minutes or less, no eamination candidate shall be permitted to enter the eamination room once the eamination has begun. 4. Eamination candidates must conduct themselves honestly and in accordance with established rules for a given eamination, which will be articulated by the eaminer or invigilator prior to the eamination commencing. Should dishonest behaviour be observed by the eaminer(s) or invigilator(s), pleas of accident or forgetfulness shall not be received. 5. Eamination candidates suspected of any of the following, or any other similar practices, may be immediately dismissed from the eamination by the eaminer/invigilator, and may be subject to disciplinary action: (i) speaking or communicating with other eamination candidates, unless otherwise authorized; (ii) (iii) (iv) (v) purposely eposing written papers to the view of other eamination candidates or imaging devices; purposely viewing the written papers of other eamination candidates; using or having visible at the place of writing any books, papers or other memory aid devices other than those authorized by the eaminer(s); and, using or operating electronic devices including but not limited to telephones, calculators, computers, or similar devices other than those authorized by the eaminer(s) (electronic devices other than those authorized by the eaminer(s) must be completely powered down if present at the place of writing). 6. Eamination candidates must not destroy or damage any eamination material, must hand in all eamination papers, and must not take any eamination material from the eamination room without permission of the eaminer or invigilator. 7. Notwithstanding the above, for any mode of eamination that does not fall into the traditional, paper-based method, eamination candidates shall adhere to any special rules for conduct as established and articulated by the eaminer. 8. Eamination candidates must follow any additional eamination rules or directions communicated by the eaminer(s) or invigilator(s).

2 MATH 00, Section 0 Midterm 2 November 4, 204 Page 2 of 7 Short-Answer Questions. Questions and 2 are short-answer questions. Put your answer in the bo provided. Full marks will be given for a correct answer placed in the bo. Show your work also, for part marks. Each part is worth 2 to 3 marks, and not all parts are of equal difficulty. Simplify your answers as much as possible in Questions and 2. 2 marks. (a) Let y = 0 2. Find dy. d Answer: dy d = 2 ln(0) 0 2 Solution: Use the chain rule dy d = ln(0) 0 2 ( 2) 2 marks (b) Evaluate tan ( tan ( )) π. 4 Answer: π 4 Solution: As done in class, one way to solve this problem is to draw a unit circle and mark down the angle of π in the second quadrant. Shifting the angle back 4 into the range of ( π, ) π 2 2 yields π because tan is negative in the second and 4 fourth quadrant. 2 marks (c) y = (sin ) ln. Find dy. d Answer: y = ( ln(sin ) Solution: Use logarithmic differentiation. ) + ln cos y sin ln y = ln ln(sin ) y y = ln(sin ) + ln sin cos Multiplying through by y gives the answer in the bo.

3 MATH 00, Section 0 Midterm 2 November 4, 204 Page 3 of 7 3 marks 2. (a) Simplify sin(sec ). Note: sec is the inverse function of sec. Answer: 2 Solution: Let y = sec then sec y = or equivalently cos y =. Drawing a triangle with angle y as done in class gives sin y = 2. 3 marks (b) Find all -coordinates at which the tangent line to the curve 2 y 2 = 6 has slope.. Solution: Implicit differentiation gives Plugging in y = and factoring gives 2y yy = 0. 2y(y ) = 0. Answer: = ±2 This implies = y since there are no points on the curve with = 0 or y = 0. The curve s equation thus becomes 4 = 6 or = ±2. 3 marks (c) Which of the following functions satisfies dp dt = p3? A) p(t) = 4 t4 B) p(t) = t+3 C) p(t) = 2t Answer: C) D) None of these. Solution: By differentiating each epression for p(t) separately, we conclude that C) is the correct answer, i.e, dp dt = 2( 2t) 3/2 ( 2) = ( 2t) 3/2 = ( ) 3 = p 3 2t

4 MATH 00, Section 0 Midterm 2 November 4, 204 Page 4 of 7 3 marks (d) Find the absolute maimum and absolute minimum values of y = + on [0.2, 4]. Answer: ma. value is y(0.2) = 5.2, min. value is y() = 2 Solution: Use the closed interval method. Start by finding critical numbers by checking where the derivative equals zero or isn t defined: y = 2 = 0 So = 0, ±. Ignore 0 and since they aren t in [0.2, 4]. We compare to get the answer in the bo. y(0.2) = y() = 2 y(4) = marks (e) Write down the degree 2 Taylor polynomial of f() = tan at = π/4. Answer: T 2 () = + 2( π/4) + 2( π/4) 2 Solution: We need to determine T 2 () = 2 c k ( π/2) k k=0 where c k = f (k) (π/2). We have k! and so f (0) () = tan f () = cos 2 which gives the answer in the bo. c 0 = c = 2 c 2 = 2 f () = 2 tan cos 2

5 MATH 00, Section 0 Midterm 2 November 4, 204 Page 5 of 7 Full-Solution Problems. In questions 3 5, justify your answers and show all your work. If a bo is provided, write your final answer there. Unless otherwise indicated, simplification of answers is not required in these questions. 3 marks 3. (a) Use a suitable linear approimation to estimate the value of Epress your answer as a single fraction. Answer: 79 9 Solution: Let f() = 2/3 and epand around = 27. Then f(27) = 9 f () = 2 3 /3 f (27) = 2 9 T () = f(27) + f (27) ( 27) = ( 27) = and thus T (26) = ( ) = marks (b) Show that the upper bound for the error in this approimation is less than 000. Solution: The error is given by E = 2! f (c)(26 27) 2 = 2 f (c) for some 26 c 27. Now f () = 2 9 4/3 has a negative power of which is decreasing but then the minus sign outside makes f () increasing. Thus the right endpoint of the interval [26, 27] makes f () largest and we obtain f () = 2 9 4/ /3 for So the error is bounded by = = as required. E = 729 < 000

6 MATH 00, Section 0 Midterm 2 November 4, 204 Page 6 of 7 4. A tanker off the coast of B.C. breaks in half spilling 0 4 m 3 of oil all at once. The oil floats on the surface and spreads. It has the shape of a cylinder whose height is becoming thinner and thinner. The height, h, of the oil slick is uniform and given by where t is time in hours. h(t) = 0 4 t 2 marks (a) Find the radius of the slick 4 hours later. Solution: Let t be time measured in hours, r be the radius of the cylinder, and V be the volume of the cylinder. Then V = πr(t) 2 h(t). Thus r(t) = t π and r(4) = π = 20, 000 π. 5 marks (b) Find how quickly the slick is spreading across the water surface 4 hours later. Solution: The quick way is to just differentiate the above epression for r(t) to get r (t) = 04 2 π t /2 so r (4) = 04 4 π = 2, 500 π. The long (and tedious) way is to differentiate the volume formula implicitely (treating V as a constant): 0 = π(2rr h + r 2 h ) ( ) We need h (t) = so that h (4) = 0 4 t Plugging t = 4 into equation ( ) and dividing by π we get and solving for r : 0 = π r π 08 r = 04 π 4π r π = 04 4π = 04 4 π ( = 2, 500 π. ) = r π 04 4π

7 MATH 00, Section 0 Midterm 2 November 4, 204 Page 7 of 7 5 marks 5. Naobi is boiling water on the stove to make a cup of coffee. The heating element is at 200 C and the water in her pot is initially at 20 C room temperature. After one minute her water has reached 40 C. How long will it take for her water to reach the boiling point? Your answer may be left in calculator-ready form. Solution: Let T be the temperature of the can in degree Celsius and t be the time in minutes. We need to fit the curve T (t) = Ae kt + E to the given data where E is the ambient temperature. In this case the ambient temperature is that of the heating element, i.e., 200 C. A = T (0) E = = 80. The other data point is t = with T () = 40. Plugging this into the equation we have thus far we get Solving for k yields k = ln 8 9. Thus 40 = 80e k T (t) = 80e ln( 8 9)t Solving 00 = 80e ln( 8 9)t for t we find t = ln 5 9 k 5 min.

Mathematics Page 1 of 9 Student-No.:

Mathematics Page 1 of 9 Student-No.: Mathematics 5-95 Page of 9 Student-No.: Midterm Duration: 8 minutes This test has 7 questions on 9 pages, for a total of 7 points. Question 7 is a bonus question. Read all the questions carefully before