Math 221 – Exam II (90 minutes) – Tuesday March 19, 2002AnswersI. (40 points.) One of the three parts of the Monotonicity Theorem sayswhat happens when the derivative of a function is positive on an interval.State and prove it. In your proof you may use (without proof) the MeanValue Theorem.Answer: Theorem. If f0(x) > 0 for all x in an interval I, then f isincreasing on that interval, i.e.x1< x2=⇒ f(x1) < f(x2)for any two points x1, x2of the interval.Proof. Choose x1and x2in the interval I with x1< x2. By the Mean ValueTheorem there is a c with x1< c < x2andf(x2) − f(x1)x2− x1= f0(c). (#)Since c is between x1and x2it lie in the interval I and hence f0(c) > 0.Hence the ratio on the left in equation (#) is positive. Since x1< x2thedenominator x2− x1is positive and hence the numerator f(x2) − f(x1) ispositive, i.e. f(x2) − f(x1) > 0. Thus f(x1) < f(x2). This theorem andproof appeared (in slightly more generality) on the handout entitled Possiblequestions for Exam II.Grader’s commentsThe responses on this question ranged from none at all to utter confusionto perfectly clear and correct. I suppose some students did not get the handoutor come to the lecture where I told the students what to expect. Some studentsproved the wrong theorem indicating (to me at least) that they were memoriz-ing rather than understanding. Other students wrote misguided things like f0is increasing rather than f is increasing; perhaps these were just dumb mis-takes and these students really understand. I did not penalize the followingtwo common errors heavily:(i) Not saying that there a number c between x1and x2satisfying (#), i.e.leaving the impression that the Mean Value Theorem holds for all c.(ii) Saying that the interval was a ≤ x ≤ b and proving that f(a) < f(b).This mistake could be defended by pointing out that since the proofworks for all intervals, the result is the same.1I don’t know if asking students to produce proofs on an exam has anyvalue. My opinion is that students have trouble reading and learning mathe-matics on their own and this kind of question will force them to read and writecarefully. These are two of the principal objectives of a college eduction.II. (20 points.) Evaluate the following limits. If the limit does not existwrite DNE and say why. Distinguish between a limit which is infinite andone which does not exist.(i) limθ→πsin θπ − θ.Answer: Plugging in gives the indeterminate form 0/0 so by l’Hˆopital’sRulelimθ→πsin θπ − θ= limθ→πcos θ−1= 1.This also follows directly from the definition of the derivative. Since sin π = 0we havelimθ→πsin θπ − θ= − limθ→πsin θ − sin πθ − π= −sin0(π) = −cos(π) = 1.(This is Problem 11 on page 165.)(ii) limx→0µ1sin x−1x¶.Answer: Plugging in gives the indeterminate form ∞−∞ so we need to dosome high school algebra first.limx→0µ1sin x−1x¶= limx→0x − sin xx sin x.Now plugging in gives the indeterminate form 0/0 so by l’Hˆopital’s Rule twicelimx→0x − sin xx sin x= limx→01 − cos xsin x + x cos x= limx→0sin x2 cos x − x sin x=02 − 0= 0.(This is Example 7 on page 165.)Grader’s commentsThe most common mistakes I saw were 1) the limit of a quotient equalsthe quotient of the limits, despite the fact that both the numerator and denom-inator tend to zero; 2) not connecting statements with equal signs; 3) saying0/2 is infinite.2III. (30 points.) A point moves along the curve y2= x3in such a waythat its distance from the origin increases at a constant rate of two units persecond. Find dx/dt at the point (x, y) = (2, 2√2).Answer: Call the distance from the origin r. Then r =px2+ y2by thePythagorean Theorem and we are given that dr/dt = 2 for all t. Sincey = x3/2near the point (x, y) = (2, 2√2) we haver =√x2+ x3.By the Chain Ruledrdt=(2x + 3x2)2√x2+ x3·dxdtsodxdt=2√x2+ x3(2x + 3x2)·drdtand hencedxdt¯¯¯¯x=2=2√4 + 8(4 + 12)· 2 =√32.(This is problem 23 on page 172.)Grader’s commentsOnly few students solved this problem correctly. Most students did notuse the distance formula and started differentiating both sides of the equationy2= x3. As a result, the average score on this problem was very low.Professors’s ResponseSome students complained that they did not know I would take problemsfrom the miscellaneous problems at the end of each chapter, but this is clearlystated in the policy statement of the syllabus.3IV. (20 points.) Sketch the graph y = f(x) of a function f which is twicecontinuously differentiable, and has the following characteristics: f0(x) < 0for |x| < 2, f0(x) > 0 for |x| > 2, f00(x) < 0 for x < 0, f00(0) > 0 for x > 0,f(−2) = 8, f (0) = 4, f(2) = 0. Draw the tangent line at each point ofinflection. (That f is twice continuously differentiable means that the secondderivative f00(x) is defined for all x and is continuous.)Answer:2468–3 –2–1 12 3The function plotted is f(x) = 4(x3/3 − 4x + 16/3)/3 which has all theabove properties, but the only three points that we know for sure are on thegraph of the problem as stated are the three given points (−2, 8), (0, 4), and(2, 0). (This is problem 19 from page 138.)4V. (30 points.) An isosceles triangle is drawn with its vertex at the originand its other two vertices symmetrically placed on the curve 12y = 36 − x2and above the x-axis. Determine the area of the largest such triangle.Answer: The vertices of the triangle are at the points(x, y) = (0, 0),µx,36 − x212¶,µ−x,36 − x212¶.Hence the base is 2x, the altitude is (36 − x2)/12, so the area isA =x(36 − x2)12= 3x −x312.Clearly A(0) = A(6) = 0 and the problem asks us to consider only thosetriangles with (36 − x2)/12 ≥ 0. i.e. 0 ≤ x ≤ 6. Hence A(x) ≥ 0 so themaximum occurs at an interior point so at the maximum0 =dAdx= 3 −x24.Therefore the maximum occurs at x =√12 and the maximum value isA(√12) =√12(36 − 12)12.(This is problem 40 page 172.)Grader’s commentsThe problem most students had was to set up a (useful) formula for thearea of the triangle. Many students didn’t get to this point, but the oneswho did were able (usually) to get the relevant critical point and the maximalarea asked for (a common mistake was to get just half the area, because theyset up the problem like that). What very few students did was explain whythe maximum (exists and) occurs when the derivative was zero. I took off 5points for this lack of explanation, which I think serious.Professors’s