Solutions to Math 41 Final Exam December 9, 2013

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1 Solutions to Math 4 Final Eam December 9,. points In each part below, use the method of your choice, but show the steps in your computations. a Find f if: f = arctane csc 5 + log 5 points Using the Chain Rule, we get: f = + e e 5 csc 4 csc cot + ln 4 = e + e + 5 csc5 cot + ln. b Find dy d in terms of and y if: y + y = 6 5 points We use implicit differentiation: dy d + y + dy = = dy + y d d = y y y = dy d = y y y + y = y y +.

2 Math 4, Autumn Solutions to Final Eam December 9, Page of 4. 9 points A tall, open pot has the shape of a cylinder, with a circular base of radius 4 in. A marble with radius r inches, where < r < 4, is placed in the pot, and the pot is filled with just enough water to cover the marble completely. What radius of marble requires the greatest amount of water to accomplish this? Show all your reasoning. We wish to compute the volume of water V w as a function of the radius r of the marble. The marble, combined with the water that encloses it, form a cylinder of height r and radius R = 4. Therefore the volume of water will be the volume of the cylinder minus the volume of the marble: V w r = rπ4 4π r = πr 4π r We want to find the maimum of this function on the interval, 4, so we find the critical points of V w : V wr = π 4πr = 4π8 r V wr is defined everywhere and has zeroes in the domain, 4 only at r = 8. Note that V wr > if < r < 8 and V wr < if 8 < r < 4, therefore by the first derivative test for global etrema, r = 8 will be a global maimum. The greatest amount of water is needed for r = 8 in.

3 Math 4, Autumn Solutions to Final Eam December 9, Page of 4. points Find each of the following its, with justification. If there is an infinite it, then eplain whether it is or. cos a sin 5 points Direct substitution gives us. So we can use L Hopital s rule cos sin = cos sin sin + cos Direct substitution for the it on the right hand side gives. So we can again use L Hopital s rule cos sin sin + cos = sin + cos cos cos + cos sin = sin + cos cos cos + cos sin = b / 5 points Let y = / Since hand side. Then ln y = ln / = ln ln/ = ln = and = we can use L Hopital s rule to find the it on the right Thus / = y = e =. ln = = =

4 Math 4, Autumn Solutions to Final Eam December 9, Page 4 of points Let f = e /. a Determine, with complete reasoning, whether f has any asymptotes horizontal or vertical, and give their equations. Compute both one-sided its for any vertical asymptotes. 5 points To find the horizontal asymptotes, we need to compute We have = and =. Hence e / = e = and Similarly we have Since neither f nor = and e / =. e / =. e / and =, so e / = and e /. f is finite, it follows that f has no horizontal asymptotes. To find the vertical asymptotes of f, first note that the domain of f is { }, and that since f is the product of a polynomial and an eponential, it follows that f is continuous at every point of this domain. So the only point we need to look for a vertical asymptote of f is at =, and this depends on finding the two one-sided its e / and + e /. We have that + =, so e/ =, + and =. + e/ Since also =, it follows that + e / = =. + Now we need to find e /. Since =, it follows that e / =. So e / is a it of the indeterminate type. substitution t =, so as, we have t and e / = et t t = e t t t. = and We make the Since this is a it of the indeterminate type, we can use l Hospital and get that e t e / = t t e t = t =. Since e / =, it follows that = is a vertical asymptote of f, and hence the only vertical asymptote.

5 Math 4, Autumn Solutions to Final Eam December 9, Page 5 of 4 b On what intervals is f increasing? decreasing? Eplain completely. Note: it is a fact that f = + e /, which you do not have to prove. 5 points The intervals of increase and decrease are determined by the sign of f. Since e / > for all, it follows that the sign of f is the sane as the sign of +. Notice that + = for =. When <, we have >, so + > and f >. When < <, we have <, so + < and f <. When >, we have >, so + > >, so f >. This means that f is increasing on,, and f is decreasing on the interval,. c On what intervals is f concave upward? downward? Eplain completely. 4 points The intervals of concavity are determined by the sign of f. Using the product rule to find f and the fact that e / = e / and + =, we get that f = e / + + e /, so f = e / + + = e / = e /. Since e / > for all, it follows that the sign of f is the same as the sign of. Hence, for >, we have that f > and for < we have f <. This means that f is concave up on, and concave down on,. d Using the information you ve found, sketch the graph y = f. Label and provide, y coordinates of any local etrema and inflection points. 4 points Note that, is not an inflection point, because is not in the domain of f there is a hole in the graph here, as f =. +

6 Math 4, Autumn Solutions to Final Eam December 9, Page 6 of 4 5. points For this problem, we try to solve the equation + =. a Show that the equation has a solution in the interval,. Eplain your reasoning completely; however, you don t have to find the eact value. 5 points Since f = + is a polynomial it is continuous on [, ]. So we can use Intermediate Value Theorem. Since f = < < = f, we deduce that there is c in, such that fc = i.e. + = has a solution in,. b Use Newton s method with initial guess = to produce the net approimation to the solution. Show all of your steps and simplify completely. points By Newton s method n+ = n fn f n for all n. Since f = 6, for all n. So if =, then n+ = n n n + n 6 n = + 6 = = c Now use your result from b to produce the net approimation. You do not need to simplify your answer. points = + 6 = + 6 d Eplain what happens if Newton s method is instead used with the initial guess =. points If = then f =. This means it is impossible to compute. Thus Newton s method is inapplicable in this case.

7 Math 4, Autumn Solutions to Final Eam December 9, Page 7 of 4 6. points A thin, horizontal copper rod is being subjected to various sources of heating and cooling along its length. At each position, where is measured in inches from the left end of the rod, suppose h represents the rate, in Celsius degrees per inch, at which the rod s temperature is changing with respect to changes in position. Approimate values of h were obtained from a heat flu analysis of the rod and listed in the chart below: h a Without calculating it, what does the quantity h d represent? Epress your answer in terms relevant to this situation, and make it understandable to someone who does not know any calculus. Be sure to use any units that are appropriate, and also eplain what the sign of this quantity would signify. 5 points The quantity hd represents the difference between the temperature of the point on the rod inches to the right of the left end and the temperature of the left end of the rod; it is measured in Celsius degrees. If the quantity is positive, then the temperature of the point on the rod inches to the right of the left end is larger than the temperature of the left end of the rod. If the quantity is negative, the temperature of the point on the rod inches to the right of the left end is less than the temperature of the left end of the rod. b Use the Right Endpoint Rule with n = to estimate 9 h d; give your answer as an epression in terms of numbers alone, but you do not have to simplify it. points We have n = so = 9 =. We use the Right Endpoint Rule, so = =, = = 6, and = = 9. 9 hd h + h + h = h + h6 + h9 = + + c Use the Midpoint Rule with n = to estimate h d; again give your final answer in terms of numbers alone. points We have n = so = = 4. We use the Midpoint Rule, so = + and = + = 6+ = 8. hd h + h = h4 + h8 = = +6 = 4

8 Math 4, Autumn Solutions to Final Eam December 9, Page 8 of points a Suppose g = 4. Let R be the region in the y-plane bounded by the -ais and the curve y = g, for 4. Find the area of R by evaluating a it of a Riemann sum that uses the Right Endpoint Rule; show all reasoning. 7 points By definition, the area of R is the it of the Riemann sum: Area of R = g i. Since we are using right endpoints, we take i = i. By graphing y = g, one sees that the bounds of integration are a = and b = 4. Thus, using the formulas = b a n and i = a + i, we find Therefore, so i= = b a n = 4 n, i = a + i = + i 4 n = 4i n. g i = i 4 i = 4i 4 4i = 6i i, n n n n Area of R = = g i i= 6i n i i= = 64 = 64 n i= i i= = 64 i= i 4 n n i n n i n i n i= i n i = 64 n i n i= i= nn + = 64 n nn + n + n 6 n + n = 64 n n + n + n 6n n = 64 n + n n n 6n n 6n = = 64 6 =. n 6n

9 Math 4, Autumn Solutions to Final Eam December 9, Page 9 of 4 πi πi π b Epress the it n cos as a definite integral, and then compute its value using n n i= the Evaluation Theorem. Show all the steps of your reasoning. 8 points Let s write the given it as the it of a Riemann sum with right endpoints. That is, we ll write i= πi πi π n cos n n in the form f i. i= To do this, let s set f i = πi πi n cos, = π n n = π/ n Since = b a n b = π. Then so, we see that b a = π. For simplicity, we may choose a =, which forces i = a + i = + i π n = πi n, f i = πi πi n cos = i cos i. n Thus, we take f = cos, so our it becomes: i= πi πi π n cos n n = π/ cos d. To evaluate this integral, we integrate by parts. Setting u = and dv = cos d, we get du = d and v = sin, so our integral becomes: π/ cos d = sin π/ = π/ [ π sin π sin = π. sin d ] [ cos ] π/ Remark: Note that other choices of a, b, and f are possible. For eample, if we take a = and b = + π, we get f = cos. As another eample, one could take a =, b = π, and f = cos. Of course, the final answer of π will be the same in all cases.

10 Math 4, Autumn Solutions to Final Eam December 9, Page of points List the following quantities in increasing order from smallest to largest. No justification is necessary. A B C D E F 5 4e 4 5 e5 e d e + e + e + e4 4 e + e + e4 4 + e5 5 e e + e 4 + e e + e e e + e e4 e + e5 9 Each of the quantities B,..., F can be viewed as a Riemann sum approimation of the integral A using a particular choice of sample points and number of rectangles n: B: using the Left Endpoint Rule with n = C: using the Right Endpoint Rule with n = D: using the Left Endpoint Rule with n = 4 E: using the Right Endpoint Rule with n = 4 F: using the Right Endpoint Rule with n = 8 since e e = e/ /, etc., where = Now the function f = e is increasing for > we can check that f is positive here, so any use of the Left Endpoint Rule gives an underestimate for A because the corresponding rectangles will lie under the graph of the positive function f, and any use of the Right Endpoint Rule gives an overestimate. Thus B and D are less than A; and C, E, and F are greater than A. Determining the relative rankings among {B, D} and {C, E, F } can be done either by inspection for eample, 4e = e + e + e + e < e + e + e + e4 4, since f is increasing, or by reasoning with Riemann sum rectangles for eample, the approimate area using 4 rectangles placed under f will cover more area than using just one rectangle under f, and so on. We find our final ranking is: B < D < A < F < E < C

11 Math 4, Autumn Solutions to Final Eam December 9, Page of 4 9. points Show all reasoning when solving each of the problems below; your final answers should not involve integral symbols. t + t + 5 if t < a If ft = 7e t, determine a formula for g = ft dt. + t if t 7 points First note that, using the substitution u = t, du = dt, t dt = du = ln u = ln t, u and, using the substitution u = t, du = dt, 7 7e t dt = eu du = 7 eu = 7 et. It was acceptable to do these in your head. For < : g = = [ t = = t + t + 5 dt ] ln t + 5t ln + 5 ln ln ln + 5 For : g = = t + t + 5dt + [ t ln t + 5t ] 7e t + t dt [ ] 7 + et arctan t = ln ln + 7 et arctan t = 9 + ln + 7 et arctan t 7 e arctan ln ln if <, So g = 9 + ln + 7 et arctan t if. It s also correct to write ln instead of ln, since, for <, is always positive.

12 Math 4, Autumn Solutions to Final Eam December 9, Page of 4 b cot θ dθ 4 points We have cot θ = cos θ sin θ. Use the substitution u = sin θ, so du = cos θdθ: cos θ cot θdθ = sin θ dθ = = cos θ u u du du cos θ = ln u + C = ln sin θ + C c / arcsin z dz 6 points Use integration by parts: let u = arcsin z, dv = dz; so du = z, v = z. / / arcsin z dz = [z arcsin z] / z dz z = [z arcsin z] / + = [z arcsin z] / + = [z arcsin z] / + / /4 [ = arcsin + 4 / / z z dz d * ] /4 / = π + * using the substitution = z, so d = dz, and = =, = = 4. d d 5 points Use the substitution u =, so = u +, and du = d: d = d = u + udu = u / + u / du = u 5/ 5/ + u/ + C = 5/ + / + C / 5

13 Math 4, Autumn Solutions to Final Eam December 9, Page of 4 + cos t. points Let g = dt. t a Find g and show that g 4 for every >. 6 points Let F t be an antiderivative of +cos t t. By the Fundamental Theorem of Calculus, so g = + cos t t = F F g = d d F F = F = + cos = + cos if >. We know cos so + cos 4. This implies g 4 when >. b Give a complete statement of the Mean Value Theorem. points If f is a differentiable function on the closed interval [a, b], then there eists c a, b such that: f fb fa c = b a NOTE! In some places, the property f needs to satisfy is listed as differentiable on a, b and continuous on [a, b]. This is a more general version of the Mean Value Theorem, but we don t need to worry about it in this class. 9 + cos t c Show that dt 8. Justify all steps. Hint: think about this integral in terms of the t function g, and use the Mean Value Theorem. You can apply the result of part a even if you did not prove it. 4 points Notice that 9 +cos t t dt = g and that g = +cos t t dt =. We want to show that g 8, so we apply the Mean Value Theorem for the function g on the interval [, ]. g is differentiable on that interval; we computed its derivative in part a. By the Mean Value Theorem, there eists a number c, such that: g c = g g = g This implies that g = g c 4 = 8 by part a. NOTE! We can also give a solution where we don t use the Mean Value Theorem: g = 9 + cos t 9 + dt dt = t t 9 t dt = [t ] 9 = = 8

14 Math 4, Autumn Solutions to Final Eam December 9, Page 4 of 4. 8 points Mark each statement below as true or false by circling T or F. No justification is necessary. T F d =. False; not integrable due to asymptote = and any reasonable etension of the notion of integral, such as improper integrals, wouldn t allow this integral to be, since > for all. T F a a arctan 5 d = for all a >. True; the function arctan 5 is odd. T F ln d <. False; although ln < for < <, the order of the its on the definite integral means that ln d = ln d >. T F If f is even and continuous at all, then h = ftdt is odd. True; we have h = ftdt = ftdt = ftdt = h, where the net-to-last equality uses the fact that f is even. The following four questions refer to the function g = { sin if, if =. T F The function g provided above is continuous at =. True. We have g = = g by the Squeeze Theorem, since g. T F The function g provided above is differentiable at every. True; for we have g = sin + cos = sin cos. T F The function g provided above is differentiable at =. True; we have g gh g = = h sin h h h h, and this it can be shown to be by another Squeeze Theorem argument using h h sin h h for h. T F For g provided above, the function g is continuous at =. False; using the formula found above for g when, it is clear that g does not eist.