MAS111 Strand 1: Solutions to Problems Spring Semester

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1 MAS Strand : Solutions to Problems Spring Semester 4-5 Week Review. Let A B to get cos cos sin. So cos cos + sin sin sin and cos cos cos + sin cos. Rearranging, we find cos + cos and sin cos. Then π/3 cos d π/3 + cos 4 d [ ] π/3 sin 4 + π δy y + δy y u + δu v + δv u v vu + δu uv + δv vδu uδv vv + δv vv + δv. Dividing through by δ gives and letting δ gives δy vδu/δ uδv/δ δ vv + δv dy d v du d u dv d v Letting u sin and v cos, we find d d sin cos cos sin sin tan d d cos cos cos sec.

2 3. We have y tan, so tan y. Differentiating with respect to using the Chain Rule gives sec y dy d. Thus dy d sec y + tan y d d e e. So Using Integration by Parts, 3 e d e d.e d [ e ] [ ] e [ ] e e. e d e d e e. 5. [ ] + d ln + ln. + d + + d + d [ tan ] π 4. Since 3 +, 3 + d ln. [ + d ] ln +

3 Week. a 5 < < < 5 4/5 <. The solution set is the open interval 4/5,. b t + 5 < 6 6 < t + 5 < 6 < t <. The solution set is the open interval,. c 3y 9 < 4 4 < 3y 9 < 4 5 < 3y < 3 5/3 < y < 3/3. The solution set is the open interval 5/3, 3/3. d 6 < 3 + < 3 < and + >, i.e. < < 3, between the roots of the quadratic. The solution set is the open interval, 3. e + 3 < < + 3 < 3 < < 3 +. The solution set is the open interval 3, 3 +. f 3 /t < / / < 3 /t < / 7/ < /t < 5/ 5/ < /t < 7/ /7 < t/ < /5 4/7 < t < 4/5. The solution set is the open interval 4/7, 4/5.. a cos > sin for nπ 3π/4 < < nπ + π/4, for all n Z. 3

4 cos > sin for nπ π/4 < < nπ + π/4, for all n Z. b Using cos sin cos and sin cos sin, the given inequality is equivalent to cos sin >, i.e cos > sin. We have seen that this is true for nπ 3π/4 < < nπ + π/4, i.e. for nπ 3π/8 < < nπ + π/8, for all n Z. 3. The triangle inequality gives us directly that a + b a + b. It remains to show that a b a + b, i.e that a + b a b a + b. Rearranging the terms in the two inequalities, this is equivalent to b a + a + b and a a + b + b. For the first of these, 4. and for the second b a + a + b a + a + b a + a + b, a a + b + b a + b + b a + b + b. 4

5 We see that > + 4 for < < and for > 4. Let s check this algebraically. > > 8 > + 4 > > and + 4 > or < and + 4 <. If > then + > and we must have 4 >, i.e. > 4, while if < then 4 < so we must have + >, i.e. < <. Week 3. f The limit appears to be 3. We can check as follows f The limit appears to be.6. We can check as follows / 5 + 3/ / as. as. 5

6 The limit appears to be / 5. The limit appears to be e. 6. If then and d cos + h cos cos lim d h h cos cos h sin sin h cos lim h h cos h cos lim h h sin lim h sin h h cos sin sin , 3 + sin 3 sin 3, using sin. Hence sin For a better bound, if then in fact sin sin.84. So and + 3+sin 3 + sin ,

7 7. The slope of the chord joining, and h, h is h h h h It looks like the limit may be ln h h. This limit is the slope of the tangent line to the curve y at the point,. This is the derivative of f at. So it appears that f ln. Week 4. a Since sin, we have sin for all. Since lim lim, it follows from the Sandwich Theorem that lim sin. b We have lim sin lim lim sin.. sin7 7 sin7 sin7. a lim lim 7 7 lim b We check that both numerator and denominator are at, so that we can apply l Hopital s rule: lim sin7 7 lim 7 cos7 7.

8 3. a cos sin + cos + cos cos cos sin cos + sin. sin So the required identity is equivalent to a standard one. So cos lim lim sin + cos sin lim lim sin.. + cos b We check that both numerator and denominator are at, so cos sin that we can apply l Hopital s rule: lim lim. 4. a We check that both numerator and denominator are at. As long as this continues to be the case, we keep differentiating top and bottom. lim lim lim b We check that both numerator and denominator are at. As long as this continues to be the case, we keep differentiating top and bottom. cos sin lim lim lim cos. c We check that both numerator and denominator are at. e lim lim e. ln / lim ln +, and we check that both numerator and denominator d This time, we write lim + ln lim + ln / lim + go to infinity as goes to zero. Then ln lim ln lim + + / lim / + / lim. + / Hence, lim lim e ln e

9 5. We check that both numerator and denominator go to infinity as goes to infinity. So we can apply l Hopital s Rule. However if we differentiate top and bottom, in the first step we get + / and in the second step we get back to where we started. On the other hand, we have lim lim + / + /. / + / 6. We consider the three cases k <, k and k > separately. Firstly, if k <, then k and e as, so k /e. Secondly, if k, then k and k /e /e as. Thirdly, if k is a positive integer, then both k and e as, so we may apply l Hopital s Rule: k lim e lim k k e. Continuing to apply l Hopital s Rule, we find k lim e lim k k kk k k! e lim e lim e. Finally, if k >, not necessarily an integer, then < k /e < k /e, where k denotes the ceiling of k, the smallest integer not less than k. Then the Sandwich Theorem gives the desired result, since we have already seen that lim k /e. 9

10 7. The shaded area is + t dt [ ln t ] + ln +. Using the smaller and larger rectangles, we see that + ln +. Hence since >, + ln +. As,, so + +. So, by the Sandwich Theorem, ln + as. Then taking eponentials, lim + e. Week 5. f 3 y 5ye y and f y 3 y 5e y.

11 . a b c

12 3. At the point,, f appears to be increasing more rapidly in the direction. f f We can check as follows:, while y f 3y, so at,, y 3. So f f is indeed bigger than y at,. 4. z f, y, where f, y 4 + y. 5. f f. So at, y,,. So the gradient is. f e y sin and f e y cos. f y e y cos and f y e y cos. Hence f + f y. 6. There are many ways to do this, for eample, as below. The curves in the picture are labelled by values of f at intervals of.

13 Week 6.. φ 3 y + y e y, φ 6y + y 4 e y, φ y + 4 y e y. φ y 3 + ye y, φ y 3 + y + y 3 e y, u u, y, v v, y. So u, v, y Rearanging the equations to give and y in terms of u and v, we obtain u v, 5 y u + v So Thus we see that u u y, y u, v , y v / u, u, v, y. +y + v y, +y v y, y u, v, y +y + y 4 +y y y y +y y, +y + y. +y y +y + y y. Since uv + y y y and u + v 4 + y, we have 4 u + v uv 4 u + v, y uv. Then, y u, v u v 4 u v + y y y, v u and this is the reciprocal of the previous answer. 3

14 4. Let the radius be r, the height h and the volume v. Then v πr h. So v v r πrh and h πr. So Thus z 5. δv v v δr + r h δh πrhδr + πr δh. δv v δv πr h δr r + δh 3 h + 4. So the volume goes up by about.4%. by z, y y. The equation of the tangent plane at,, is given z z z, +, y, y i.e. z + y, i.e. z + y. 6. z r z r + z y y r y cos θ + 4 y + 4y 3 sin θ 4 + y cos θ + y sin θ 4r r cos θ + r sin θ 4r 3. Alternatively, z θ z θ + z y y θ y r sin θ + 4 y + 4y 3 r cos θ 4 + y ry cos θ sin θ 4r 4 sin θ cos θ cos θ sin θ. z r cos θ 4 + r cos θ r sin θ + r sin θ 4 r 4 cos θ + sin θ r 4, 7. so z r 4r3 and z θ. g + g y f + f f y y y f y, by equality of the mied partials. So g is harmonic. 4

15 If f, y e y cos, then we want g such that g e y cos, g y e y sin. So g, y e y sin works. Week 7. z u z u + z y y u z + z y, z v z v + z y y v z z y. Multiplying gives the result.. Differentiate both sides with respect to to give y + dy + cos + y + dy d d. So dy y + cos + y d. + cos + y 3. By the Chain Rule, In other words, Now z z z u u + z v z v u + y z v. z u + y z v z u + z u + z u u + y v. + y u + y v z v z u + y z u + 4 z u + 4y z u v + y z v using the product rule u + y z v v using equality of second order partials. 5

16 By symmetry, we have Adding, we get z y z u + 4y z u + 4y z u v + z v. z + z y 4 z u y z u + 8y z u v + + y z v 4. By the Chain Rule, and Then and 4 z u + 4u z u + 8v z u v + u z v. y y u u + y v v y u + y v y t y u u t + y v v t c y v c y u. y y t y u + y v u + v y u + y v y u + y u v + y v t c y v c y u c v c u c y v c y u c y v c y u v + c y u. Substituting these epressions in and simplifying, we see that the Wave Equation is equivalent to y u v, i.e. y. u v Now u y v means that y v is constant with respect to u, i.e. y v hv for some function of v. And then integrating with respect to v, we see that y gv + fu, where g is any function of v such that g h and fu is the constant of integration with respect to v. Now substituting u ct, v + ct gives the result. 6

17 Week 8. i ii S S 3y d dy + y d dy The second integral is easier. + y d dy 3y d dy 7 3 3y dy + y dy d [ ] + y dy [ 7 3 y y3 [ 3 ] 3 3y ] dy + y d dy. + y + y dy [ tan y tan y ] π 4 tan.. D 3y d dy y y 3 3y d dy 3y y dy 3 y4 [ y ] 4 y4 6. [ 3 3 3y ] y dy 7

18 Now let s do it the other way round: D 3y d dy 3y dy d 3 d [ ] 6. [ y y 3] y y d 3. From the diagram we see that it will be more efficient to make the inner variable between y and 3 y and y the outer one. The other way round, we would need to split the integral as a sum of two integrals. 8

19 R + y + + y + d dy 3 y y 3 y y + y + d dy + y + y d dy + y + by long division of polynomials in [ y ln + y + ] 3 y y dy 8 4y y ln 4 + y ln y dy 8 4y y ln + y ln + y ln y dy [ 8y y ln y + y ln y integrating by parts [ 8y y ln y + y ln y ] 4 y ln + ln ln. 4 ] y y dy 4. y e 3 3 d dy D where the region D is indicated in the diagram. e 3 3 d dy, 9

20 So y e 3 3 d dy e 3 [ 3 3 e3 ] 3 dy d e 3 3 d 3 e. Week 9. a a a + y dy d a y d dy + y a [ ] y + y d dy y dy [ ] y a a.. y y 9 6 d dy [ y y 9 6 dy d ] 6 d 5 6 d [ ] 9 6 3/ 9.

21 3. D y d dy / / / 4 [ y dy d y ] d [ ] /. d Note that d d

22 4. The region D is described by the inequalities r <, θ π. Using the change of variables r cos θ, y r sin θ, we have + y r, and we find D + y + + y d dy π r cos θ + sin θ + r 4 r dθ dr π cos θ + sin θ + sin θ cos θ dθ r 3 + r 4 dr [ ] π 4 ln + r4 + sinθ dθ [ 4 ln θ ] π cosθ π ln z R y, so z So + R y z and similarly z y y z. z z + + y z + y z z + + y z R R y. The area of the graph of z R y lying above D is an eighth of the surface area of the sphere. Therefore the surface area of the sphere is 8 D R π/ R y d dy 8 8 π R R r dr dθ R r [ R R r ] R 4πR.

23 6. u, v, y ye y + y e. e The region D corresponds to u, v. Hence + y e + y d dy u, v + + y +, y d dy D D du dv u + [lnu + ] dv ln 3 dv ln 3. 3

24 7. The region D corresponds to the region R of the u, v-plane described by the inequalities u and v. u, v, y So,y u,v 3. Hence lny d dy D /y /y y / y/ 3. lnuv du dv 3 R ln u + ln v du dv [u ln u u + u ln v] dv ln + ln v dv 3 [ ln v + v ln v v] ln. 3 4

25 8. The triangle D in the, y-plane corresponds to the triangle T in the u, v-plane bounded by the lines v u a.k.a. y /, u a.k.a. y and v u + a.k.a. 3y +. So,y u,v. Then u, v, y. D y y + e y d dy uu + e uv du dv T u+ +u uu + e uv dv du [u + e uv ] u+ u+ du u + e u +u e u +u du [ eu +u e u +u ]. 5

26 9. The region D corresponds to the region of the u, v-plane described by the inequalities u and 6 v 8. u, v, y y y y. So u, v, y y, since > y on D. Hence du dv y d dy. Then y + y + y d dy R v + u du dv 8 [ ] dv 8 6 v + u 6 [ ] 8 v + ln v ln 36 7 ln. 35 D 8 6 v + u v + + v + ln9/ ln7/8 du dv dv Week. Let s N N n n. Let s work to 4 decimal places. N s N Thus s 7 is.5 to d.p. The net term to be added is /64.565, which won t change the first decimal place. But the net few partial sums, to 4 decimal places, are s 8.574, s , s.5498, s.558. So s.6 to d.p., and the first decimal place has changed. In fact, n n π /

27 . Since the harmonic series /n is divergent, the series in question is not absolutely convergent. Here is a table of partial sums, to 4 d.p. N s N s, s 3, s 5 is a decreasing sequence, since / > /3, /4 > /5, etc. And s, s 4, s 6 is an increasing sequence, since /3 > /4, /5 > /6, etc. The odd partial sums are all greater than all the even partial sums. Hence s, s 4, s 6 is an increasing sequence, bounded above by any odd partial sum, so converges to a limit. Likewise, s, s 3, s 5 is a decreasing sequence, bounded below by any even partial sum, so converges to a limit. These limits are the same, since s n s n a n. This common limit is then the sum of the series, which converges. The limit is in between successive partial sums s n and s n, within a n /n of each of them. If we keep going until n <.5, i.e. until n >,, then we will have 4 d.p. correct. So, the convergence is slow. 3. We know that n /n converges. The series cos n n n has positive terms, so its partial sums form an increasing sequence. They are bounded by the sum n /n cos n since n n, so the sequence of partial sums of cos n n n converges. Thus n cos n n converges absolutely and we know that this implies it converges. 4. f ln +, f +, f + f + 3,..., f n n n! + n, for all n >. The coefficient of n in the Maclaurin series is f n /n! n n!/n! n /n, for n >. And f ln. So the Maclaurin series is / + 3 /3 4 /4 + The radius of convergence is lim n a n /a n+ lim n n+/n. For, the series is / /3 /4, which diverges because it is minus the harmonic series. For, the series is / + /3 /4 +, which converges see problem. 5. a n a n+ n! n +! n + n + n!n +! n + + /n + /n + /n 4 7

28 as n. So the radius of convergence is 4. Looking more closely, a n a n+ n+ n+ < 4, so a n 4 n < a n+ 4 n+. Therefore, a n 4 n and the series cannot converge for ±4. 6. I cannot calculate sin d. Make the approimation Then sin d sin 3 /6 + 5 /. /6 + 4 / d [ 3 /8 + 5 /6] /8 + /6 73/9.89. Using one more term, Then sin sin 3 /6 + 5 / 7 /54. d /6 + 4 / 6 /54 d [ 3 /8 + 5 /6 7 /358] /8 + /6 / Assuming that term-by-term integration of the series is justified, the infinite series for sin d has terms of alternating sign and strictly decreasing absolute value. Therefore, as in problem, the sum of the series is between any two successive partial sums. Therefore, it must be.89 to 4 d.p. 8