2u 2 + u 4) du. = u 2 3 u u5 + C. = sin θ 2 3 sin3 θ sin5 θ + C. For a different solution see the section on reduction formulas.

Transcription

1 Last updated on November, 3. I.a A x + C, I.b B t + C, I.c C tx + C, I.d I xt + C, J x t + C. I4.3 sin x cos x dx 4 sin x dx 8 cos 4x dx x 8 3 sin 4x+C. I4.4 Rewrite the integral as cos 5 θ dθ and substitute u sin θ. We get cos 5 θ dθ u du cos 4 θ cos θ dθ }{{} d sin θ u + u 4 du u 3 u3 + 5 u5 + C sin θ 3 sin3 θ + 5 sin5 θ + C. For a different solution see the section on reduction formulas. I4.5 Hopefully you remembered that cos θ + sin θ. The answer is θ + C. I4.6 Use sin A sin B cosa B cosa + B to rewrite the integrand as sin x sin x cosx cos3x. You then get sin x sin x cosx cos3x dx sinx sinx + C. 6 I7. I7. I7.3 x n ln x dx xn+ ln x n + e ax sin bx dx e ax cos bx dx I7.6 π sin4 xdx π I7.7b π sin5 xdx π sin6 xdx π π sin7 xdx xn+ n + + C. eax a a sin bx b cos bx + C. + b eax a a cos bx + b sin bx + C. + b

2 I7.8 cos n xdx n sin x cosn x + n n cos n xdx; cos 4 x dx 3 8 x + 4 cos3 x sin x + 3 cos x sin x + C 8 Setting x and x π/4 leads to π/4 cos 4 xdx π I7.9 Hint: first integrate x m. I7. x ln x x + C I7. xln x x ln x + x + C I7.3 Substitute u ln x. I7.4 π/4 tan 5 xdx π/4 tan xdx 4 + ln I7.7 Substitute u + x. I9.a + 4 x 3 4 I9.b + x+4 x 3 4 I9.c x +x+ x 3 4 I9.d x3 x x + x + x x. You can simplify this further: x 3 x x+. I9.a x + 6x + 8 x + 3 x + 4x + so dx x + 6x + 8 lnx + lnx C. I9.b I9.c 5 dx x arctanx C. + 6x + dx x + 4x + 5 arctanx + + C 5 I9.3 Multiply both sides of the equation x + 3 xx + x A x + with xx + x and expand B x + + x + x x x +6x+8 / x+ + / x+4 and C x x + 3 Ax + x + Bxx + Cxx + Ax + Bx x + Cx + x A + B + Cx + C Bx A.

3 3 Comparing coefficients of like powers of x tells us that A + B + C C B A 3 Therefore A 3 and B C, i.e. x + 3 xx + x 3 x + x + + x and hence x + 3 dx 3 ln x + ln x + + ln x + constant. xx + x I9.4 To solve multiply by x: x + 3 xx + x A x + B x + + C x, x + 3 x + x A + Bx x + + Cx x and set x or rather, take the limit for x to get A 3; then multiply by x + : x + 3 Ax + Cx + + B + xx x x and set x or rather, take the limit for x to get B ; finally multiply by x : x + 3 Ax Bx + + C, xx + x x + and set x or take the limit for x to get C. I9.5 Apply the method of equating coefficients to the form x + 3 x x A x + B x + C x. In this problem, the Heaviside trick can still be used to find C and B; we get B 3 and C 4. Then A x 3 x + 4 Axx + 3x + 4x x x x so A 3. Hence x + 3 x x dx 3 ln x ln x + constant. x I9.6 Hint: look at the numerator. I

4 4 I9. x + ln x + C I9. 5 ln x 3 ln x. I9.3 x ln x + 5 ln x + C I9.4 4 ln ex 4 ln ex + + arctanex + C I9.5 Let e x z. Then e x dx dz. Now we can substitute: e x dx dz + e x + z Since we see a + z the natural substitution is z tan θ. This yields dz sec θdθ, and we get sec θdθ + tan θ sec θdθ sec θ sec θdθ ln sec θ + tan θ + C Now tan θ z and so sec θ + tan θ + z. ln + z + z + C. I9.6 arctane x + + C I9.7 x ln + e x + C I9. ln x + x + ln x + C I3. arcsin x + C I3. Use the substitution x sin θ and dx cos θdθ to obtain: dx cos θdθ 4 x 4 4 sin θ cos θdθ cos θ dθ θ + C arcsin x + C

5 5 I3.3 There are several ways to solve this problem. The following solution involves trigonometric substitution. We set x tan θ and so dx sec θdθ and we get + x dx + tan θ sec θdθ sec 3 θdθ Now we need to integrate sec 3 θ, which is a tricky problem in its own right! cos 3 θ dθ cos θ cos 4 θ dθ cos θ sin θ dθ Now let u sin θ so du cos θdθ and we get: du u Now we have a partial fractions problem. du + u u Au + B + u + Cu + D u du

6 6 We want to rewrite +u u Au+B +u + Cu+D u. We need to use equating coefficients. After expanding and then equating the numerators, we obtain A + Cu 3 + B A + C + Du + A B + C + A + C B A + C + D Du + B + D. Equating coefficients yields A B + C + D B + D So C A and D { B. Subsituting these into the two middle B A + A + B equations we get: The first equation A B + A + B simplifies to 4A + so A 4 and so C 4. The second equation to 4B so B so D. We now complete the integral computation: 4 u u + 4 u u du 4 + u + + u + u + u du [ ] ln + u 4 + u ln u + + C u [ ln + u 4 u u ] u + C 4 ln + u u u u + C Let s substitute back in with u sin θ to get: 4 ln + sin θ sin θ sin θ sin θ + C 4 ln + sin θ sin θ sin θ cos θ + C 4 ln + sin θ sin θ tan θ sec θ + C Finally, since x tan θ, we have sin θ yielding: 4 ln + x +x x +x x +x, and sec θ + x x + x + C We remark that, using rules of logarithms and some other algebraic manipulations, this can be transformed into the formula appearing in the book: x + x + ln x + + x + C

7 7 I3.4 First we complete the square to get x x x x x x. So we now return to the integral: dx dx x x x Sub x sin θ and so dx cos θdθ and we get cos θdθ sin θ cos θdθ cos θ dθ θ + C arcsinx + C. I3.7 We substitute x sin θ so that dx cos θ and get: dx x 4 x x x x x x cos θ 4 4 sin θ cos θ cos θ dθ dθ [θ] x x [ arcsin x ] arcsin arcsin π/6 π/6 π/3. I3.3 3 arctanx + + C I3.4 3x + 6x x + x x + x Therefore substitute u x+.

8 8 I5. We use integration by parts with f x and g sin x. Then f and g cos x. So we get: a x sin x a fg [fg] a a [ x cos x] a f g a cos a + a a cos xdx cos xdx a cos a + [sin x] a a cos a + sin a sin a cos a + sin a. I5. a x cos xdx a + sin a + a cos a I x dx [ x x ] I5.4 Use u x so that du xdx and so du xdx. /3 /4 x dx x/3 du x u x/4 x/3 x/4 [u /] x/3 u / du x/4 [ x /] /3 /4 [ /3 / /4 ] / [ / / ] I5.5 same as previous problem after substituting x /t I5.6 Complete the square and substitute u x + /4. See the next problem for details. ln x + x + 7 x+ 4 arctan 4 + C

9 9 I5.6 Completing the square leads to x dx x + x + 7 x dx x x dx x This suggests the substitution u x+ 4 x dx x + x The first integral is 4u 4du u + u du u + u + +C,, i.e. x 4u : u du 4 u + du u +. which one can find by substituting v u +. The second integral is in the list??. Thus we end up with x dx x + { x + x + } x + x ln C { x + x + } x + x + 7 ln C. 4 4 This can be further simplified to { x + x + x + 7 ln + } x C 4 4 x + x + 7 ln x + + x C 4 x + x + 7 ln { x + + x + x + 7 } + ln 4 + C x + x + 7 ln { x + + x + x + 7 } + Ĉ. I5. x + ln x ln x + + C I5.7 x lnx + x + x lnx + + C I5.3 Substitute x t and integrate by parts. Answer: x arctan x x + arctan x + C I5.33 tanx secx + C I x+ ln x + + arctanx + C

10 II4. dx M + x lim dx M + x [ ] M lim M + x x lim M to [ + M + ]. II4. The integrand becomes infinite as x integral. so this is indeed an improper / x 3 dx lim a a x 3 dx lim [ a 4 x ] a lim a 4 a + 4. II4.3 The integral is improper at x 3: 3 dx 3 x a lim a 3 dx 3 x [ lim 3 x a 3 ] xa x lim 3 a + 3 a 3 3. II4.4 II4.5 Substitute u x to get the antiderivative xe x dx e x + C. The improper integral then is: xe x dx lim M M xe x dx lim M e M + e. II4.8 7 x / dx lim a 7 a x / dx lim a [ x / ] 7 xa 7. II4.9 This integral is infinite. II4. e II4.3

11 II4.4. II4.5 To do the integral, substitute x u. The answer is. II4. πr dx π x dx π II6.3 x x + x x + < x for all x >, so False. II6.4 True II6.5 False. II6.7 True because II6.8 False. II6.9 True. II6. False. II6. True. therefore x < x+ x x + x+. For x > it is always true that x + > x and is true for all x >. II6. The integrand is fu u u +, which is continuous at all u. Therefore the integral is improper at u, but nowhere else. To determine if the integral converges we may therefore look at the tail, i.e. we can look at I u u + du instead of the integral starting at u. For all u we have + u u, and therefore Therefore u + u u u u. u + u du du u. This implies for the integral in the problem that u + u du <. In other words the integral converges. To get an actual estimate for how big the integral is we already have, so we need to estimate the integral for < u <. On this interval we have u + u u u,

12 so that u + u du u du 3. Therefore u + u du u + u du + u + u du II6.3 For large values of u the integrand is approximately given by u 3 u + u3 u u. Since the integral du u diverges, we expect the integral I to diverge too. To confirm this we try to estimate the tail from below. For u > we have which implies Therefore It follow that u 3 u + du + u < u, u 3 u + > u3 u 4u. u 3 u + du du 4u. u 3 u + du + u 3 u + du In other words the integral in the problem diverges. du 4u. III4. General solution yx Ce x ; initial conditions imply C e. III4. Implicit form of the solution tan y x Solution yx arctan 3 x /3 + C, so C tan π/3 3. III4.3 Implicit form of the solution: y + y + x + x A + A. If we solve for y we get y ± A + A + x x Whether we need the + or depends on A. III4.4 Implicit form of the solution 3 y3 + 4 x4 C; C 3 A3. Solution is y 3 A x4.

13 III4.5 Integration gives ln y y + ±e C e x. 3 x+c. Solve for y to get y y+ ±ex+c Let B ±e C be the new constant and we get y y + Bex whence y + Bex Be x. The initial value y A tells us that B A A+ solution with initial value y A is y III4.6 yx tan arctana x. III4.7 y x x3 3 + III6.5 y Ce x 3 ex III6.6 y xe sin x + Ae sin x III6.8 mx cos x, yx tan x + C cos x, C. A + + A ex A + A e x. III6.9 mx cos x, yx + sin x cos x ln + C cos x, C. sin x III6. Rewrite as dy dx + y cos x N cos x. mx e tan x, yx N + Ce tan x, C N. III6. Rewrite as dy dx y x. mx x, yx x ln x + Cx, C ln. III6.5 y Ce x3 /3, C 5e /3 III6.6 y Ce x x3, C e, and therefore the III9. a ln y t + C or y, y ec e t / Since e C can be any positive value, and e C can be any negative value, we can lump both together with the y solution to obtain: y A e t / Plugging in our initial value of y.., we get. A e /, so A e / Solving for y3., y3. e / e 3. / e 4.83 b There is only step between y. and y3.. t., y, m, so y3. +.

14 4 c If h., then it takes n steps to reach y3., so t., y, m y y. +. y, t., m., y y3. +. In tabular form: t k y k m k ft k, y k y k+ y k + m k h t. y +. t. y. +. t 3. y Now we can do the same thing with steps of h /3 and h.5 h/3 t k y k m k ft k, y k y k+ y k + m k h t. y. + /3 /3 t 5/3 y /3 5/3 /3 5/9 /3 + 5/9/3 /7 t 7/3 y /7 7/3 /7 7/8 /7 + 7/8/3 5/43 t 3 3. y 3 5/43 h/ t k y k m k ft k, y k y k+ y k + m k h t. y. + / 5 t 3/ y 5 3/ 5 5/ 5 + 5// 5/4 t y 5/4 5/4 5/8 5/4 + 5/8/ 5/6 t 3 5/ y 3 5/6 5/ 5/6 75/3 5/6 + 75/3/ 5/64 t 4 3 y 4 5/64 III9. a If h, then there is only one step between y and y, therefore: x, y, m, and e y + h/ If h /, then x, y, and m are the same, and y + / 3/ x /, y 3/, m 3/, and e y 3/ + 3/ / 9/4 h/3 If h /3, then x, y, and m are the same, and y + /3 4/3 x /3, y 4/3, m 4/3, and y 4/3 + 4/3 /3 6/9 x /3, y 6/9, m 6/9, and e y 3 6/9 + 6/9 /3 64/7.37 h x k y k m k fx k, y k y k+ y k + m k h x y + x y h/ x k y k m k fx k, y k y k+ y k + m k h x y + / 3/ x / y 3/ 3/ 3/ + 3// 9/4 x y 9/4

15 5 h/3 x k y k m k fx k, y k y k+ y k + m k h x y + /3 4/3 x /3 y 4/3 4/3 4/3 + 4/3/3 6/9 x /3 y 6/9 6/9 6/9 + 6/9/3 64/7 x 3 y 3 64/7 b. Hint: 9/4 3 / and 64/7 4 3 /3 3 III.3a Let Xt be the rabbit population size at time t. The rate at which this population grows is dx/dt rabbits per year. 5 X from growth at 5% per year X from death at % per year car accidents +7 immigration from Sun Prairie Together we get dx dt 3 X 3. This equation is both separable and first order linear, so we can choose from two methods to find the general solution, which is If X99 then Xt, + Ce.3t., + Ce.3 99, C, e.3 99 don t simplify yet! Hence X994, +, e.3 99 e.3 994, +, e , +, e.9, III.4a sec. III.4b T s t T r + Ce Kt. C has the same units as T r and T s, so C is a temperature: degree Fahrenheit. lim t T s t T r : in the long run the soup will approach room temperature. III.4c ii Given T s 8, T r 75, and T s 5 5. This gives the following equations: T r +C 8, T r +Ce 5K 5 C 5, 5K ln 3 5 ln 7 ln 7. When is T s 9? Solve T s t 9 for t using the values for T r, C, K found above K is a bit ugly so we substitute it at the end of the problem: T s t T r + Ce Kt e Kt 9 e Kt Hence ln /7 t K ln 7 K ln 7 ln 7/.

16 6 III.6 a Let yt be the amount of retaw in gallons in the tank at time t. Then dy dt 5 y }{{} growth }{{ 3 }. removal b yt 6 + Ce t/ 6 + y 6e t/. c If y then yt 6 + 4e t/ so that lim t yt +. d y 6. III.7 Finding the equation is the hard part. Let At be the volume of acid in the vat at time t. Then A 5% of 5gallons. A t the volume of acid that gets pumped in minus the volume that gets extracted per minute. Per minute 4% of gallons, i.e. 8 gallons of acid get added. The vat is well mixed, and At out of the gallons are acid, so if gallons get extracted, then A da dt 8 A 8 A 5. of those are acid. Hence The solution is At 4 + Ce t/5 4 + A 4e t/5 4 5e t/5. The concentration at time t is At 4 5e t/5 concentration.4.5r t/5. total volume If we wait for very long the concentration becomes At concentration lim t.4. III.8c P is the volume of polluted water in the lake at time t. At any time the fraction of the lake water that is polluted is P/V, so if 4 cubic feet are drained then P V 4 of those are polluted. Here V 9 ; for simplicity we ll just write V until the end of the problem. We get dp dt in minus out 3 P V 4 whose solution is P t 4 8V + Ke V t. Here K is an arbitrary constant which we can t call C because in this problem C is the concentration. The concentration at time t is Ct P t V 8 + K 4 e V t V 8 + C e 4 V t. 8 No matter what C is we always have lim Ct t because lim t e 4 V t. If C 8 then the concentration of polluted water remains constant: Ct 8.

17 7 IV4. Use Taylor s formula : Qx x 7 + x 7. A different, correct, but more laborious clumsy solution is to say that Qx Ax + Bx + C,, compute Q x Ax + B and Q x A. Then Q7 49A + 7B + C 43, Q 7 4A + B 9, Q 7 A. This implies A /, B 9 4A , and C 43 7B 49A 79. IV4. px 3 + 8x x IV4.7 T e t + t +! t + + n! tn + IV4.8 T e αt + αt + α! t + + αn n! tn + IV4.9 T sin3t 3t 33 3! t ! t5 + + k 3 k+ k+! t k+ + IV4. T sinh t t + 3! t3 + + k+! tk+ + IV4. T cosh t +! t + + k! tk + IV4. T +t t + t + n n t n + IV4.3 T 3 t t t t n+ n+ t n + note the cancellation of factorials IV4.4 T ln + t t t + 3 t3 + + n+ n t n + IV4.5 T ln + t T ln[ + t] ln + ln + t ln + t t + 3 t3 + + n+ n t n + IV4.6 T ln + t T ln + t t 4 t + 6 t3 + + n+ n t n + IV4.7 T ln + t t t t3 + + n+ n n t n + IV4.8 T ln [ +t t T ln + t ln t] t + 3 t3 + 5 t5 + + k+ tk+ + [ ] IV4.9 T t T / t + / +t + t + t t k + we could also substitute x t in the geometric series / + x x + x +, later in this chapter we will use little-oh to justify this point of view. [ ] t IV4.3 T t T / t / +t t + t 3 + t t k+ + note that this function is t times the previous function so we would think its Taylor series is just t times the Taylor series of the previous function. Again, little-oh justifies this.

18 8 IV4.3 The pattern for the n th derivative repeats every time we increase n by 4. So we indicate the the general terms for n 4m, 4m +, 4m + and 4m + 3: T sin t + cos t +t! t 3! t3 + 4! t4 + + t4m 4m! + t4m+ 4m +! t4m+ 4m +! t4m+3 4m + 3! + IV4.3 Use a double angle formula T sin t cos t sin t t 3 3! t m+ 4m +! t4m+ 4m+3 4m + 3! t4m+3 + IV4.33 T 3 tan t t + 3 t3. There is no simple general formula for the n th term in the Taylor series for tan x. IV4.34 T [ + t 3 t4] + t 3 t4 IV4.35 T [ + t 5 ] + 5t + t + t 3 + 5t 4 + t 5 IV4.36 T 3 + t + /3! t + /3/3! t + + /3/3 /3 /3 n+ n! t n + IV4.37! 6 IV4.38 Because of the addition formula sinα + β sin α cos β + sin β cos α we should get the same answer for f and g, since they are the same function! The solution is T sinx + a sin a + cosax sin a! + sin a 4n! x4n + x cos a x 3 + 3! cos a 4n +! x4n+ sin a 4n +! x4n+ cos a 4n + 3! x4n+3 + IV7. so fx f 4 x cos x, f x f 5 x sin x, f x cos x, f 3 x sin x, f f 4, f f 3, f. and hence the fourth degree Taylor polynomial is 4 f k T 4 {cos x} x k x k!! + x4 4!. The error is k R 4 {cos x} f 5 ξ x 5 5! sin ξ x 5 5!

19 9 for some ξ between and x. As sin ξ we have cos x x! + x4 R 4 x x5 < 4! 5! 5! for x <. Remark: Since the fourth and fifth order Taylor polynomial for the cosine are the same, it must be that R 4 x R 5 x. It follows that 6! is also an upper bound. IV7.3a The polynomial is px + x 9 3 x. Then p.7986 and the error satisfies: 3 9 p ! The 3 9 according to a computer is: IV7.3b The polynomial is px + x 9 3 x x3. Then p and the error satisfies: 3 9 p ! The 3 9 according to a computer is: IV7.4 The polynomial is px x 4 7 x. Then and the error satisfies: p p ! The according to a computer is: IV9. Notice that + x x + x as x. Thus, + x is not ox.

20 IV9.3 Recall that + x + x + ox and similarly that x + x + o x x + ox therefore + x x x + ox x x as x. Thus, + x x is not ox. IV9.4 Suppose that fx and gx are both ox. This means that fx lim x x lim gx x x Then we know that fx + gx fx lim lim x x x x + lim gx x x Thus, fx+gx is also ox. Since these are arbitrary ox functions, we can say that ox + ox ox. IV9.8 Notice that the functions x 3 and x 3 are both ox, since x 3 lim x x lim x 3 x x But, x 3 x 3 x 3 is not ox 3, since x 3 lim x x 3 IV9.9 Suppose that fx ox. This means that fx lim x x But then fx lim x x lim fx x x lim fx x x so that fx ox. IV9.4a We know from the Taylor expansion for + x that + x + x + ox From this, we can see that + x +ox k exactly when x is ox k, which is true when k <.

21 IV9.4c We know from the Taylor expansion of cosx that so that cosx x4 + ox 4 cosx x4 + ox 4 which is ox k exactly when k < 4. IV9.5d The PFD of g is gx x x. gx + x + 3 x + + n+ x n +. So g n / n+ and g n is n! times that. IV9.6 You could repeat the computations from problem??, and this would get us the right answer with the same amount of work. In this case we could instead note that hx xgx so that hx x + x + 3 x n+ x n+ + Therefore h n / n. The PFD of kx is kx x cancel! x x x, the Taylor series of k is just the Geometric series. IV9.8 T e at + at + a! t + + an n! tn +. IV9.9 e +t e e t so T e +t e + et + e! t + + e n! tn + IV9. Substitute u t in the Taylor series for e u. T e t t +! t4 3! t6 + + n t n + n! IV9. PFD! The PFD of +t t Series we get +t is t + t. Remembering the Geometric T + t t + t + t + t t n + IV9. Substitute u t in the Geometric Series / u. You get T + t t + t 3 t n n t n + IV9.3 fx x 3! + x4 5! n n n+! xn

22 IV9.4 T ln + x x x x + 3 x3 + + n n xn + x x + 3 x + + n n xn + IV9.5 T e t t +t+ ++! t + ++! + 3! t ! + + n! t n + IV9.6 / t t / so T + t t + t + 3 t3 + be careful with minus signs when you compute the derivatives of t /. You can make this look nicer if you multiply top and bottom in the n th term with n : T t + t t t n t n + 4 n IV9.7 T t + t t t n t n + 4 n IV9.8 T arcsin t t + t t t n 4 n t n+ n + + IV9.9 T 4 [e t cos t] t + 3 t3 6 t4. IV9.3 T 4 [e t sin t] t t + 3 t3 + ot 4 the t 4 terms cancel. IV9.3 PFD of / t t +t t 3 +t + 3 t. Use the geometric series. IV t + t 3 + t + t /3. This is very similar to problem??. The answer follows from Newton s binomial formula.

23 3 IV.b The Taylor series is sint t t 3 /6 + and the order one and two Taylor polynomial is the same pt t. For any t there is a ζ between and t with sint pt f 3 ζ t 3 3! When ft sint, f n ζ for any n and ζ. Consequently, for nonnegative t. Hence sinx dx px dx Since px dx 3 A ɛ sint pt t3 3! sinx px dx x 6 3! dx 7 3! 7 ɛ 3 A the approximate value we have that sinx dx A + ɛ IV.c b 43 3 c 3 6! 3 V3. / lim n n n 3 lim n V3. Does not exist or + n n n 3 n V3.3 / lim n n n +n 3 lim n n V3.4 + V3.5 lim n + n 3 lim n + n n 3 n lim n 3 n n +n 3 3 n 3 n lim n n n lim n + n 3 n 3 n + 3 n + 3 n V3.6 Does not exist or because e >. lim n e n + lim n e n + n n lim n an indeterminate n form. e n + n n + Note: n is not V3.7. For any polynomial n p, p >, and any exponential a n, a >, the exponential will go to much faster than the polynomial as n goes to, regardless the size of p or a. Recall L Hop Rule the exponential does not change in essence while the polynomial keeps decreasing in power V3.8. For any factorial n!, and any exponential a n, a >, the factorial will go to much faster than the exponential as n goes to, regardless the size of a. See final example in the section.

24 4 n!+ V3.9 write the limit as lim n n+! lim n n! n+! + lim n n+! lim n n+ + lim n n+!. n! n+n+ n+n+ V3. Let a n n! n!. Then a n+ n+! n+! n! n+ a n n+ < an. Every increase in n cuts a n at least in half. Then lim n a n. V3. Use the explicit formula?? from Example??. The answer is the Golden Ratio φ. V6. The remainder term R n x is equal to f n ζ n n! x n for some ζ n. For either the cosine or sine and any n and ζ we have f n ζ. So R n x x n x n!. But we know lim n n n! and hence lim n R n x. V6. The k th derivative of gx sinx is g k x ± k socx. Here socθ is either sin θ or cos θ, depending on k. Therefore k th remainder term is bounded by R k [sin x] gk+ c k +! x k+ k+ x k+ socx x k+ k +! k +!. x Since lim k+ k k+! we can use the Sandwich Theorem and conclude that lim k R k [gx], so the Taylor series of g converges for every x. V6.6 Read the example in??. V6.7 We know that the Taylor s series of fx x is T fx + x + x + x 3 N n xn lim N n xn Using the formula N n arn a rn+. r N n xn. xn+ xn+ x x For x, fx is undefined. For x, N n xn xn+ oscillates about / as N. So the limit does not exist, ie. the Taylor series does not converge. For < x <, N x n+ lim N n xn lim N ie. the Taylor series converges. V6.8 Let x u. Then by the previous question the Taylor series converges when < u <, ie. < x <, ie. when < x <. V6.9 < x <. V6. 3 < x < 3. Write fx as fx 3 Series. 3 x and use the Geometric

25 5 V6. x < /5 VI9.3e a 3 b 4 4 x VI9.6c a Since + x c 36 d e x the number x would have to satisfy both + x + x and + x. That s impossible, so there is no such x. b No drawing, but # x p + +x is the parametric x representation of a straight line through the points, when x and, 3 when x. x + y c x and y must satisfy. Solve x+y, x+y x + y to get x, y 3. VI9.7 Every vector is a position vector. To see of which point it is the position vector translate it so its initial point is the origin. Here AB # 3, so AB # is the position vector of the point 3, 3. 3 VI9.8 One always labels the vertices of a parallelogram counterclockwise see??. ABCD is a parallelogram if AB+ # AD # AC. # AB #, # AC, 3 # 3 AD. So AB # + AD # AC, # and ABCD is not a parallelogram. VI9.9a As in the previous problem, we want AB # + AD # AC. # If D is the point d, d, d 3 then AB #, AD # d d, AC # 4, so that d 3 3 # AB + AD # AC # will hold if d 4, d and d 3 3. VI9.9b Now we want AB # + AC # AD, # so d 4, d, d 3 5. VI.3b a x # 3 + t 3 t t. + t b Intersection with xy plane when z, i.e. when t, at 4,,. Intersection with xz plane when y, when t, at 3,, i.e. at A. Intersection with yz plane when x, when t 3, at, 3, 4. VI.4b a L[t] b t

26 6 VI.5c a # p # b + # c /, # q a # + # c /, # r a # + # b /. b m # a # + 3 # p a # See Figure??, with AX twice as long as XB. Simplify to get m # # 3 a + # 3 b + # 3 c. c Hint : find the point N on the line segment BQ which is twice as far from B as it is from Q. If you compute this carefully you will find that M N. VI. To decompose # b set # b # b + # b /, with # b / t # a for some number t. Take the dot product with # a on both sides and you get # a # b t # a, whence 3 4t and t 3 4. Therefore # b / 3 a, # 4 # b # b 3 a. # 4 To find # b / and # b you now substitute the given values for a # and # b. The same procedure leads to a # and a # / : a # / 3 # b, a # a # 3 # b. VI. This problem is of the same type as the previous one, namely we have to decompose one vector as the sum of a vector perpendicular and a vector parallel to the hill s surface. The only difference is that we are not given the normal to the hill so we have to find it ourselves. The equation of the hill is x + 5x 3 so the vector n # 5 is a normal. The problem now asks us to write f # grav f # + f # /, where f # t n # is perpendicular to the surface of the hill, and # f / id parallel to the surface. Take the dot product with n, # and you find t n # n # f # grav 69t 5mg t 5 69mg. Therefore # f mg 69 mg # mg, f / f # grav f # 6 69 mg mg, VI.c # a # b # a # a # b + # b ; # a # b 4 # a 4 # a # b + # b ; # a + # b 54, # a # b 6 and # a # b 3. VI.3 Compute AB # BA # 3, # BC CB #,. Hence # AB, # BC 8, # AC. # AC CA # And also AB # AC # AB cos A # AC # AB # AC # 5. A similar calculation gives cos B so we have a right triangle; and cos C 5. VI.4 AB #, AC # t # t, BC t 3 t. If the right angle is at A then AB # AC #, so that we must solve t + t. Solution: t, and C, 3. If the right angle is at B then AB # BC #, so that we must solve t 3 + t. Solution: t 5, and C 5,.

27 If the right angle is at C then AC # BC #, so that we must solve t t 3 + t t. Note that this case is different in that we get a quadratic equation, and in that there are two solutions, t, t 5. This is a complete solution of the problem, but it turns out that there is a nice picture of the solution, and that the four different points C we find are connected with the circle whose diameter is the line segment AB: 7 VI.5a l has defining equation x + y which is of the form n # x # constant if you choose n # /. VI.5b The distance to the point D with position vector # d from the line l is n # # d a # n # where a # is the position vector of any point on the line. In our case # d # and the point A,, a # OA #, is on the line. So the distance to the origin from the line is n # a # n # / + / 5. VI.5c 3x + y, normal vector is # m 3. VI.5d Angle between l and m is the angle θ between their normals, whose cosine is cos θ 5/4 5. # n # m # n # m / VI3.3a # the cross product of any vector with itself is the zero vector. VI3.3c # a + # b # a # b # a # a + # b # a # a # b # b # b # a # b. VI3.4 Not true. For instance, the vector # c could be # c a # + # b, and a # # b would be the same as # c # b. VI3.5a A possible normal vector is n # AB # AC # 4 4. Any non zero 4 multiple of this vector is also a valid normal. The nicest would be # 4 n. VI3.5b # n # x # a, or # n # x # n # a. Using # n and # a from the first part we get 4x + 4x 4x 3 8. Here you could replace # a by either # b or # c. Make sure you understand why; if you don t think about it, then ask someone. VI3.5c Distance from D to P is # n # d # a # n 4/ There are many valid choices of normal # n in part i of this problem, but they all give the same answer here. Distance from O to P is # n # # a # n 3 3. VI3.5d Since # n # # a and # n # d # a have the same sign the point D and the origin lie on the same side of the plane P.

28 8 VI3.5e The area of the triangle is # AB # AC 3. VI3.5f Intersection with x axis is A, the intersection with y-axis occurs at,, and the intersection with the z-axis is B. VI3.6a Since # n # AB # AC 3 the plane through A, B, C has defining equation 3x + y + z 3. The coordinates,, 3 of D do not satisfy this equation, so D is not on the plane ABC. VI3.6b If E is on the plane through A, B, C then the coordinates of E satisfy the defining equation of this plane, so that α 3. This implies α 5. VI3.7a If ABCD is a parallelogram then the vertices of the parallelogram are labeled A, B, C, D as you go around the parallelogram in a counterclockwise fashion. See the figure in 43.. Then AB # + AD # AC. # Starting from this equation there are now two ways to solve this problem. first solution If D is the point d, d, d 3 then AD # while AB # and # 3 AC 3, and thus d, d and d 3. d d + d 3 d d +, d 3. Hence # AB + # AD # AC implies second solution Let a, # # b, # c, # d be the position vectors of A, B, C, D. Then AB # # b a, # etc. and AB # + AD # AC # is equivalent to # b a # + # d a # # c a. # Since we know a, # # b, # c we can solve for # # d and we get d # # c b + a # +. VI3.7b The area of the parallelogram ABCD is # AB # AD VI3.7c In the previous part we computed # AB # AD 3., so this is a normal 3 to the plane containing A, B, D. The defining equation for that plane is x + y + 3z. Since ABCD is a parallelogram any plane containing ABD automatically contains C. VI3.7d,,,,,,,, 3. VI3.8a Here is the picture of the parallelepiped which you can also find on page 3: Knowing the points A, B, D we get AB # #, AD. EF GH ABCD Also, since is a parallelepiped, we know that all its faces are parallelogram, and thus EF # AB, # etc. Hence: we find these coordinates for the points A, B,... A,,, given; B,,, given; C,,, since AC # AB # + # 3 AD ; D,,, given; E,,, given F,,, since we know E and EF # AB #

29 G 3,, 3, since we know F and F # G EH # # H,, 3, since we know E and # EH # AD VI3.8b The area of ABCD is # AB # AD. AD. 9 VI3.8c The volume of P is the product of its height and the area of its base, which we compute in the previous and next problems. So height volume area base 6 7. VI3.8d The volume of the parallelepiped is # AE # AB # AD 6.