MATH spring 2008 lecture 3 Answers to selected problems. 0 sin14 xdx = x dx. ; (iv) x +

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1 MATH - sprig 008 lecture Aswers to selected problems INTEGRALS. f =? For atiderivatives i geeral see the itegrals website at (5-vi (0 i ( ( i ( π ; (v π a. This is example 8. i the otes. (0 iii π 0 si xdx = π cos xdx = si x cos x + cos xdx; π/ cos xdx = π Hit: first itegrate x m. ( ii x l x + C; (iii x(l x x l x + x + C π/ ta 5 xdx = 0 ( ( + π/ ta xdx = 0 + l (6 i + x+ x ; (ii + x ; (iii x +x+ x x ; (iv x + x = x + x+ (7 i 6x(x (x (x + (x + + (8 i x + 6x + 8 = (x + = (x + (x + so (8 ii dx x +6x+8 = l(x + l(x + + C. (8 iii (6 i 5 a 0 dx x +6x+0 = arcta(x + + C. dx x +x+5 = 5 arcta(x + + C. x si xdx = si a a cos a. x +6x+8 = / x+ + / x+ ad (6 ii a 0 x cos xdx = (a + si a + a cos a. (6 iii (6 iv (6 v / / x dx = [ x x ] = 5 8. x dx = [ x x ] / = / 5 8. same as (iv after substitutig x = /t. TAYLOR S FORMULA (5 i T e t = + t +! t + +! t + (5 ii T e αt = + αt + α! t + + α! t + (5 iii T si(t = t! t + 5 5! t5 + + k+ (k+! tk+ + (5 iv T sih t = t +! t + + (k+! tk+ + (5 v T cosh t = +! t + + (k! tk + (5 vi T +t = t + t + ( t + (5 vii T ( t = + t + t + t + + (+ 5 t + (ote the cacellatio of factorials + (5 viii T l( + t = t t + t + + ( + t +

2 (5 ix T l( + t = l[ ( + t] = l + l( + t = l + t t + t + + ( + t + (5 x T l + t = l( + t = t t + 6 t + + ( + t + (5 xi T l( + t = t t + t + + ( + t + (5 xii T l[(+t(+t] = l(+t+l(+t = (+t + t + + t + + ( + (+ t + (5 xiii T l ( +t t = l( + t l( t = t + t + 5 t5 + + k+ tk+ + [ ] (5 xiv T t = T / t + / +t = + t + t + + t k + (you could also substitute x = t i the geometric series /( + x = x + x +, later i this chapter we will use little-oh to justify this poit of [ view. ] t (5 xv T t = T / t / +t = t + t + t t k+ + (ote that this fuctio is t times the previous fuctio so you would thik its Taylor series is just t times the taylor series of the previous fuctio. Agai, little-oh justifies this. (5 xvi The patter for the th derivative repeats every time you icrease by. So we idicate the the geeral terms for = m, m +, m + ad m + : (5 xvii T (si t + cos t = + t! t! t +! t + + Use a double agle formula tm (m! + tm+ (m +! tm+ (m +! tm+ (m +! + T ( si t cos t = si t = t! t + + m+ (m +! tm+ m+ (m +! tm+ + (5 xviii T ta t = t + t. There is o simple geeral formula for the th term i the Taylor series for ta x. [ (5 xix T + t t] = + t t (5 xx T [( + t 5 ] = + 5t + 0t + 0t + 5t + t 5 (5 xxi T + t = + /! t + (/(/! t + + (/(/ (/ (/ +! t + DIFFERENTIAL EQUATIONS (9 y(t = Aet + Ae t (9 i y = Ce x /, C = 5e / (9 ii y = Ce x x, C = e (9 iii Implicit form of the solutio ta y = x + C, so C = ta π/ =. Solutio y(x = arcta ( x / (9 iv Implicit form of the solutio: y + y + x + x = A + A. If you solve for y you get y = ± A + A + x x Whether you eed the + or depeds o A. (9 v Itegratio gives l + y y = x + C. Hece +A A = ±ec, ad y = ( + Aex + A ( + Ae x + A. (9 vi y(x = ta ( arcta(a x.

3 (9 vii y = xe si x + Ae si x (9 viii Implicit form of the solutio y + x = C. C = A ; Solutio is y = A x. (97 y(t = Ae t cos t + Be t si t. (98 Solutio with give iitial values has A = 7, B = 0. (99 i y = Ae t + Be t + C cos t + D si t (99 ii The characteristic roots are r = ± ±, so the geeral solutio is y = Ae t cos t + Be t si t + Ce t cos t + De t si t. (99 iii y = A + Bt + C cos t + D si t (99 iv y = A + Bt + C cos t + D si t. (99 v The characteristic equatio is r + = 0, so we must solve The characteristic roots are r = = e (π+kπi. r = e ( π + kπi where k is a iteger. The roots for k = 0,, are differet, ad all other choices of k lead to oe of these roots. They are k = 0 : r = e πi/ = cos π + i si π = + i k = : r = e πi = cos π + i si π = k = : r = e 5πi/ = cos 5π + i si 5π = i + i i

4 The real form of the geeral solutio of the differetial equatio is therefore y = Ae t + Be t cos t + Ce t si t (99 vi y = Ae t + Be t cos t + Ce t si t (00 Problem Geeral Solutio Solutio with give iitial values (i y(t = A cos t + B si t y(t = si t (ii y(t = A cos t + B si t y(t = cos t (iii y(t = Ae t + Be t y(t = e t e t (iv y(t = Ae t + Be t y(t = e t e t (v y(t = Ae t + Be t y(t = e t e t (vi y(t = Ae t + Be 5t y(t = 5 et e5t (vii y(t = Ae t + Be 5t y(t = (e 5t e t / (viii y(t = Ae t + Be 5t y(t = 5 e t ( e 5t (ix y(t = Ae t + Be 5t y(t = e t e 5t (x y(t = e t( A cos t + B si t y(t = e t( cos t si t (xi y(t = e t( A cos t + B si t y(t = e t si t (xii y(t = e t( A cos t + B si t y(t = e t( cos t + si t (xiii y(t = e t( A cos t + B si t y(t = e t si t (xiv y(t = Ae t + Be t y(t = e t e t (0-i (0-ii (0-iii (0-iv (0-v (0-vi y = + Ae t + Be t y = Ae t + Be t + te t y = A cos t + B si t + 6 t si t y = A cos t + B si t + 8 cos t y = A cos t + B si t + t si t y = A cos t + B si t 8 cos t VECTORS (-i No, x would have to satisfy both x + = ad + x = at the same time, which is impossible. (-iii Solve x + y = ad x + y = to get x =, y =. ( AB = ( is the positio vector of the poit C(,. (5 The order of the letters i ABCD tells you how the corers of the parallelogram are labelled. If ABCD is a parallelogram the the corers of the parallelogram are A, B, C ad D i clockwise order.

5 5 C D D C B B A A ABCD ABCD is a parallelogram if AB + AD = AC. AB = (, AD = (, ( t t +t ABDC AC = (, so AB + AD AC Hece ABCD is ot a parallelogram. (6-i Agai use AB + AD = AC to fid D(, 0, (6-ii Now use AB + AC = AD to get D(,, 5. (8 a = v w, b = v + w. Solve to get v = 6 a + b, w = a + 6 b. (-i Lie l is give by x =. ( (-ii l itersects the x x plae whe x = 0, i.e. whe + t = 0, i.e. whe t =, so at x = ( 0 0 ( 0 Other itersectios: x = ad x =. (-i p = ( b + c/, q = ( a + c/, r = ( a + b/. (-ii m = p + a = ( a + b + c/. (7-i, ii a b = a a b + b ; a b = a a b + b. (7-iii a + b = + ( + 7 ; a b = ( + 7 ; a b = ( + 7. (8 ( a + b ( a b = a vb. (9 AB = (, ( BC = (, AC =. Hece AB =, BC = ad AC = 0. cos A = AB AC 0; cos B = 0 (so this is a right triagle, ad B is the right agle!; = AB AC 0 cos C = 5 0. (0 If A is the right agle, the AB AC so AB AC = 0. Compute AB = (, AC = ( t t, so we wat (t + ( t = 0, ad thus t = 0. I this case C is the poit (0,. If B is the right agle the t = 5 ad C is (5,. If C is the right agle, the the same approach works, but i this case the equatio for t turs out to be quadratic, ad there are two solutios: AC BC = (t (t 5 = t = or t = 5. ( Defiig equatio for l: x + y = (there are may aswers, all of which are obtaied by multiplyig this equatio with a costat. This equatio is of the form x = a where = ( / ad where a is the positio vector for ay poit o l. Normal for l: = ( / (agai, ay multiple of these vectos is also a ormal, ad there are o others. Distace from ay poit x to l is d = ± ( x a, so for the origi set x = 0 ad you get d = = / 5. +.

6 6 Equatio for m: x + y =. Normal vector for m: ay multiple of m = (. m Agle betwee l ad m is the agle betwee their ormals, so it s cos (l, m = m = / 6. ( l is parallel to v = ( ad m is parallel to w = (. Agle betwee the lies is thus the agle (5 (i betwee the vectors v ad w, whihc you fid from cos ( v, w = ; ( v w v w = 65 (ii 0: ( Do t compute aythig for this is the cross product of a vector with itself! 00 (iii. ( 00 (iv. (v k: Same as (iii i i, j, k otatio. (vi k: Same as (iv. (vii i j + k = (. (viii After some pleasat cacellatios you get: j = ( 0 (6 (i, ii Both 0 (cross product of a vector with itself. (iii ( a + b ( a b = a b. (7 False! If a ad b are ay two vectors the with c = a + b you have 0. c b = ( a + b b = a b + b b = a b eve though a c. (8 (i To fid a ormal vector to a plae whe you are give three poits A, B, C i that plae form two vectors i the plae, e.g. AB ad BC, ad compute their cross product. Here we have AB = ( 0, ( BC = 0 = m = ( AB BC = is a ormal. Ay multiple of a ormal is agai a ormal, so there are may correct aswers. (ii Defiig equatio is m ( x a = 0, or, m = m a. So you get x + x x = m a = 8. Here you could replace a with either b or c. You should get the same defiig equatio. If you had computed a differet ormal vector the you would get a differet defiig equatio. But whatever equatio you get, it must be the same as x + x x = 8 after multiplyig with a costat. (iii The distace from ay poit with positio vector x to the plae is ±( m ( x a/ m. For D ad O you get m ( d a m =, m ( 0 a m =. Therefore the distace from D to P is, ad the distace from the origi to P is +. Note that distaces are positive umbers, so we must discard the mius sig! (iv Sice m ( d a/ m ad m ( 0 a/ m have opposite sigs, the poits D ad O (the origi lie o opposite sides of the plae P.

7 (v The area of a triagle ABC is AB BC. We have already computed the cross product, so we fid that the area is. (vi The plae P itersects the axes i the poits (, 0, 0, (0,, 0 ad (0, 0,. (9 (i m = ( AB BC = is a ormal to the plae through A, B, C. Sice m ( d b = 5 0 the poit D does ot lie o the plae. (ii m ( e b = α 5, so the poit E lies o the plae if α = 5 ad ot otherwise. (50 (i AD + AB = AC leads to D(0,, 0. (ii Area of the parallelogram ABCD is AD AB =. (iii The plae P cotais the poits A, B, C, D. Pick ay three of those, for istace A, B, D, ad compute the cross product m = ( AB AD = to fid a ormal vector to the plae P. The defiig equatio is the m ( x a = 0, i.e. x x x =. (iv P itersects the axes i the poits (, 0, 0, (0,, 0 ad (0, 0,. 7

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