Exam 1 Spring 23 Answer

Exam 1 Spring 2023 constant 1(a) sin() f(x) f(x) + cos.crx = 2 sin + = ↓ 0+ sinixn sin() sinxx = X simplify Needto Do Not 1(b) f(x) cos(sinx) = f'(x) sin(sinx).cOSX = (c) f(x) COSX.siux = *Cos2x-sin2x f'(x) 10sX.CosX sinx(sinx) = + = s crep & 1(d) f(x) tan8 tan = xs 7 f(x) 8 (e) f(x) cox (20sx) 5 6 - = If f(x) f'(x) (gy sex cos - sinX. E - E = cos(6x) - - (3x) sin + (4x) + - 1 sin + bsin(6x) 3c0s(3x) +4cos(4x) sins - 6 sinb I 3c0s3 + ⑧ z 0 0 + I - 7 2π I 4c0s5 + 4. above + 4cos + ⑧ Tsin?- bsin++ 3c0s see = = sin().E + Tsin(7x) = secx.tanx = c0SX = - = = = . y = = sincosysin f(x) secx-taux => = = f(x) cos(ix) f'(x) or /CosX---sink - = ·,f(x) f - -20x - 4x = = f'(x) 2(a) is sect. tan = *1s see 5. above - 2 1 = - 2 1 = - z= Match 2(b) H(x) c0s"(2x) (6x) 2sinx sin = + + (cos(2x)) +sin (6x) 2sinx + = Hx zcos(2X).(zsin(2x))-z 6c0s(6x) + = H'5 = - 4c0s 2. .sin2 E3 z 1 =- = = y(t)(z) 15 3(a)(8X x6 y(z prep(E 5.5" = + y 2 + 1 * . sinx 3(c)(X- 1"yX2 5x 1- + c 4++ " yy yi c + + secxtanxx .secx + c + 1 - - + + 6X" 5x + + dx 6x + tanx-8sinX -cosX + Match! + x = 6 x 5 yx + + + + yEax - + j,x px = 3(b)/sec x -8cosX -- 5 ((6 x-+x- 1 y.- - + + + + +2cos 18 bcoST +2cos 6( 1 + 13 6 - 2cosX + - - 6c0s6. - .sin 4cos + + zdx "(x+ 1 + (XY = (xy = = 1 - + 1 - - - 1dx X47 - - x 4dX + c (5+5y3 = 1 - c + xyxx PIit(*t 3(d)(X 5x2t - x+1dx (X - = 5 x - Diagram 7 Track given - (x (x- + - dx + z- z c + 1 - c + Train x P y x sx 1 1) 4. - 7x + + = dX XM 1 = x+xX2e - ⑦ ( ↑ ④ 10 z Station Variables Let x Distance between train Distance between = z = O Angle between = Given = E Equation train track station and line and connecting the train the station ft sec X6 Find pointP and ?when z 20 = = tanf= Differentiate attanE g) = 2 sect >se-8. E - = 10x) 2 10X e ExtraSolvable Information ? 1(2012 (10) - = = 100 100 = 3002055 1055 = = ⑦ 10 ↓" Substitute 2 20 500 20 . 10.6 It(800) < dE - = set 0 = E = = = L at (20)2 300 = - 0 s00 (0001" d0 =-60. Solve F= = per second Radians negative, decreasing > Answer angle This -3 Radians 20 5. f(x) 33 = Antidifferentiate k - decreasing is this secondat every f(4) 2 + and = - (f'(x)dx (93 5 f(x) - (32X - 6 + 5 3 = - Moment. 2dx = = of rate a at makes sense x 2dx + 322 X, 16 = - = - 16 - 2/x + 2x 2 + 2x C + + X2 Test Initial Condition 2 12 - f(x) = - 2, 2.y (- + + 6 16 - 1 - 4 8 C + + 3 c - = = - + ↳ Finally, f(x) Max 1 2 6 6 c = 25x + 2x - = - - q q Height a(t) 1 6. 80 ft. building s(0) 80 = v(t) V V = 32 - ft. V(0) 64f +1se + 32z = - 64 = = Find tmax=? FindMaxHeightLe S(tmax) =? SLA) - = = Findtimpact =? V(timpact) =? Find 32 + - 64 + 1bt" 16t3 vS6 5. 7 + + 64t 80 + + 80 (a) MaxHeightis when v(t) = - V(t) 327 64 + Heightoccurs Max (b) Max Heightoccurs set 0 = = 7 => = tmax2 = sec. seconds Imax when 327 64 Solve = t=2 at s(tmax) s(2) => 0 = =2 64 16(2)" 64(2) + 80 - + = 64 128 + feet 80 144 + = Max Heightis 144feet (c) strikes ground s(t) = - when 16+" S(t) = 64t 80 + + 0 = = - 16t2 16(t 4+ - - 5 set 5)(t+1) 0 = - - t 5 0t 1 = + - t t5 = Hits (1) groundafter Velocity = - 1 ignore timpactseconds impactv(timpact) at Impactvelocity is 0 = - 90 v(3) = feetper = 32.(5) 64 + = - 160 64 + = - 1 second 96ft(s negative sense. makes DOWN @ impact.