Exam 1 Spring 2022

Exam 1 Spring 2022 constant Ila ) FIX )= f- 1lb ) cost cost cost + = 1×1=0+1-2 ( iosrx )¥ f- sink ) ' - % Do Not - simplify Need to flxkcoslsinx ) f 'lH= Kc ) cost + fix)= - sinlsinx ) cosx . cosx . sin 0% f- ' 4) cosx.com/tsinx(-sinx)--cosZx-sin2x = prep / (d) costs FIX)= ¥6 f- 1×1=5 0¥ #) -4in 1¥ Cos 210 = cos 1¥ ) " sin - (7×-6) ' -142×-7 # 7 10517×7 78 fix )= tank) - 1×1=18 ' . . ✗ f- cos = " ' 114 5 ¥6 cos = cos 17×7 "8 tanbx.tsinGXD-7-coshxtsec43xj.is . tank) (tan 13×1/2 - OR , 2cal 1-1×1 f- 13×1 . ' ' ' tan / 2×7 sink) -_ f- 4) tan = tan / 2×1 . cos 12×1.2 Ttg 1¥)= tan / ¥) 2- - / It ) cos 2. V3 =tan) = 1- sin / 2×1 - -2 + Ez + g / ¥1.2 53.1-44 Ey Sec . + 42×1.2 sink .tt/.sec2(2.Ig).2 Ik cos - sin (2)21 a see 4.4=53+453--553 "É=c¥z sec 2 -2 = = 2 < 2lb) f- 1×1=2 sink ) ( ' sinx 2 = f- 1×1=2.2 sinx " cos ✗ - 214 /¥ ) ' HIX / H' 1×1 H = H' (F) 3 (a) /§ ✗ St - = - % ✗ 14×1 ' 4. ¥) = ¥ t f. 5-6×5 % fseix = E. ✗ + .4= ¥ + + + É! ¥ - % ✗ + 6 ✗ + % + secx - 8 cosx tanx "2= 2 . 2 - 3141×2+1×-2 8 + sinxt sinx - rx + ¥ dx DX . ¥ ✗ + kg ✗ " , = -65¥ -23×4 + - + c +C DX tjsecx ¥ { E. tanx 7 cos ✗ + } 4- -4 G- +5-6×-6-6 Éˢdx ±Y÷ ¥± . - 65×6 ✗ + 35-6×6+6,7 = 2% 4=-14%1.4%-253 . 141¥ sin , PEP 3lb) sin - % ' cos × . 14×1 cos = sinx 45*14) .ws#-)=4/Ey.Fy---TzTz-- = = 4 E E f = / J +C ✗Txt % ✗ % + ¥e+¥r×d✗ + ✗ - + ✗ %+ ' % ✗ DX =¥÷±×÷±i÷±i÷+c = 3- % ✗ + § % ✗ 2 _ ✗% 2 _ % 3✗ + C ' = 2 " split Hd ) g. 2 ×dx= 1¥ ¥ ✗ = Diagram > , , " ✗ h ✗ + "" given 15/14 =a¥ ¥1,4 I + ° +c Train ✗ P ax + 181, DX Fg ✗ 1%+1,43×1%1 Track =) dx + 34 =) 4. ✗ 0-1 n < 10 z Station Variables let ✗ = Distance = Distance 2- ① = Given 0¥ Equation + an -0 and point P train and station between Angle between track line and connecting the Train the station 6 1- +1sec dd¥= Find between train = ? when f- 2sec . ✗ = 6ft ' / sef 1244=12 feet = Differentiate ¥ tano =¥ ' seco > 10×-1 seio.ae#t=-l0x-?dI-t Extra Solvable Information 12 8 z=fÉ=FF4=F44 10 Substitute f z F44 12 > ? - . = ¥4 - , 6 see -0 __%s¥,=¥=F¥ :# ¥¥¥+=-÷y save . I¥= -60 - = zyy ¥ Radians negative decreasing > Answer angle This Radians 1% , f- 5. ' 1×1=2 sinx - of rate a at this Moment . (1-1)=-6 f- 5 cosx at second every makes sense , decreasing is per second Anti differentiate , fly )= / ' f- A) dx given Use =) Zsinx to value ft)= finally , f- 1×1=-2 Max 6. 48ft building . 2 | for solve I - - Scosxdx - -5 iosx Ssinx - - Zcosx - Ssinxtc tc 0 cost = sinttc set - -6 = > - 2 V10)= 32ft / see find tmax = vltt Max Max Height Height ↳ -32++32 sltk -16T sltmax) set at t -4 second occurs when tmax sltmax)=sH)= Max Velocity at Height is time t= V01 - 72 + V. t + sits -16-12+32++48 vlt )=O when occurs + ? ult 1=-321-+32--0 Height 32T - = Max c- -8 " . = is -6 at -4=-32 Slo)= 48ft Height = -8 Height Find Max Otc - > Solve 321=32 tmax =\ second . =L -16+32+48=16+48--64 feet 64 feet 2 Makes sense v12)= -32 (2) +32=-641-32 ↳ Ball is travelling Down so = -32ft / see velocity negative .