Applications of exponential growth and decay



ME-V1 Introduction to vectorsThis document contains sample questions and solutions from the sample unit ME-V1 Introduction to vectors.Identifying scalars and vectorsA scalar is a quantity that has magnitude only. A vector is a quantity that has both magnitude and direction.Identify each quantity as a scalar or vector: length, displacement, area, volume, velocity, acceleration, speed, time, mass, weight, density, force, temperatureScalarVectorLengthDisplacementAreaVelocityVolumeAccelerationSpeedWeightTimeForceMassDensityTemperatureScalarVectorLengthDisplacementAreaVelocityVolumeAccelerationSpeedWeightTimeForceMassDensityTemperatureVector notation Write vector OA as a column vector, an ordered pair and in component form.Column vector: OA=42Ordered pair: OA=(4, 2)Component form: OA=4i~+2j~On a Cartesian plane:Plot two points A and BPlot two additional points C and D and draw vectors AC and DB such that AC=3i~+j~ and DB=-5i~+3j~Multiplying a vector by a scalar If u=5i-2j calculate 4u. 4u=45i-2j4u=20i-8jGeometrically, this can be shown as: On a blank page draw a vector v~. Use this to draw vectors 3v~, -2v~ and 12v~.If u~=3i~+7j~, find 3u~, -2u~ and 12u~.3u~=33i~+7j~=9i~+21j~-2u~=-23i~+7j~=-6i~-14j~12u~=123i~+7j~=32i~+72j~Adding and subtracting vectors Given OA= 8i~-3j~ and OB=3i~+2j~: graph vectors OA and OB on a Cartesian planeFind the vector AB.AB=-5i~+5j~Explain the relationship between AB, OA and OB. AB=?AB=-OA+OB=OB-OAConfirm the relationship numerically.OB-OA=3i~+2j~-8i~-3j~=3i~+2j~-8i~+3j~=-5i~+5j~=ABIf v~=3i~+7j~ and u=-5i~-3j~, calculate v?+uv?+u=3i~+7j~+-5i~-3j~=3i~+7j~-5i~-3j~=-2i~+4j~Consider the vectors u~=3i~-2j~ and v~=i~+4j. Find:u~+ v~u?+v=3i~-2j~+i~+4j=4i~+2j~u~- v~u-v=3i~-2j~-i~+4j=3i~-2j~-i~-4j~=2i~-6j~v~- u~v-u=i~+4j-3i~-2j~=i~+4j-3i~+2j~=-2i~+6j~Confirm the results for a. to c. by geometrically representing these on a Cartesian plane.u~+ v~u~- v~v~- u~If v~=5i~+4j~, u~=-3i~-7j~ and w~=5i~-4j~, match each statement to its component form.StatementComponent form expressionsu~+v~25i~-20j~v~-w~2i~-3j~v~+w~-u~13i~+7j~5w~0i~+8j~-v~15i~+4j~2v~+w~-5i~-4j~StatementComponent form expressions – solutions u~+v~2i~-3j~v~-w~0i~+8j~v~+w~-u~13i~+7j~5w~25i~-20j~-v~-5i~-4j~2v~+w~15i~+4j~Magnitude of a vectorIf u~=8i~-6j~, find the:magnitude of the vector u~u~=82+-62=64+36=100=10unit vector uu=u~u~=8i~-6j~10=810i~-610j~=45i~-35j~If v~=i~+7j~, find the:magnitude vector v~.u~=1+72=1+49=50=52unit vector vv=v~v~=i~+7j~52=152i~+752j~=210i~-7210j~Given OA= 4i~-3j~ and OB=-2i~+3j~, find the magnitude of ABAB=OB-OA=-2i~+3j~-4i~-3j~=-2i~+3j~-4i~+3j~=-6i~+6j~Direction of a vectorA particle is projected from the origin with an initial velocity of 5i~+3j~?ms-1. What is the angle of projection?tanθ=35θ=tan-135=30.96…°A particle is projected from the origin with an initial velocity of -2i~+3j~?ms-1. What is the angle of projection?tanα=32α=tan-132θ=180-tan-132=123.69…°Scalar (dot) productu~=-3i~-7j~, If v~=5i~+4j~ and w~=5i~-4j~:Calculate u~?v~u~?v~=x1x2+y1y2u~?v~=-3?5+-7?4u~?v~=-15-28u~?v~=-43Find the angle between v~ and u~u~.v~=|u~||v~|cosθcosθ=u~.v~|u~||v~|u~=-32+(-7)2=9+49=58v~=52+42=25+16=41cosθ=-4358?41θ=cos-1-4358?41=151.85…°Show the distributive property holds, i.e. v~?u~+w~=v~?u~+v~?w~u~+w~=-3i~-7j~+5i~-4j~=2i~-11j~v~?u~+w~=5?2+4?-11=10-44=-34v~?u~+v~?w~=5?-3+4?-7+5?5+4?-4=-15-28+25-16=-34A, B and C are points defined by the position vectors a~=i~+3j~, b~=2i~+j~ and c~=i~-2j~ respectively. Find the size of ∠ABC.∠ABC is the angle between vectors BA and BCu~.v~=|u~||v~|cosθcosθ=u~.v~|u~||v~|, or in our case, letting ∠ABC=θ, cosθ=BA .BC|BA ||BC|BA= a~- b~=i~+3j~-2i~+j~=-i~+2j~BC= c~- b~=i~-2j~-2i~+j~=-i~-3j~BA .BC=-1×-1+2×-3=-5|BA |=-12+22=5|BC |=-12+(-3)2=10cosθ=BA .BC|BA ||BC|cosθ=-55×10cosθ=-55×5×2cosθ=-55×2cosθ=-12θ=cos-1-12=180°-cos-112=180°-45°=135°∴∠ABC=135°If v=i~+7j~ and u=8i~-6j~, find u~?v~ and hence find the angle between u~ and v~ (to the nearest degree). u~?v~=x1x2+y1y2u~?v~=1×8+7×-6u~?v~=-34u~?v~=u~v~cosθ-34=50×100cosθ-34=502cosθ-17252=cosθθ=119oParallel and perpendicular vectors If v~=6i~+4j~, u~=2i~-3j~, w~=12i~-18j~Show u~?and v~ are perpendicularu~.v~=x1x2+y1y2=2×6+-3×4=12-12=0∴u?and v are perpendicularShow u~ and w~ are parallelShow w~=λu~ or if show x1x2=y1y2x1x2=122=6y1y2=-18-3=6∴u~?and ?w~ are parallel.Alternate method: Show u~?w~=|u~||w|?or?u~?w~=-|u~||w~|If a~=xi~+2j~ is perpendicular to vector b~=2i~-j~, find x.a~?b~=0 as they are perpendicular.x×2+2×-1=02x-2=02x=2x=1If a~=-6i~+yj~ is parallel to vector b~=2i~-4j~, find y.Two vectors are parallel if they are scalar multiples of one another.y-4=-62y-4=-3∴y=12Vector projection Let a~=i~+3j~ and b~=i~-j~, find the scalar projection of a~ onto b~.Scalar projection of a~ onto b~=|a~|cosθNote: The scalar projection will be negative as a~ opposes b~cosθ=a~.b~|a~||b|a~cosθ=a~.b~|b~|a~cosθ=1×1+3×(-1)12+-12a~cosθ=-22a~cosθ=-2find the vector projection of a~ in the direction of b~.projb~a~= scalar projection of a~ onto b~×the unit vector of b~=-2×i~-j~2=-i~+j~Let a~=i~+3j~ and b=i~-j~, find the vector projection of a~ in the direction b~. a~?b~=i~+3j~ i~-j~a~?b~=1×1+3×-1a~?b~=-2b~?b~= i~-j~ i~-j~b~?b~=1×1+(-1)×-1b~?b~=2projb~a~=a~?b~b~?b~b~projb~a~=-22 i~-j~projb~a~=-i~+j~Proving geometric resultsThe diagonals of a parallelogram meet at right angles if and only if it is a rhombus (ACMSM039)u+v?u-v=u. u-v.v=u2-v2If the diagonals meet at a right angle then u+v?u-v=0∴u2-v2=0u2=v2u=v∴ The parallelogram is a rhombusThe midpoints of the sides of a quadrilateral join to form a parallelogram (ACMSM040)Part 1: First show the line joining the midpoints of two sides of a triangle is parallel to the third side and is half its length.LM=LJ+JM=12IJ+12JK=12IJ+JK=12IKPart 2: Consider quadrilateral ABCD with midpoints F, G, E and H.FE=12BD and GH=12BDFG=12AC and EH=12AC∴ FGHE is a parallelogram as opposite sides are equal in length.The sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sides (ACMSM041)Proof using vectors:u+v2+u-v2=u+v?u+v+u-v?u-v=u?u+u?v+v?u+v?v+u?u-u?v-v?u+v?v=2u?u+2v?vu+v2+u-v2=2u2+2v2 i.e. the sum of the squares of the length of the sides∴ The sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sidesProof without vectors:Consider the cosine rule for angle C in triangle DBC:cosC=s22+s32-d122s2s3Consider the cosine rule for angle D in triangle DAC:cosD=s32+s42-d222s3s4cosD=cosπ-C=-cosC (Cointerior angles are supplementary when BC||AB)-cosC=s32+s42-d222s3s4cosC=-s32-s42+d222s3s4s22+s32-d122s2s3=-s32-s42+d222s3s4sub s2=s4 (Opposite sides of a parallelogram are equal)s22+s32-d122s4s3=-s32-s42+d222s3s4Multiply both sides by 2s3s4s22+s32-d12=-s32-s42+d22s22+s32+s32+s42=d12+d22sub s1=s3 (Opposite sides of a parallelogram are equal)s22+s12+s32+s42=d12+d22s12+s22+s32+s42=d12+d22∴ The sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sides.Modelling motionIn still water, Jacques can swim at 3 m/s. Jacques is at point A on the edge of a canal, and considers point B directly opposite. A current is flowing from the left at a constant speed of 1 m/s. If Jacques dives in straight towards B, and swims without allowing for thecurrent, what will his actual speed and direction be? (answers correct to 1 decimal place)Direction is the deviation of the desired path.tanθ=13θ=tan-113=18.4349…°≈18.4° correct to 1 decimal placeLet S=speed of travel. Using Pythagoras’ Theorem,S2=32+12S=10=3.1622…≈3.2mscorrect to 1 decimal placeJacques wants to swim directly across the canal to point B. At what angle should Jacques aim to swim in order that the current will correct his direction? And what will Jacques’ actual speed be? (answers correct to 1 decimal place) Direction Jacques should aim to swim:sinθ=13θ=sin-113θ=19.4712…°θ≈19.4° (correct to 1 decimal place)Let S=speed of travel. Using Pythagoras’ Theorem,32=S2+129=S2+18=S2S=8=2.824…≈2.8 m/s (correct to 1 decimal place)Projectile motionA projectile is launched off a cliff 15m above the ground at an angle of 30° to the horizontal with an initial velocity of 12 m/s. Find:Initial velocitiessin30=uy12uy=12sin30=6 m/scos30=ux12ux=12cos30=12×32 =63 m/sHorizontal componentsx=0x=0?dt=c1When t=0, x=63 63=c1∴x=63x=63?dt=63t+c2When t=0, x=00=63×0+c2c2=0∴x=63tVertical componentsy=-9.8y=-9.8?dt=-9.8t+c3When t=0, y=6 6=-9.8×0+c3c3=6∴y=-9.8t+6y=-9.8t+6?dty=-4.9t2+6t+c4When t=0, y=1515=-4.9×02+6×0+c4c4=15∴y=-4.9t2+6t+15Note: The projectile motion Geogebra applet can be used to check solutions for a, b and c.The time of the flightThe time of the flight is the time taken for the projectile to hit the ground, i.e. the vertical displacement is zero.Solve y=-4.9t2+6t+15 for t when y=00=-4.9t2+6t+15Use the quadratic formula with a=-4.9, b=6, c=15t=-6±62-4×-4.9×152(-4.9)=-6±330-9.8t=-6-330-9.8, -6+330-9.8t=-1.2414…, 2.4659… ∴ the time of the flight is 2.47 seconds (correct to 2 decimal places)The solution can be checked by graphing y=-4.9t2+6t+15 using graphing software. (use x in place of t in the graphing software)The range of the projectileThe range of the projectile is the horizontal displacement when the projectile strikes the ground.Solve x=63t for x when t=-6-330-9.8x=63×-6-330-9.8x=25.6264…∴ the range of the projectile is 25.63m (correct to 2 decimal places)The final velocity of the projectileTo find the final velocity we need to obtain the x and y velocity components when t=-6-330-9.8, the time when the projectile hits the ground.The x component of the velocity is a constant: x=63.To find the y component of the final velocity, solve y=-9.8t+6 for y when t=-6-330-9.8.y=-9.8×-6-330-9.8+6y=-330 m/s or -18.1659 m/sNote: The negative signifies the object is travelling downwards.Final velocityvfinal2=632+3302=438vfinal=438=20.9284…tanθ=33063θ=tan-133063=60.2271…∴ the final velocity is 20.93m/s at an angle of 60.23° to the horizontal. (correct to 2 decimal places)The maximum height the projectile reachesThe maximum height of the projectile is the vertical displacement when y=0Step 1: Solve y=-9.8t+6 for t when y=0.0=-9.8t+6-6=-9.8t-6-9.8=tt=3049or 0.6122…Step 2: Solve y=-4.9t2+6t+15 for y when t=3049y=-4.930492+63049+15=16.8376…≈16.84 (correct to the nearest centimetre)Derive the Cartesian equation of the trajectory (y as a function of x)x=63tt=x63Substitute t=x63 into y=-4.9t2+6t+15y=-4.9x632+6x63+15y=-4.9x2108+3x3+15A projectile is launched from the ground at an angle of θ° to the horizontal with an initial velocity of u m/s. Assuming no air resistance, show the maximum range is achieved whenθ=45° orπ4.Initial velocitiessinθ=uyuuy=usinθcosθ=uxuux=ucosθHorizontal componentsx=0x=0?dt=c1When t=0, x=ucosθucosθ=c1∴x=ucosθx=ucosθ?dtx=ucosθt+c2When t=0, x=00=ucosθ×0+c2c2=0∴x=utcosθt=xucosθVertical componentsy=-9.8y=-9.8?dt=-9.8t+c3When t=0, y=usinθusinθ=-9.8×0+c3c3=usinθ∴y=-9.8t+usinθy=-9.8t+usinθ?dty=-4.9t2+usinθt+c4When t=0, y=00=-4.9×02+6×0+c4c4=0∴y=utsinθ-4.9t2Derive the Cartesian equation of the trajectory by substituting t=xucosθy=uxucosθsinθ-4.9xucosθ2=xtanθ-4.9x2u2cos2θThe range of the projectile is the horizontal displacement, x, when the vertical displacement, y, is 0.0=xtanθ-4.9x2u2cos2θ0=xtanθ-4.9xu2cos2θSolve tanθ-4.9xu2cos2θ=0u2cos2θtanθ-4.9x=0u2sinθcosθ-4.9x=0u2sinθcosθ=4.9xx=u2sinθcosθ4.9x=u22sinθcosθ9.8=u2sin2θ9.8The maximum value of the functions is when sin2θ=1 as -1≤sin2θ≤1sin2θ=12θ=90°θ=45° orπ42011 Mathematics Extension 1 HSC, question 6, part bThe diagram shows the trajectory of a ball thrown horizontally, at speed v ms–1, from the top of a tower h metres above ground level.The ball strikes the ground at an angle of 45°, d metres from the base of the tower, as shown in the diagram. The equations describing the trajectory of the ball are x=vt and y=h-12gt2, (Do NOT prove this.) where g is the acceleration due to gravity, and t is time in seconds.Prove that the ball strikes the ground at time t=2hg seconds.The ball will strike the ground when y=0Solve y=h-12gt2 for t when y=00=h-12gt212gt2=hgt2=2ht2=2hgt=±2hg ∴The ball strikes the ground at time, t=2hg (t must be positive)Hence, or otherwise, show that d=2h.d is the horizontal displacement when , t=2hg, x=vtThe ball strikes the ground at 45° so the horizontal and vertical compoants of velocity are equal in magnitude.tan135=yx-1=yxx=yHorizontal component of velocityx=ddtx=ddtvt=vVertical component of velocityy=ddty=ddth-12gt2=-gtx=-y, ∴v=gtSubstitute v=gt into (1)x=gt2Substitute t=2hg to find the horizontal displacement when the ball strikes the ground, d.x=g2hg2=g×2hg=2h, ∴d=2h ................
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