Chapter6 - Chapter 6 Applications of Integration MAT186H1F...

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Chapter 6: Applications of IntegrationMAT186H1F Lec0103 BurbullaChapter 6 Lecture NotesNine LecturesFall 2014Chapter 6 Lecture NotesNine LecturesMAT186H1F Lec0103 BurbullaChapter 6: Applications of IntegrationChapter 6: Applications of Integration6.1 Velocity and Net Change6.2 Regions Between Curves6.3 Volume by Slicing6.4 Volume by Shells6.5 Length of Curves6.6 Surface Area6.7 Physical Applications6.8 Logarithmic and Exponential Functions Revisited6.9 Exponential Models6.10 Hyperbolic Functions: An OverviewChapter 6 Lecture NotesNine LecturesMAT186H1F Lec0103 Burbulla
Chapter 6: Applications of Integration6.1 Velocity and Net Change6.2 Regions Between Curves6.3 Volume by Slicing6.4 Volume by Shells6.5 Length of Curves6.6 Surface Area6.7 Physical Applications6.8 Logarithmic and Exponential Functions Revisited6.9 Exponential Models6.10 Hyperbolic Functions: An OverviewNet Distance and Total Distance TravelledSupposesis the position of a particle at timetfort[a,b].Thenbav dt=bas(t)dt=s(b)-s(a).s(b)-s(a) is called the displacement, or net distance travelled, bythe particle over the time interval [a,b].This is in contrast to thetotal distance travelled, which is given by the integral of speed.That is, the total distance travelled by the particle fort[a,b],isba|v|dt.Chapter 6 Lecture NotesNine LecturesMAT186H1F Lec0103 BurbullaChapter 6: Applications of Integration6.1 Velocity and Net Change6.2 Regions Between Curves6.3 Volume by Slicing6.4 Volume by Shells6.5 Length of Curves6.6 Surface Area6.7 Physical Applications6.8 Logarithmic and Exponential Functions Revisited6.9 Exponential Models6.10 Hyperbolic Functions: An OverviewDerivation of Total DistanceIfv*iis the velocity of the particle at some timet*iin the timeinterval [ti-1,ti],then the speed|v*i|is approximately constant onthe time interval. So on this time interval the distance travelled isapproximately|v*i| ×Δt,sincedistance = speed×time.So the total distance travelled islimn→∞ni=1|v*i| ×Δt=ba|v|dt.Chapter 6 Lecture NotesNine LecturesMAT186H1F Lec0103 Burbulla
Chapter 6: Applications of Integration6.1 Velocity and Net Change6.2 Regions Between Curves6.3 Volume by Slicing6.4 Volume by Shells6.5 Length of Curves6.6 Surface Area6.7 Physical Applications6.8 Logarithmic and Exponential Functions Revisited6.9 Exponential Models6.10 Hyperbolic Functions: An OverviewExample 1Supposev=t2-11t+ 24 fort[0,8].Then80v dt=80(t2-11t+ 24)dt=t33-112t2+ 24t80=323This means that after 8 seconds the particle is 32/3 units to theright of where it started from.Chapter 6 Lecture NotesNine LecturesMAT186H1F Lec0103 BurbullaChapter 6: Applications of Integration6.1 Velocity and Net Change6.2 Regions Between Curves6.3 Volume by Slicing6.4 Volume by Shells6.5 Length of Curves6.6 Surface Area6.7 Physical Applications6.8 Logarithmic and Exponential Functions Revisited6.9 Exponential Models6.10 Hyperbolic Functions: An OverviewThe total distance is a much more complicated calculation becausev>0 fort<3; butv<0 fort>3.Thus the total distancetravelled is80|v|dt=30v dt-83v dt=t33-

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