Unit 8 Day 1 Integral as Net Change

Transcription

1 Unit 8 Day 1 Integral as Net Change

2 Free Response Practice Packet p.1

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4 Terminology Displacement of an object is the distance from an arbitrary starting point at the end of some time interval. Actual Distance Traveled is how far an object travels regardless of direction.

5 Review: Position and Velocity If s(t) is the position of an object then the velocity of the object will be s (t). The area between the velocity curve and the -ais represents the total displacement or change in position. b Displacement s '( t) dt v t dt Eample Sketch: a b a

6 Review: Position and Velocity To find the actual distance traveled integrate the absolute value of the velocity function. Actual Distance b a v t dt Eample Sketch:

7 Velocity measured in feet per second: 3 v( t) t 10t 4 t on t [0, 6] When will the object be stopped? Use a sign line to determine when the object is moving with positive displacement and negative displacement.

8 Eample: 3 v( t) t 10t 4 t on t [0, 6] Use integration to determine the net or total displacement.

9 Eample: 3 v( t) t 10t 4 t on t [0, 6] Draw the path of the particle on a number line. Then determine the total distance traveled. Use integration to find the total distance traveled.

10 Generalization t t 1 ' s ( t) dt s( t ) s( t ) 1 Integral of the rate of change Net, or total, change over [t 1,t ]

11 Accumulation Eample Rate of Change Function N (t) Real Life Meaning Industry Rate at which pollutants enter a lake, measured in pounds per month b a Interpret rate of change function dt N'(t) dt 0 Total number of pounds of pollutants that enter the lake over a period of months The rate at which pollutants enter a lake from a factory is 3/ N '( t) 80t where N is the total number of pounds of pollutants in the lake at time t. How many pounds of pollutants enter the lake in 16 months? What is the average rate pollutants enter the lake over the 16 month time period? Include units.

12 Eample Problem Starting at 7:00 water leaks from a tank at a rate of (+0.5t) gal/hr. How much water leaks out between 9:00 and 11:00?

13 Eample Problem # Using data table and trapezoidal approimation. The number of cars passing an observation point is called the rate of traffic flow in cars/hour. The table below represents the traffic flow for a road. Use the data to estimate the number of cars using the road from 7:00-9:00. Time 7:00 7:15 7:30 7:45 8:00 8:15 8:30 8:45 9:00 Rate r(t) 1,044 1,97 1,478 1,844 1,451 1,378 1,

14 Eample Problem # Using data table and trapezoidal approimation. Time 7:00 7:15 7:30 7:45 8:00 8:15 8:30 8:45 9:00 Rate r(t) 1,044 1,97 1,478 1,844 1,451 1,378 1, Solution:

15 Practice (Recommend Doing at Some Point) Packet p. Moving right=positive displacement Moving left=negative displacement BE SURE to TRY this PAGE!!!

16 Answers: Packet p. 1a. Stopped at t = 0, π/3 seconds 1b. Moving right (0, π/3) 1c. Moving left (π/3, π/]. Distance moved right 3. Distance move left 4. Displacement from origin = m 5. Total distance = 6 m

17 Practice packet p.3 1a. Stopped at t = π/, 3π/ seconds 1b. Displacement = 0 m 1c. Total distance = 0 m a. Stopped at t = 0, π/, 3π/, π seconds b. Displacement = 0 m c. Total distance = 0/3 m 3a. Stopped at t = 0,, π seconds 3b. Displacement = m 3c. Total distance = m

18 Applications of Integrals AREA BETWEEN CURVES

19 Recall: Integration is used to find the area between the curve and the -ais. S b a f d

20 Today... finding the area between two curves f ( ) g( ) 0 [ a, b] Given for then the area of region A is: A = b A y y a b a top bottom d f g d

21 NOT really any different.... Area between curve and -ais can be thought of as area between curve f() and curve g()=0. = b A y y d b a b a top bottom f g d = f ( ) 0 d f ( ) d a a b

22 Find the area between these two equations. First consider a very thin vertical strip. ytop The length of the strip is: ytop ybottom or ybottom Since the width of the strip is a very small change in, we could call it d.

23 ytop d y top y bottom ybottom Repeat with more thin strips then add them together: Since the strip is a long thin rectangle, the area of the strip is: length width d d 9 1 Note: The boundaries of integration are the -coordinates of the intersection points of the equations.

24 Eample Problem #1: Calculate the area of the region between f ( ) 4 10 and g( ) 4 over the interval [1,3] Step #1 Determine if f()>g() or g()>f(). For this problem we will use the calculator!!

25 Eample Problem #: Calculate the area of the region between g ( ) 4 and f ( ) Step #1 Determine whether f()>g() or g()>f(). For this problem we will draw the sketch by hand!!! Hint: You will need to find where the two functions intersect to find the integration boundaries!!

26 E: Calculate the area between f ( ) 5 7 and on [-,5] g( ) 1 NO CALCULATOR! Trick, tricky! A. Graph f() find and y-intercepts, minimum value (there is one because the function is an upwards facing parabola!) B. Graph g() find the and y-intercepts C. Find the intersection of the two curves by setting them equal to each other.

27 E: Calculate the area between on [-,5] and A. Graph f() -intercepts: (-1.14,0) and (6.14, y-intercept: 0), (0,-7) minimum: (.5, -13.5) B. Graph g() -intercept: (1,0) y-intercept: (0,-1) C. Find intersection points : (5,-7) and (1, -11) SKETCH: g( ) 1 f ( ) 5 7

28 Why should always have all the fun? Let s integrate with respect to y b dy a a b d value value function in d y value y value function in y dy

29 Find the area bounded by y by the line y, the -ais, and y y

30 y If we try vertical strips, we have to integrate in two parts: d d y 4 d d 0

31 y d d y yleft We can find the same area using a horizontal strip. dy yright

32 yleft dy yright Since the width of the strip is dy, we find the length of the strip by solving y y for in terms of y. y y 0 y length of strip y dy 10 3 NOTE: The boundaries of integration are the y-coordinates of the intersection points of the equations. width of strip

33 Sometimes dy Integration Works Best ending y starting y ( right left) dy Both equations need to be in terms of y. This means solve both equations for. 3 y = = 3 y y = + = y -

34 Using the Calculator as a Tool Watch the demo and try on your calculator y ( y ) dy 4.15

35 Summary Use the way that gives you the easiest integral to evaluate ending ending y ( top bottom) d ( right left) dy starting starting y Both equations need to be in terms of. This means solve both equations for y. Both equations need to be in terms of y. This means solve both equations for.

36 General Strategy for Area Between Curves: 1 Sketch the curves. Decide on vertical or horizontal strips. (Pick whichever is easier to write formulas for the length of the strip, and/or whichever will let you integrate fewer times.) 3 Write an epression for the area of the strip. If the width is d, the length must be in terms of. If the width is dy, the length must be in terms of y. 4 Find the limits of integration. (If using d, the limits are values; if using dy, the limits are y values.) 5 Integrate to find area.