Modeling Accumulations: Introduction to the Issues /07/0
The purpose of calculus is twofold:. to find how something is changing, given what it s doing;. to find what something is doing, given how it s changing. We did () geometrically and algebraically. We did () algebraically. Let s do () geometrically!
If you travel at mph for 4 hours, how far have you gone?
If you travel at mph for 4 hours, how far have you gone? Answer: 8 miles.
If you travel at mph for 4 hours, how far have you gone? Answer: 8 miles. Another way: 3 4 (graph of speed, i.e. graph of derivative)
If you travel at mph for 4 hours, how far have you gone? Answer: 8 miles. Another way: Area = 8 3 4 (graph of speed, i.e. graph of derivative)
If you travel at mph for hours, and mph for hours, how far have you gone? Area = +4= 6 3 4 (graph of speed, i.e. graph of derivative)
If you travel at.5 mph for hour, mph for hour,.5 mph for hour, mph for hour, how far have you gone? Area =.5++.5+= 5 3 4 (graph of speed, i.e. graph of derivative)
If you travel at.75 mph for /4 hour,.5 mph for /4 hour,... mph for /4 hour, how far have you gone? Area =.75.5 +.5.5 + +.5 = 4.5 3 4 (graph of speed, i.e. graph of derivative)
If you travel at t mph for hours, how far have you gone? Area = 4 (it s a triangle) 3 4 (graph of speed, i.e. graph of derivative)
Choose another sequence of speeds: 3 4
Choose another sequence of speeds: 3 4
Choose another sequence of speeds: 3 4
Choose another sequence of speeds: 3 4
Choose another sequence of speeds: 3 4
Choose another sequence of speeds: 3 4
Choose another sequence of speeds: y = 8 x,area=??? 3 4
Estimate the area under the curve y = 8 x between x =0andx =4: Area =??? 3 4
Estimate the area under the curve y = 8 x between x =0andx =4: Estimate : pick the highest point Area 8 3 4
Estimate the area under the curve y = 8 x between x =0andx =4: Estimate : pick two points Area +4 = 5 3 4
Estimate the area under the curve y = 8 x between x =0andx =4: Estimate 3: pick four points Area 8 + + 9 8 + = 3.75 3 4
Estimate the area under the curve y = 8 x between x =0andx =4: Estimate 4: pick eight points Area 3 + 8 + + = 3.875 3 4
Estimate the area under the curve y = 8 x between x =0andx =4: Estimate 5: pick sixteen points Area.9875 3 4
Estimate the area under the curve y = 8 x between x =0andx =4: Estimate 6: pick thirty two points Area.7996875 3 4
Estimating the Area of a Circle with r = - 0 -
Estimating the Area of a Circle with r = Divide it up into rectangles: - 0 -
Estimating the Area of a Circle with r = Divide it up into rectangles: - 0 -
Estimating the Area of a Circle with r = Divide it up into rectangles: Estimate area of the half circle (f (x) = p x ) and mult. by. - 0
Estimating the Area of a Circle with r = Divide it up into rectangles: Estimate area of the half circle (f (x) = p x ) and mult. by. height = f(0) = A= - 0 height = f() = 0 base= #rect. Area 4 * = 4* 4*3 4*4 4*5 base=
Estimating the Area of a Circle with r = Divide it up into rectangles: Estimate area of the half circle (f (x) = p x ) and mult. by. - 0 #rect. Area 4 * = 4* 4*3 4*4 4*5
The Method of Accumulations Big idea: Estimating, and then taking a limit. Let the number of pieces go to i.e. let the base of the rectangle for to 0. This not only gives us a way to calculate, but gives us a proper definition of what we mean by area! Also good for volumes and lengths...
A small dam breaks on a river. The average flow out of the stream is given by the following: hours m 3 /s hours m 3 /s hours m 3 /s 0 50 4.5 460 8.5 43 0.5 30 4.5 350 8.5 390 0.5 30 4.75 70 8.75 365 0.75 430 5 50 9 35 550 5.5 030 9.5 300.5 750 5.5 950 9.5 80.5 950 5.75 89 9.75 60.75 50 6 837 0 33 350 6.5 770 0.5 0.5 550 6.5 75 0.5 99.5 700 6.75 658 0.75 88.75 745 7 60 80 3 750 7.5 579.5 75 3.5 740 7.5 535.5 68 3.5 700 7.75 500.75 55 3.75 630 8 460 50 4 550
A small dam breaks on a river. The average flow out of the stream is given by the following: 500 000 500 0 3 4 5 6 7 8 9 0 3
Over each time interval, we estimate the volume of water by Average rate 900 s 500 V = 500m 3 /s*900s 000 500.5.5.7
Over each time interval, we estimate the volume of water by Average rate 900 s 500 000 500 0 3 4 5 6 7 8 9 0 3
Over each time interval, we estimate the volume of water by Average rate 900 s hours m 3 hours m 3 hours m 3 0 35000 4.5 34000 8.5 380700 0.5 07000 4.5 5000 8.5 35000 0.5 79000 4.75 43000 8.75 38500 0.75 387000 5 035000 9 9500 495000 5.5 97000 9.5 70000.5 675000 5.5 855000 9.5 5000.5 855000 5.75 80800 9.75 34000.75 035000 6 753300 0 09700 5000 6.5 693000 0.5 98000.5 395000 6.5 65500 0.5 7900.5 530000 6.75 5900 0.75 6900.75 570500 7 549000 6000 3 575000 7.5 500.5 57500 3.5 566000 7.5 48500.5 500 3.5 530000 7.75 450000.75 39500 3.75 467000 8 44000 35000 4 395000 total=33,39,800
A tent is raised and has height given by xy over the grid where 0 < x < and0< y <. What is the volume of the tent?
A tent is raised and has height given by xy over the grid where 0 < x < and0< y <. What is the volume of the tent? Estimate via boxes! Volume = base *height. x y height = xy volume 0 0 0 0* 0 0 0* 0 0 0* * total volume 0
A tent is raised and has height given by xy over the grid where 0 < x < and0< y <. What is the volume of the tent? Estimate via boxes! Volume = base *height. x y height = xy volume * * * 4 4* total volume 9 0
A tent is raised and has height given by xy over the grid where 0 < x < and0< y <. What is the volume of the tent? Estimate via boxes! Volume = base *height. x y height = xy volume.5.5.5.5 *.5.5.75.75 *.5.5.75.75 *.5.5.5.5 * total volume 4.5 0