section one
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Section One. Precalc Review, Average Value, Exponential Growth & Decay, and Slope Fields. By: RE, Rusty, Matthew, and D Money. Unit Circle. Geometrical figure relating trig functions to specific angles and coordinates. Unit Circle. - PowerPoint PPT PresentationTRANSCRIPT
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Section One
Precalc Review, Average Value, Exponential Growth & Decay, and
Slope Fields
By: RE, Rusty, Matthew, and D Money
![Page 2: Section One](https://reader036.vdocuments.mx/reader036/viewer/2022062520/56815f5a550346895dce3b9f/html5/thumbnails/2.jpg)
Unit Circle
• Geometrical figure relating trig functions to specific angles and coordinates
![Page 3: Section One](https://reader036.vdocuments.mx/reader036/viewer/2022062520/56815f5a550346895dce3b9f/html5/thumbnails/3.jpg)
Unit Circle
Angle is in radians and measured counterclockwise from 0 radians.
Coordinates are always fractions of 1; 1 is the radius of the unit circle
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Practice Problem 1
• Graph the following equation using only skills from pre-calc:
• Y=4/(x^2-9) • Vertical asymptotes are found by setting
denominator equal to 0, then solving for x• (x^2-9)=0 • (x+3)(x-3)=0• X=3,-3
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Practice Problem 1
• Y=4/(x^2-9)• Vertical asymptotes at x=3,-3• Parent graph of 1/(x^2) – Stretched vertically fourfold
• Test values on either side ofeach asymptote• X=0, y=?; x=-4, y=?; x=4, y=?• (0, -4/9) negative; (-4, 4/5) positive; (4, 4/5)
positive
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Solution 1
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Practice Problem 2
• Ratio of donkeys to people in Kazakhstan – 1:3• Donkey population of Kazakhstan in 1900 – 7• Growth rate of people in Kazakhstan - .23857• How many donkeys were in Kazakhstan in
1999?– Assume that the donkey-people ratio remains
constant, as it generally does
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Practice Problem 2
• Determine human population of Kazakhstan in 1900– Donkeys/People=1/3– 7/P=1/3– P=21 people
• Use growth equation with variables– y=Ye^kt– y is people in 1999; Y is people in 1900; k is growth
rate of people; t is time in years
0y
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Practice Problem 2
• y=(21)e^(.23857*99)• y=379,815,876,900 people– They reproduce rather quickly
• How many donkeys?– Donkeys/People=1/3– D/379,815,876,900=1/3
• D=126,605,292,300 donkeys in Kazakhstan in 1999
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126,605,292,300 DONKEYS!!!
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Practice Problem 3
• Find the average value of f(x) on the interval [7,20]
• Given:
2017)(
33
12)(
7
dxxf
dxxf
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Practice Problem 3
20
7
20
3
7
3
)()()( dxxfdxxfdxxf
• Find the value of the integral of f(x) from 7 to 20 in terms of known values
20
3
3
20
)()( dxxfdxxf
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Practice Problem 3
20
7
3
20
7
3
)()()( dxxfdxxfdxxf
• Find the value of the integral of f(x) from 7 to 20 in terms of known values
=-(-17)-(-12)=29
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Practice Problem 3
• Use average value formula:
• Plug in variables:
• Now, solve:
=1/13(29)=29/13
20
7
)(720
1 dxxf
b
a
dxxfab
)(1
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Practice Problem 4
• Graph the slope field of the differential equation dy/dx=y-x
• Make a table of values
• X: 0 0 0 0 0 0 0• Y: 0 1 2 3 -1 -2 -3• Dy/dx: 0 1 2 3 -1 -2 -3
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Practice Problem 4
• X: 1 1 1 1 1 1 1• Y: 0 1 2 3 -1 -2 -3• Dy/dx: -1 0 1 2 -2 -3 -4• X: 2 2 2 2 2 2 2• Y: 0 1 2 3 -1 -2 -3• Dy/dx: -2 -1 0 1 -3 -4 -5• X: 3 3 3 3 3 3 3• Y: 0 1 2 3 -1 -2 -3• Dy/dx: -3 -2 -1 0 -4 -5 -6
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Practice Problem 4
• X: -1 -1 -1 -1 -1 -1 -1• Y: 0 1 2 3 -1 -2 -3• Dy/dx: 1 2 3 4 0 -1 -2• X: -2 -2 -2 -2 -2 -2 -2• Y: 0 1 2 3 -1 -2 -3• Dy/dx: 2 3 4 5 1 0 -1• X: -3 -3 -3 -3 -3 -3 -3• Y: 0 1 2 3 -1 -2 -3• Dy/dx: 3 4 5 6 2 1 0
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Practice Problem 4
• Plotting all points with slope dy/dx on a graph with domain and range both of [-3,3] should look like this: