analytic geometry & calculus

8/11/2019 Analytic Geometry & Calculus

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Analytic Geometry

Kristina S. Camsol


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Distance between two points in a plane

d= x2x12

+ y2y12

Slope

m=y

x=

y2

y1

x2x1

Parallel Lines

m1=m2

Perpendicular Lines

m1= 1m2


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Angle between two lines

tan =m2m1

1+m1m2

m1= slope of equation 1

m2= slope of equation 2

Example:

Find the angle (in degrees) between the two lines:

6x-y+8=0

-3x-11y+10=0


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Solution:6x-y+8=0 eq. 1

-3x-11y+10=0 eq. 2

Soln:

eq. 1 y=6x+8 ; m1=6

eq. 2 y=311

x+10 ; m2=311

tan =

311

61+ 311 6

= 84.2


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Distance between a point and a

line

d= Ax1+By1+C

A2+B2

Example:

Find the distance from the point (-2,-1) to the line 3x-4y=12.

Gen. equation of a line:

Ax + By + C = 0


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Solution:

3x-4y-12=0 ; A = 3, B = -4, C = -12

d= 3 2 + 4 1 + 12

32+ 42= 2.8

d = 2.8


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Distance between two parallel lines

d=C1C2

A2+B2

Example:

Find the distance between 3x+4y+5=0 and 3x+4y+2=0.

Soln:

d=

5

232+42

d = 0.6


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Area of a polygon using the coordinates of

its vertices

Coordinates must be arranged in the matrix in a counter-clockwise

direction based on the Cartesian plane.

x1

x2

x3y1 y2 y3

A=1

2

x1y2+x2y3+x3y1 y1x2+y2x3+y3x1


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Example:(4,10), (9,7), (11,2), (2,2)

Arrange counter-clockwise direction on the cartesian plane:

(11,2), (9,7), (4,10), (2,2)

A = 0.5{[(11)(7)+(9)(10)+(4)(2)+(2)(2)]-

[(2)(9)+(7)(4)+(10)(2)+(2)(11)]}

A = 45.5


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Circle

General EquationAx2+Cy2+Dx+Ey+F=0

Standard Equations

x2+y2=r2

xh2+ yk2=r2

h=

D

2A k=

E

2A


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Circle

Example:What is the distance between the center of the circles x2+y2+2x+4y-3=0 and x2+y2-8x-6y+7=0?

Soln:

A1=1, D1=2, E1=4 A2=1, D2=-8, E2=-6

C1(-1,-2) Using the distance formula, the answerC2(4,3) would be approximately 7.07 units.

h1=2

2(1)= 1 k1= 42(1) = 2 h2=

(8)

2(1)=4 k2=

(6)2(1)

= 3


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Parabola

Ax2+Dx+Ey+F=0Cy2+Dx+Ey+F=0

Standard Equations

y2=4ax

yk2=4a xh

x2=4ay

xh2=4a yk

Vertex is at the point (h,k).

Focus is the point (inside the parabola) aunits from the vertex.

Directrix is the line (outside the parabola) aunits from the vertex.


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Parabola

Example:What is the equation for the parabola with focus (-2,-2) and

directrix y=4?

Soln:midpoint= y1y22 =

2+4

2 =1

V(-2,1)

a = 4-1 = 3

(x-h)2= -4a(y-k)

(y-1) = ()(x+2)2y1 =

1

12 x+22


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Parabola

Eccentricity is the ratio of the distance to thefocusto the distance to the directrix

e=f

d

, since f = d

e=1

Latus rectum is a line that passes through the

focus and perpendicular to the axis of the conic

LR=4a


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Parabola

For horizontal axis: For vertical axis:

Cy2+Dx+Ey+F=0

h= E2

4CF4CD

k=E

2C

a=E

4A

Ay2+Dx+Ey+F=0

h= D2A

k=D24AF

4AE

a=E

4A


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Ellipse

Ellipse General EquationAx2+Cy2+Dx+Ey+F=0

b = semi-minora = semi-major

e < 1

d1+ d2= 2a

e=c

a

LR

LR=2b

2

a


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Ellipse Standard Equations

MA is horizontal MA is vertical

x2

a2+

y2

b2=1

x2

b2+y2

a2=1

xh2

a2 +

yk2

b2 =1

xh2

b2 +

yk2

a2 =1


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Find the length of the latus rectum of the following ellipse:25x2+9y2-300x-144y+1251=0

Soln:

25x2-300x+9y2-144y+1251=0

25(x2-12x)+9(y2-16y)+1251=0

25(x2-12x)+9(y2-16y)=-1251

25(x2-12x+36)+9(y2-16y+64)=-1251+900+576

25(x-6)2+9(y-8)2=225

A

ax62

3 + y82

5 =1 LR= 2 32

5 =3.6


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Hyperbola

Hyperbola General EquationAx2Cy2+Dx+Ey+F=0 verticalCy2Ax2+Dx+Ey+F=0 horizontal


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Hyperbola

Hyperbola Standard Equations

xh2

a2

yk2

b2

=1

yk 2a2

xh 2b

2 =1

From the illustration:

e=c

a

e > 1

c


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Hyperbola

Example:

From the hyperbola y2/25x2/9 = 1. Give the

vertices, foci and equations of asymptotic lines.


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Hyperbola

Soln:

a = 5 b = 3 c = 5.8

V1(0,-5) V2(0,5)

F1(0,-5.8) F2(0,5.8)

y = -(5/3)x y = (5/3)x


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Polar Coordinates

Polar coordinates (r,)

Rectangular coordinates (x,y)Relationship:

x = rcos y = rsin

r2 = x2+ y2


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Polar Coordinates

Example:

What is (13, 22.6) in Cartesian Coordinates?

r = 13, = 22.6

x = rcos y = rsin

x = (13)(cos22.6) y = (13)(sin22.6)

x ~ 12 y ~ 5

(12,5)


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Differential & Integral

Calculus

Kristina S. Camsol


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CalculusLatin calxstone, and Greek chalislimestone

Gottfried Wilhelm von Leibniz & Isaac Newtonconsidered as the

founders of todayscalculus.


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Limit Theorems

1. If limxa

f x =Lexists, then it is unique.

2. limxa

f x +g x = limxa

f x + limxa

g x =L+K

3. limxaf x g x = limxa f x limxa g x =LK4. lim

xaf x g x = lim

xaf x lim

xag x =L K

5. lim = , 06. lim = , 07. lim

xa = lim

xa

8. limxa

= limxa

f x

limxa f x =L limxa g x =K


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Limit Theorems

9. limxa = limxa =

10. If f(x) is a polynomial function, lim

xa =

11. lim = lim , > 1

12. If f(x) h(x) g(x), and if

limxa

f x =L= limxa

g(x) , then limxa

h x =L


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Limit Theorems

If f(x) is a polynomial, and lim+ = = lim , then thelimit exists at L.

But if lim+ = lim , then the limit does not exist.


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Let h be defined by

h x = 4x2 if x12+x2 if x>1

limx1

4x2 =3 limx1+

2+x2 =3

limx1h(x) = limx1+h(x)

The limit if h(x) exists at 3.


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h(x)=4-x2

h(x)=2+x2


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Let f(x) be defined by

f x = 6x2 if 0x


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Special Limits

1. lim = 1

2. lim = 0

3. lim 1 =

4. lim 1 =


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LHpitalsRule

If f(x) and g(x) are two functions such thatlim = 0 and lim = 0

Then lim()()has the indeterminate form 0/0 at a.

LHpitalsRules states that

lim()() = l i m

()()


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LHpitalsRule

Let f(x) be defined by sinx and g(x) be x.f(x)= cosx and g(x)= 1.

lim()() =lim

sin =l im

cos1 = 1


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LHpitalsRule

Let f(x) be defined by x2

-1 and g(x) be x2

+3x-4.f(1)= 12-1 = 0

g(1)= 12+3(1)-4 = 0

lim()() =l im 1 3 4

lim(1)(1)(4)(1) =l im

1 4 =0.4


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LHpitalsRule

f(x) = x2

-1 f(x)=2xg(x) = x2+3x-4 g(x)=2x+3

lim

x1f(x)

g(x)

=

limx12x

2x+3

=0.4


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Derivatives

B. Exponential Functions

9.d

dx au =au ln a

dudx

10.d

dx eu =eu

dudx

C. Logarithmic Functions

11.d

dx logau =

logaedudx

u

12.d

dx log10 u =

log10 edudxu

13.d

dx lnu =

dudxu

D. Trigonometric Functions

14.ddx

sinu =cosududx

15.ddx

cosu =sinududx

16.ddx

tan u = sec2 ududx

17.ddx

cot u = csc2 ududx

18. ddx sec u = sec u tan u dudx

19.ddx

cscu =cscucotududx


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Maxima & Minima

First

Derivative

Second

Derivative

Behavior

Maximum

Point

0 Negative Concave

DownwardMinimum

Point

0 Positive Concave

Upward

Point of

Inflection

0


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Maxima & Minima

Find two nonnegative numbers whose sum is 12 such that theirproduct is an absolute maximum.

Soln:

Let x and y be the two nonnegative numbers

x+y=12

f(x)=xy

f(x)=x(12-x)

f(x)=12-2x

Let f(x)=0

12-2x=0

x=6


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Maxima & Minima

Find the area of the largest rectangle having two vertices on the x-axis and two

vertices on or above the x-axis and on the parabola y = 9x2.

Soln:

A = 2xy

y = 9x2

A = f(x) = 2x(9x2)

f(x) = 18x2x3

f(x)= 186x2

f(x)= 0

6x

2

= 18x = 1.732

A = 2(x)(9x2)

A = 2(1.732)(91.7322)

A = 20.785


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Implicit Differentiation

x2+ y2= 9 2xdx + 2ydy = 0

y= 9x2= 9x212 2ydy = -2xdx

y=1

2 9x2

12 2x

2ydy

2ydx=

2xdx

2ydx

y= x

9x2

dydx

= xy

dydx

= x

9x2


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Implicit Differentiation

Find the first derivative of x3+ y3= 8xy.

x3+ y3= 8xy

3x2dx + 3y2dy = 8[xdy+ydx]

3x2dx + 3y2dy = 8xdy + 8ydx

3y2dx8xdy = 8ydx3x2dx

(3y28x)dy = (8y3y2)dx

3y28x dy

3y28x dx=

8y3y2 dx

3y28x dx

dy

dx =

8y3y2

3y28x


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Related Rates

An airplane is flying parallel to the ground at an altitude of 2 km and at a speed

of 4.5 km/min. If the plane flies directly over the Statue of Liberty, at what rate is

the line-of-sight distance between the plane and the statue changing 20 sec later?

Soln:

use Pythagorean theoremc2=a2+b2

a = 2

b = (4.5 km/min)(1 min/60 s)(20 s)

= 1.5 km

c = 2.5


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Related Rates

From Pythagorean theorem,

c2= a2+ b2

The first derivative would be 2cdcdt

=2adadt

+2bdbdt

dadt

=0

dbdt=4.5

dcdt

=?

Using the first derivative of the theorem

(2.5)dcdt

= (2)(0) + (1.5)(4.5)

dcdt

=2.7 km/min


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Related Rates

A ladder is 25 ft long and leaning against a vertical wall. The bottom of the ladder is

pulled horizontally away from the wall at 3 ft/s. We wish to determine how fast thetop of the ladder is sliding down the wall when the bottom is 15 ft from the wall.

Soln:

From Pythagorean theorem,

c = 25, b = 15 cdc

dt

=ada

dt

+bdb

dt

a = 20

dcdt

=0 (25)(0) = (20)(dadt

)+(15)(3)

dbdt

=3dadt

=2.25 ft/s

dadt

=?

The negative sign denotes the direction to which the top of the ladder is sliding.


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Additional Applications

If a closed tin can of volume 60 in3is to be in the form of a right circular cylinder, find the base

radius of the can if the least amount of tin is to be used in its manufacture.Soln:

SA = 2rh + 2r2

V = r2h = 60

h=60

r2

S(r) = 2r 60

r2 +2r2

S(r) =120

r +2r2

S(r)= 120

r2+4r

S(r)= 0

120

r2 +4r=0

r= 30

3~2.12


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Integration

A. Basic Integrals

1. du = u+C2. adu= au+C3. undu = un+1n+1+C , n14.

du

u

= ln u +C

B. Exponential & Logarithmic Functions

5. eudu = eu+C6. audu =auln a +C7. ueudu =eu u1 +C8. ln u du = u ln u u+C9. duulnu = ln ln u +CBy parts

udv = uv vdu


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Integration

Integration by parts

udv =uv vduLet u = f(x) , v = g(x)du = f(x),dv = g(x)

Example:

x ln x dxu = lnx, dv = xdx

du =dxx

, v =x2

2+C

ln x xdx = ln x x2

2+C x2

2+C

dx

x

=12

x2 lnx+Clnx14

x2Clnx+C

=12

x2 lnx14

x2+C


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Integration

Integration by other substitution techniques

x5 x2+4dxLet z = x2+4, z2= x2 + 4, 2zdz = 2xdx

x5 x2+4dx =

x2

2x2+4 (xdx)

= z24 2 (z)(zdz)= z68z4+16z2 dz=

17

z785

z5+83

z3+C

= 1105 z3 15z4168z2+560 +C

= 4 15 4 1 6 8 4 560

=1

105 x2+4

32 15x448x2+128 +C


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Application of Integrals

Area

Centroid

Volume

Work

=

= = = 2

= = 12

=

= ()

= () ()

Vertical strip Horizontal strip Polar coordinates


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Area

Example:

y2= 2x2 & y = x5Let f(x) = (2x-2)1/2

g(x) = x5

Point/s of intersection

2x212=x5

Square both sides

2x2 = x210x + 25

x212x + 27 = 0

x1= 3 x2= 9

y1= - 2 y2= 4

A= ydxx2x1

A2= 2x2 x5 dx93 A2=

383

A1

A2


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Area

Using horizontal strip

A= xdyy2y1 f(y) = (y2+ 2)

g(y) = y + 5

Using the limits from the previous

solution:

A= y+5 12 y2+2

42 dy

A = 18


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Centroid

Find the centroid of the first quadrant region bounded by the curve y2= 4x, the x

axis, and the lines x = 1 and x = 4.

A= 4x dx41 A =

x= x 4x dx41283

=9335

y= 12 4x 4x dx41 283

=4528

Centroid =9335

,4528

,

id


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Centroid

2. Find the centroid of the region bounded by the curves y = x2and y = 2x+3.

A= 2x+3 x231 dxA =

323

x= x 2x+3 x2

dx31

323

=1

y

=

12

2x+3 x22dx31323

=17

5

Centroid = 1,175

,

l


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Volume

Volume of a Solid

Volume of a disk

Volume of a Washer

=

= ()

= () ()

V l


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Volume

Volume of a Solid

Example: Find the volume of

A(x) = r2

V= r2dxh0 V=r2 x0

V= r2h

=

V l


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Volume

Volume of a disk

Determine the volume of the solid obtained by rotating the region bounded by

y = x2-4x+5, x=1, x=4, and the x axis

about the x axis.

V= x24x+5 2 dxba V=

785

= ()

V l


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Volume

Volume of a washer

Determine the volume of the solid obtained by rotating the portion of the region

bounded by y3= x and y =x4about the y axis.

Soln:

f(y) = 4y

g(y) = y3

V= 4y2 y32 dy20 V = 51221

= () ()

W k


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Work

=

A particle is moving along the x axis under the action of a force of f(x) pounds when the

particle is x feet from the origin. If f(x) = x2+ 4, find the work done as the particle moves

from the point x = 2 to the point x = 4.

W=

f x dx4

2

W= (x2+4)dx42

W=

1

3 x

3

+4x 4

2

W = 26


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M lti l I t l


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Multiple Integrals

f x,y dA = f x,y dyy2

y1

dx

x2

x1

Evaluate 3y2x2 dAR if R is the region consisting of all points (x,y) for which-1 x 2 and 1 y 3.

3y2x2 dAR = 3y2x2 dx21 dy31 = 3yx 23 x3 21 dy31 = 3y 2 1 23 23 13 dy31 = 9y6 dy

3

1

= 92

y26y 31=

92

3212 6 31

= 24

M lti l I t l


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Multiple Integrals

V= f r, rdrdR

Find the volume of the solid in the first octant bounded by the cone z = r and the cylinder r =3sin.

z = f(r,) = r

V= r2drdR V= r2dr3sin0 d

20

V= 13

r3 3sin

0 d20

V=9 sin3 d

20

V= 9cos+3cos3

20

V = 6

sin3

= 1cos2 sin


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E d f Di i


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End of Discussion

Thank You!

analytic geometry & calculus

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