In chapter 1, we talked about parametric equations. Parametric equations can be used to describe motion that is not a function. If f and g have derivatives

In chapter 1, we talked about parametric equations.Parametric equations can be used to describe motion that is not a function.

x f t y g t

If f and g have derivatives at t, then the parametrized curve also has a derivative at t.

10.1 Parametric Functions

The formula for finding the slope of a parametrized curve is:

dy

dy dtdxdxdt

This makes sense if we think about canceling dt.


The formula for finding the slope of a parametrized curve is:

dy

dy dtdxdxdt

We assume that the denominator is not zero.


To find the second derivative of a parametrized curve, we find the derivative of the first derivative:

dydtdxdt

2

2

d y

dx d

ydx

1. Find the first derivative (dy/dx).

2. Find the derivative of dy/dx with respect to t.

3. Divide by dx/dt.


22 3

2Find as a function of if and .

d yt x t t y t t

dx


22 3

2Find as a function of if and .

d yt x t t y t t

dx

1. Find the first derivative (dy/dx).

dy

dy dtydxdxdt

21 3

1 2

t

t


2. Find the derivative of dy/dx with respect to t.

21 3

1 2

dy d t

dt dt t

2

2

2 6 6

1 2

t t

t

Quotient Rule


3. Divide by dx/dt.

2

2

2 6 6

1 2

1 2

t t

t

t

2

3

2 6 6

1 2

t t

t


dtdxdtdy

dx

yd'

2

2

The equation for the length of a parametrized curve is similar to our previous “length of curve” equation:


dtdt

dy

dt

dxL

b

a

22

1. Revolution about the -axis 0x y 2 2

2b

a

dx dyS y dt

dt dt

2. Revolution about the -axis 0y x

2 2

2b

a

dx dyS x dt

dt dt


This curve is:

sin 2

2cos 5

x t t t

y t t t


Quantities that we measure that have magnitude but not direction are called scalars.

Quantities such as force, displacement or velocity that have direction as well as magnitude are represented by directed line segments.

A

B

initialpoint

terminalpoint

AB��

The length is AB��

10.2 Vectors in the Plane

A

B

initialpoint

terminalpoint

AB��

A vector is represented by a directed line segment.

Vectors are equal if they have the same length and direction (same slope).


A vector is in standard position if the initial point is at the origin.

x

y

1 2,v v

The component form of this vector is: 1 2,v vv


A vector is in standard position if the initial point is at the origin.

x

y

1 2,v v

The component form of this vector is: 1 2,v vv

The magnitude (length) of 1 2,v vv is:2 2

1 2v v v


P

Q

(-3,4)

(-5,2)

The component form of

PQ��

is: 2, 2 v

v(-2,-2) 2 2

2 2 v

8 2 2


If 1v Then v is a unit vector.

0,0 is the zero vector and has no direction.


Vector Operations:

1 2 1 2Let , , , , a scalar (real number).u u v v k u v

1 2 1 2 1 1 2 2, , ,u u v v u v u v u v

(Add the components.)

1 2 1 2 1 1 2 2, , ,u u v v u v u v u v

(Subtract the components.)


Vector Operations:

Scalar Multiplication:1 2,k ku kuu

Negative (opposite): 1 21 ,u u u u


Let 2, 1 and 5,3 . Find 3 . u v u v

3 3 2 , 3 1 = 6, 3

3 = 6, 3 5,3 6 5, 3 3 11,0

u

u v


A vector with | | 1 is a . If is not the zero vector

10,0 , then the vector is a

| | | |

.

unit vector

unit vector in the direction

of

u u v

vu v

v v

v


Find a unit vector in the direction of 2, 3 . v

222, 3 2 3 13, so

1 2 32, 3 ,

13 13 13

v

v

v


v

vu

u

u+vu + v is the resultant vector.

(Parallelogram law of addition)


The angle between two vectors is given by:

1 1 1 2 2cosu v u v

u v

This comes from the law of cosines.See page 524 for the proof if you are interested.


The dot product (also called inner product) is defined as:

1 1 2 2cos u v u v u v u v

Read “u dot v”

This could be substituted in the formula for the angle between vectors to give:

1cos

u v

u v


Find the dot product.

4,3 1, 2

4,3 1, 2 (4)( 1) (3)( 2) 10


Find the angle between the vectors 3,2 and 1,0 . u v

1

-1

-1

cos

3,2 1,0cos

3,2 1,0

3 cos

13 1

33.7

u v

u v


Application: Example 7

A Boeing 727 airplane, flying due east at 500 mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?


N

Eu


N

E

v

u

60o


N

E

v

u

We need to find the magnitude and direction of the resultant vector u + v.

u+v


N

E

v

u

The component forms of u and v are:

u+v

500,0u

70cos60 ,70sin 60v

500

7035,35 3v

Therefore: 535,35 3 u v

538.4 22535 35 3 u v

1 35 3tan

535

6.5


N

E

The new ground speed of the airplane is about 538.4 mph, and its new direction is about 6.5o north of east.

538.4

6.5o


Any vector can be written as a linear

combination of two standard unit vectors.

,a bv

1,0i 0,1j

,a bv

,0 0,a b

1,0 0,1a b

a b i j

The vector v is a linear combination

of the vectors i and j.

The scalar a is the horizontal

component of v and the scalar b is

the vertical component of v.

10.3 Vector-valued Functions

tr

If we separate r(t) into horizontal and vertical components, we can express r(t) as a linear combination of standard unit vectors i and j.

t f t g t r i j f t i

g t j



Let A = (-2,3) and B = (4,6)

Find AB in terms of i and j

AB = <6,3> = 6i + 3j

In three dimensions the component form becomes:

f g ht t t t r i j k


Most of the rules for the calculus of vectors are the same as we have used, except:

Speed v t

velocity vectorDirection

speed

t

t

v

v

“Absolute value” means “distance from the origin” so we must use the Pythagorean theorem.


3cos 3sint t t r i j

a) Find the velocity and acceleration vectors.

3sin 3cosd

t tdt

r

v i j

3cos 3sind

t tdt

v

a i j

b) Find the velocity, acceleration, speed and direction of motion at ./ 4t


3cos 3sint t t r i j

3sin 3cosd

t tdt

r

v i j 3cos 3sind

t tdt

v

a i j


velocity: 3sin 3cos4 4 4

v i j3 3

2 2 i j

acceleration: 3cos 3sin4 4 4

a i j 3 3

2 2 i j



3 3

4 2 2

v i j3 3

4 2 2

a i j

speed:4

v

2 23 3

2 2

9 9

2 2 3

direction:

/ 4

/ 4

v

v3/ 2 3/ 2

3 3

i j

1 1

2 2 i j


3 2 32 3 12t t t t t r i j 2 26 6 3 12d

t t t tdt

r

v i j

a) Write the equation of the tangent where .1t

At :1t 1 5 11 r i j 1 12 9 v i j

position: 5,11 slope:9

12

tangent: 1 1y y m x x

311 5

4y x

3 29

4 4y x

3

4


The horizontal component of the velocity is .26 6t t

b) Find the coordinates of each point on the path where the horizontal component of the velocity is 0.

26 6 0t t 2 0t t

1 0t t 0, 1t

0 0 0 r i j

1 2 3 1 12 r i j

1 1 11 r i j

0,0

1, 11


a) Write the equation of the tangent where t = 1.

position: slope:

tangent: 1 1y y m x x


r(t) = (2t2- 3t)i + (3t3- 2t)j v(t) = (4t - 3)i + (9t2 - 2)j

At t = 1 r(1) = -1i + 1j v(1) = 1i + 7j

(-1,1)1

7

)1(71 xy 87 xy

The horizontal component of the velocity is 4t - 3 .

b) Find the coordinates of each point on the path where the horizontal component of the velocity is 0.


4t – 3 = 0

4

3t

jir

4

32

4

33

4

33

4

32

4

332

jir

64

15

8

9

4

3

64

15,

8

9


j ir

r(0) j,)(cos)sin( ttdt

dFind r

j)(cos)sin( tt ir Cj )(sin)(cos tt ir

Cj )0(sin)0(cos ij

C ij ijC

i ir jj)(sin)(cos tt

j)1(sin)1(cos tt ir

One early use of calculus was to study projectile motion.

In this section we assume ideal projectile motion:

Constant force of gravity in a downward direction

Flat surface

No air resistance (usually)

10.4 Projectile Motion

We assume that the projectile is launched from the origin at time t =0 with initial velocity vo.

ov

Let o ov v

then cos sino o ov v v i j

The initial position is: r 0 0 0o i j


ov

Newton’s second law of motion:

Vertical acceleration

f ma2

2f

d rm

dt


ov


The force of gravity is:

Force is in the downward direction

f ma f mg j2

2f

d rm

dt


ov



f ma f mg j2

2f

d rm

dt

mg j2

2

d rm

dt


ov



f ma f mg j2

2f

d rm

dt

mg j2

2

d rm

dt


2

2

d rg

dt j

Initial conditions:

r r v when o o

drt o

dt

2o

1r v r

2 ogt t j

21r

2gt j 0 cos sin o ov t v t i j

o vdr

gtdt

j


21r cos sin 0

2 o ogt v t v t j i j

21r cos sin

2o ov t v t gt

i j

Vector equation for ideal projectile motion:


21r cos sin

2o ov t v t gt

i j

Vector equation for ideal projectile motion:

Parametric equations for ideal projectile motion:

21cos sin

2o ox v t y v t gt


Example 1:A projectile is fired at 60o and 500 m/sec.Where will it be 10 seconds later?

500 cos 60 10x

2500x

21500sin 60 10 9.8 10

2y

3840.13y

Note: The speed of sound is 331.29 meters/secOr 741.1 miles/hr at sea level.

The projectile will be 2.5 kilometers downrange and at an altitude of 3.84 kilometers.


The maximum height of a projectile occurs when the vertical velocity equals zero.

sin 0o

dyv gt

dt

sinov gt

sinovt

g

time at maximum height


The maximum height of a projectile occurs when the vertical velocity equals zero.

sin 0o

dyv gt

dt sinov gt

sinovt

g

We can substitute this expression into the formula for height to get the maximum height.


2

max

sin sin1sin

2o o

o

v vy v g

g g

21sin

2oy v t gt

2 2

max

si2

2

n sin

2o ov v

yg g

2

max

sin

2ov

yg

maximumheight


210 sin

2ov t gt When the height is zero:

10 sin

2ot v gt

0t time at launch:


210 sin

2ov t gt When the height is zero:

10 sin

2ot v gt

0t time at launch:

1sin 0

2ov gt

1sin

2ov gt

2 sinovt

g

time at impact

(flight time)


If we take the expression for flight time and substitute it into the equation for x, we can find the range.

cos ox v t

cos 2 sin

oov

x vg


If we take the expression for flight time and substitute it into the equation for x, we can find the range.

cos ox v t 2 sincos o

o

vx v

g

2

2cos sinovx

g

2

sin 2ovx

g Range


The range is maximum when

2

sin 2ovx

g Range

sin 2 is maximum.

sin 2 1

2 90o

45o

Range is maximum when the launch angle is 45o.


If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x.

cosox v t

coso

xt

v

21sin

2oy v t gt


If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x.

cosox v t

coso

xt

v

21sin

2oy v t gt

2

1sin

2cos cosoo o

x x

vy g

vv

This simplifies to: 22 2

tan2 coso

gy x x

v

which is the equation

of a parabola.


If we start somewhere besides the origin, the equations become:

coso ox x v t 21sin

2o oy y v t gt


A baseball is hit from 3 feet above the ground with an initial velocity of 152 ft/sec at an angle of 20o from the horizontal. A gust of wind adds a component of -8.8 ft/sec in the horizontal direction to the initial velocity.

The parametric equations become:

152cos 20 8.8ox t 23 152sin 20 16oy t t

ov oy 1

2g

These equations can be graphed on the TI-83 to model the path of the ball:

the calculator must be in degrees.

Max height about 45 ft

Distance traveled about 442 ft

Timeabout

3.3 sec

Usingthe

tracefunction:

In real life, there are other forces on the object. The most obvious is air resistance.

If the drag due to air resistance is proportional to the velocity:

dragF kv (Drag is in the opposite direction as velocity.)

Equations for the motion of a projectile with linear drag force are given on page 546.


Polar graphing is like the second method of giving directions. Each point is determined by a distance and an angle.

Initial ray

r A polar coordinate pair

determines the location of a point.

,r

10.5 Polar Graphing

1 2 02

r

r a

o

(Circle centered at the origin)

(Line through the origin)

10.5 Polar Graphing

30o

2

More than one coordinate pair can refer to the same point.

2,30o

2,210o

2, 150o

210o

150o

All of the polar coordinates of this point are:

2,30 360

2, 150 360 0, 1, 2 ...

o o

o o

n

n n

10.5 Polar Graphing

10.5 Polar Graphing

Equations Relating Polar and Cartesian Coordinates

θx

yr

Polar Rectangular

sincos ryrx

Rectangular Polar

x

yryx tan222

10.5 Polar Graphing

Change from rectangular to polar.

)3,1( )300,2(

)2,2( )4

3,22(

)3,4( )323,5(

10.5 Polar Graphing

Change from polar to rectangular.

6,3

2

3,

2

33

3

2,4

)32,2(

)315,1(

2

2,

2

2

10.5 Polar Graphing

Write as a Cartesian Equationr = 2 cos θ

r

xyx 222

22

22 2yx

xyx

xyx 222

02 22 yxx +1 +1

1)1( 22 yx

10.5 Polar Graphing

Write as a Cartesian Equation

r = 4 tan θ sec θ

x

r

x

yyx 422

222 4

x

ryyx

2

2222 4

x

yxyyx

241

x

y

4

2xy

10.5 Polar Graphing

Write as a Polar Equationxy = 2

2sincos rr

2cossin2 r

seccsc22 r

seccsc2r

10.5 Polar Graphing

Write as a Polar Equationx - y = 6

6sincos rr

6)sin(cos r

sincos

6

r

Tests for Symmetry:

x-axis: If (r, q) is on the graph, so is (r, -q).

r

2cosr

r

10.5 Polar Graphing

y-axis: If (r, q) is on the graph,

r

2sinr

r

so is (r, p-q)

or (-r, -q).

10.5 Polar Graphing

origin: If (r, q) is on the graph,

r

r

so is (-r, q)

or (r, q+p) .

tan

cosr

10.5 Polar Graphing

If a graph has two symmetries, then it has all three:

2cos 2r

10.5 Polar Graphing

Try graphing this on the TI-83.

2sin 2.15

0 16

r

10.6 Calculus of Polar Curves

To find the slope of a polar curve:

dy

dy ddxdxd

sin

cos

dr

dd

rd

sin cos

cos sin

r r

r r

We use the product rule here.


To find the slope of a polar curve:

dy

dy ddxdxd

sin

cos

dr

dd

rd

sin cos

cos sin

r r

r r

sin cos

cos sin

dy r r

dx r r


Example: 1 cosr sinr

sin sin 1 cos cosSlope

sin cos 1 cos sin

2 2sin cos cos

sin cos sin sin cos

2 2sin cos cos

2sin cos sin

cos 2 cos

sin 2 sin


The length of an arc (in a circle) is given by r . q when q is given in radians.For a very small q, the curve could be approximated by a straight line and the area could be found using the triangle formula:

1

2A bh

r dr

21 1

2 2dA rd r r d


We can use this to find the area inside a polar graph.

21

2dA r d

21

2dA r d

21

2A r d


Example: Find the area enclosed by: 2 1 cosr 2 2

0

1

2r d

2 2

0

14 1 cos

2d

2 2

02 1 2cos cos d

2

0

1 cos 22 4cos 2

2d


2

0

1 cos 22 4cos 2

2d

2

03 4cos cos 2 d

2

0

13 4sin sin 2

2

6 0 6


Notes:

To find the area between curves, subtract:

2 21

2A R r d

Just like finding the areas between Cartesian curves, establish limits of integration where the curves cross.


When finding area, negative values of r cancel out:

2sin 2r

22

0

14 2sin 2

2A d

Area of one leaf times 4:

2A

Area of four leaves:

2 2

0

12sin 2

2A d

2A


To find the length of a curve:Remember: 2 2ds dx dy

For polar graphs: cos sinx r y r 2

2 drds r d

d

So: 22Length

drr d

d


There is also a surface area equation similar to the others we are already familiar with:

22S 2

dry r d

d

When rotated about the x-axis:

22S 2 sin

drr r d

d


Documents

In chapter 1, we talked about parametric equations. Parametric equations can be used to describe motion that is not a function. If f and g have derivatives