in chapter 1, we talked about parametric equations. parametric equations can be used to describe...
TRANSCRIPT
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.
10.1 Parametric Functions
The formula for finding the slope of a parametrized curve is:
dy
dy dtdxdxdt
We assume that the denominator is not zero.
10.1 Parametric Functions
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.
10.1 Parametric Functions
22 3
2Find as a function of if and .
d yt x t t y t t
dx
10.1 Parametric Functions
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
10.1 Parametric Functions
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
10.1 Parametric Functions
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
10.1 Parametric Functions
dtdxdtdy
dx
yd'
2
2
The equation for the length of a parametrized curve is similar to our previous “length of curve” equation:
10.1 Parametric Functions
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
10.1 Parametric Functions
This curve is:
sin 2
2cos 5
x t t t
y t t t
10.1 Parametric Functions
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).
10.2 Vectors in the Plane
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
10.2 Vectors in the Plane
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
10.2 Vectors in the Plane
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
10.2 Vectors in the Plane
If 1v Then v is a unit vector.
0,0 is the zero vector and has no direction.
10.2 Vectors in the Plane
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.)
10.2 Vectors in the Plane
Vector Operations:
Scalar Multiplication:1 2,k ku kuu
Negative (opposite): 1 21 ,u u u u
10.2 Vectors in the Plane
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
10.2 Vectors in the Plane
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
10.2 Vectors in the Plane
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
10.2 Vectors in the Plane
v
vu
u
u+vu + v is the resultant vector.
(Parallelogram law of addition)
10.2 Vectors in the Plane
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.
10.2 Vectors in the Plane
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
10.2 Vectors in the Plane
Find the dot product.
4,3 1, 2
4,3 1, 2 (4)( 1) (3)( 2) 10
10.2 Vectors in the Plane
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
10.2 Vectors in the Plane
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?
10.2 Vectors in the Plane
N
Eu
10.2 Vectors in the Plane
N
E
v
u
60o
10.2 Vectors in the Plane
N
E
v
u
We need to find the magnitude and direction of the resultant vector u + v.
u+v
10.2 Vectors in the Plane
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
10.2 Vectors in the Plane
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
10.2 Vectors in the Plane
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
10.3 Vector-valued Functions
10.3 Vector-valued Functions
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
10.3 Vector-valued Functions
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.
10.3 Vector-valued Functions
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
10.3 Vector-valued Functions
3cos 3sint t t r i j
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
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
10.3 Vector-valued Functions
b) Find the velocity, acceleration, speed and direction of motion at ./ 4t
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
10.3 Vector-valued Functions
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
10.3 Vector-valued Functions
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
10.3 Vector-valued Functions
a) Write the equation of the tangent where t = 1.
position: slope:
tangent: 1 1y y m x x
10.3 Vector-valued Functions
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.
10.3 Vector-valued Functions
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
10.3 Vector-valued Functions
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
10.4 Projectile Motion
ov
Newton’s second law of motion:
Vertical acceleration
f ma2
2f
d rm
dt
10.4 Projectile Motion
ov
Newton’s second law of motion:
The force of gravity is:
Force is in the downward direction
f ma f mg j2
2f
d rm
dt
10.4 Projectile Motion
ov
Newton’s second law of motion:
The force of gravity is:
f ma f mg j2
2f
d rm
dt
mg j2
2
d rm
dt
10.4 Projectile Motion
ov
Newton’s second law of motion:
The force of gravity is:
f ma f mg j2
2f
d rm
dt
mg j2
2
d rm
dt
10.4 Projectile Motion
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
10.4 Projectile Motion
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:
10.4 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
10.4 Projectile Motion
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.
10.4 Projectile Motion
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
10.4 Projectile Motion
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.
10.4 Projectile Motion
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
10.4 Projectile Motion
210 sin
2ov t gt When the height is zero:
10 sin
2ot v gt
0t time at launch:
10.4 Projectile Motion
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)
10.4 Projectile Motion
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
10.4 Projectile Motion
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
10.4 Projectile Motion
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.
10.4 Projectile Motion
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
10.4 Projectile Motion
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.
10.4 Projectile Motion
If we start somewhere besides the origin, the equations become:
coso ox x v t 21sin
2o oy y v t gt
10.4 Projectile Motion
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.
10.4 Projectile Motion
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.
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
sin cos
cos sin
dy r r
dx r r
10.6 Calculus of Polar Curves
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
10.6 Calculus of Polar Curves
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
10.6 Calculus of Polar Curves
We can use this to find the area inside a polar graph.
21
2dA r d
21
2dA r d
21
2A r d
10.6 Calculus of Polar Curves
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
10.6 Calculus of Polar Curves
2
0
1 cos 22 4cos 2
2d
2
03 4cos cos 2 d
2
0
13 4sin sin 2
2
6 0 6
10.6 Calculus of Polar Curves
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.
10.6 Calculus of Polar Curves
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
10.6 Calculus of Polar Curves
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
10.6 Calculus of Polar Curves
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
10.6 Calculus of Polar Curves