Transcript Ch.2 Limits and derivatives

Parametric equations
Parametric equation: x and y expressed in terms of a
parameter t, for example, x  a cos t , y  b sin t
 A curve can be described by parametric equations x=x(t),
y=y(t). Each value of t determines a point (x,y). So the
parametric equations define a function.
 Typical parametric equations:
circle: x  r cos t , y  r sin t , 0  t  2 .
ellipse: x  a cos t , y  b sin t , 0  t  2

Example
Ex. Sketch the curve with parametric equations
x  sin t, y  sin 2 t.
2
 Sol. Observe that y  x and 1  x  1 so the curve is part
of the parabola. Since sint is periodic, the point (x,y) moves
back and forth infinitely often along the parabola, as t
changes.


Ex. A cycloid is defined by
x  r (  sin  ), y  r (1  cos ),   R
Derivative of functions defined
by parametric equations
Suppose y=y(x) is defined by the parametric equation
dy
x   (t ), y   (t ). Then
dy dt  (t )
y  

.
dx
dx
 (t )
dt
Ex. Find an equation of the tangent line to the curve

 x  ln(1  t 2 )
at the point (ln 2,1  ).

4
 y  1  arctan t
1
Sol. dy y(t ) 
2
1
1
1

t


   y(1)   .
2t
dx x(t )
2t
2
1 t2
Question
dy
 x  ln(sin t )
Suppose 
, find
.
y
dx
 y  e sin t  1
y
dy
dy
dy
e
cos t
y
y
 (sin t )e
 e cos t  0  
.
Sol.
y
dt
dt
dt 1  e sin t
dy
e y cos t
dy dt 1  e y sin t
e y sin t
y 1
 


.
y
cos t
dx dx
1  e sin t 2  y
dt
sin t
Question
d 2 y  x  a(t  sin t )
Find 2 if 
.
dx
 y  a(1  cos t )
dy
dy dt
a sin t
t
Sol.


 cot ,
dx dx a(1  cos t )
2
dt
d dy
1 2t
( )  csc
2
d y d dy
1
dt
dx
2
2
 ( )


.
2
dx
dx
dx dx
a(1  cos t )
4 t
4a sin
dt
2
Area formula

When a function is in parametric form:
 x  x(t )
(  t   ),

 y  y(t )
Then the area of the region bounded by the curve and x=a,
x=b is
b


a


A   ydx   y(t )dx(t )   y(t ) x(t )dt.
Remark: in the formula, ( x( ), y( ))
is the left endpoint
Example

x2 y 2
Ex. Find the area of the region bounded by 2  2  1.
a b

Sol.
x  a cos t , y  b sin t
a
0
 A  4 ydx  4 b sin t (a sin t )dt   ab.
0
2
Example

Find the area under one arc of the cycloid
x  a(t  sin t ), y  a(1  cos t ).
2 a

Sol. A  0
a
2
ydx   a(1  cos t )d[a(t  sin t )]
0
2

2
0
(1  cos t )2 dt  3 a 2 .
Example: volume
Ex. Find the volume of the solid obtained by rotating about
y-axis the region bounded by the cycloid x  a(t  sin t ),
y  a(1  cos t ), (0  t  2 ) and y=0.
2 a
2
 Sol. V  2
xydx  2 a(t  sin t )  a(1  cos t )d[a(t  sin t )]


0
 2 a
3

2
0

0
(t  sin t )(1  cos t )2 dt 
 6 3a3.
Arc length formula

If a smooth curve is defined by the parametric equation
x  x(t ), y  y(t ) (  t   )
we have ds  (dx) 2  (dy ) 2  [ x(t )]2  [ y(t )]2 dt
s


[ x(t )]2  [ y(t )]2 dt.
Example


Ex. Find the length of one arc of the cycloid
x  a(t  sin t ), y  a(1  cos t ) (0  t  2 ).
Sol. s  
2
0
[ x(t )]  [ y(t )] dt  a 
2
2
2
0
2
t
 2a  sin dt  8a.
0
2
2(1  cos t )dt
Example: surface area
Ex. Find the area of the surface obtained by rotating the
cycloid x  a(t  sin t ), y  a(1  cos t ) (0  t  2 ) about y-axis.


Sol.
S 
2 a
0
 4 a
2

2
0
2
2 xds   2 x(t ) [ x(t )]2  [ y(t )]2 dt
0
2
2
t
t
t 
2
(t  sin t )sin dt  4 a   t sin dt   sin t sin dt 
0
2
2
2 
0
2
t
t 4 3 t
2
 4 a  2t cos  4sin  sin   16 2 a 2
2
2 3
2 0

Polar coordinates



A coordinate system represents a point in the plane by an
ordered pair of numbers called coordinates.
The usual rectangular coordinate system, also called
Cartesian coordinates, uses (x,y) to locate a point.
In many situation, the Polar coordinate system is more
convenient for some purposes. A point P is represented by
an ordered pair (r ,  ) where r is the distance from O to P,
 is the angle between the polar axis and the line OP.
Polar coordinates



The polar coordinate for a point may not be unique: r can
be negative, (r , ) and (r,   ) represent the same point.
 is not unique, (r ,  ) can be represented by (r,  2n )
An angle is positive if measured in the counterclockwise
direction from the polar axis and negative clockwise.
Ex. Plot the points whose polar coordinates are given
(a) (1,5 / 4)
(b) (2,3 )
(c) (2, 2 / 3) (d) (3,3 / 4)

Polar coordinates

Relationship between polar and Cartesian coordinates
x  r cos , y  r sin 
r 2  x2  y2 tan   y / x


Polar curves: r  f ( )
Simple examples: r=a is a circle centered origin with
radius a;    is a line passing through origin with slope
tan 
Polar curves


Ex. Sketch the curve with polar equation r  2 R cos 
Sol.
x  2R cos2  , y  2R cos sin 
 ( x  R)2  y 2  R2
r  2R cos  2Rx / r  r 2  2Rx  ( x  R)2  y 2  R2

Question: r  2 R sin 
Calculus in polar coordinate


tangent line to a polar curve r  f ( )
x  f ( )cos , y  f ( )sin 
dy y( ) r ( )sin   r cos 


dx x( ) r ( ) cos   r sin 
Ex. For the cardioid r  1  sin  , (a) find the slope of the
tangent line when    / 3
(b) find the points on the
cardioid where the tangent line is horizontal or vertical.
Area formula

Area problem: the boundary of a region is given in polar
coordinates: r  r ( ) (     ). find the area A.
Use differential element method. In the total interval[ ,  ],
take any element [ ,  d ], then the sub-area corresponding
1 2

A

r ( )d  dA, because d is small, r ( )
to this element is
2
is approximately a constant and thus the sub-region a sector.

1  2
Therefore, A   dA   r ( )d .

2 

Area formula
Alternatively, by the relationship between polar coordinate
and Cartesian coordinate, the polar coordinate equation can
be converted to parametric equation:

x  r ( )cos , y  r ( )sin  .
Example

Find the area of the region bounded by the cardioid
r  a(1  cos  ).
Sol I Sketch the graph first. By symmetry,

1  2
2
A  2   r ( )d  a  (1  cos  ) 2 d
0
2 0
2

1

cos
2

3
a
 a 2  (1  2cos  
)d 
.
0
2
2
 Sol II The parametric equation is x  a(1  cos  )cos  ,
y  a(1  cos  )sin  . So
2
2a
0
3
a
A  2 ydx  2 a 2 (1  cos  )(1  2cos  )sin 2  d 

0

2

Example


Ex. Find the area of the region bounded by the two-leaved
rose r 2  a 2 cos 2 .
Sol Sketch the graph first. By symmetry,
1  /2 2
A  4   a cos 2 d  a 2 .
2 0
Arc length


To find the length of a polar curve r  f ( ) (     ), we
regard  as a parameter and write the parametric equation
of the curve as
x  f ( )cos , y  f ( )sin  .
So the arc length is computed as follows
dx
dy
 f ( ) cos   f ( ) sin  ,
 f ( ) sin   f ( ) cos 
d
d
s


[ x( )]  [ y( )] d  
2
2


[ f ( )]2  [ f ( )]2 d .
Example


Ex. Find the length of the cardioid r  1  sin  .
Sol.
2
2
2
2
s   (1  sin  )  (cos  ) d   2  2sin  d
0
0
2
2cos 
(I)

d  2 2  2sin   0
0
0
2  2sin 
2 2 | cos  |

d  8
0
2  2sin 



 2
2
2
(II) 2  2sin   2sin  cos  2sin cos  2(sin  cos )
2
2
2
2
2
2
2