dy dy / dt y '  t 


(a) An equation of the tangent line at t = 3 dx dx / dt x '  t 
Let c (t) = (t2 + 1, t3 − 4t). Find:
(b) The points where the tangent is horizontal.
dy
23
dy 3t  4



dx t 3 6
dx
2t
2
 2   7 16 
c
 3,

3  3 3

23
c  3  10,15   y  15   x  10  c  2    7 ,  16 

 3

6
3 3
dy
 3 
 0  y ' t   0 & x ' t   0
dx
2
y '  t   3t  4  0  t  
3
`
2






`
3
2 
8
8
8
24
16
 2 

4











3
3
27
3
3 3 3 3 3 3

3
2 
8
8
8
24
16
 2 

4









3
27
3 3 3 3 3
3 3
 3
`
We need a point and a slope
THEOREM 1 Arc Length Let c(t) = (x(t), y(t)), where
x (t) and y (t) exist and are continuous. Then the arc
length s of c(t) for a ≤ t ≤ b is equal to
b
s   x '  t   y '  t  dt
2
a
2
The simplest parametrization of y = f (x) is c (t) = (t, f (t)).
x ' t   y ' t   1  y ' t 
2
2
2
Which leads to the arc length formula derived in Section 9.1.
b
s   1  y '  t  dt
2
a
Arc Length:
but (in a parametric curve)
b
s   x '  t   y '  t  dt
2
a
2
x  t  will not always equal t.
C  2 r
The arc length integral can be evaluated explicitly only in
special cases. The circle and the cycloid are two such cases.
Use THM 1 to calculate the arc length of a circle of radius R.
x  R cos  ,
y  R sin 
x '    y '      R sin     R cos    R
2
2
s
2
2
 Rd  R  
2
0
0
2
2
 2 R
b
s   x '    y '   d
2
a
2
In the parametric equation for a circle, x and y are usually in terms of  .
Centered at  0, 0  ...
Calculate the length s of one arch of the cycloid generated by
a circle of radius R = 2.
x  Rt  R sin t ,
y  R  R cos t
x  2  t  sin t  ,
y  2 1  cos t 
x '  t   2 1  cos t  ,
y '  t   2sin t
t 0
 x '  t   y '  t   4 1  cos t   4sin 2 t
2
2
2
 4 1  2 cos t  cos t  sin t 
2
2
 1  cos t 
 8 1  cos t   16 

2


2 t
Square Root
 16sin
2
1  cos t
2 t
Trig Identities...
 sin
2
2
Throughout this chapter,
I'll give you a handful of new trig identities.
2

t
s  4  sin dt  8 sin udu
2
0
0
 8   cos u 0  8  cos u   16

0
u  t / 2  du  dt / 2
b
s   x '  t   y '  t  dt
2
a
2
Speed is defined as the rate of change of distance traveled
with respect to time, so by the 2nd Fundamental Theorem of
Calculus,
t
ds d
2
2
2
2
Speed 
  x '  u   y '  u  du  x '  t   y '  t 
dt dt t0
THEOREM 2 Speed Along a Parametrized Path
The speed of c (t) = (x (t), y (t)) is
ds
2
2
Speed 
 x ' t   y ' t 
dt
To determine speed with the 2nd FTC, let's write our
bounds in terms of t and our integrand in terms of u.
b
s   x '  t   y '  t  dt
2
a
2
The next example illustrates the difference between distance
traveled along a path and displacement (also called net
change in position). The displacement along a path is the
distance between the initial point c (t0) and the endpoint
c (t1). The distance traveled is greater than the displacement
unless the particle happens to move in a straight line.
A particle travels along the path c (t) = (2t, 1 + t3/2).
Find:
(a) The particle’s speed at t = 1 (assume units of meters and
minutes).
3 1/2
x '  t   2, y '  t   t
2
ds
9

 4   2.5 m/min
dt
4
ds
2
2
Speed 
 x ' t   y ' t 
dt
A particle travels along the path c (t) = (2t, 1 + t3/2).
Find:
(a) The particle’s speed at t = 1 (assume units of meters and
minutes).
(b) The distance traveled s and displacement d during the
interval 0 ≤ t ≤ 4. u  4  9 t  du  9 dt 
3
1/2
x
'
t

2,
y
'
t

t




4
4
2
13
9
4 1/2
8 3/2 13 8
u  
4  tdt   u du 
133/2  8   11.518 m

4
4
94
27
27

4
s
0

c  0    0,1 , c  4    8,9 
d  64  64  8 2  11.314 m
b
s   x '  t   y '  t  dt
2
a
2
In physics, we often describe the path of a particle moving
with constant speed along a circle of radius R in terms of a
constant ω (lowercase Greek omega) as follows:
c (t) = (R cos ωt, R sin ωt)
The constant ω, called the angular velocity, is the rate of
change with respect to time of the particle’s angle θ.
Next Slide...
A particle moving on a
circle of radius R with
angular velocity ω has
speed |ω|R.
Angular Velocity Calculate the speed of the circular path of
radius R and angular velocity ω. What is the speed if
R = 3 m and ω = 4 rad/s?
c (t) = (R cos ωt, R sin ωt)
x '  t    R sin t , y '  t    R cos t
ds


dt
  R sin t    R cos t 
2
2
  2 R 2 sin 2 t   2 R 2 cos 2 t   R
Thus, the speed is constant with
value |ω|R. If R = 3 m and ω = 4
rad/s, then the speed is |ω|R = 3(4)
= 12 m/s.
ds
2
2
Speed 
 x ' t   y ' t 
dt
~ 13, 23, 25