Reminder from class discussion:
The speed of an object traveling along a
parametric curve is given by:
2
 dx   dy 
   
 dt   dt 
(1)
2
Let x  cos  t  and y  sin  2t 
for 0  t  2.
a) Sketch this parametric curve.
dy 2 cos  2t 

dx
 sin  t 
dy
.
dx
b)
Find
c)
Find the ordered pairs (x, y) of all points where the tangent line is
i)
Horizontal
ii) Vertical
dy
dx
 0.
 0.
Tangent line will be horizontal if
Tangent line will be vertical if
dt
dt
2 cos  2t   0
 sin  t   0
2t 
t
t
d)

2

4
t   k, k 
  k, k 


2
t  0 :  cos  0  ,sin  0   or 1, 0 
t   :  cos   ,sin    or  1, 0 
k, k 
 2 
 
  
:  cos   ,sin    or 
,1
4 
4
 2 
 2 
 

 3 
 3
:  cos 
 ,sin 
 4 
 2



2

, 1
  or  

 2

t
3
4
t

5 
2 
 5 
 5  
:  cos 
,1
 ,sin 
  or  
4 
 4 
 2 
 2 
t
7
4

 7
:  cos 
 4


 7
 ,sin 

 2
 2


, 1
  or 

 2

Assuming these equations represent the position in meters of an object at time t seconds, find the


speed of the object at time t  and then again at time t  .
4
2
Speed 
t
t
  sin  t   2   2 cos  2t   2 
sin 2  t   4 cos 2  2t 

1
1
 
 
: sin 2    4cos 2   
0 
meters per second
4
2
2
4
2

 
: sin 2    4cos 2    1  4  5 meters per second
2
2
e)
Eliminate the parameter t and write the Cartesian (x-y) equation of the curve. (Try!)
y  sin  2t 
y  2sin  t  cos  t 
y 2  4sin 2  t  cos 2  t 


y 2  4 1 x 2 x 2
(2) Let x  8cos3  t  and y  8sin 3  t 
a)
Find each of the following.
dx
dy
i)
ii)
 24 cos 2  t  sin  t 
 24sin 2  t  cos  t 
dt
dt
2
24sin  t  cos  t 
sin  t 
dy
iii)


  tan  t 
2
dx 24 cos  t  sin  t 
cos  t 
b)
Write the equation of the tangent line when t 
dy
dx
t 
6
d)
6
.
3
 
  tan  
6
 3
 
 
x    8cos 3    8 
  3 3
6
6
 2 
3

3
 
 
1
y    8sin 3    8    1
6
6
2
y 1  
c)


3

3
3
x  3 3 or y  
x4
3
3
Sketch this parametric curve in the space at the
right and label the value of t at each cusp.
t
At each cusp, what is true about the value
dx
dy
of both
and
?
dt
dt
Find each of the following:
dy
lim
0
i.
t 0 dx
2
t 0
t 
Both are equal to 0.
e)

t
ii.
lim
t

2
dy
 
dx
3
2
f)
If these equations represent the position in meters of a particle in motion at time t seconds, find the
speed of the particle at time:

i)
t
4
2
2
2


1 2 
1 2
   
  
2 
2 

24
cos
sin

24sin
cos


24



24






    
 
 


2 2  
2 2 
 4   4  
4
 4 


 72  72
 12 m/s
ii) t  2
 24cos 2  2 sin  2   24sin 2  2 cos  2
2
(3)
2
 9.08 m/s
Let x  2cos  t  +cos  2t  and y  2sin t   sin  2t  . (Note: This curve is called a deltoid.)
a)
Sketch this parametric curve.
b)
Find
dy
dx
dy 2 cos  t   2 cos  2t 

dx 2sin  t   2sin  2t 

cos  t   cos  2t 
sin  t   sin  2t 
2
(4)
A balloon is filled with air and then
released. As the air escapes, the
balloon flies around the room for 5
seconds following the path given by
the parametric equations
x  0.7t  3sin  2t  and
y  0.5t  4  4cos  3t  , where the
position is measured in feet.
a)
Sketch the graph of this path for
time 0 ≤ t ≤ 5 seconds.
b)
Find the speed of the balloon at
time t = 1.5 seconds.
dy
dx
 0.5  12sin  3t 
 0.7  6 cos  2t 
dt
dt
 0.7  6 cos  3    0.5  12sin  4.5 
2
2
 13.05 ft/sec
c)
Find the slope of the tangent line to the path at t = 1.5 seconds.
0.5  12sin  4.5 
dy 0.5  12sin  3t  dy


 1.69
dx 0.7  6 cos  2t  dx t 1.5
0.7  6 cos  3
d)
Find the first time that the balloon is moving horizontally and find its speed at this time.
dy
dx
 0 and
 0.
Balloon will be moving horizontally when
dt
dt
1
0.5  12sin  3t   0  sin  3t   
24
First time that this occurs is when t  1.061 sec. The speed at this instant is:
2
 0.7  6 cos  2 1.061    0 2  3.84 ft/sec
e)
Find the first time that the balloon is moving vertically and find its speed at this time.
dx
dy
 0 and
 0.
Balloon will be moving vertically when
dt
dt
1
0.7  6 cos  2t   0  cos  2t  
42
First time that this occurs is when t  0.727 sec. The speed at this instant is:
 0 2   0.5  12sin  3  0.727  
2
 10.34 ft/sec
(5)
d2y
To find the second derivative
with parametric equations, we can use the following formula:
dx 2
d  dy 
2
d y dt  dx 

dx
dx 2
dt
Let's return to a simple ellipse from Parametrics 1: x  5cos  t  and y  2sin  t 
a)
Find
d2y
.
dx 2
dy
 2 cos  t 
dt
dx
 5sin  t 
dt
d  dy  2
   csc  t 
dt  dx  5
b)
Find
dy
2
  cot  t 
dx
5
2
csc  t 
d y 5
2

  csc3  t 
2
5sin  t 
25
dx
2
d2y
d2y
and
.
dx 2 t 
dx 2 t 
6
6
What would these values typically tell us about the concavity of the graph at these points?
d2y
2
2 3
16
d2y
2
2
16
3
 
3 


csc



2


  csc3       2  
 
2
2
25
25
25
25
25
25
dx t 
dx t 
6
 6
6
6
A negative value of the second derivative typically indicates that the graph is concave down at the
point in question, while a positive value typically indicates that the graph is concave up.
c)
Based on the results to part b, what special type of point would you expect the graph of the ellipse to


have between t   and t  ? Does the graph appear to have such a point (in the same way that
6
6
we have encountered these points when graphing functions)?
Based on the results of part b, one would expect the graph of the ellipse to have an inflection point


between t   and t  . The graph does appear to change concavity at t = 0, but the inflection
6
6
point there is not quite the same as ones we have encountered previously because the graph is not
that of a function and “comes back up itself” after the inflection point..