10.1 Parametric functions
Photo by Vickie Kelly, 2008
Mark Twain’s Boyhood Home
Hannibal, Missouri
Greg Kelly, Hanford High School, Richland, Washington
Photo by Vickie Kelly, 2008
Mark Twain’s Home
Hartford, Connecticut
Greg Kelly, Hanford High School, Richland, Washington
In chapter 1, we talked about
parametric equations.
Parametric equations can be
used to describe motion that is
not a function.
6
4
2
-6
x  f t 
y  g t 
-4
-2
0
2
4
6
-2
-4
-6
If f and g have derivatives at t, then the parametrized
curve also has a derivative at t.

The formula for finding the slope of a parametrized
curve is:
dy
dy
 dt
dx
dx
dt
This makes sense
if we think about
canceling dt.
The formula for finding the slope of a parametrized
curve is:
dy
dy
 dt
dx
dx
dt
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:
dy
2
d
d y
dt


y

 
2
dx
dx
dx
dt
1. Find the first derivative (dy/dx).
2. Find the derivative of dy/dx with respect to t.
3. Divide by dx/dt.

Example:
d2y
2
3
Find
as
a
function
of
t
if
x

t

t
and
y

t

t
.
2
dx
0.5
-2
-1.5
-1
-0.5
0.5
0
-0.5
-1

Example:
d2y
2
3
Find
as
a
function
of
t
if
x

t

t
and
y

t

t
.
2
dx
1. Find the first derivative (dy/dx).
dy
dy
y 
 dt
dx
dx
dt
1  3t

1  2t
2

2. Find the derivative of dy/dx with respect to t.
dy d  1  3t  2  6t  6t
 
 
2
dt dt  1  2t 
1  2t 
2
2
Quotient Rule

2  6t  6t
3. Divide by dx/dt.
dy
2
d y
dt

2
dx
dx
dt
1  2t 

2
2
1  2t

2  6t  6t
2
1  2t 
3

The equation for the length of a parametrized
curve is similar to our previous “length of curve”
equation:
2
2
 dx   dy 
L       dt
 dt   dt 
(Notice the use of the Pythagorean Theorem.)

Likewise, the equations for the surface area of a
parametrized curve are similar to our previous
“surface area” equations:
Revolution about the x-axis
2
 y  0
2
 dx   dy 
S   2 y      dt
a
 dt   dt 
b
Revolution about the y-axis
2
 x  0
2
 dx   dy 
S   2 x      dt
a
 dt   dt 
b

6
4
2
-6
-4
-2
0
2
4
6
This curve is:
-2
-4
x  t   t  sin 2t
-6
y  t   t  2 cos  5t 
