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Consider the parametric curve given by the equations x(t)=t^2 + 6t - 26 y(t)=t^2 + 6t - 28 How many units of distance are covered by the point P(t) = (x(t),y(t)) between t=0, and t=5 ?

• We reviewed parametrically defined functions in Section 1.4. Instead of defining the
points (x, y) on a planar curve by relating y directly to x, we can define both coordinates as
functions of a parameter t. The resulting set of points may or may not define y as a function of x (that is, the parametric curve might fail the vertical line test).
10.1
• Parametric Curves in the Plane
• Slope and Concavity
• Arc Length
• Cycloids
. . . and why
Parametric equations enable us
to define some interesting and
important curves that would be
difficult or impossible to define in
the form y f(x).
EXAMPLE 1 Reviewing Some Parametric Curves
Sketch the parametric curves and identify those which define y as a function of x. In
each case, eliminate the parameter to find an equation that relates x and y directly.
(a) x cos t and y sin t for t in the interval [0, 2)
(b) x 3 cos t and y 2 sin t for t in the interval [0, 4]
(c) x t and y t 2 for t in the interval [0, 4]
SOLUTION
(a) This is probably the best-known parametrization of all. The curve is the unit circle
(Figure 10.1a), and it does not define y as a function of x. To eliminate the parameter, we
use the identity cos t
2
sin t
2
1 to write x
2
y
2
1.
(b) This parametrization stretches the unit circle by a factor of 3 horizontally and by a factor
of 2 vertically. The result is an ellipse (Figure 10.1b), which is traced twice as t covers the interval [0, 4]. (In fact, the point (3, 0) is visited three times.) It does not define y as a
function of x. We use the same identity as in part (a) to write

3
x

2

2
y

2
1.
(c) This parametrization produces a segment of a parabola (Figure 10.1c). It does define
y as a function of x. Since t x
2
, we write y x
2
2. Now try Exercise 1.
Figure 10.1 A collection of parametric curves (Example 1). Each point (x, y) is determined by
parametric functions of t, but only the parametrization in graph (c) determines y as a function of x.
3
–3
–2
(a)
y
x
(b)
4
1
–4 –1
–3
3
1
• eliminating t we have,
x - y = 2
this is an equation of line,

at t=0 (putting values)
the inital point is => (-26,-28)

at t=5 (putting values)
the final point is => (29,27)

the distance between the 2 points,
D = [(55)^2 + (55)^2]^.5
=> D = 55√2

• We reviewed parametrically defined functions in Section 1.4. Instead of defining the
points (x, y) on a planar curve by relating y directly to x, we can define both coordinates as
functions of a parameter t. The resulting set of points may or may not define y as a function of x (that is, the parametric curve might fail the vertical line test).
10.1
• Parametric Curves in the Plane
• Slope and Concavity
• Arc Length
• Cycloids
. . . and why
Parametric equations enable us
to define some interesting and
important curves that would be
difficult or impossible to define in
the form y f(x).

eliminating t we have,
x - y = 2
this is an equation of line,

at t=0 (putting values)
the inital point is => (-26,-28)

at t=5 (putting values)
the final point is => (29,27)

the distance between the 2 points,
D = [(55)^2 + (55)^2]^.5

=> D = 55v2
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