In other words, at time t , the particle is located at the point

, x f t y g t Parametric Equations In other words, at time t, the particle is located at the point

, c t f t g t Parametric Curve

Sketch the curve with parametric equations22 4, 3x t y t

2 2 4 3t x t y t

2

0

2

4

8

4

0

4

7

3

7

19Notice, every point has a time

and we also have a direction arrow!

CONCEPTUAL INSIGHT The graph of a function y = f (x) can always be parametrized in a simple way as c (t) = (t, f (t))For example, the parabola y = x2 is parametrized by c (t) = (t, t2) and the curve y = et by c (t) = (t, et). An advantage of parametric equations is that they enable us to describe curves that are not graphs of functions. For example, the curve below is not of the form y = f (x) but it can be expressed parametrically.

Eliminating the Parameter Describe the parametric curvec (t) = (2t − 4, 3 + t2)

of the previous example in the form y = f (x).

Solve for y as a function of x:4 1

2 4 22 2

xx t t x

221 1

3 2 7 22 4

y x x x

Substitute:

in terms of !t x

in terms of !y x

217 2

4y x x

(We have in terms of

and in terms of .)

y t

t x

' 200 9.8 0

20.408 s 2 2041 0.408 m

y t t

t y

A bullet follows the trajectoryc (t) = (80t, 200t − 4.9t2)

until it hits the ground, with t in seconds and distance in meters. Find:

(a) The bullet’s height at t = 5s.(b) Its maximum height.

The height of the bullet at time t is y (t) = 200t − 4.9t2

877 m5 .5 y

The maximum height occurs at the critical point of y (t):

THEOREM 1 Parametrization of a Line

(a) The line through P = (a, b) of slope m is parametrized by

for any r and s (with r 0) such that m = s/r.

(b) The line through P = (a, b) and Q = (c, d) has parametrization

The segment from P to Q corresponds to 0 ≤ 1 ≤ t.

Solution

, in ,r sx a t y b t t

, in ,x a t y b t tc a d b

yx

y mx b

1 1y y m x x

'ly many parametrizations of a line. These are equivalent expressions for and . In (a) we are given point-slope, in (b) two points.x t y t

x ay s b m x a

rb st y b

(a) Use x = a + rt, to write t in terms of x… implies t = (x − a)/r:

THM 1

This is the equation of the line through P = (a, b) of slope m. The choice r = 1 and s = m yields the parametrization below.

(b) This parametrization defines a line that satisfies (x (0), y (0)) = (a, b) and (x (1), y (1)) = (c, d). Thus, it parametrizes the line through P and Q and traces the segment from P to Q as t varies from 0 to 1.

So how do we parametrize a line? 1 1y y m x x

, ,Q c d

Parametrization of a Line Find parametric equations for the line through P = (3, −1) of slope m = 4.

, in ,r sx a t y b t t

4 1 & s 4 , 3, 1s

m r P a br

3 , 1 4x t y t

Parametric Curve: 3 , 1 4 c t t t

'ly many parametrizations of a line.

4 5 & s 20 , 3, 1s

m r P a br

3 5 , 1 20c t t t

The circle of radius R with center (a, b) has parametrization

cos , sinx R y R

Let’s verify that a point (x, y) given by the above equation, satisfies the equation of the circle of radius R centered at (a, b):

In general, to translate a parametric curve horizontally a units and vertically b units, replace c (t) = (x (t), y (t)) by c (t) = (a + x (t), b + y (t)).

2 2

2

2 2

2 22 2

cos sin

cos sin

a R a b R b

R

x a y b

RR

Suppose we have a parametrization c (t) = (x (t), y (t)) where x (t) is an even function and y (t) is an odd function, that is, x (−t) = x (t) and y (−t) = −y (t). In this case, c (−t) is the reflection of c (t) across the x-axis:

c (−t) = (x (−t), y (−t)) = (x (t), −y (t))

The curve, therefore, is symmetric with respect to the x-axis.

Parametrization of an Ellipse Verify that the ellipse with equation

is parametrized by

Plot the case a = 4, b = 2.

Show that the equation of the ellipse is satisfied with x = a cos t, y = b sin t:

2 2

1x y

a b

, cos s or in fac t b t tt

2 2 22 2

2cos sin

+ cos 1sina t b t

t ta

y

b b

x

a

To plot the case a = 4, b = 2, we connect the points corresponding to the t-values in the table. This gives us the top half of the ellipse corresponding to 0 ≤ t ≤ π. Then we observe that x (t) = 4 cos t is even and y (t) = 2 sin t is odd. As noted earlier, this tells us that the bottom half of the ellipse is obtained by symmetry with respect to the x-axis.

c (−t) = (x (−t), y (−t)) = (x (t), −y (t))

The curve, therefore, is symmetric with respect to the x-axis.

Different Parametrizations of the Same Curve Describe the motion of a particle moving along each of the following paths.(a) c1(t) = (t3, t6)(b) c2(t) = (t2, t4)(c) c3(t) = (cos t, cos2 t)

Each of these parametrizations satisfies y = x2, so all three parametrize portions of the parabola y = x2.

3

3 6 21

(a) As varies from to , varies from to .

, traces all of

(moving left to right and passing through each point once).

t t

c t t t y x

c (t) = (t, f (t))

Different Parametrizations of the Same Curve Describe the motion of a particle moving along each of the following paths.(a) c1(t) = (t3, t6)(b) c2(t) = (t2, t4)(c) c3(t) = (cos t, cos2 t)

Each of these parametrizations satisfies y = x2, so all three parametrize portions of the parabola y = x2.

2 2 42(b) 0 the path of , traces only the right

half of the parabola. The particle comes in towards the origin as

varies from to 0, and goes back out to the right as varies

fr

x t c t t t

t t

om 0 to .

Different Parametrizations of the Same Curve Describe the motion of a particle moving along each of the following paths.(a) c1(t) = (t3, t6)(b) c2(t) = (t2, t4)(c) c3(t) = (cos t, cos2 t)

Each of these parametrizations satisfies y = x2, so all three parametrize portions of the parabola y = x2.

23

(c) As varies from to , cos oscilates between 1 and 1.

the particle following cos ,cos oscilates between

the points 1,1 and 1,1 on the parabola.

t t

c t t t