# Parametric Equations Worksheet

Parametric Equations Worksheet
• Page 1
1.
John hits a ball when it is 3 ft above the ground . The ball leaves the bat with an initial velocity 20 ft/sec, at 60° angle with the horizontal and heads towards a 9 ft wall at a distance of 8 ft. At what height does the ball hit the wall?
 a. 9 ft b. 8 ft c. 6.6 ft d. 8.6 ft

#### Solution:

The motion of the ball can be given by the parametric equations x = (20 cos 60o)t = 10t, y = - 16t2 + (20 sin 60o)t + 3 = - 16t2 + 10(3)t + 3, where x is the horizontal distance, y is the vertical distance covered by the ball in time t sec.

Time taken by the ball to reach the 9 ft wall, which is at 8 ft from the home place = 820 cos 60° = 0.8 sec

At 0.8 sec, y = - 16(0.8)2 + 10(3)(0.8) + 3 = 6.6 ft approximately.
[Substitute t = 0.8 for y.]

So, the ball will hit the 9 ft wall at a height of 6.6 ft .

2.
Find the equation obtained by eliminating the parameter $t$ from the parametric equations $x$ = $t$ + 1, $y$ = 2$t$ + 5 .
 a. $y$ = 2$x$ + 3 b. $y$ = 3$x$ + 2 c. $y$ = 3$x$ - 2 d. $y$ = 2$x$ - 3

#### Solution:

x = t + 1, y = 2t + 5

x = t + 1 t = x - 1
[Solve for t.]

Then y = 2(x - 1) + 5
[Substitute t = x - 1 in y = 2t + 5.]

So, y = 2x + 3 is the required equation obtained by eliminating the parameter from the given equations.

3.
The parametric equations of the path of a moving particle are $x$ = 8$t$ + 8, $y$ = 9$t$ - 2. Find the path of the particle.
 a. an ellipse b. a straight line c. a circle d. a parabola

#### Solution:

x = 8t + 8, y = 9t - 2
[Write the parametric equations of the path of the particle.]

Consider x = 8t + 8 t = x - 88
[Solve for t.]

y = 9(x - 8)8 - 2
[Substitute t = x - 88 in y = 9t - 2.]

So, y = (98) x + (- 11 ) is the equation obtained by eliminating the parameter t.

The equation y = (98)x + (- 11 ) is a straight line with slope 98 and y - intercept (- 11 ).

So, the path of the moving particle is a straight line.

4.
Find the equation obtained by eliminating the parameter $t$ from the equations $x$ = $\sqrt{25-{t}^{2}}$, $y$ = $t$ .
 a. $x$2 - $y$2 = 25 b. $x$2 + $y$2 = 25 c. $x$2 - $y$ = 25 d. $x$2 + $y$ = 25

#### Solution:

x = 25-t2, y = t

x = 25-y2
[Substitute t = y in x = 25-t2.]

x2 = 25 - y
[Square on both sides.]

x2 + y = 25
[Simplify.]

So, x2 + y = 25 is the required equation obtained by eliminating the parameter t.

5.
Choose the equation obtained by eliminating the parameter $t$ from the equations $x$ = 8 cos $t$, $y$ = 8 sin $t$, 0 ≤ $t$$\pi$.
 a. $x$2 + $y$2 = 64 b. $x$2 + $y$2 = 8 c. $x$$y$ = 64 d. $x$2 - $y$2 = 64

#### Solution:

x = 8 cos t, y = 8 sin t, 0 ≤ tπ

cos t = x8, sin t = y8
[Solve for cos t, sin t.]

x264 + y264 = 1
[Use the identity cos2 t + sin2 t = 1.]

x2 + y2 = 64
[Simplify.]

So, x2 + y2 = 64 is the required equation obtained by eliminating the parameter t.

6.
Choose the equation obtained by eliminating the parameter $t$ from the equations $x$ = $t$ + 5, $y$ = $\frac{16}{t}$.
 a. ($x$ + 5)$y$ = 16 b. ($x$ - 5)$y$ = 16 c. ($x$ - 16)$y$ = 8 d. ($x$ + 16)$y$ = 5

#### Solution:

x = t + 5, y = 16t

t = 16y
[Solve for t from y = 16t.]

x = (16y) + 5
[Substitute t = 16y in x = t + 5.]

(x - 5)y = 16
[Simplify.]

So, the required equation obtained by eliminating the parameter t is, (x - 5)y = 16.

7.
Identify the equation obtained by eliminating the parameter $t$ from the equations $x$ = 2$t$, $y$ = 8 - $t$2.
 a. $x$2 - 4$y$ + 32 = 0 b. $x$2 + 4$y$ = 0 c. $x$2 + 4$y$ - 32 = 0 d. $x$2 - 4$y$ = 0

#### Solution:

x = 2t, y = 8 - t2

t = x2
[Solve for t from x = 2t.]

y = 8 - (x2)2
[Substitute t = x2 in y = 8 - t2.]

x2 + 4y - 32 = 0
[Simplify.]

So, x2 + 4y - 32 = 0 is the required equation obtained by eliminating the parameter.

8.
Find the equation obtained by eliminating the parameter from the equations $x$ = $a$$t$2, $y$ = 2$a$$t$.
 a. $x$2 + 4$a$$y$ = 0 b. $y$2 = 4$a$$x$ c. $y$2 + 4$a$$x$ = 0 d. $x$2 = 4$a$$y$

#### Solution:

x = at2, y = 2at

Consider y = 2at

y2 = 4a2t2
[Square on both sides.]

y2 = 4a(at2)

y2 = 4ax
[Substitute x for at2.]

So, y2 = 4ax is the required equation obtained by eliminating the parameter t.

9.
What is the graph represented by $x$ = 3 + 5$t$, $y$ = 4 + 5$t$, 1 ≤ $t$ ≤ 3?
 a. a straight line b. a line c. a circle d. a parabola

#### Solution:

x = 3 + 5t, y = 4 + 5t, 1 ≤ t ≤ 3

The end values of t are 1, 3.
[As 1 ≤ t ≤ 3.]

At t = 1, x = 8, y = 9
[Substitute t = 1 for x, y.]

At t = 3, x = 18, y = 19
[Substitute t = 3 for x, y.]

So, the end points of the graph are (8, 9) and (18, 19).

t = x - 35
[Solve for t from x = 3 + 5t.]

y = 4 + 5(x - 35)
[Substitute t = x - 35 in y = 4 + 5 t.]

y = x + 1
[Simplify.]

So, y = x + 1 is the equation obtained by eliminating t, which represents a line with slope 1, y - intercept 1.

As the graph has end points at (8, 9), (18, 19), it is a straight line.

10.
Which of the following points corresponds to $t$ = 1 in the parametrization $x$ = $t$2 + 4, $y$ = $t$ + ($\frac{2}{t}$), where $t$ is a non zero real number?
 a. (3, 5) b. (5, - 3) c. (- 3, 5) d. (5, 3)

#### Solution:

x = t2 + 4, y = t + (2t)

At t = 1, x = (1)2 + 4 = 5
[Substitute t = 1 for x.]

At t = 1, y = 1 + (21) = 3
[Substitute t = 1 for y.]

So, (5, 3) is a point corresponding to t = 1 in the parametrization given.