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 .


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


Correct answer : (1)
 3.  
The parametric equations of the path of a moving particle are x = 8t + 8, y = 9t - 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.


Correct answer : (2)
 4.  
Find the equation obtained by eliminating the parameter t from the equations x = 25-t2, y = t .
a.
x2 - y2 = 25
b.
x2 + y2 = 25
c.
x2 - y = 25
d.
x2 + 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.


Correct answer : (4)
 5.  
Choose the equation obtained by eliminating the parameter t from the equations x = 8 cos t, y = 8 sin t, 0 ≤ tπ.
a.
x2 + y2 = 64
b.
x2 + y2 = 8
c.
xy = 64
d.
x2 - y2 = 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.


Correct answer : (1)
 6.  
Choose the equation obtained by eliminating the parameter t from the equations x = t + 5, y = 16t.
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.


Correct answer : (2)
 7.  
Identify the equation obtained by eliminating the parameter t from the equations x = 2t, y = 8 - t2.
a.
x2 - 4y + 32 = 0
b.
x2 + 4y = 0
c.
x2 + 4y - 32 = 0
d.
x2 - 4y = 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.


Correct answer : (3)
 8.  
Find the equation obtained by eliminating the parameter from the equations x = at2, y = 2at.
a.
x2 + 4ay = 0
b.
y2 = 4ax
c.
y2 + 4ax = 0
d.
x2 = 4ay


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.


Correct answer : (2)
 9.  
What is the graph represented by x = 3 + 5t, y = 4 + 5t, 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.


Correct answer : (1)
 10.  
Which of the following points corresponds to t = 1 in the parametrization x = t2 + 4, y = t + (2t), 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.


Correct answer : (4)

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