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Parametric Equations Xyz Homework

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Intro to parametric equations (video)

Intro to parametric equations

Let's say I have a cliff. Let's make this cliff, I don't know, let's say it's 50 meters high. And on this cliff, I have a car. And this car is not just sitting on the cliff, it's driving off of it. A very dramatic problem. So let's see, I have this car here. And it's driving off of this cliff at 5 meters per second. And I want to know, what is the path of this car as it falls off of the cliff? So let's set up a little coordinate axis here. So let's say that this is the y-axis right there. And then this will be my x-axis. So this is y, this is x. Let's say that this is the point-- well, we know that this is a 50 meter high cliff. Maybe y equals 0 is sea level. So this would be 50 right here. And let's say that this point right here on the cliff, that's at x is equal to 10. So this point right here's the point 10, 50. And let's say the car's right at this point, right about to drive off the cliff. And time is equal to 0. So this is that time equal to 0. So t for time. Time is equal to zero. So my question is, what happens to this car as it drives off the cliff? So this is a bit of a physics problem, and I won't go deep into the physics. And I won't prove some of the equations. And I encourage you to watch the [UNINTELLIGIBLE] videos, the projectile motion videos, if you want to know where the equations come from. But the point here is just to get the equations and see what the graph looks like. So if I want to know x as a function of time-- a suitably vibrant color-- so x as a function of time is going to be what? Well, we're going to assume that we're on a planet that has no air. We're in a vacuum. So if we start off in the x direction at 5 meters per second to the right, we won't be decelerated by air or friction or anything else. Newton's laws of motion: an object in motion stays in motion, unless it's affected by a net force. And there won't be any net force in the x direction. It's just going to keep moving to the right at 5 meters per second. And position, or distance, is just equal to velocity times time. Our velocity is 5 times time. And, of course, it didn't start at x is equal to 0. This is at time equals 0. So it started at x is equal to 10. So you want to know-- it's kind of x of 0, where it started off, so plus 10. And this should be a little intuitive for you, right? At time is equal to 0, this term cancels out, where x is equal to 10. That makes sense. At time is equal to 1, we should be a little bit-- we'll be 5 meters further out, so on and so forth. Fair enough. That's x as a function of the parameter time. As you probably realize, that this is a video on parametric equations, not physics. So it's nice to early on say the word parameter. Parameter. And time tends to be the parameter when people talk about parametric equations. Although it could be anything. It could be radius or angle or who knows what else? So let's figure out what y is as a function of time. So y as a function of time, it's going to be equal to the initial y position, or y of 0, which is 50. We're 50 meters up in the air. Plus our initial velocity in the y direction. And we don't actually have any initial velocity in the y direction. The car isn't jumping, or isn't diving. It's just moving horizontally to the right. And the cliff is supporting it. So it has no y velocity. But if you were curious, it would be the y velocity times time. But since there's no y velocity times time, at least initially, I'll put nothing there. Plus the acceleration of gravity times time squared over 2. We want to figure out the sine. And just so you know, I mean, it's nice to touch on the physics a little bit, just so you know where these formulas come from and you know the motivation behind why you would even use a parametric equation. Gravity goes downwards in this example, and downwards in this example is in the minus y direction. y is decreasing. And the real-- and you know, it's not exact-- but gravity is normally 9.8 meters per second squared in most textbooks. But for the sake of simplifying this, we'll say that it's approximately 10 meters per second squared. That's how fast everything will be accelerated downward on this planet. Since it has no air, let's assume it's a planet with a little bit more mass than earth. And since it's going downwards, its direction is negative. So in our formula up here, it's our initial position-- we had no velocity times time, so I won't put that there-- minus 10 meters per second squared times t squared over 2. And you can watch the projectile motion videos to figure out how I got these formulas right there. But that's not the point of this. The point of this is to graph what happens to the cars and learn a little bit about parametric equations. So what is the path of this car as it falls off the cliff? Let's make a table here. So x and y are a function of this third parameter, t. So we're going to set t at different values and we're going to figure out what x and y are equal to. And I'm just going to arbitrarily pick some t's. T is equal to 0. T is equal to 1, 2, and 3. At time is equal to 0, what is x? x of 0-- this is 0-- x is equal to 10 meters. At time is equal to 1, what is x? And this is x of 1, right? If I wanted to write that notation. So 5 times 1 is 5, plus 10 is 15. x of 2? 5 times 2 plus 10, that's 20. And it makes sense. Every second we're getting 5 meters more to the right. Or x is increasing by 5 meters. So when t is equal to 3, 15 plus 10 is 25. Easy enough. The y is a little bit more complicated. And just to simplify this, it's the same thing as 5, right? 10 divided by 2. So 50 minus 5t squared. So time is equal to 0, this term cancels out. We just have 50 meters up in the air. At time is equal to 1, 1 squared is 1 times 5, is five. 50 minus 5 is 45 five meters in the air. Is that right? Right, yeah, time 1, 50. Right. And then at time is equal to 2, 2 squared is 4. 4 times 5 is 20. 50 minus 20 is 30. And then finally, at time is equal to 3-- and I just say finally because that's the last number we picked-- time is equal to 3. 3 squared is 9. 9 times 5 is 45. 50 minus 45 is 5. So let's plot these points. So the time is equal to 0. That's what we got right there. At time is equal to 1, we're at x is equal to 15. That's roughly, you see this is 5, 10, 15-- let me do all of them-- 15, 20, 25. And then the y-axis-- let me label that while we're at it-- this is roughly 10, 20, 30, 40, 50. So at time is equal to 0, we're at 10, 50. That's that point right there. At time is equal to 1, we're at 15, 45. So x is 15, y is 45, which is right about there. So this is t is equal to 1. At time is equal to 2, or at the coordinate 20, 30, it's right about there. So this is at time is equal to 2. And then at time is equal to 3, we're at 25, 5. So we're right there. And if we kept going on, at some point we're going to hit the ground. And you can figure out, actually-- set this equal to 0 and you figure out the exact time you hit the ground. Actually, let's do that. If this is equal to 0, 50, you get t is equal to the square root of 10. Which is a little over 3 seconds. Which makes sense, right? A little over 3 seconds, we're going to be hitting the ground. But anyway, what's the path of this car? Well, it's going to look something like this. Ooh, it starts getting accelerated downwards, and then plunk! It hits the ground at 3 point something seconds. Now what was interesting here is that by setting the parameter, not only did we get the curve-- right? We got this curve, which is kind of half of a parabola, half of a downward shaping parabola-- and we could actually eliminate the t, and just get the equation for that parabola. And we'll do that in future videos. But what was interesting, by making it a parametric equation, we know the direction of the car. If you just saw this graph without the car and everything else I drew, you wouldn't know which way the car was falling. But now we know that as t is increasing, we're going in that direction. So we can draw some arrows here. So because it's a parametric equation, we can draw some arrows. And then the most important thing is we know exactly where the car is at any time t. You can substitute t is equal to 1.25 seconds and you'll know exactly where the car is. So you can plot these points and you can kind of get a sense that as time goes on, we're getting accelerated downwards. And that's why, for every second further, especially the y distance gets further and further apart. Anyway, I just wanted to give you this example. Although this was a good physics problem, the intention wasn't to teach you physics. The intention is to give you the motivation behind why parametric equations even exist. These two things are parametric equations. We defined x and y as a function of a third parameter, t, instead of defining y in terms of x or x in terms of y like we've done every other time since then. And this is super useful. I mean, you could imagine when you have really hard physics problems where you want to figure out the three dimensional position of something, then you'll have x as a function of t, y as a function of t, z as a function of t. All sorts of interesting problems come out of using parametric equations, not just in physics. But anyway, I thought a good place to start is the motivation. Because the first time I learned parametric equations I was like, why mess up my nice and simple world of x's and y's by introducing a third parameter, t? This is why. Because you can figure out the path of things. You can figure out the direction of something as it moves along a curve, and you can figure out its exact position at any, in this case, time.

Removing the parameter in parametric equations

Removing the parameter in parametric equations

Other articles

Parametric equations theory and problems

Parametric equations

Equation which except the unknown quantity contains another letter which can take different values from some multitude is called parametric equation. This letter taking part in the equation is called parameter. Actually with every parametric equation is written a multitude of equations. We will observe the solutions of simple parametric equations and module parametric equations.

Problem 1 Solve the equation in reference to x
A) x + a = 7
B) 2x + 8a = 4
C) x + a = 2a – x
D) ax = 5
E) a – x = x + b
F) ax = 3a

A) x + a = 7 x = 7 – a, a solution to the given equation is found.
In different values of the parameter, the solutions are x = 7 – a

B) 2x + 8a = 4 2x = 4 - 8a x = 2 – 4a

C) x + a = 2a – x x + x = 2a – a 2x = a x = a/2

D) ax = 5, when a is different from 0 we can divide both sides by a and we get x = 5
If a = 0, we get equation from the kind 0.x = 5, which has no solution;

E) a – x = x + b a – b = x + x 2x = a – b x = (a – b)/2

F) When a = 0 the equation ax = 3a is equal to 0.x = 0
Therefore every x is a solution. If a is different from 0, then
ax = 3a x = 3a/a x = 3

Problem 2 If a is a parameter solve the equation:
A) (a + 1)x = 2a + 3
B) 2a + x = ax + 4
C) a 2 x – x = a
D) a 2 x + x = a

A) If a + 1 is different from 0, i.e. a ≠ -1,
then x = (2a + 3)/(a + 1);
if a + 1 = 0, i.e. a = - 1
the equation get this look 0.x = (2).(-1) + 3
0.x = 1, which has no solution;

B) 2a + x = ax + 4
x – ax = 4 - 2a
(1 – a).x = 2(2 – a)
If (1 – a) ≠ 0, i.e. a ≠ 1; the solution is
x = 2(2 - a) / (1 - a);
if a = 1 the equation is 0.x = 2(2 - 1)
0.x = 2, which has no solution

C) a 2 x – x = a
x(a 2 -1) = a
(a - 1)(a + 1)x = a
If a - 1 ≠ 0 and a + 1 ≠ 0 i.e. a ≠ 1, -1,
the solution is x = a/(a - 1)(a + 1)
If a = 1 or a = -1, the equation is 0.x = ±1, which has no solution

Problem 3 If a and b are parameters solve the equation:
A) ax + b = 0
B) ax + 2b = x
C) (b - 1)y = 1 – a
D) (b 2 + 1)y = a + 2

A) ax + b = 0 ax = -b
If a ≠ 0, the solution is x = -b/a.
If a = 0, b ≠ 0, the equation get the look 0.x = -b and has no solution.
If a = 0 and b = 0, the equation is 0.x = 0 and every x is solution;

B) ax + 2b = x ax – x = -2b (a - 1)x = -2b
If a - 1 ≠ 0, i.e. a ≠ 1, the solution is x = -2b/a - 1
If a - 1 = 0, i.e. a = 1, but b ≠ 0, the equation is 0.x = - 2b and has no solution

C) if b - 1 ≠ 0, i.e. b ≠ 1,
the solution is y = (1 – a)/(b - 1)
If b - 1 = 0, i.e. b = 1, but 1 – a ≠ 0,
i.e. a ≠ 1, the equation is 0.y = 1 – a and has no solution.
If b = 1 and a = 1 the equation is 0.y = 0 and every y is solution

D) b 2 + 1 ≠ 0 for every b(why?), therefore
y = (a + 2)/(b 2 + 1) is solution to the equation.

Problem 4 For which values of x the following expressions have equal values :
A) 5x + a and 3ax + 4
B) 2x - 2 and 4x + 5a

To have equal values we must find the solutions of the equations
5x + a = 3ax + 4 and 2x – 2 = 4x + 5a

A) 5x + a = 3ax + 4
5x - 3ax = 4 – a
(5 - 3a)x = 4 – a
If 5 - 3a ≠ 0, i.e. a ≠ 5/3, the solution is x = (4 – a)/(5 - 3a)
If 5 - 3a = 0, i.e. a = 5/3, the equation is 0.x = 4 – 5/3
0.x = 7/3, which has no solution

B) 2x - 2 = 4x + 5a
-2 - 5a = 4x - 2x
2x = - 2 - 5a
x = -(2 + 5a)/2

Problem 5 Solve the parametric equation:
A) |ax + 2| = 4
B) |2x + 1| = 3a
C) |ax + 2a| = 3

A) |ax + 2| = 4 ax + 2 = 4 or ax + 2 = -4
ax = 2 or ax = - 6
If a ≠ 0, the equations are x = 2/a or x = -6/a
If a = 0, the equations have no solutions

B) If a 0 it is equivalent to 2x + 1 = 3a
or 2x + 1 = -3a 2x = 3a - 1 x = (3a - 1)/2 or
2x = -3a - 1 x = (3a - 1)/2 = -(3a + 1)/2

C) |ax + 2a| = 3 ax + 2a = 3 or ax + 2a = - 3,
and we find ax = 3 - 2a or ax = -3 - 2a
If a = 0 there are no solutions, if a ≠ 0
they are x = (3 - 2a)/a and x = -(3 + 2a)/a

Problem 6 Solve the equation 2 – x = 2b – 2ax, where a and b are real parameters. Find for which values of a the equation has for solution a natural number, if b = 7

We write the given equation like this (2a - 1)x = 2(b - 1)
The following cases are possible:
If 2a - 1 ≠ 0, i.e. a ≠ 1/2, the equation has single solution
x = 2(b - 1)/(2a - 1)
If a = 1/2 and b = 1, the equation gets the kind 0.x = 0 and every x is solution
If a = 1/2 and b ≠ 1, we get 0.x = 2(b - 1), where 2(b - 1) ≠ 0
In this case the equation has no solution.
If b = 7 and a ≠ 1/2 the single solution is
x = 2(7 - 1)/(2a - 1) = 12/(2a - 1)
If a is whole number, then 2a - 1 is also whole number and the solution
x = 12/(2a - 1) is natural number when
2a - 1 is positive divisor of 12.
To have a as whole number, the divisor of 12 mu be odd.But the only whole positive odd numbers divisible by 12 are 1 and 3.
Therefore 2a - 1 = 3 a = 2 or 2a - 1 = 1
a = 1 a = 2 or 2a - 1 = 1 a = 1

From the definition for module we get
|ax - 2 – x| = 4 ax - 2 – x = 4 or ax - 2 – x = - 4
From the first equality we get x(a - 1) - 2 = 4
(a - 1)x = 4 + 2 (a - 1)x = 6
From the second we have (a - 1)x = -2
If a - 1 = 0, i.e. a = 1, the last equations have no solutions.
If a ≠ 1 we find x = 6/(a - 1) or x = -2/(a - 1)
To have these roots as whole negative numbers must:
For the first equality a - 1 to be negative divisors of 6, and for the second positive divisors of 2
So a - 1 = -1; -2; -3; - 6 or a - 1 = 1; 2
We get a - 1 = -1 a = 0; a - 1 = -2
a = -1; a - 1 = -3 a = -2; a - 1 = -6 a = -5
or a - 1 = 1 a = 2; a - 1 = 2 a = 3
So a = -5; -2; -1; 0; 2; 3 are solutions of the problem.

Problem 8 Solve the equation:
A) 3ax – a = 1 – x, where a is parameter;
B) 2ax + b = 2 + x, where a and b are parameters

A) 3ax + x = 1 + a (3a + 1)x = 1 + a.
If 3a + 1 ≠ 0, i.e. a ≠ -11 /3 /3. the solution is
x = (1 + a)/(3a + 1)
If a = - 1/3 the equations gets the look 0.x = 1.1/3, which has no equation.

B) 2ax – x = 2 – b (2a - 1)x = 2 – b
If 2a - 1 ≠ 0, i.e. a ≠ 1/2, x = (2 – b)/(2a - 1) is the solution.
If a = 1/2 the equation gets the look 0.x = 2 – b
Then, if b = 2, every x is solution, if b ≠ 2, the equation has no solution.

We remake the equation to 6kx - 36 + 24 = 5kx kx = 12

A) If x = - 4/3, for k we get the equation - 4/3k = 12 k = - 9

B) The equation kx = 12 has no solutions when k = 0

C) When k ≠ 0 the root is x = 12/k and he is a natural number, if k is whole positive number, dividing 12, i.e. k = 1, 2, 3, 4, 6, 12

A) 2ax + 1 = x + a 2ax – x = a - 1
(2a - 1)x = a - 1
If 2a - 1 ≠ 0, i.e. a ≠ 1/2, the only solution of the equation is
x = (a - 1)/(2a - 1)
If 2a - 1 = 0, i.e. a = 1/2, the equation take the look
0.x = 1/2 - 1 0.x = - 1/2, which has no solution

B) 2ax + 1 = x + b
2ax – x = b - 1
(2a - 1)x = b - 1
If 2a - 1 ≠ 0, i.e. a ≠ 1/2, the solution is
x = (b - 1)/(2a - 1)
If a = 1/2, the equation is equivalent to 0.x = b - 1
If b = 1 every x is solution, if b ≠ 1 there is no solution.

A) If x = -2/3 is solution to the equation it is valid that
3[a(-2/3) - 4] + 4 = 2a(-2/3)
-2a - 12 + 4 = -4a/3
4a/3 - 2a = 8 (4a - 6a)/3 = 8
-2a/3 = 8 a = -12

B) 3(ax - 4) + 4 = 2ax 3ax - 2ax = 12 - 4 ax = 8
If a ≠ 0 the solution of the equation is x = 8/a, it is whole number if it is divisor of 8.
Therefore; ±2; ±4; ±8
If a=0 the equation has no solution

C) To have natural(whole positive) number for the solution x=8/a the number must a=1, 2, 4, 8

Problem 12 The equation 2 – x = 2b – 2ax is given, where a and b are parameters. Find for which values of a the equation has for solution a natural number, if b = 7

To be able to cross the graphic of a function the abscissa axis, needs the equation
(3a - 1).x -2a + 1 = 0 to have solution and will not cross it if there is no solution.
From the given equation we get (3a - 1)x = 2a - 1
If 3a - 1 ≠ 0, i.e. a ≠ 1/3, the equation have solution
x = (2a – 1)/(3a - 1), therefore the graphic of the function crosses the abscissa axis.
If a = 1/3 we get 0.x = 2/3 - 1 0.x = -1/3, which has no solution .
Therefore if a = 1/3 the graphic do not cross the abscissa axis.

Problem 14 Solve the parametric equation:
A) |x -2| = a
B) |ax -1| = 3
C) |ax - 1| = a - 2

A) if a 0 we get:
|x - 2| = a x - 2 = a or x - 2 = -a
From x - 2 = a => x = a + 2, and from
x - 2 = -a => x = 2 – a
If a = 0, then x - 2 = 0 or x = 2

B) |ax - 1| = 3 ax - 1 = 3 or ax - 1 = -3
from where ax = 4 or ax = - 2
If a ≠ 0 the solutions are x = 4/a or x = - 2/a
If a = 0 there is no solution

A) We will solve the second equation. We remake it in this way
(-x - 1) 2 - 1 = x 2
[(-1)(x + 1) ] 2 - 1 = x 2
x 2 + 2x + 1 - 1 = x 2
2x = 0 x = 0
For the first one we get
(x + m)/2 = 1 – m x + m = 2 - 2m x = 2 - 3m
The two equations are equivalent if they have the same roots, i.e.
2 - 3m = 0 m = 2/3

C) Because x > 3, 3 – x 2 – 5x + 3 = 0
x 2 - 4x – 0 x(x - 4) = 0
x = 0 or x = 4
By condition x > 3, therefore only x = 4 is solution. For the second equation we get
ax – x = 1 - 2a (a - 1)x = 1 - 2a
If a - 1 = 0 there is no solution(Why?), if a - 1 ≠ 0, i.e. a ≠ 1, the solution is
x = (1 - 2a)/(a - 1) The two equations will be equivalent if 4 = (1 - 2a)/(a - 1) 4(a - 1) = 1 - 2a 4a + 2a = 1 + 4 6a = 5 a = 5/6

Parametric equations in the forum

Learn Parametric Equations of a Circle with Experts

Parametric Equation of a Circle

Learn about parametric equations based on circle here and understand the concept better with solved examples provided. For more help, students can connect to an online tutor anytime and get the required help in the concept.

In mathematics, parametric equations are a method of defining a relation using parameters. A simple kinematical example is when one uses a time parameter to determine the position, velocity, and other information about a body in motion.

Abstractly, a Parametric Equation defines a relation as a set of equations. It is therefore somewhat more accurately defined as a parametric representation. It is part of regular parametric representation.

What is Parametric Equation

A parametric equation of circle is the coordinates of a point on the circle in terms of a single variable θ. These single variables are called as parameter. The parametric equations of a circle with radius (r ≥ 0) and center (h, k) are given by

x = h + r cos θ 0 ≤ θ ≤ 2 `pi`

Parametric equation of a circle centered at the origin with the radius r.

Equation of a circle with single variable formula

Each formula gives a portion of a circle

y= `sqrt(r^2 - x^2)` (Top)

y= - `sqrt(r^2 - x^2)` (Bottom)

x= `sqrt(r^2 - y^2)` (Right side)

x = - `sqrt(r^2 - y^2)` (Left side)

Parametric Equation Examples

Below are provided parametric equation examples for a better understanding:

Find the radius of parametric equation of the circle for the given x = 5 sin t and y = 5 cos t where 0 < t < `pi`

Formula for the parametric equation of a circle

(5 sin t) 2 + (5 cos t) 2 = r 2

5 2 (sin 2 t + cos 2 t) = r 2

25 (sin 2 t + cos 2 t) = r 2 Hence (sin 2 t + cos 2 t = 1)

Radius of the parametric circle is 5.

Find the radius of parametric equation of the circle for the given x = 2 sin t and y = 2 cos t where 0 < t < `pi`

Formula for the parametric equation of a circle

(2 sin t) 2 + (2 cos t) 2 = r 2

2 2 (sin 2 t + cos 2 t) = r 2

4 (sin 2 t + cos 2 t) = r 2 Hence (sin 2 t + cos 2 t = 1)

Radius of the parametric circle is 2.

Find the parametric equation of a circle with radius= 6 and center (2, 4).

x = h + r cos `theta`

y = k + r sin `theta`

Parametric equation of a circle with radius 6 and center (2, 4)

x = 2 + 6 cos `theta`

y = 4 + 6 sin `theta`

Literature review pecking order theory, Parametric equations xyz homework

Literature review pecking order theory, Parametric equations xyz homework

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