Parametric equations

[Probability]

Hi would just like a second opinion please to check if I have done this correctly?

I have

x = - 3t - 3 and y = 2 - t

I Have said

t = - 1/3x - 3

Then I said

y = 2 - t = - 1/3(x) - 3

then I said

y = 2 - t = 2( - 1/3)x + 3

y = 2 - t = - 6 + 1 + 6

y = 2 - t = - 5 + 6

y = 1

A second opinion would be appreciated.

 

[lookagain]

Hi would just like a second opinion please to check if I have done this correctly?

I have

x = - 3t - 3 and y = 2 - t

I Have said

t = - 1/3x - 3 \(\displaystyle \ \ \ \) X

Then I said

y = 2 - t = - 1/3(x) - 3 \(\displaystyle \ \ \ \) X, etc.

then I said

y = 2 - t = 2( - 1/3)x + 3

y = 2 - t = - 6 + 1 + 6

y = 2 - t = - 5 + 6

y = 1

A second opinion would be appreciated.

No, you'll have to start from scratch.

You need instructions stated here for the problem, such as

"Eliminate t and write one equation with y in terms of x," or something similar.


For now, what I would state is:


1) Have another attempt and solve (x = -3t - 3) for t in terms of x.

2) Solve (y = 2 - t) for t in terms of y.

3) Set the t expressions above equal to each other and manipulate the resulting equation
until it is an equation where y is equal to an expression in terms of x.

 

[stapel]

Hi would just like a second opinion please to check if I have done this correctly?

What were the instructions? What, exactly, are you supposed to be doing?

I have

x = - 3t - 3 and y = 2 - t

Are you maybe supposed to "express the parametric equations as a single Cartesian equation"?

I Have said

t = - 1/3x - 3

Your formatting is ambiguous. I think you meant to say the following:

. . . . .\(\displaystyle t\, =\, \left(\frac{1}{3}\right)(-x\, -\, 3)\)

y = 2 - t = - 1/3(x) - 3

I'm sorry, but I don't follow...?

y = 2 - t = 2( - 1/3)x + 3

y = 2 - t = - 6 + 1 + 6

Where did these lines come from?

I think you maybe were attempting a substitution. Assuming I guessed the instructions correctly, then try writing things out clearly, so fractions, etc, don't get confused.

You have y = 2 - t, so then t = 2 - y. Try plugging "2 - y" in for "t" in the equation for x, and simplifying. This will give you a (non-constant) equation for y in terms of x.

 

[Probability]

This is the very first time I have come across these types of equations and am working blind with them. The information given in the course book in my opinion is significantly over the top for a beginner in this subject. I think that some people with many years experience have seriously forgot that people just can't start from a level of Phd, but must start from basic first principles.

In the work I am creating a circle, I have line AB and BC, I have worked out the gradient of AB and the midpoint, and the equation of the perpendicular line. Having put my value of x into the equation I get the result of my y value so I assume that my equation is correct at this point.

Then I am asked to use the parametric equations to prove this line is the same, which I am assuming is talking about the bisector of AB?

In my understanding there is only two ways to solve this problem

x = - 3t - 3

3t = x + 3

y = 2 - t

t = 2 + y

My problem here is which variable am I suppose to select?

Taking this one step at a time.

If

t = 2 + y = 2(x + 3) = 2x + 6

and my previous value of x =- 6

then

2(-6) + 6 = - 6

This suggests that t = y = - 6

This is not correct because I know that y = 1

I am not sure I know any other way to solve this?

 

[pka]

I have
x = - 3t - 3 and y = 2 - t.

Solving for t we get
\(\displaystyle \\\dfrac{{x + 3}}{{ - 3}} = \dfrac{{y - 2}}{{ - 1}}\\x+3=3y-6\\x-3y=-9\)

 

[Probability]

Solving for t we get
\(\displaystyle \\\dfrac{{x + 3}}{{ - 3}} = \dfrac{{y - 2}}{{ - 1}}\\x+3=3y-6\\x-3y=-9\)

Thanks for your help

 

[lookagain]

I am not sure I know any other way to solve this?

Probability, why don't you *go back* and read what I already posted to you in post #2 and try that!?

Next time, either 1) don't miss my post or 2) don't ignore my post. Either way, your particular recent attempt

shouldn't even have been happened, because I gave you the guidelines to follow back in post #2.

 

[Probability]

Probability, why don't you *go back* and read what I already posted to you in post #2 and try that!?

Next time, either 1) don't miss my post or 2) don't ignore my post. Either way, your particular recent attempt

shouldn't even have been happened, because I gave you the guidelines to follow back in post #2.

Sorry don't mean to upset anyone, but try to see it from my point of view. I have never worked with parametric equations before and I am struggling not only to understand how to work with them, but at the same time apply something I don't understand to a problem that I have also not yet fully understood.

I am also trying to understand poorly worded questions at the same time.

I don't fully understand pka post either, if the denominator - 3 is multiplied across to give -3y,why then when - 3 is multiplied by - 2 to give - 6 is that not +6

I am confused with the whole issue.

 

[pka]

I don't fully understand pka post either, if the denominator - 3 is multiplied across to give -3y,why then when - 3 is multiplied by - 2 to give - 6 is that not +6

I wish you had asked earlier. It is after all simple algebra.

\(\displaystyle \\\dfrac{{x + 3}}{{ - 3}} = \dfrac{{y - 2}}{{ - 1}}\\x+3=3y-6\\x-3y=-9\)

Are you clear on where \(\displaystyle \dfrac{{x + 3}}{{ - 3}} = \dfrac{{y - 2}}{{ - 1}}\) comes from?

Is it clear to you that is the same as:
\(\displaystyle \\(x+3)\left(\frac{{-3}}{{ - 3}}\right) =(y-2)\left( \frac{-3}{- 1}\right)\\(x+3)(1)=(y-2)(3)~?\)

So we get \(\displaystyle x+3=3y-6\).

 

[Probability]

I wish you had asked earlier. It is after all simple algebra.

\(\displaystyle \\\dfrac{{x + 3}}{{ - 3}} = \dfrac{{y - 2}}{{ - 1}}\\x+3=3y-6\\x-3y=-9\)

Are you clear on where \(\displaystyle \dfrac{{x + 3}}{{ - 3}} = \dfrac{{y - 2}}{{ - 1}}\) comes from?

Is it clear to you that is the same as:
\(\displaystyle \\(x+3)\left(\frac{{-3}}{{ - 3}}\right) =(y-2)\left( \frac{-3}{- 1}\right)\\(x+3)(1)=(y-2)(3)~?\)

So we get \(\displaystyle x+3=3y-6\).

Thank you pka I understand it now after some brain ache, but it will still take some getting used to.