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#1 2006-03-08 03:15:54

lakitu
Member
Registered: 2006-03-08
Posts: 5

parametric equations as a cartesian equation help

hi apologise if this is in the wrong forum smile

my lecturer has told me that i need to be able to express parametric equations as a cartesian equation and then Simplify it into the form y = a x² + b x + c in my exam later this month. my mind boggles !

here is an example i have found.

Express the parametric equations x = t + 1 and y = -3 t² + 3 t as a Cartesian equation in just x and y.

any help would be great!

kind regards lakitu smile

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#2 2006-03-08 03:48:52

lakitu
Member
Registered: 2006-03-08
Posts: 5

Re: parametric equations as a cartesian equation help

hi again, i have re written my question in a better way, sorry for the confusion.


i have figured out how to do a simple parametric - cartesian equation in the form of x and y here is what i have already done.

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

multiply (1) by 3 and (2) by 2.

3x = 6 t - 6 -----------------(3)
2y = 6 t - 4 -----------------(4)

eliminate t from the two eqns by subtracting (4) from (3)

3x - 2y = -6 + 4
3x - 2y = -2


in my exam i am assured i will need to know harder ones like this, but i cannot figure out how to do these.

Express the parametric equations x = t + 1 and y = -3 t² + 3 t as a Cartesian equation in just x and y.

i also need to simpify this in the form y = a x² + b x + c.

Any help would be great.

kind regards lakitu smile

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#3 2006-03-08 03:51:19

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,775

Re: parametric equations as a cartesian equation help

Since x=t+1, t=x-1.
Given
y=-3t²+3t
Therefore,
y=-3t(t+1)+6t
y=-3(x-1)(x)+6(x-1)
y=-3x²+3x+6x-6
y=-3x²+9x-6.

You can verify this by substituting x=t+1 in the Cartesian equation.
y=-3(t+1)²+9(t+1)-6
y=-3t²-6t-3+9t+9-6
y=-3t²+3t
This is the parametric equation of y given.
Hence, the Cartesian equation is correct.

smile cool


Character is who you are when no one is looking.

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#4 2006-03-08 03:51:24

ashwil
Member
Registered: 2006-02-27
Posts: 121

Re: parametric equations as a cartesian equation help

If x = t+1, then rearranging gives t = x-1

We can then substitute (x-1) for t in the equation y = -3t² + 3t so that:

y = -3(x-1)² + 3(x-1)

Multiply out to give:

y = -3(x²-2x+1) + 3x -3

and further to give:

y = -3x² +6x -3 + 3x -3 = -3x² + 9x - 6

Factorising -3x² + 9x - 6 gives:

y = (-3x + 3)(x-2) OR -3(x-1)(x-2)

Solving for y=0 gives x= 1 or 2

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#5 2006-03-08 03:52:47

ashwil
Member
Registered: 2006-02-27
Posts: 121

Re: parametric equations as a cartesian equation help

Shucks! Beaten to it by 5 seconds!!

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#6 2006-03-08 04:40:59

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: parametric equations as a cartesian equation help

Hehe, don't worry, it happens to all of us. I was beaten by 3 seconds not too long ago.

But eventually, you're going to beat someone by a tiny amount as well, and the universe will be restored. smile


Why did the vector cross the road?
It wanted to be normal.

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#7 2006-03-08 04:48:06

ashwil
Member
Registered: 2006-02-27
Posts: 121

Re: parametric equations as a cartesian equation help

mathsy, calculate the following:

What is the probability that ashwil will not only be able to answer an algebra/calculus question correctly, but will also be able to type & post his answer before ganesh, Ricky, krassi_holmz, yourself, mathisfun etc etc.?

I want my mummy!

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#8 2006-03-08 05:29:32

lakitu
Member
Registered: 2006-03-08
Posts: 5

Re: parametric equations as a cartesian equation help

thank you to both of you ! big help smile i understand it all smile

regards

lakitu

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#9 2006-03-08 05:50:31

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: parametric equations as a cartesian equation help

for ashwil:
the probability is 5 seconds less than ganesh's. smile


IPBLE:  Increasing Performance By Lowering Expectations.

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