# Core 4 Cartesian Equation Q

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Thread starter 2 years ago
#1
Please can someone help with part a of this question! Not really done these too much in class, trying to make T the subject of one equation but struggling. Thanks!

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2 years ago
#2
(Original post by jazz_xox_)
Please can someone help with part a of this question! Not really done these too much in class, trying to make T the subject of one equation but struggling. Thanks!
You don't always need to make the subject, sometimes there are less obvious but quicker and neater approaches.

For example, note that if you take the left equation and say then get that

which we can square and rearrange to get

Also note that you can obtain entirely in terms of by rearranging for it in
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Thread starter 2 years ago
#3
(Original post by RDKGames)
You don't always need to make the subject, sometimes there are less obvious but quicker and neater approaches.

For example, note that if you take the left equation and say then get that

which we can square and rearrange to get

Also note that you can obtain entirely in terms of by rearranging for it in
Okay thank you, I understand what you've said but still unsure of how to get the equation in the form they want? Where does the y^3 come from?

And also regarding your last point for rearranging the equation for t^2, how do you go about doing this?
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2 years ago
#4
(Original post by jazz_xox_)
Okay thank you, I understand what you've said but still unsure of how to get the equation in the form they want? Where does the y^3 come from?

And also regarding your last point for rearranging the equation for t^2, how do you go about doing this?
If you work through it then it will become apparent where it comes from.

To rearrange for , just start by getting rid off the denominator so that hence then factor out in the next step or so before getting it on its own.
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Thread starter 2 years ago
#5
(Original post by RDKGames)
If you work through it then it will become apparent where it comes from.

To rearrange for , just start by getting rid off the denominator so that hence then factor out in the next step or so before getting it on its own.
Got it, thanks! On the first step you did, , by dividing both sides by T would you always divide the numerator by the T? I didn't think of this to do, never really done it before.

And one more thing (sorry for a million questions), I'm doing a question on implicit differentiation- and set dy/dx=0 and ended up with x + y - 1 = 0, all I have is the original equation of the curve, and no point- how can I go about solving this for the stationary points?

Thanks again
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2 years ago
#6
(Original post by jazz_xox_)
Got it, thanks! On the first step you did, , by dividing both sides by T would you always divide the numerator by the T? I didn't think of this to do, never really done it before.
Well you don't "always" in these situations do this approach, but in cases like this where it's acceptable, then yes you divide both sides by whatever and that means the numerator gets divided through. This should follow from simple GCSE/AS-Level algebra.

And one more thing (sorry for a million questions), I'm doing a question on implicit differentiation- and set dy/dx=0 and ended up with x + y - 1 = 0, all I have is the original equation of the curve, and no point- how can I go about solving this for the stationary points?

Thanks again
If you end up with then that means all the stationary points of your implicit curve lie on this straight line. To find them, solve the system of the two equations since you're only looking for the intersections between the two.
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