Parametric equation to cartesian Exam question

OP

Original post by mqb2766

The square root and the divide by 1-x, should be red flags?
Also the domain for t limits the values of the curve as does the definition.
What interval for x do these suggest?

I just wasn't sure as the t limits gave me an x value of 3 and then the denominator gives an x value of 1 but then I don't know how to represent the domain

Reply 3

You should get a range of values from the t domain?

Reply 4

ghostwalker

17

For info: Here is the markscheme from Edexcel Practice Paper C, 9MA0.

The final line is gibberish, IMO.

Reply 5

Interesting model solution.

Reply 6

OP

Original post by mqb2766

You should get a range of values from the t domain?

Yeah although both ends of the range of t give the same x value

Reply 7

Original post by BartGR3

Yeah although both ends of the range of t give the same x value

But what happens in the middle? It may not always be essential, but should be known in general.

(edited 3 years ago)

Reply 8

OP

Original post by mqb2766

But what happens in the middle? It may not always be essential, but should be known in general.

What do you mean by what happens in the middle?

Reply 9

x(pi/4) = x(-pi/4) = 3
But x(t) is not constant. What is its actual range?

Reply 10

OP

Original post by mqb2766

x(pi/4) = x(-pi/4) = 3
But x(t) is not constant. What is its actual range?

Oh does the middle of the sec^2x graph go to a minimum below those points? How would you find this minimum?

Reply 11

OP

Okay now I got x=2 in the middle of that range, so would that mean that the domain would range from 2 to 3? I think the mark scheme above confused me, I wasn’t sure why they put x<1 as the domain and then x...2

Reply 12

ghostwalker

17

Original post by BartGR3

Okay now I got x=2 in the middle of that range, so would that mean that the domain would range from 2 to 3? I think the mark scheme above confused me, I wasn’t sure why they put x<1 as the domain and then x...2

In red - that's partly why I said the last line is gibberish (in the mark scheme).

And, yes, the range of x as a function of t is 2 to 3, and that becomes the domain of the cartesian equation.

The markscheme's answer is also incorrect in that it takes the positive square root to get y, whereas y can be negative if t is < 0.

It would have been better if they'd left it as y^2=..., which would have been correct.

If you've any graphing software that can handle parametric definitions, I'd recommend plotting the curve and you'll be able to see what's going on; if you can't visualise it already. I can run one off, if not.

Reply 13