The Student Room Group

Parametric Equations

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Reply 20
Original post by TeeEm
bottom is

from 2pi/3 to 5pi/3

or

-4pi/3 to -pi/3


Ah that makes sense. Is there a reason how you knew that the limits in the question would be associated with the upper half of the curve and not the bottom?
Reply 21
Original post by PhyM23
Ah that makes sense. Is there a reason how you knew that the limits in the question would be associated with the upper half of the curve and not the bottom?


absolutely
the values of theta which produce the top half lie between -pi/3 and 2pi/3
They are even marked in the picture
Reply 22
Original post by TeeEm
absolutely
the values of theta which produce the top half lie between -pi/3 and 2pi/3
They are even marked in the picture


I can see that they are marked in the picture, but what I'm asking is how you knew that, when you were working the limits out, they would be associated with the top half?
Reply 23
Original post by PhyM23
I can see that they are marked in the picture, but what I'm asking is how you knew that, when you were working the limits out, they would be associated with the top half?


because pi/2 which also is marked in the picture lies on the top half and happens to lie between -pi/3 and 2pi/3
Reply 24
Original post by TeeEm
because pi/2 which also is marked in the picture lies on the top half and happens to lie between -pi/3 and 2pi/3


Ah of course. This all seems perfectly clear now. Thank you for bearing with me; this topic has never clicked in my head until now.

Thank you very much for your help :smile:
Reply 25
Original post by PhyM23
Ah of course. This all seems perfectly clear now. Thank you for bearing with me; this topic has never clicked in my head until now.

Thank you very much for your help :smile:


no worries
Original post by TeeEm
I have not revised parametrics


What about your exam?????

I expect more of you Tee.
Reply 27
Original post by Louisb19
What about your exam?????

I expect more of you Tee.


I am sorry about that ... Hopefully nobody noticed
Reply 28
Why do you integrate the curve between A and B? Wouldn't you cut off part of the curve that is between B and E?
Reply 29
Original post by Ano123
Why do you integrate the curve between A and B? Wouldn't you cut off part of the curve that is between B and E?


you can do it this way too if you are not sure how parametrics work
look at post 6
Reply 30
Original post by TeeEm
you can do it this way too if you are not sure how parametrics work
look at post 6


How does parametric integration work exactly?
Reply 31
Original post by Ano123
How does parametric integration work exactly?


read the posts and look at the question and its solution and it should make more sense, assuming you have seen parametric before.
Reply 32
Original post by TeeEm
read the posts and look at the question and its solution and it should make more sense, assuming you have seen parametric before.


So what area would you be calculating if you integrated between A and E parametrically?
Reply 33
Original post by Ano123
So what area would you be calculating if you integrated between A and E parametrically?


the area between the curve and the x axis including the little section on the bottom right which you do not want
Reply 34
What I'm saying is why isn't the area calculated this image.jpg
when you integrate between A and E. If you integrated this using Cartesian coordinates (ignoring the bottom half of the ellipse) between A and E you would calculate the area shown above wouldn't you. But when switching to parametric integration it doesn't seem to be the same. I thought that the only difference between parametric and normal Cartesian integration was that you use the t value for the corresponding c values as the limits. And you integrate with respect to t (and of course multiplying brought by dx/dt). It seems here that there is something fundamentally flawed about how I'm thinking of it.
(edited 8 years ago)
Reply 35
Original post by Ano123
What I'm saying is why isn't the area calculated this image.jpg
when you integrate between A and E. If you integrated this using Cartesian coordinates (ignoring the bottom half of the ellipse) between A and E you would calculate the area shown above wouldn't you. But when switching to parametric integration it doesn't seem to be the same. I thought that the only difference between parametric and normal Cartesian integration was that you use the t value for the corresponding c values as the limits. And you integrate with respect to t (and of course multiplying brought by dx/dt). It seems here that there is something fundamentally flawed about how I'm thinking of it.


...k2.gif
Reply 36
Original post by TeeEm
...k2.gif


I know.
It would be fairly long winded to read at the best of times.
Reply 37
Original post by Ano123
I know.
It would be fairly long winded to read at the best of times.


I am off to bed too now
I am sure tomorrow many eager people here will help you follow this.
Goodnight.
Reply 38
What area would you be calculating if you we're integrating from θ=-π/3 to say θ=π.

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