Integration with parametic equation

Original post by otrivine

ok

what would you integrate then can you show me :redface:

I gave you the requisite formula in my first reply post - just substitute the expressions for y and x into there.

Reply 21

OP

Original post by Smaug123

I gave you the requisite formula in my first reply post - just substitute the expressions for y and x into there.

oh so we integrate both x and y ! so is d(x) there to mean you integrate x also right

Reply 22

davros

16

Original post by otrivine

ok

what would you integrate then can you show me :redface:

Smaug123 gave you the formula in post 4 - just plug the theta limits into this.

Reply 23

TenOfThem

18

Original post by otrivine

oh so we integrate both x and y ! so is d(x) there to mean you integrate x also right

Stop guessing and look at the formula that you will have in your book (with examples) and from Smaug earlier on

Reply 24

Original post by otrivine

oh so we integrate both x and y ! so is d(x) there to mean you integrate x also right

Er, sort of. What you have currently is y and x both in terms of some parameter

\theta

. In order to integrate y with respect to x (which is what you want to find) rather than with respect to theta, you need to use what amounts to the chain rule in order to change variables from theta to x.

Reply 25

OP

Original post by Smaug123

Er, sort of. What you have currently is y and x both in terms of some parameter

\theta

. In order to integrate y with respect to x (which is what you want to find) rather than with respect to theta, you need to use what amounts to the chain rule in order to change variables from theta to x.

Oh, right cause in the forumula you put on post 4, you put the sub method right, cause I thought that my forumla could work

Reply 26

TenOfThem

18

Original post by otrivine

Oh, right cause in the forumula you put on post 4, you put the sub method right, cause I thought that my forumla could work

It is the same "formula"

You need to understand what "integration with respect to a variable" means so that you do not think that you can just pretend that x is theta

Reply 27

Original post by otrivine

Oh, right cause in the forumula you put on post 4, you put the sub method right, cause I thought that my forumla could work

Your formula would only work if y were in terms of x, because you're integrating with respect to x. There are two things you can do to fix it: you can actually put y in terms of x (by eliminating theta from the equations) or you can use my post-4 formula, which essentially turns dx into d(theta) and means you integrate with respect to theta.

Reply 28

OP

Original post by Smaug123

Your formula would only work if y were in terms of x, because you're integrating with respect to x. There are two things you can do to fix it: you can actually put y in terms of x (by eliminating theta from the equations) or you can use my post-4 formula, which essentially turns dx into d(theta) and means you integrate with respect to theta.

Bingo! will it be possible to get a cartesian equation to combine x and y and then integrate with the area formula, or you know what Ill stick with the sub method , sounds easier

Reply 29

davros

16

Original post by otrivine

Oh, right cause in the forumula you put on post 4, you put the sub method right, cause I thought that my forumla could work

Your formula does work - but you have to keep everything in terms of one independent variable. You either work out the Cartesian equation of the curve and write y in terms of x and then integrate y(x) with respect to x, using the x-limits; OR you write y in terms of theta, convert dx to d(theta) using dx/d(theta) and change your limits to the appropriate theta-limits.

You should have some examples of this technique in your book :smile:

Reply 30

Original post by otrivine

Bingo! will it be possible to get a cartesian equation to combine x and y and then integrate with the area formula, or you know what Ill stick with the sub method , sounds easier

Both are possible and, indeed, not hard if you know what sin(arccos(x)) is. The method I called neater is neater, and more generally applicable.

Reply 31

TenOfThem

18

Original post by otrivine

Bingo! will it be possible to get a cartesian equation

If it had wanted you to take this approach it would probably have asked you to find the cartesian equation earlier in the question

Reply 32

OP

Original post by davros

Your formula does work - but you have to keep everything in terms of one independent variable. You either work out the Cartesian equation of the curve and write y in terms of x and then integrate y(x) with respect to x, using the x-limits; OR you write y in terms of theta, convert dx to d(theta) using dx/d(theta) and change your limits to the appropriate theta-limits.

You should have some examples of this technique in your book :smile:

Thank you so much Davros, yes sorry I got it now!

Reply 33

OP

Original post by TenOfThem

If it had wanted you to take this approach it would probably have asked you to find the cartesian equation earlier in the question

I am so awful at maths, I just want to get it done, in Jan 2013 I got 75/100 for C3!

Reply 34

OP

Original post by Smaug123

Both are possible and, indeed, not hard if you know what sin(arccos(x)) is. The method I called neater is neater, and more generally applicable.

Thank you and for the limits! Can you just say how you did cause I did get one of the limit which was 5 root2

Reply 35

posthumus

20

Okay a lot of you would probably do it differently but I did this to find the intercept of the oval thing...

y=4sin(angle)
y=0 a the x intercept
angle = sin^-1(0)
angle = 0

now I can find x... sub the angle into the other formula x=5cos(angle)
x=5cos(0)
x=5

Guys, is there actually an alternative or better way of doing this ?? :smile:

I know this isn't the most reliable method :tongue:

Now you can find area of triangle (well you could do that before without finding the x intercept value):

area of triangle = 0.5 x 2root(2) x [5root(2) - 5root(2)/2]

integrate the parametric equations ....

I would then integrate y dx/d(angle) d(angle)
I would change the limits by subbing in 5root(2)/2 & 5 into x=5cos(angle)
This will give me limits

Finally when I get the area, I will subtract from the area of the triangle found previously :smile:

Okay now I don't know if that's the best way of doing it ! Or if I've just made it sounds complicated :tongue:

Sorry I really need to learn how to use latex :smile:

Reply 36

Original post by otrivine

Thank you and for the limits! Can you just say how you did cause I did get one of the limit which was 5 root2

Not if you're integrating with respect to theta… if you're integrating with respect to x, then that sounds plausible.

Reply 37

davros

16

Original post by otrivine

Thank you so much Davros, yes sorry I got it now!

No problem! Once you've done a few of these you'll probably recognize that the curve you're integrating is actually an ellipse...and you've probably seen the cartesian equation for that!

Reply 38

Original post by posthumus

Guys, is there actually an alternative or better way of doing this ?? :smile:

I know this isn't the most reliable method :tongue:

Yep, there is; the way we've been prodding OP to do is outlined in broad brushstrokes in post 4 and then elaborated on a bit later (chiefly by davros and me).

Reply 39

OP