The Student Room Group

Polar Graph - Where did I go wrong?

Untitled.png

Attachment not found
Reply 1
Original post by creativebuzz
Untitled.png

Attachment not found


Your working do not match the question in any shape or form ...
Have you taken anything you should not have taken?
Original post by TeeEm
Your working do not match the question in any shape or form ...
Have you taken anything you should not have taken?


Oh sorry, I posted the wrong question

Untitled.png
Original post by creativebuzz
Oh sorry, I posted the wrong question

Untitled.png


Your second term of dy/d theta shouldn't contain sin theta
Original post by tiny hobbit
Your second term of dy/d theta shouldn't contain sin theta


WIN_20151011_210133.JPG

Ah yes, thank you! But now I have two answers and the question only asks for one.
Original post by creativebuzz
WIN_20151011_210133.JPG

Ah yes, thank you! But now I have two answers and the question only asks for one.


The range of values of theta only goes up to pi/4 (presumably to avoid problems with values of theta where the graph doesn't exist).
(edited 8 years ago)
Original post by tiny hobbit
The range of values of theta only goes up to pi/4 (presumably to avoid problems with values of theta where the graph doesn't exist).


EDIT: Oh I see! Thank you :biggrin:
Original post by tiny hobbit
The range of values of theta only goes up to pi/4 (presumably to avoid problems with values of theta where the graph doesn't exist).


Last quick question. Why is the limit 0, I thought it would be pi as if you make r equal to zero in r = a(1+costheta) then you get theta = pi

Untitled.png
Original post by creativebuzz
Last quick question. Why is the limit 0, I thought it would be pi as if you make r equal to zero in r = a(1+costheta) then you get theta = pi

Untitled.png


The red curve is the boundary from the intial line upto the radius OP (at neither end of this chunk of area does r = 0). At the initial line, theta = 0. At P, theta = pi/3.

Then the blue curve is the boundary from OP to where the curve meets O, which is when r = 0. When r = 0 for this curve, theta = pi/2 so that you gives you the limits on the second integral.

If you let me know your email address by PM, I'll send you my notes on polar coordinates.
(edited 8 years ago)
Original post by tiny hobbit
The red curve is the boundary from the intial line upto the radius OP (at neither end of this chunk of area does r = 0). At the initial line, theta = 0. At P, theta = pi/3.

Then the blue curve is the boundary from OP to where the curve meets O, which is when r = 0. When r = 0 for this curve, theta = pi/2 so that you gives you the limits on the second integral.

If you let me know your email address by PM, I'll send you my notes on polar coordinates.


I've attached a picture showing the two questions I have, other than that I've inboxed you my email :smile:

Untitled.png
Original post by creativebuzz
I've attached a picture showing the two questions I have, other than that I've inboxed you my email :smile:

Untitled.png


The part of the red curve that goes to 0 is not a boundary of the shaded area. The part of the blue curve that goes to 0 is a boundary of the shaded area.
Original post by tiny hobbit
The part of the red curve that goes to 0 is not a boundary of the shaded area. The part of the blue curve that goes to 0 is a boundary of the shaded area.


What makes you think that 0 was the boundary for the blue curve and not for the red curve? They both touch at that point after all
Original post by creativebuzz
What makes you think that 0 was the boundary for the blue curve and not for the red curve? They both touch at that point after all


The parts of the red curve that end at O are the ones that go left of the x-axis. These parts do not go along the edge of the shaded area.

The only part of the red curve that goes along the edge of the shaded area is the bit from the theta = 0 line round to P.
Original post by tiny hobbit
The parts of the red curve that end at O are the ones that go left of the x-axis. These parts do not go along the edge of the shaded area.

The only part of the red curve that goes along the edge of the shaded area is the bit from the theta = 0 line round to P.


Why is r=0 counted for the blue area even though we just clarified that r=0 doesn't go along the edge of the shaded area?
Original post by creativebuzz
Why is r=0 counted for the blue area even though we just clarified that r=0 doesn't go along the edge of the shaded area?


Have we? r = 0 isn't a part of a curve, it is a single point at the end of a section of curve.

I'm not at all sure that you've really got the idea of what r is. It is the distance of a point on the curve from O. So if r = 0, the curve passes through O.

For the blue curve, the part of the curve which goes along the edge of the shaded area is for theta = pi/3 (at P) to theta = pi/2 (when the curve goes through O).
Original post by tiny hobbit
Have we? r = 0 isn't a part of a curve, it is a single point at the end of a section of curve.

I'm not at all sure that you've really got the idea of what r is. It is the distance of a point on the curve from O. So if r = 0, the curve passes through O.

For the blue curve, the part of the curve which goes along the edge of the shaded area is for theta = pi/3 (at P) to theta = pi/2 (when the curve goes through O).


But the both intersect at the origin, which is equivalent to r=0

Quick Reply

Latest