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# Polar coordinate question watch

1. If I'm sketching the line rcos t = 2,

is this just a straight line going in the +ve x direction by 2?

i.e

O _______________> l
2
2. no man, it's vertical.

rcost = x, rsint=y
3. okay - silly error on my part, but then is it just a straight line upwards by 2?
4. what do you mean 'by 2'.

in cartesian coordinates, it's the line x=2.
5. Okay, the question says:

Sketch the line (or half0line) with polar equation:
rcos t = 2

So am I meant to be sketching them on a cartesian or polar graph? I presumed polar.

In the last question which said sketch t = -PI/3, I simply drew an acute angle of 60 degrees going down the page - was this not correct?
6. (Original post by studienka)
Okay, the question says:

Sketch the line (or half0line) with polar equation:
rcos t = 2

So am I meant to be sketching them on a cartesian or polar graph? I presumed polar.

In the last question which said sketch t = -PI/3, I simply drew an acute angle of 60 degrees going down the page - was this not correct?
Of course you're meant to be sketching them on a polar graph, but "a straight line upwards by 2" doesn't mean anything. It's the straight line equivalent to the line x=2 in cartesian coordinates.
7. hmmm - I thought I had this, now I'm getting a bit lost.

If I'm drawing the polar version of the cartesian eq x = 2, isn't that just this:
Attached Images
8. img128.pdf (3.5 KB, 95 views)
9. if it were, what would rcost = 3 look like?
10. ok - then i've not got this yet, can i have a bit of help with this
11. a bit longer?????
12. why? why oh why would the value of r be limited to a maximum of 3? and why would the value of t always be pi/2?
13. Fair enough point, but then I'm lost now
14. what is r when t=0?

what about for other values of t?
15. Okay - I did this, and while it's a rough sketch, is it these pair of parabola's?
Attached Images
16. img129.pdf (4.7 KB, 64 views)
17. (Original post by studienka)
Okay - I did this, and while it's a rough sketch, is it these pair of parabola's?

I'm quite sure we've told you what it is. It's a line. Specifically, it's the cartesian line x = 2. If r cos theta = 2 - imagine it - then drawing a right angled triangle between the pole, the initial line and any point on this line should make it obvious. r cos theta = 2 means the 'adjacent' side will always be 2. That is, the horizontal distance from the pole will always be 2.
18. See attached diagram.

The horizontal distance from the pole is fixed at 2. So it must be a vertical line passing through the point (2, 0) (in polar coordinates and in cartesian coordinates). So r cos theta = 2 is equivalent to x = 2.
Attached Images

19. I follow the triangle explanation pretty well - so it's just the line x = 2 thoughout the diagram?

I'm now concerned about my answer to the next one, which said:

rcos(t - pi/6) = 2

See my diagram and I wait for you to say I've gone wrong on this one as well.
Attached Images
20. img126.pdf (6.2 KB, 80 views)
21. thnx for the diagram, but should I just draw this?

If I'm saying that the distance from the polar point is always 2 (and t can take any value between pi and -pi, should I not keep this line continuing (as you would do with x=2 on a cartesian graph)
22. (Original post by studienka)
thnx for the diagram, but should I just draw this?

If I'm saying that the distance from the polar point is always 2 (and t can take any value between pi and -pi, should I not keep this line continuing (as you would do with x=2 on a cartesian graph)
No, my diagram was just an explanation. It'll be an infinitely long vertical line through (2, 0).
23. (Original post by studienka)
I follow the triangle explanation pretty well - so it's just the line x = 2 thoughout the diagram?

I'm now concerned about my answer to the next one, which said:

rcos(t - pi/6) = 2

See my diagram and I wait for you to say I've gone wrong on this one as well.
Can't work out what you were trying to draw. Try making the substitution s = t - pi/6. Then r cos s = 2 is the same line we had a second ago, relative to the initial line s = 0. But this is the line t - pi/6 = 0, or t = pi/6. So where's the initial line t = 0?

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