# Help with Further Mathematics Polar Coordinates

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#1
Hi,
I'm struggling with graphing polar curves, I think my approach is what the problem is. I just cant get my thought process in line, I've been struggling to graph this one in particular below:
r = sec( theta - pi/4 )

Thanks!
0
1 year ago
#2
(Original post by Ethan7702)
Hi,
I'm struggling with graphing polar curves, I think my approach is what the problem is. I just cant get my thought process in line, I've been struggling to graph this one in particular below:
r = sec( theta - pi/4 )

Thanks!
The approach that is recommended in most text books is to make a table of values for theta across the range of values you normally use (most often -pi to pi, but some prefer 0 to 2pi - either is fine), using sufficiently many divisions so that you can see what is going on (every pi/12, say), then use a calculator or your general knowledge to find the corresponding values for r. You can then plot these points as accurately as you like, and the shape of the curve should be clear. There isn't really a need to plot the points super-accurately using a protractor, its only the general idea that you need.

I prefer to do things a little differently. In this case, I would sketch a graph of y = sec(theta - pi/4) first, which shouldn't be too tricky, as we're expected to know what a graph of sec looks like, and the subtraction of pi/4 is a simple translation of this graph. Then I would imagine what it would look like if instead of using the y values as the height of the curve for a particular angle, these values were used as the length of the radial vector pointing in the direction of that angle. I find this a bit less time consuming than working out loads of values with a calculator, and it keeps me thinking about my general trigonometry and transformations knowledge.

Most courses stick to the convention that r must be non-negative, so whichever method you use, if it leads to negative values of r at any point, you need to treat these as zero values of r (strictly speaking, null values as the curve doesn't exist at these points). But be careful - if you are checking your sketches against a graphic calculator or online graph plotter, many of these do use negative values for r, and so they may include portions of the curve that you are supposed to discard.
0
1 year ago
#3
(Original post by Ethan7702)
Hi,
I'm struggling with graphing polar curves, I think my approach is what the problem is. I just cant get my thought process in line, I've been struggling to graph this one in particular below:
r = sec( theta - pi/4 )

Thanks!
Right, so from what I remember (I dropped Further Maths last year), you plot a table of coordinates for values of ϴ being 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π. You use the equation to work out the values of r corresponding to each ϴ value. Now, I don't think the graphing needs to be exactly to scale but once you see the general trend in the values of r (increasing, decreasing, constant, fluctuating), you can approximately plot the r values against the ϴ values in a 4-quadrant grid. Hope this helped in some way 1
1 year ago
#4
(Original post by Ethan7702)
Hi,
I'm struggling with graphing polar curves, I think my approach is what the problem is. I just cant get my thought process in line, I've been struggling to graph this one in particular below:
r = sec( theta - pi/4 )

Thanks!
Further to my first reply, which is perfectly OK advice in general, I've looked at your particular example and it's an interesting one. I would start off just by thinking about r = sec(theta). If you multiply through by cos(theta), you get r.cos(theta) = 1. Now - what do you recognise about the LHS? Does this make it very easy indeed to draw the curve? If so, what is the effect of replacing the theta by theta - pi/4?
1
1 year ago
#5
so r = 1/cos ( Θ - 45º )

rcos ( Θ - 45º ) = 1

rcosΘsin45º + rsinΘcos45º = 1

rcosΘ and rsinΘ have special meanings in polar regions....
1
1 year ago
#6
(Original post by Ethan7702)
Hi,
I'm struggling with graphing polar curves, I think my approach is what the problem is. I just cant get my thought process in line, I've been struggling to graph this one in particular below:
r = sec( theta - pi/4 )

Thanks!
The "table of values" is a good approach in general to polar plotting and works especially well for plots that end up as a series of "lobes" or "petals". However, there is another way of tackling cases involving sec(theta) or cosec(theta).
First, note that r = 1 / cos(theta - pi/4) then use the cos(a - b) formula and a little rearranging to give you:
r(cos(theta) + sin(theta)) = sqrt(2)
From there you can substitute in cos(theta) = x/r and sin(theta) = y/r to give you the cartesian line that this "polar" plot actually consists of.
1
1 year ago
#7
also you can think of the effect of having Θ - 45º instead of just Θ .... it is a rotation of 45º of the graph of secΘ in the anticlockwise direction
1
#8
(Original post by Pangol)
Further to my first reply, which is perfectly OK advice in general, I've looked at your particular example and it's an interesting one. I would start off just by thinking about r = sec(theta). If you multiply through by cos(theta), you get r.cos(theta) = 1. Now - what do you recognise about the LHS? Does this make it very easy indeed to draw the curve? If so, what is the effect of replacing the theta by theta - pi/4?
Thanks

I've tried out graphing it by sketching sec(theta) by the side, and then translating it by pi/4. This makes it much easier for me to graph it.
About the other approach in the quoted post, I recognize rcos(theta) as x, of course. So, in Cartesian form, its just x=1 I guess, but how do I implement replacing theta with theta - pi/4 now? I read that a translation in Cartesian plane is a rotation in the polar plane, but how do I actually do it?
Last edited by Ethan7702; 1 year ago
0
1 year ago
#9
https://www.wolframalpha.com/input/?...x+-+pi%2F4+%29

Ooh that's a very cool shaped graph 0
1 year ago
#10
(Original post by Ethan7702)
Thanks

I've tried out graphing it by sketching sec(theta) by the side, and then translating it by pi/4. This makes it much easier for me to graph it.
About the other approach quoted post, I recognize rcos(theta) as x, of course. So, in Cartesian form, its just x=1 I guess, but how do I implement replacing theta with theta - pi/4 now? I read that a translation in Cartesian plane is a rotation in the polar plane, but how do I actually do it?
I suggested in my earlier post that you can rearrange the polar equation to be r(cos(theta) + sin(theta)) = sqrt(2). From there you can substitute in cos(theta) = x/r and sin(theta) = y/r. From there it's a small step to y = mx + c.
0
#11
(Original post by the bear)
also you can think of the effect of having Θ - 45º instead of just Θ .... it is a rotation of 45º of the graph of secΘ in the anticlockwise direction
Thanks
I got stuck rotating it! How do I know I have rotated it 45 degrees, or pi/4?
Last edited by Ethan7702; 1 year ago
0
#12
(Original post by old_engineer)
I suggested in my earlier post that you can rearrange the polar equation to be r(cos(theta) + sin(theta)) = sqrt(2). From there you can substitute in cos(theta) = x/r and sin(theta) = y/r. From there it's a small step to y = mx + c.
Im actually not familiar with the cos(a-b) identity as I haven't taken P3 for my As levels, but I'll look into it.

Edit: I think this is the only way I could've reached the conclusion that its linear. So for other polar graphs, should I tend to convert them to Cartesian form so that way its easier? Or do you recommend this approach for tackling sec and cosec in particular?
Last edited by Ethan7702; 1 year ago
0
1 year ago
#13
(Original post by Deggs_14)
https://www.wolframalpha.com/input/?...x+-+pi%2F4+%29

Ooh that's a very cool shaped graph That's not a polar plot unfortunately... By using the variables x and y you are implicitly telling Wolrfram Alpha that you want a Cartesian plot.

If we use r,theta instead we get the correct Polar plot... which is not as exciting in this case: https://www.wolframalpha.com/input/?...a+-+pi%2F4+%29
0
1 year ago
#14
(Original post by Ethan7702)
Im actually not familiar with the cos(a-b) identity as I haven't taken P3 for my As levels, but I'll look into it.

Edit: I think this is the only way I could've reached the conclusion that its linear. So for other polar graphs, should I tend to convert them to Cartesian form so that way its easier? Or do you recommend this approach for tackling sec and cosec in particular?
I would recommend looking at conversion to cartesian form where sec and cosec are involved. Otherwise a table of values and a sketch are usually best.

By the way cos(a - b) = cosa.cosb + sina.sinb, which comes out quite simple when b = pi/4.
0
1 year ago
#15
(Original post by Ethan7702)
Im actually not familiar with the cos(a-b) identity as I haven't taken P3 for my As levels, but I'll look into it.

Edit: I think this is the only way I could've reached the conclusion that its linear. So for other polar graphs, should I tend to convert them to Cartesian form so that way its easier? Or do you recommend this approach for tackling sec and cosec in particular?
I wouldn't recommend converting to cartesian coordinates in all cases. Sometimes this can be quite hard or time-consuming. This is a bit of a special case.

In terms of replacing theta by theta - pi/4, think about what this means on a point-to-point basis.

- to work out the r to use with an angle of 0, you would see what the r was for 0 - pi/4 = -pi/4. You can think of this as taking the -pi/4 r and rotating it by pi/4 to use it for an angle of 0.
- to work out the r to use with an angle of pi/2, you would see what the r was for pi/2 - pi/4 = pi/4. You can think of this as taking the pi/4 r and rotating it by pi/4 to use it for an angle of pi/2.
- to work out the r to use with any angle, you would see what the r was for that angle - pi/4. You can think of this as taking the angle - pi/4 r and rotating it by pi/4 to use it for the new angle.

In other words, replacing theta by theta - a is equivalent to rotating the original curve through an angle of a about the pole. This applies to all curves, not just this straight line.
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#16
(Original post by Pangol)
I wouldn't recommend converting to cartesian coordinates in all cases. Sometimes this can be quite hard or time-consuming. This is a bit of a special case.

In terms of replacing theta by theta - pi/4, think about what this means on a point-to-point basis.

- to work out the r to use with an angle of 0, you would see what the r was for 0 - pi/4 = -pi/4. You can think of this as taking the -pi/4 r and rotating it by pi/4 to use it for an angle of 0.
- to work out the r to use with an angle of pi/2, you would see what the r was for pi/2 - pi/4 = pi/4. You can think of this as taking the pi/4 r and rotating it by pi/4 to use it for an angle of pi/2.
- to work out the r to use with any angle, you would see what the r was for that angle - pi/4. You can think of this as taking the angle - pi/4 r and rotating it by pi/4 to use it for the new angle.

In other words, replacing theta by theta - a is equivalent to rotating the original curve through an angle of a about the pole. This applies to all curves, not just this straight line.
I see.
Returning to the part where you simplified it to rcos (theta) = 1, so is that x = 1 ? also, how should I rotate now?
0
1 year ago
#17
(Original post by Ethan7702)
Thanks
I got stuck rotating it! How do I know I have rotated it 45 degrees, or pi/4?
they are the same angle expressed in different ways.
0
1 year ago
#18
(Original post by Ethan7702)
I see.
Returning to the part where you simplified it to rcos (theta) = 1, so is that x = 1 ? also, how should I rotate now?
In just the way described in the previous post - by pi/4 about the pole (origin), using the normal convention that rotations by positive amounts are snticlockwise.
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#19
(Original post by the bear)
they are the same angle expressed in different ways.
My bad, I phrased it wrongly. I meant how do I generally rotate these polar graphs(by whatever angle), I find it difficult to do.
0
#20
(Original post by Pangol)
In just the way described in the previous post - by pi/4 about the pole (origin), using the normal convention that rotations by positive amounts are snticlockwise.
I see. Thanks so much for giving your time!
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