# Sketch the graph - Polar curves

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#1
Sketch the graph of r = sec(theta - pi/4)

Used a table of values from theta = 0 to 2 pi but didn't get the straight line graph required.

0
1 month ago
#2
(Original post by golgiapparatus31)
Sketch the graph of r = sec(theta - pi/4)

Used a table of values from theta = 0 to 2 pi but didn't get the straight line graph required.

https://www.desmos.com/calculator/2y6nel3jpw
What did you get for the table?
Last edited by mqb2766; 1 month ago
0
#3
(Original post by mqb2766)
https://www.desmos.com/calculator/r1umtv4egb
What did you get for the table?
θ = 0, pi/4, pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4, 2pi
r = 1.41, 1, 1.41, infinity, -1.41, -1, -1.41, infinity, 1.41

I found it difficult to plot these points (r, theta). Couldn't handle the negatives and infinity very well.

But I found another way to get the answer:
I wrote rcos(theta - pi/4 ) = 1
then used the cos(A-B) formula and the relations x = r cos theta and y = r sin theta to get x+y=sqrt 2 which is the line
0
1 month ago
#4
(Original post by golgiapparatus31)
θ = 0, pi/4, pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4, 2pi
r = 1.41, 1, 1.41, infinity, -1.41, -1, -1.41, infinity, 1.41

I found it difficult to plot these points (r, theta). Couldn't handle the negatives and infinity very well.

But I found another way to get the answer:
I wrote rcos(theta - pi/4 ) = 1
then used the cos(A-B) formula and the relations x = r cos theta and y = r sin theta to get x+y=sqrt 2 which is the line
Yes the simple trig is
So for an adjacent side of 1, the hypotenuse is r hence it's a straight line. When theta=pi/4, r=1.
Last edited by mqb2766; 1 month ago
1
#5
(Original post by mqb2766)
Yes the simple trig is
So for an adjacent side of 1, the hypotenuse is r hence it's a straight line. When theta=pi/4, r=1.
Thank you.

I don't understand the part "for an adjacent side of 1, the hypotenuse is r hence it's a straight line". How do you know? I had to do the algebraic manipulation to get it in the form x +y = k to know it's a straight line

Also, do you have general advice on how to plot polar curves?

Thanks a lot for your help
0
1 month ago
#6
(Original post by golgiapparatus31)
Thank you.

I don't understand the part "for an adjacent side of 1, the hypotenuse is r hence it's a straight line". How do you know? I had to do the algebraic manipulation to get it in the form x +y = k to know it's a straight line

Also, do you have general advice on how to plot polar curves?

Thanks a lot for your help
I've updated the diagram a bit
https://www.desmos.com/calculator/m1phxncbv7

Note if it was just
r = sec(theta)
This would be a vertical line passing through the x-axis at x=1. This is because it corresponds to a right angled triangle where r is the hypotenuse, x is the adjacent base and theta the angle. The base 1 is fixed and as theta varies, the hypotenuse traces out a vertical line. For simple trig functions, it's worth thinking about the geometry as well as algebra.

Yours is similar, except the "base" is rotated by pi/4.
0
#7
(Original post by mqb2766)
I've updated the diagram a bit
https://www.desmos.com/calculator/m1phxncbv7

Note if it was just
r = sec(theta)
This would be a vertical line passing through the x-axis at x=1. This is because it corresponds to a right angled triangle where r is the hypotenuse, x is the adjacent base and theta the angle. The base 1 is fixed and as theta varies, the hypotenuse traces out a vertical line. For simple trig functions, it's worth thinking about the geometry as well as algebra.

Yours is similar, except the "base" is rotated by pi/4.
Thanks a lot, I have understood what you have said.

The diagram also made it clear to me that my points that I was plotting were just not accurate enough.

Thank you
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