# parametric equations, trigonometry, oscilloscope, .....urgh?

Discussion in 'Math' started by ninjaman, Aug 30, 2016.

1. ### ninjaman Thread Starter Member

May 18, 2013
306
1
Hello,

I am reading "Use of dual-trace oscilloscopes" by Charlie Roth.
It has in it this question:
"If x = 2 + 3 cos wt and y = 1 + 3 sin wt what is:
maximum x deflection =
minimum x deflection =
max y deflect =
min y deflect =
now sketch a rectangle on the screen showing the limiting values of the x and y deflections. The trace must lie within this rectangle."

I managed to get the rectangle correct, but getting the circle is a little tricky. I am not using a calculator and there is only a grid of (10,8) squares like an oscilloscope screen. I looked on line by entering the main formula above and found parametric equations of an ellipse. this explained the first numbers (2 and 1) 3 is the amplitude so the circle will have a max of x = 3 and y = 3 and a min of x = -1 and y = -2. i think this is correct. Problem I am having is that this is not in the trig books that I have, or explained in the oscilloscope book. Is there a good site that will explain this?

2. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
What did you get for the questions asked (min/max x/y deflections)? Because the numbers you mention for max x and y aren't correct. How do you get max x = 3 and max y = 3?

Try picking about ten values for wt spread out between 0 and 2pi to plot what the scope is going to plot.

3. ### ninjaman Thread Starter Member

May 18, 2013
306
1
max x = 5
min x = -1
max y = 4
min y =-2
this created a rectangle on the grid. i then had to create a trace with y = 3 sin and x = 3 cos
this is the formula, "x = 2 + 3 cos wt and y = 1 + 3 sin wt"
i assumed that the "3" in the above was the same as the amplitude on a regular sine wave where you would have a max of 3 and min of -3. however, this is a circle so i think that i have gotten confused somewhere.

4. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
So is y = 3 sin or is y = 1 + 3 sin wt ?

You can't have it both ways.

orrza1 likes this.
5. ### ninjaman Thread Starter Member

May 18, 2013
306
1
y = 1 + 3 sin wt
x = 2 + 3 cos wt

6. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Okay, and you've identified that the min/max for x is -1/+5 and for y it is -2/+4.

Have you plotted a handful of points to see what you get? A spreadsheet makes this trivially easy to do (and also lets you play around with the equations).

When you say, "however, this is a circle," is that because you plotted points and got a circle? If not, what led you to you to the conclusion that it was a circle? And on what basis do you think that this is not correct?

7. ### ninjaman Thread Starter Member

May 18, 2013
306
1

I managed to get this, with a lot of values some research and a little mucking about. I think I can proceed now with the book but I will have to use excel a lot to create the graphs.
Simon

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8. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Good. Being able to visualize things using Excel (or whatever tool is available) is extremely useful. But you don't want to become a slave to what it says. So now let's consider the original equations just a bit and see if we can spot how they match up with the plot.

First, since sin() and cos() both average to zero, the center of the circle (or whatever it turned out to be) is determined by the other terms, namely x = 2 and y = 1. If you look at the plot, sure enough, that's where it is centered.

We know that the equation for a circle of radius R centered at the origin is

x² + y² = R²

What about for a circle centered at (x, y) = (Cx, Cy)?

We can shift the graph by using

(x - Cx)² + (y - Cy)² = R²

Play with this to understand why this is true.

Now let's look at our parametric equations

x = 2 + 3 cos wt
y = 1 + 3 sin wt

And see if we can get them into the form of a circle with an offset origin.

(x - 2) = 3 cos(wt)
(y - 1) = 3 sin(wt)

Square both sides of each equation

(x - 2)² = 3² cos²(wt)
(y - 1)² = 3² sin²(wt)

Now add the two equations together

(x - 2)² + (y - 1)² = 3² sin²(wt) + 3² cos²(wt)
(x - 2)² + (y - 1)² = 3² [sin²(wt) + cos²(wt)]

And we know that

sin²(wt) + cos²(wt) = 1

so we have

(x - 2)² + (y - 1)² = 3²

Matching up terms with

(x - Cx)² + (y - Cy)² = R²

We see that we have a circle of radius 3 centered at (2, 1).

bogosort likes this.
9. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
sin()/cos() has a maximum value of 1 and minimum value of -1.

that's all it takes.

10. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
if you need to convert that to x/y space, remember that sin^2 + cos^2 = 1.

sin(wt) = (y-1)/3 and cos(wt) = (x-2)/3.

So (y-1)^2 + (x-2)^2 = 9.

that's a circle with a radius of 3, and center at (2,1).

11. ### MrAl Distinguished Member

Jun 17, 2014
2,554
515
Hi,

You can recognize a circle in a number of ways. With the two equations:
"x = 2 + 3 cos wt and y = 1 + 3 sin wt"

you can visualize them as a circle because:
x=r*cos(A)
y=r*sin(A)

is a circle centered at the origin (0,0) (and A=angle as long as w is the same for both).

The other numbers just shift the center of the circle.
If you add 1 to x you shift it over one unit because now x=1+r*cos(A).
If you add 1 to y you shift it up one unit because now y=1+r*cos(A).

If you square both equations:
x^2=r^2*cos(A)^2
y^2=r^2*sin(A)^2

and add them together you get:
x^2+y^2=r^2*(cos(A)^2+sin(A)^2)

and the trig forms reduce to 1, so we end up with:
x^2+y^2=r^2

which is also the equation for a circle.

If you call the offset in x 'h' and the offset in y 'k' then we have:
x-h=r*cos(A)
y-k=r*sin(A)

and doing the same thing we end up with:
(y-k)^2+(x-h)^2=r^2

which is the equation of a circle with center at (h,k).

It helps to study at least the basic plane curves so that we know what we have when we see them at a later date. There are a lot of interesting curves out there that we seldom ever see.