Drawing with trigonometric function

Rotation Matrix for 45 Degrees:I can just model the equation with sines. So y_1 = 300 * Sin(pi*x / 1000) (Assuming 300 is the highest point in the curve)y_2 = (1000/pi) * sin^-1(x/300)
  • #1
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Homework Statement


Come up with a set of equations to draw the following picture:

http://img197.imageshack.us/img197/7428/equation.jpg [Broken]

Homework Equations





The Attempt at a Solution



The circle is easy.

1000 = sqrt(x^2+y^2)

I can just model the equation with sines. So

y_1 = 300 * Sin(pi * x / 1000) (Assuming 300 is the highest point in the curve)

The sine graph on the y-axis can be easily tilted by switching x and y. Therefore:

x = 300 * Sin(pi*y_2 / 1000)

sin^-1 (x/300) = pi * y_2 / 1000

y_2 = (1000/pi) * sin^-1(x/300)


Is my y_2 correct?

And I'm kinda stumbling on how to translate the axis in 45 degree angle to draw the sine graph like that. Any help would be appreciated.
 
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  • #2
l46kok said:
I can just model the equation with sines. So

y_1 = 300 * Sin(pi * x / 1000) (Assuming 300 is the highest point in the curve)
You should probably add a restriction for x, like -1000 ≤ x ≤ 1000.

l46kok said:
The sine graph on the y-axis can be easily tilted by switching x and y. Therefore:

x = 300 * Sin(pi*y_2 / 1000)

sin^-1 (x/300) = pi * y_2 / 1000

y_2 = (1000/pi) * sin^-1(x/300)


Is my y_2 correct?
Technically, no, because x has the domain restriction -π/2 ≤ x ≤ π/2. Must all equations be y as functions of x? If not, I would just leave it in terms of x. Also, I think you need a negative in front of the R.H.S.:
x = -300 * Sin(pi*y_2 / 1000)
 
  • #3
No it can be in terms of X too.

How should I modelize the lines which goes at 45 degree angle?
 
  • #4
l46kok said:
The circle is easy.

1000 = sqrt(x^2+y^2)
Oops! The radius for that equation is about 31.6227766.

l46kok said:
No it can be in terms of X too.

How should I modelize the lines which goes at 45 degree angle?
Are you familiar with rotation of axes? You'll need to make use of it, I think.
 
  • #5
eumyang said:
Oops! The radius for that equation is about 31.6227766.
No it's not,

l46kok said:
The circle is easy.

1000 = sqrt(x^2+y^2)

:tongue:
 
  • #6
eumyang said:
Oops! The radius for that equation is about 31.6227766.


Are you familiar with rotation of axes? You'll need to make use of it, I think.

Yeah, translating the axes by defining a new set of axes such as x' and y', then rotating it via trigonometric function but I was struggling big time about that. Any suggestions?
 
  • #7
Mentallic said:
No it's not,
:tongue:
Wow... that's was embarrassing. :redface: Let's try that again.
l46kok said:
The circle is easy.

1000 = sqrt(x^2+y^2)
This only gives you the upper semicircle. Define this implicitly instead:
x2+y2=10002

l46kok said:
Yeah, translating the axes by defining a new set of axes such as x' and y', then rotating it via trigonometric function but I was struggling big time about that. Any suggestions?
Do you know the equations that define x' and y' in terms of x, y, and θ? Take the equation
y' = 300 * sin(pi*x'/1000)
and replace with what x' and y' equals. Then find a new set of equations for x' and y', in terms of a different angle θ, and replace into the equation above.
 
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  • #8
Call your original function y = f(x). The easiest way to rotate is to use the rotation matrix:

[tex]R(x) = \left[ \begin{array}{cc}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{array}\right]
\left[
\begin{array}{c}
x\\
f(x)
\end{array}\right][/tex]

Pick any θ you wish and plot it parametrically for x from 0 to 1000.
 

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