Drawing with trigonometric function

[asd1249jf]

Homework Statement

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

http://img197.imageshack.us/img197/7428/equation.jpg

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.

 

[eumyang]

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.

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)

 

[asd1249jf]

No it can be in terms of X too.

How should I modelize the lines which goes at 45 degree angle?

 

[eumyang]

The circle is easy.

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

Oops! The radius for that equation is about 31.6227766.

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.

 

[Mentallic]

Oops! The radius for that equation is about 31.6227766.

No it's not,

The circle is easy.

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

 

[asd1249jf]

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?

 

[eumyang]

No it's not,

Wow... that's was embarrassing. Let's try that again.

The circle is easy.

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

This only gives you the upper semicircle. Define this implicitly instead:
x²+y²=1000²

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.

 

[LCKurtz]

Call your original function y = f(x). The easiest way to rotate is to use the rotation matrix:

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

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