MATLAB Plotting a Curve in Polar Coordinates Using MATLAB

AI Thread Summary
The discussion focuses on plotting a curve in polar coordinates using MATLAB, specifically an Archimedean spiral defined by r = theta. The user initially struggles with MATLAB syntax, particularly with defining the range for theta and using the correct operators. After receiving guidance, they learn to use the parameterization x(t) = r*cos(theta) and y(t) = r*sin(theta) for plotting. Additionally, they seek assistance with line integrals, which leads to a discussion on integrating the arc length formula. The user expresses gratitude for the help and indicates they have resolved their issues.
morry
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Ok, so I've been given a curve in polar coordinates.

I came up with a parameterisation:
x(t)=rcos(theta)
y(t)=rsin(theta)

But now I have to plot the graph using MATLAB and I have no idea. Theta lies between 0 and 2pi.

This is what I put in and got back in matlab:
>> t=[0:pi/50:2pi]
? t=[0:pi/50:2pi]
|
Error: Missing MATLAB operator.

Cheers.
 
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Should that be 2*pi?
 
Thanks Tide. I must still be doing something wrong.

The graph is archimedes spiral. r= theta

Heres what I've tried to do:

>>syms x y r theta
>>x=r*cos(theta)
>>y=r*sin(theta)
>>ezpolar(x,y)

I only need to plot from theta= 0-->2pi

Im really lost.

Any info on how to do line integrals would also be appreciated.

Thanks again.
 
You need to find the element of arc length along the curve. This should help:

ds = \sqrt {dr^2 + r^2 d\theta^2}

which you can easily integrate.
 
Thanks Tide. I am not having trouble doing the maths by hand. I can find line integrals no worries by hand its just that I have no idea how to use matlab.
 
morry,

Since r = \theta you just need to evaluate the integral

\int_{0}^{2\pi} \sqrt{1+\theta^2} d\theta

so look for a MatLab command resembling "int(sqrt(1+x^2), x = 0..2pi)"
 
Thanks for the help again Tide, I think I've got it all out! Finally!
 

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