for the polar graph of
\[r=\frac{2}{3\sin(\theta)-\cos(\theta)} \] find the restrictions on theta and describe what this will mean in terms of the shape of the graph
since dividing by zero is always a bad idea ....
3sin = cos 3tan = 1 tan = 1/3
\[\tan(\theta)=1/3 \] is not allowed yeah
how will this affect hape of graph ?
i cant be sure, i want to say its asymptotic; but ill have to graph it to be sure
the cartesian form of the equation graphs y=1/3x+2/3 straight line with m=1/3
yes, and that creates a useful alternative, but does not excuse the original defects
so y=1/3x+2/3 would not show the restriction?
it shouldnt. just like:\(\large \frac{2x^2}{x}\) would translate to the line 2x, but it wouldnt show us the restriction at x=0 since its an algebraic equivalent
when theta hits these angles http://www.wolframalpha.com/input/?i=3sinx-cosx%3D0 the value of the polar is undefined and therefore cannot be mapped
so just discontinuity at that hpoint
its been awhile, but i do believe that at those intervals we have a jump discontinuity in the cartesian form
and since the limit from left and right are equal, the unbroken straight line is equivalent, but not equal to it
i could very easily be wrong, but i think this does all its business from \(\theta=0\) to \(\theta=2\pi\) it is not a line, but rather a line segment
yeah i am probably wrong
the solution says the line has gradient = 1/3 and so at tan(theta)=1/3, r has infinite solutions? is that correct
the 100 ss would indicate that his wrongest solutions is far better than any correct solutions i may muster ;)
tangent is defined as y/x gradient is defined as y/x hmmm
http://www.wolframalpha.com/input/?i=y%3D2%2F%283sinx-cosx%29 if we plot this on a \(\theta r\) plane we can clearly see the asymptotic behavior each point can be mapped onto the \(xy\) plane, except for the places where the asyptotes exist
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