Lecture 13, 14

Lecture 13, 14 - MATH 118 LECTURES 13& 14 POLAR...

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Unformatted text preview: MATH 118, LECTURES 13 & 14: POLAR EQUATIONS 1 Polar Equations We now know how to equate Cartesian coordinates with polar coordinates, so that we can represents points in either form and understand what we are talking about. Furthermore, we know how to go back and forth between the two representations without losing any information. From a Calculus point of view, however, we are not interested in distinct sets of points—we are interested in connected curves of points (i.e. func- tions). Our next logical question, therefore, is to ask how we can represent functions y = f ( x ) and relations f ( x, y ) = 0 in terms of polar coordinates. We will also be interested in returning from polar coordinates to Cartesian coordinates, just as we were for single points. Consider the equation x 2 + y 2- 4 y = 0 . We want to represent this equation in polar coordinates ( r, θ ), but how can we go about doing this? Well, we know from switching from Cartesian to polar coordinates for a single point that we have the relationship x = r cos( θ ) , and y = r sin( θ ) . If we plug these relations into the equation, we have x 2 + y 2- 4 y = r 2 cos 2 ( θ ) + r 2 sin 2 ( θ )- 4 r sin( θ ) = 0 = ⇒ r 2 = 4 r sin( θ ) = ⇒ r = 4sin( θ ) . So we have transformed the problem from a relationship between y and x into a relationship between θ and r . We notice also that we can properly say r is a function of θ —that is to say, for each value of θ we can assign a unique point value of r . This was not a property of the original equations since x did not uniquely determine values of y (i.e. the graph fails the vertical line test). In that sense, the polar equations are a more natural approach to this equation. (We also notice that r can be negative for some values of 1 θ , meaning that as we plot along the axis at angle θ , we plot backwards a magnitude of r .) Consider, however, representating the following equation in polar coor- dinates y = x 2 + 1 . Using our relations above we get r sin( θ ) = r 2 cos 2 ( θ ) + 1 = ⇒ r 2 cos 2 ( θ )- r sin( θ ) + 1 = 0 . Unlike the previous example, we cannot solve for r uniquely as a function of θ . We are left with the relation f ( r, θ ) = 0 in the form above. This means that some values of θ may correspond to multiple values of r . Other values of θ may not correspond to any value of r at all. These examples raise an important point about how we should choose to represent functions. For the first example, we could not represent y as a function of x , but we could represent r as a function of θ ; conversely, for the second example, we could represent y as a function of x , but not represent r as a function of θ (at least not choosing (0 , 0) as our pole) (see Figure 1)....
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Lecture 13, 14 - MATH 118 LECTURES 13& 14 POLAR...

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