In the previous section, we learnt how to trace the curves, when equations are in Cartesian form. In this section, we will learn to trace polar curves.
General Observations (For Equations in Polar Coordinates)
I) The equation in polar form is generally , where and .
II) The polar coordinate system has a pole and an initial line. The directed distance is measured from pole and the angle is measured w.r.t. the initial line.
So, initial line is analogous to axis and the pole is analogous to the origin.
IV) If for , , then the pole lies on the curve.
V) Slope of tangent at a point is equal to
V) Polar Curves Nomenclature:
Cardioid : ,
Parabola : ,
Rose Curves : ,
Lemniscate of Bernoulli :
Spirals : ,
Q.1) Trace the curve (June 2016)
1) is an even function, so function will have same value for and .
2) is bounded between and , hence will be bounded between and .
3) Intersection with the initial line will be at , because at , .
4) When becomes , the curve will pass through the pole, because . Hence, at , .
5) The next value of when becomes will be , because at , and .
6) As value of varies from to , value of will reduce from to . Since the curve is symmetric about the initial line, we will get a loop (or a petal) of the curve.
7) Considering the periodicity of function, same values of will be obtained in the intervals , and .
8) Since at , and again at , becomes maximum, we’ve a good idea about the curve in the interval .
9) Let’s repeat this in the intervals and .
In the figure below, I’ve plotted 2 such curves, one with , the violet curve and second with , the orange curve. The red line is , the green line is and the blue line is .
Q.2) Trace the curve . (December 2015)
This is similar to Q.1.
Q.3) Trace the curve (June 2015)
Remarks : (Assuming )
1) is an even function, so the curve will be symmetric about the initial line .
2) The curve will not have multiple loops, because the argument of is and not as it was in Q.1.
3) Intersection with initial line will be at the point and intersection with will be at the point , i.e. the pole. Intersection with the line will be the point .
4) The equation of tangent at the pole is obtained by equating to , which gives , or . Since the curve is symmetrical about and value of reduces from to , there will be a cusp at the pole.
5) Slope of tangent at the point is obtained by . From the equation, . So, = for . Hence the tangent will be parallel to at .
With this information, the curve will look like this :
Value of that I took is . Tangent at is the orange line. The curve is in green colour.