Curve Tracing II

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 r=f(\theta), where r^2 = x^2+y^2 and tan (\theta) = \frac y x.

II) The polar coordinate system has a pole and an initial line. The directed distance r is measured from pole and the angle \theta is measured w.r.t. the initial line.

So, initial line is analogous to X axis and the pole is analogous to the origin.

III) Symmetry

curve_tracing_table_2.jpg

IV) If for {\theta=0}, {r=0}, then the pole lies on the curve.

V) Slope of tangent at a point {(r_1, \theta_1)} is equal to { r \frac {d \theta}{dr}}

V) Polar Curves Nomenclature:

Cardioid : {r = a (1 \pm cos \ \theta)}, {r = a (1 \pm sin \ \theta)}

Parabola : {r (1 \pm cos \ \theta) = 2a}, {r (1 \pm sin \ \theta) = 2a}

Rose Curves : {r = a sin (n \theta)}, {r = a cos (n \theta)}

Lemniscate of Bernoulli : {r^2 = a^2 cos (2 \theta)}

Spirals : {r = a \theta}, {r = ae^{m \theta}, \ a,m > 0}


  • Illustrative Examples

Q.1) Trace the curve {r = a cos (3 \theta)} (June 2016)

Remarks :

1) {Cosine} is an even function, so function will have same value for {\theta} and {- \theta}.

2) {Cosine} is bounded between {-1} and {1}, hence {r} will be bounded between {-a} and {a}.

3) Intersection with the initial line will be at {r=a}, because at {\theta = 0}, {cos \ \theta = 1}.

4) When {3 \theta} becomes {\frac {\pi}{2}}, the curve will pass through the pole, because {cos \frac \pi 2 =0}. Hence, at {\theta = \frac {pi}{6}}, {r=0}.

5) The next value of {\theta} when {r} becomes {0} will be {\frac \pi 2}, because at {\theta = \frac {\pi}{2}}, {3 \theta = \frac {3 \pi}{2}} and {cos \ \frac {3 \pi}{2}= 0}.

6) As value of {\theta} varies from {0} to {\frac \pi 6}, value of {r} will reduce from {a} to {0}. 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 {cosine} function, same values of {r} will be obtained in the intervals {(0, \frac {2 \pi} {3})}, {(\frac {2 \pi}{3}, \frac {4 \pi}{3})} and {(\frac {4 \pi}{3}, 2 \pi)}.

8) Since at {\theta = \frac {\pi}{2}}, {r=0} and again at {\theta = \frac {2 \pi} {3}}, {r} becomes maximum, we’ve a good idea about the curve in the interval {0 \le \theta \le \frac {2 \pi}{3}}.

9) Let’s repeat this in the intervals {\frac {2 \pi}{3} \le \theta \le \frac {4 \pi}{3}} and {\frac {4 \pi}{3} \le \theta \le \frac {6 \pi}{3}= 2 {\pi}}.

In the figure below, I’ve plotted 2 such curves, one with {a=1}, the violet curve and second with {a=2}, the orange curve. The red line is {\theta = \frac \pi 6}, the green line is {\theta = \frac {2 \pi}{3}} and the blue line is {\theta = \frac \pi 2}.

rose_curve_1.jpg


Q.2) Trace the curve {r = a \ sin (2 \theta)}. (December 2015)

This is similar to Q.1.


Q.3) Trace the curve {r = a (1 + cos \theta)} (June 2015)

Remarks : (Assuming {a>0})

1) {Cosine} is an even function, so the curve will be symmetric about the initial line {\theta = 0}.

2) The curve will not have multiple loops, because the argument of {cosine} is {\theta} and not {n \theta, n \ne 1} as it was in Q.1.

3) Intersection with initial line will be at the point {(2a,0)} and intersection with {\theta = \pi} will be at the point {(0, \pi)}, i.e. the pole. Intersection with the line {\theta = \frac \pi 2} will be the point {(a,0)}.

4) The equation of tangent at the pole is obtained by equating {r} to {0}, which gives {1 + cos (\theta)= 0}, or {\theta = \pi}. Since the curve is symmetrical about {\theta =0} and value of {r} reduces from {\theta = 0} to {\theta = \pi}, there will be a cusp at the pole.

5) Slope of tangent at the point {(2a,0)} is obtained by {r \frac {d \theta}{dr}}. From the equation, {\frac {dr}{d \theta} = -a \ sin \ \theta}. So, {\frac {d \theta}{dr} = \frac {-1}{a \ sin \ \theta}} = {- \infty} for {\theta = 0}. Hence the tangent will be parallel to {\theta = \frac {\pi}{2}} at {(2a,0)}.

With this information, the curve will look like this :

Value of {a} that I took is {5}. Tangent at {(2a,0)} is the orange line. The curve is in green colour.

cardioid_1.jpg

 

 

 

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