# 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

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 ${\theta = \frac \pi 2}$.

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.