Thursday, 20 March 2014



Let we will discuss about Cardioide

r=a(1+cos theta)

-----  If we put  -theta  for  theta  in the equation of the curve we find that;

r=a{1+cos (-theta)}=a(1+cos theta)

i.e. the equation of the curve does not change. Therefore the given curve is symmetrical about the initial line.

-----  Now    r=0  when;

1+cos theta = 0 

i.e.  cos theta = -1 

therefore    theta = pi 

Hence the curve passes through the origin and the equation of the tangent at the pole is

theta = pi  i.e. the initial line.

-----  Now we plot some of the points on the curve.

When  theta=0,  then  r=2a;

when  theta=pi/3,  then  r=3a/2

when  theta=pi/2,  then  r=a

when  theta=pi,  then  r=0.

-----  From the given equation

r=a(1+cos theta),  when we have

(dr/d theta)=-a sin theta 

That is, when the value of  theta  increases from

0  to  pi,  then the value of   r  decreases and as been stated earlier decreases from

2a   to  0

Again, since the given curve is symmetrical about the initial line, therefore when the value of    theta   increases from    pi  to    2pi,

Then the value of   r    increases from   0  to  2a.

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