Question

How do you find the length of cardioid \[r = 1 + \sin \theta \]?

Answer 1

Hint: Cardioid is one of the curves shaped like heart. Equation of the cardioid is given above whose length is to be found. We will use the formula given below to find the length. Also we will take the limits from 0 to 2pi. But we need to use rationalisation, substitution and limits division in the problem to solve to find the length. So, let’s solve it!
Formula used:
\[L = \int {\sqrt {{r^2} + {{\left( {\dfrac{{dr}}{{d\theta }}} \right)}^2}} d\theta } \]

Complete step-by-step solution:
Given that,
\[r = 1 + \sin \theta \]
This is the equation of the cardioid so given. The formula needs to find the differentiation also. So we will find the derivative first.
\[\dfrac{{dr}}{{d\theta }} = \cos \theta \]
The limits are from 0 to 2pi.
Now substitute these values in the formula above we get,
\[L = \int\limits_0^{2\pi } {\sqrt {{{\left( {1 + \sin \theta } \right)}^2} + {{\left( {\cos \theta } \right)}^2}} d\theta } \]
Taking the squares and expanding the first bracket,
\[L = \int\limits_0^{2\pi } {\sqrt {1 + 2\sin \theta + {{\sin }^2}\theta + {{\cos }^2}\theta } d\theta } \]
On adding the square terms of sin and cos we get,
\[L = \int\limits_0^{2\pi } {\sqrt {1 + 2\sin \theta + 1} d\theta } \]
On adding the numbers we get,
\[L = \int\limits_0^{2\pi } {\sqrt {2 + 2\sin \theta } d\theta } \]
Taking 2 common outside,
\[L = \int\limits_0^{2\pi } {\sqrt {2\left( {1 + \sin \theta } \right)} d\theta } \]
\[L = \sqrt 2 \int\limits_0^{2\pi } {\sqrt {\left( {1 + \sin \theta } \right)} d\theta } \]
On rationalising the root,
\[L = \sqrt 2 \int\limits_0^{2\pi } {\sqrt {1 + \sin \theta } \times \dfrac{{\sqrt {1 - \sin \theta } }}{{\sqrt {1 - \sin \theta } }}d\theta } \]
On multiplying the numerators, we get,
\[L = \sqrt 2 \int\limits_0^{2\pi } {\dfrac{{\sqrt {\left( {1 - {{\sin }^2}\theta } \right)} }}{{\sqrt {1 - \sin \theta } }}d\theta } \]
On solving the numerator we get,
\[L = \sqrt 2 \int\limits_0^{2\pi } {\dfrac{{\sqrt {\left( {{{\cos }^2}\theta } \right)} }}{{\sqrt {1 - \sin \theta } }}d\theta } \]
Cancelling the square and root,
\[L = \sqrt 2 \int\limits_0^{2\pi } {\dfrac{{\cos \theta }}{{\sqrt {1 - \sin \theta } }}d\theta } \]
Now we will use substitution as,
\[1 - \sin \theta = u\]
Taking the derivative,
\[ - \cos \theta d\theta = du\]
Now we will shift the minus sign,
\[ - du = \cos \theta d\theta \]
Now substitute to take the integration,
\[L = - \sqrt 2 \int {\dfrac{{du}}{{\sqrt u }}} \]
Writing the denominator in the form of power,
\[L = - \sqrt 2 \int {{u^{\dfrac{{ - 1}}{2}}}du} \]
Now taking the integration we get,
\[L = - \sqrt 2 \left[ {\dfrac{{{u^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right] + C\]
Now taking the denominator of the denominator in the numerator,
\[L = - 2\sqrt 2 \left[ {{u^{\dfrac{1}{2}}}} \right] + C\]
Now we will substitute the original value of u and will split the limits as per the convenience,
\[L = - 2\sqrt 2 \left( {\left[ {\sqrt {\left( {1 - \sin \theta } \right)} } \right]_0^{\dfrac{\pi }{2}} - \left[ {\sqrt {\left( {1 - \sin \theta } \right)} } \right]_{\dfrac{\pi }{2}}^{\dfrac{\pi }{2} + \pi } + \left[ {\sqrt {\left( {1 - \sin \theta } \right)} } \right]_{\dfrac{\pi }{2} + \pi }^{2\pi }} \right)\]
On applying the upper and lower limits we get,
\[L = - 2\sqrt 2 \left( {\left[ {0 - 1} \right] - \left[ {\sqrt 2 - 0} \right] + \left[ {1 - \sqrt 2 } \right]} \right)\]
On calculating we get,
\[L = - 2\sqrt 2 \left( { - 1 - \sqrt 2 + 1 - \sqrt 2 } \right)\]
1 will be cancelled,
\[L = - 2\sqrt 2 \left( { - 2\sqrt 2 } \right)\]
On multiplying we get,
\[L = 8\]
This is the final answer. This is the length of the cardioid.

Note: Here note that the limit is splitted because if we place the upper and lower limits we will get the answer totally as zero. So we have converted it into three breaks. Also note that, we have used rationalisation as well as substitution to reach the answer. Since the equation before rationalisation has no output to solve.