Question

How do you find the exact length of the polar curve\[r = 1 + \sin \left( \theta \right)\] ?

Answer 1

Hint: To deal with the polar coordinates we have to solve using the trigonometric rules, here for this question we have to get the periodic intervals of the given trigonometric identity and accordingly by using properties we can solve it.

Complete step by step solution:
The given equation is: \[r = 1 + \sin \left( \theta \right)\]
This is the equation of cardioid. The equation here is in a periodic function form. Therefore, we will find out the period so that we can get the length after that. We know that the time period of \[1 + \sin \left( \theta \right)\] is \[2\pi \]. This says that \[T = 2\pi \] for \[1 + \sin \left( \theta \right)\].
We can find the length of the polar curve by doing simple integration of the arc length on a particular interval. The internal here is denoted as \[I\]. On integrating, we get:
\[l = \int\limits_I {ds} \]
Here, \[ds = \sqrt {{r^2} + {{\left( {\dfrac{{dr}}{{d\theta }}} \right)}^2}d\theta } \]

Now, we will calculate the derivative, and then we get:
\[ \Rightarrow \dfrac{{dr}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {1 + \sin \theta } \right) = \cos \theta \]
Now, we can put the value of \[\dfrac{{dr}}{{d\theta }}\] in the equation\[ds = \sqrt {{r^2} + {{\left( {\dfrac{{dr}}{{d\theta }}} \right)}^2}d\theta } \]:
\[ \Rightarrow ds = \sqrt {{{(1 + \sin \theta )}^2} + {{(\cos \theta )}^2}d\theta } \]
\[ \Rightarrow ds = \sqrt {(1 + 2\sin \theta + {{\sin }^2}\theta + {{\cos }^2}\theta d\theta } \]
Now, we can apply the trigonometric formula \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] in the equation, and we get:
\[ \Rightarrow ds = \sqrt {(2 + 2\sin \theta d\theta } )\]
\[ \Rightarrow ds = \sqrt {2(1 + \sin \theta )d\theta } \]

Now, we have to take out the square root from the equation. Here, the square root is multiplied with a coefficient that is \[2\]. Now, we know the trigonometric formula:
\[{\cos ^2}\left( {\dfrac{\alpha }{2}} \right) = \dfrac{{1 + \cos \alpha }}{2}\]
If \[\alpha = \theta - \dfrac{\pi }{2}\] then \[\cos \alpha = \cos \left( {\theta - \dfrac{\pi }{2}} \right)\]
\[ \Rightarrow \cos \alpha = \sin \theta \]
Therefore, when we now substitute, we get:
\[ \Rightarrow {\cos ^2}\left( {\dfrac{{\theta - \dfrac{\pi }{2}}}{2}} \right) = \dfrac{{1 + \cos \left( {\theta - \dfrac{\pi }{2}} \right)}}{2}\]
\[ \Rightarrow 2{\cos ^2}\left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right) = 1 + \sin \theta \]
When we calculate the last part of \[ds\], we get:
\[ds = \sqrt {4{{\cos }^2}\left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)d\theta } \]
\[ \Rightarrow ds = 2\left| {\cos \left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)} \right|d\theta \]
So, we have simplified and taken out the square root part. Now, we will integrate this part to get the time interval of the length of the curve. To integrate, first we have to pull out the absolute value. To do that we can first plot a graph and find out the coordinates. Here, the coordinates are, \[ - 10,10, - 5,5\].

Now, we know that:
\[\cos \left( {\dfrac{\pi }{2} - \dfrac{\pi }{4}} \right) \geqslant 0 \Leftrightarrow - \dfrac{\pi }{2} + 2k\pi \leqslant \dfrac{x}{2} - \dfrac{\pi }{4} \leqslant \dfrac{\pi }{2} + 2k\pi \]
Where “k” is any integer value. So, we get:
\[- \dfrac{\pi }{2}\pi + 4k\pi \leqslant x \leqslant \dfrac{3}{2}\pi + 4k\pi \]
\[\Rightarrow - \dfrac{\pi }{2}\pi \leqslant x \leqslant \dfrac{3}{2}\pi \]
Now, we will do the integration on the arc of the length(“I”):
\[I = \left( { - \dfrac{\pi }{2},\dfrac{3}{2}\pi } \right)\]
We are removing the absolute value from here because \[\cos \left( {\dfrac{\pi }{2} - \dfrac{\pi }{4}} \right) \geqslant 0\] on \[I\].
\[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{3}{2}\pi } {2\cos \left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)d\theta } = 4\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{3}{2}\pi } {\dfrac{1}{2}\cos } \left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)d\theta \]
\[ \Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{3}{2}\pi } {2\cos \left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)d\theta } = \left[ {4\left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)} \right]_{ - \dfrac{\pi }{2}}^{\dfrac{3}{2}\pi }\]
\[ \Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{3}{2}\pi } {2\cos \left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)d\theta } = 4\sin \left( {\dfrac{\pi }{2}} \right) - 4\sin \left( { - \dfrac{\pi }{2}} \right)\]
\[ \Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{3}{2}\pi } {2\cos \left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)d\theta } = 4 + 4\]
\[ \therefore\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{3}{2}\pi } {2\cos \left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)d\theta } = 8\]

Note: Here we have used integration because on solving for the given question it was needed to integrate the expression to get the desired length and by integrating in a within range we can get the length of the whole curve assuming for the single unit.