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

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

Explanation

Solution

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(θ)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(θ)1 + \sin \left( \theta \right) is 2π2\pi . This says that T=2πT = 2\pi for 1+sin(θ)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 II. On integrating, we get:
l=Idsl = \int\limits_I {ds}
Here, ds=r2+(drdθ)2dθds = \sqrt {{r^2} + {{\left( {\dfrac{{dr}}{{d\theta }}} \right)}^2}d\theta }

Now, we will calculate the derivative, and then we get:
drdθ=ddθ(1+sinθ)=cosθ\Rightarrow \dfrac{{dr}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {1 + \sin \theta } \right) = \cos \theta
Now, we can put the value of drdθ\dfrac{{dr}}{{d\theta }} in the equationds=r2+(drdθ)2dθds = \sqrt {{r^2} + {{\left( {\dfrac{{dr}}{{d\theta }}} \right)}^2}d\theta } :
ds=(1+sinθ)2+(cosθ)2dθ\Rightarrow ds = \sqrt {{{(1 + \sin \theta )}^2} + {{(\cos \theta )}^2}d\theta }
ds=(1+2sinθ+sin2θ+cos2θdθ\Rightarrow ds = \sqrt {(1 + 2\sin \theta + {{\sin }^2}\theta + {{\cos }^2}\theta d\theta }
Now, we can apply the trigonometric formula cos2θ+sin2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 in the equation, and we get:
ds=(2+2sinθdθ)\Rightarrow ds = \sqrt {(2 + 2\sin \theta d\theta } )
ds=2(1+sinθ)dθ\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 22. Now, we know the trigonometric formula:
cos2(α2)=1+cosα2{\cos ^2}\left( {\dfrac{\alpha }{2}} \right) = \dfrac{{1 + \cos \alpha }}{2}
If α=θπ2\alpha = \theta - \dfrac{\pi }{2} then cosα=cos(θπ2)\cos \alpha = \cos \left( {\theta - \dfrac{\pi }{2}} \right)
cosα=sinθ\Rightarrow \cos \alpha = \sin \theta
Therefore, when we now substitute, we get:
cos2(θπ22)=1+cos(θπ2)2\Rightarrow {\cos ^2}\left( {\dfrac{{\theta - \dfrac{\pi }{2}}}{2}} \right) = \dfrac{{1 + \cos \left( {\theta - \dfrac{\pi }{2}} \right)}}{2}
2cos2(θ2π4)=1+sinθ\Rightarrow 2{\cos ^2}\left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right) = 1 + \sin \theta
When we calculate the last part of dsds, we get:
ds=4cos2(θ2π4)dθds = \sqrt {4{{\cos }^2}\left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)d\theta }
ds=2cos(θ2π4)dθ\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 - 10,10, - 5,5.

Now, we know that:
cos(π2π4)0π2+2kπx2π4π2+2kπ\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:
π2π+4kπx32π+4kπ- \dfrac{\pi }{2}\pi + 4k\pi \leqslant x \leqslant \dfrac{3}{2}\pi + 4k\pi
π2πx32π\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=(π2,32π)I = \left( { - \dfrac{\pi }{2},\dfrac{3}{2}\pi } \right)
We are removing the absolute value from here because cos(π2π4)0\cos \left( {\dfrac{\pi }{2} - \dfrac{\pi }{4}} \right) \geqslant 0 on II.
π232π2cos(θ2π4)dθ=4π232π12cos(θ2π4)dθ\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
π232π2cos(θ2π4)dθ=[4(θ2π4)]π232π\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 }
π232π2cos(θ2π4)dθ=4sin(π2)4sin(π2)\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)
π232π2cos(θ2π4)dθ=4+4\Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{3}{2}\pi } {2\cos \left( {\dfrac{\theta }{2} - \dfrac{\pi }{4}} \right)d\theta } = 4 + 4
π232π2cos(θ2π4)dθ=8\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.