Question
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) ?
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)
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) is 2π. This says that T=2π for 1+sin(θ).
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=I∫ds
Here, ds=r2+(dθdr)2dθ
Now, we will calculate the derivative, and then we get:
⇒dθdr=dθd(1+sinθ)=cosθ
Now, we can put the value of dθdr in the equationds=r2+(dθdr)2dθ:
⇒ds=(1+sinθ)2+(cosθ)2dθ
⇒ds=(1+2sinθ+sin2θ+cos2θdθ
Now, we can apply the trigonometric formula cos2θ+sin2θ=1 in the equation, and we get:
⇒ds=(2+2sinθdθ)
⇒ds=2(1+sinθ)dθ
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:
cos2(2α)=21+cosα
If α=θ−2π then cosα=cos(θ−2π)
⇒cosα=sinθ
Therefore, when we now substitute, we get:
⇒cos22θ−2π=21+cos(θ−2π)
⇒2cos2(2θ−4π)=1+sinθ
When we calculate the last part of ds, we get:
ds=4cos2(2θ−4π)dθ
⇒ds=2cos(2θ−4π)dθ
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(2π−4π)⩾0⇔−2π+2kπ⩽2x−4π⩽2π+2kπ
Where “k” is any integer value. So, we get:
−2ππ+4kπ⩽x⩽23π+4kπ
⇒−2ππ⩽x⩽23π
Now, we will do the integration on the arc of the length(“I”):
I=(−2π,23π)
We are removing the absolute value from here because cos(2π−4π)⩾0 on I.
−2π∫23π2cos(2θ−4π)dθ=4−2π∫23π21cos(2θ−4π)dθ
⇒−2π∫23π2cos(2θ−4π)dθ=[4(2θ−4π)]−2π23π
⇒−2π∫23π2cos(2θ−4π)dθ=4sin(2π)−4sin(−2π)
⇒−2π∫23π2cos(2θ−4π)dθ=4+4
∴−2π∫23π2cos(2θ−4π)dθ=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.