Question

How do you find the arc length of the cardioid $r=1+\cos \theta $ from $0$ to $2\pi $ ?

Answer 1

Hint:We have been given the equation of a cardioid whose arc length we have to find in a fixed interval. We shall use the formula of arc length as $L=\int\limits_{a}^{b}{\sqrt{{{r}^{2}}+{{\left( \dfrac{dr}{d\theta } \right)}^{2}}}.d\theta }$. Thus, to apply this formula, we will first find the derivative of the given function and then substitute it in the formula of arc length to integrate it within the provided interval.

Complete step by step solution:
Given that $r=1+\cos \theta $.
We shall first differentiate the function $r=1+\cos \theta $ with respect to angle, $\theta $ and find the derivative, $\dfrac{dr}{d\theta }$.
$\Rightarrow \dfrac{dr}{d\theta }=\dfrac{d}{d\theta }\left( 1+\cos \theta \right)$
$\Rightarrow \dfrac{dr}{d\theta }=\dfrac{d}{d\theta }\left( 1 \right)+\dfrac{d}{d\theta }\cos \theta $
Here, we shall use a property of differentials $\dfrac{d}{dx}c=0$and $\dfrac{d}{dx}\cos x=-\sin x$
$\Rightarrow \dfrac{dr}{d\theta }=-\sin \theta $ …………………….. equation (1)
The formula of arc length, $L$ is given as:
$L=\int\limits_{a}^{b}{\sqrt{{{r}^{2}}+{{\left( \dfrac{dr}{d\theta } \right)}^{2}}}.d\theta }$
Where,
$a=$ lower limit of integral
$b=$ upper limit of integral
$\dfrac{dr}{d\theta }=$derivative of the function
$r=$the equation of cardioid
Substituting the value of derivative of function, we get
$L=\int\limits_{0}^{2\pi }{\sqrt{{{\left( 1+\cos \theta \right)}^{2}}+{{\left( -\sin \theta \right)}^{2}}}.d\theta }$
The above function, L has the time period $=\pi $ and thus we can use the property of definite integration $\int\limits_{0}^{nT}{f\left( x \right).dx}=n\int\limits_{0}^{T}{f\left( x \right).dx}$ where $T$ is the time period of $f\left( x \right)$.
We will expand the terms within the square, ${{\left( 1+\cos \theta \right)}^{2}}$using the algebraic property ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.
$\Rightarrow L=2\int\limits_{0}^{\pi }{\sqrt{\left( 1+{{\cos }^{2}}\theta +2\cos \theta \right)+\left( {{\sin }^{2}}\theta \right)}.d\theta }$
Applying ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get
$\Rightarrow L=2\int\limits_{0}^{\pi }{\sqrt{1+2\cos \theta +1}.d\theta }$
$\Rightarrow L=2\int\limits_{0}^{\pi }{\sqrt{2+2\cos \theta }.d\theta }$
$\Rightarrow L=2\int\limits_{0}^{\pi }{\sqrt{2\left( 1+\cos \theta \right)}.d\theta }$
We will use the property of trigonometric function here according to which $1+\cos \theta =2{{\sin }^{2}}\theta $.
 $\Rightarrow L=2\int\limits_{0}^{\pi }{\sqrt{2\left( 2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right) \right)}.d\theta }$
$\Rightarrow L=2\int\limits_{0}^{\pi }{\sqrt{4{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)}.d\theta }$
We know that $2\times 2=4$ or ${{2}^{2}}=4$. Substituting this value, we get
 $\Rightarrow L=2\int\limits_{0}^{\pi }{\sqrt{\left( {{2}^{2}} \right){{\cos }^{2}}\dfrac{\theta }{2}}.d\theta }$
$\Rightarrow L=2\int\limits_{0}^{\pi }{2\cos \dfrac{\theta }{2}.d\theta }$
Taking the constant 2, outside the integral, we get
$\Rightarrow L=4\int\limits_{0}^{\pi }{\cos \dfrac{\theta }{2}.d\theta }$
$\Rightarrow L=4\left[ 2\sin \dfrac{\theta }{2} \right]_{0}^{\pi }$
$\Rightarrow L=8\left[ \sin \dfrac{\pi }{2}-\sin 0 \right]$
$\Rightarrow L=8\left[ 1-0 \right]$
$\Rightarrow L=8$
Therefore, the arc length of the cardioid $r=1+\cos \theta $ from $0$ to $2\pi $ is 8 units.

Note: We must have prior knowledge of the various formulae of integration and in order to proceed with such problems. The property of integration used in this problem is $\int{\cos x.dx=\sin x+C}$ but since the limits of integration were already provided, thus we removed the constant of integration, C and substituted the upper and lower limits as 0 and $2\pi $ respectively.