# Find the maximum height (above the x-axis) of the cardioid r = 2(1 + \cos\theta).

## Question:

Find the maximum height (above the {eq}x {/eq}-axis) of the cardioid

{eq}r = 2\left(1 + \cos\theta\right) {/eq}.

## Maximum of a Function:

Let us consider a real-value function {eq}y=f(x). {/eq}

A point {eq}x_0 {/eq} is a relative maximum for the function if the first derivative of the function is equal to zero at this point

and the second derivative at this point is negative, i.e.

{eq}f'(x_0)=0 \\ f''(x_0) < 0. {/eq}

We are given the cardiodi in polar form

{eq}r = 2\left(1 + \cos\theta\right). {/eq}

The height (above the {eq}x {/eq}-axis) of the cardioid is the y coordinate, that is related to polar coordinates {eq}(r,\theta) {/eq} as

{eq}y = r \sin \theta = 2(1+\cos\theta)\sin\theta \\ = 2\sin\theta + 2\sin\theta\cos\theta \\ = 2\sin\theta + \sin 2\theta. {/eq}

The maximum of this function is found setting the first derivative to zero

{eq}y'(\theta) = 2\cos\theta + 2\cos 2\theta = 0 \\ \Rightarrow \cos 2\theta + \cos \theta = 0 \\ \Rightarrow \cos^2 \theta - \sin^2\theta + \cos \theta = 0 \\ \Rightarrow 2\cos^2 \theta + \cos \theta -1 = 0. {/eq}

Upon setting {eq}\displaystyle \cos \theta = u \Rightarrow \theta = \cos^{-1}(u) {/eq}

we rewrite and solve former equation

{eq}\displaystyle \Rightarrow 2u^2 + u -1 = 0 \Rightarrow u=\frac{1}{2},\:u=-1 \\ \displaystyle u = \frac{1}{2} \Rightarrow \theta = \cos^{-1}(\frac{1}{2} ) = \frac{\pi}{3}, \; \theta = \frac{5\pi}{3} \\ \displaystyle u = -1 \Rightarrow \theta = \pi. {/eq}

Therefore we have three critical points

{eq}\displaystyle \theta = \frac{\pi}{3}, \; \frac{5\pi}{3}. \; \pi {/eq}

and, since the function y is the sum of two sines function, its maximum is achieved when

{eq}\displaystyle \theta = \frac{\pi}{3} \Rightarrow y(\frac{\pi}{3}) = 2.598. {/eq}