# Thread: Maximum Y-Coordinate Values Of A Cardioid

1. ## Maximum Y-Coordinate Values Of A Cardioid

Algebraically determine the maximum value of the y-coordinate of points on the cardioid $r = 1 + cos \theta$ for 0 θ π.

I used the formula:

$\frac{dy}{dx}=\frac{\frac{dr}{d \theta} \sin \theta +r \cos \theta}{\frac{dr}{d \theta} \cos \theta-r \sin \theta}$, which yielded $\frac{-sin\theta(sin\theta+cos\theta+cos^{2}\theta)}{-sin\theta(cos\theta-sin\theta-sin\theta cos\theta)}$.

If I solve $-sin\theta(sin\theta+cos\theta+cos^{2}\theta)$, setting it to zero, it will give the value of theta at which horizontal tangents exist. The problem is that I don't know how to solve it.

After graphing it, I know that the max value occurs when $\theta = \frac{\pi}{3}$ and $r=1.5$, but I'm not sure how to get to those values.

2. ## Re: Maximum Y-Coordinate Values Of A Cardioid

All you need is that

$y=r\cos(\theta)=(1+\cos(\theta))\sin(\theta)=\sin( \theta)+\sin(\theta)\cos(\theta)$

Now we have

$\frac{dy}{d\theta}=\cos(\theta)+\cos^2(\theta)-\sin^2(\theta)=2\cos^2(\theta)+\cos(\theta)-1$

Setting equal to zero and factoring gives

$(2\cos(\theta)-1)(\cos(\theta)+1)=0$

This will give you both the max and min values.

3. ## Re: Maximum Y-Coordinate Values Of A Cardioid

Wow, that's so much better than my inchoate solution.

Thanks a lot!

4. ## Re: Maximum Y-Coordinate Values Of A Cardioid

Hello, Algebrah!

Determine the maximum value of the y-coordinate of points on the cardioid: $r \:=\: 1 + \cos \theta\;\text{ for }0\,\le\,\theta\,\le\,\pi$

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We have: . $r' \:=\:\text{-}\sin\theta$

$\text{Formula: }\:\frac{dy}{dx} \;=\;\frac{r\cos\theta + r'\sin\theta}{\text{-}r\sin\theta + r'\cos\theta}$

$\text{Then: }\:\frac{dy}{dx} \;=\;\frac{(1+\cos\theta)\cos\theta + (\text{-}\sin\theta)\sin\theta}{\text{-}(1+\cos\theta)\sin\theta + (\text{-}\sin\theta)\cos\theta} \;=\;\frac{\cos\theta + \cos^2\theta - \sin^2\theta}{\text{-}\sin\theta - 2\sin\theta\cos\theta}$

We have: . $\cos\theta + \cos^2\theta - \sin^2\theta \;=\;0$

. . . . $\cos\theta + \cos^2\theta - (1 - \cos^2\theta) \;=\;0$

. . . . . $\cos\theta + \cos^2\theta - 1 + \cos^2\theta \;=\;0$

. . . . . . . . . . $2\cos^2\theta + \cos\theta - 1 \;=\;0$

n . . . . . . $(\cos\theta + 1)(2\cos\theta - 1) \;=\;0$

. . $\begin{array}{cccccccccccc}\cos\theta + 1 \:=\: 0 &\Rightarrow& \cos\theta \:=\: \text{-}1 & \Rightarrow & \theta \:=\: \pi & {\color{red}**}\\ \\[-3mm] 2\cos\theta - 1 \:=\: 0 & \Rightarrow & \cos\theta \:=\: \frac{1}{2} &\Rightarrow & \theta \:=\: \frac{\pi}{3} \end{array}$

** There is a horizontal tangent at $\theta = \pi$
. . .but it gives a minimum value for y.

Hence, the maximum y occurs at $\theta \,=\,\tfrac{\pi}{3}$

$\text{Therefore: }\:y \;=\;r\sin\theta \;=\;\left(1 + \cos\tfrac{\pi}{3}\right)\sin\tfrac{\pi}{3}\;=\; \left(\frac{3}{2}\right)\left(\frac{\sqrt{3}}{2} \right)\;=\;\frac{3\sqrt{3}}{4}$