Solutions for a polar equation

Question

A polar curve has the equation $$ r=a(1+ \cos \theta)$$

The point on the curve with polar coordinates (r,$\theta$) has Cartesian coordinates (x,y).

Find the minimum value of y.

Attempt I found the $ \frac{dy}{d\theta} $and equated it to 0. However I found two solutions; $\cos \theta=1/2 $and $\cos \theta=-1$

$\cos \theta=1/2 $gives the correct value for y. Can somebody tell me what the second solution mean?

Well, what are the other possibilities at a point where the derivative of a function is equal to zero? — amd, May 17 '17 at 00:18
dtheta is approaching infinite? But I doubt that makes sense. — mathnoob123, May 17 '17 at 00:19
Maybe I got it. $ \frac{dy}{d\theta} = \frac{dy}{dr} \times \frac{dr}{d\theta} $ since $\frac{dy}{dr}$ is 0 when $\theta=\pi$,$ \frac{dy}{d\theta}=0$ — mathnoob123, May 17 '17 at 00:34
a rel. min can occur where $\frac{dy}{d\theta} = 0$ and $ \frac{dx}{d\theta} \neq 0$ — Dando18, May 17 '17 at 00:36
Can somebody please look at this question too? https://math.stackexchange.com/q/2283827/335742 — mathnoob123, May 17 '17 at 00:38
$y = r \sin(\theta) = a(1+\cos(\theta))\sin (\theta) $. If $\cos (\theta) = -1$, $\sin(\theta) = 0$, so this cannot be a min, as $ y < 0$ wherever $\sin ( \theta) < 0$ and $\cos (\theta) \ne 1$ (I assumed $a > 0$ ) — infinitylord, May 17 '17 at 01:05

score 1 · Accepted Answer · answered May 17 '17 at 00:49

1

Relative extrema on polars occur where $\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \neq 0$. This second part is important.

Think about where extrema occur when $y$ is a function of $x$ ($y(x)$ - "a normal equation"). Extrema occur where $\frac{dy}{dx} = 0$ . Now note that $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$, which is zero when the numerator ($\frac{dy}{d\theta}$) is zero and is undefined when the denominator ($\frac{dx}{d\theta}$) is zero.

In your question $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos(\theta) - r \sin(\theta)$, which is zero at $\theta = \pi$ so not the location of a min.

answered May 17 '17 at 00:49

Dando18

5,368

@Faiq In fact, for $\theta=\pi$ we have a cusp. – imranfat May 17 '17 at 01:05
Is this correct? $ \frac{dy}{d\theta} = \frac{dy}{dr} \times \frac{dr}{d\theta} $ since $\frac{dy}{dr}$ is 0 when $\theta=\pi$,$ \frac{dy}{d\theta}=0$ – mathnoob123 May 17 '17 at 15:37
@FaiqRaees yes $\frac{dy}{d\theta}(\pi)=0$, but also $\frac{dx}{d\theta}(\pi)=0$ so $\frac{dy}{dx}(\pi)$ is undefined. Note that $\frac{dy}{d\theta}(\pi/3) = \frac{dy}{d\theta}(5\pi/3) = 0$ where $\frac{dx}{d\theta} \neq 0$ at those points so they are the locations of extrema. – Dando18 May 17 '17 at 17:21

Solutions for a polar equation

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