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?
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.