Slope of a cardioid on the origin (0,0) I just started calculus and got this problem stuck in my head. It seems, in my case, a problem of bad understanding or definition.
The cardioid:
$r=1+\cos(θ)$
Derivative:
$\frac{dy}{dx}=\frac{\cosθ+\cos^2θ−\sin^2θ}{−\sinθ−2\sinθ\cosθ}$
When trying to calculate the derivative of the cardioid on the origin ($θ=\pi$), an indetermination shows
¿Does that means that the derivative does not exist on that point? I read in some site that this is not conclusive; instead, we should try to do the limit of the derivative function on that point.
$L'Hopital:$
$$\lim_{θ\to \pi}\frac{d(\cosθ+\cos^2θ−\sin^2θ)/dθ}{d(−\sinθ−2\sinθ\cosθ)/dθ}=\lim_{θ\to \pi}\frac{-\sinθ-2\sin2θ}{−\cosθ−2\cos2θ}=0$$
Following that procedure, the answer should be an horizontal line, slope $0$, which seems reasonable if you remember the graph of a cardioid.
The solution seems to imply (for me) that the derivative function is, actually, like this (something I could not find anywhere, so I assume I am somehow wrong):
Derivative of $f(x)$ at the x point:
$$\lim_{h\to x}f'(h)$$
Is "$0$" the right answer to this problem, or the cardioid is not differentiable on the origin point?
 A: The principle "differentiable if and only if has a tangent line" holds for Cartesian graphs. There is, however, a subtlety between graphs and parametric paths, including polar graphs.
Throughout, let's suppose $f$ is a continuously-differentiable real-valued function ($f'$ exists and is continuous) in some interval.

*

*The Cartesian graph $y = f(x)$ has a tangent line at each point $(x_{0}, f(x_{0}))$, with Cartesian equation $y = f(x_{0}) + f'(x_{0})(x - x_{0})$.

*By contrast, the polar graph $r = f(\theta)$ can fail to have a tangent line at the origin if $f(\theta_{0}) = f'(\theta_{0}) = 0$ for some $\theta_{0}$. (!)

Generally, we say a parametric path $(u(t), v(t))$ is differentiable if $u$ and $v$ are individually differentiable. We say the path is regular or an immersion at $t_{0}$ if $u'$ and $v'$ are continuous in a neighborhood of $t_{0}$ and the velocity $(u'(t_{0}), v'(t_{0}))$ is non-zero.
The curve traced by a parametric path has a tangent line at $(u(t_{0}), v(t_{0}))$ if the path is regular there. Otherwise the path may or may not have a tangent line. For instance:

*

*The path $(u, v) = (t^{3}, t^{2})$ smoothly (but non-regularly) traces the cusp $y = x^{2/3}$, which has no tangent to the origin.

*The path $(u, v) = (t^{3}, t^{3})$ smoothly (but non-regularly) traces the line $y = x$.

Exercises:

*

*The path $(u(t), v(t)) = (t, f(t))$, which traces the Cartesian graph $y = f(x)$, is regular everywhere. (Recall we're assuming $f$ is continuously-differentiable.)

*The path $(u(t), v(t)) = (f(t)\cos t, f(t)\sin t)$, which traces the polar graph $r = f(\theta)$, is regular if and only if $f(\theta) \neq 0$ or $f'(\theta) \neq 0$. (Hint: It may help to calculate the speed, the magnitude of the velocity.)

In the example at hand, $f(\theta) = 1 + \cos\theta$. Thus $f(\pi) = f'(\pi) = 0$ (N.B. $\theta = \pi$, not $2\pi$), and we have no guarantee of a tangent line at $(0, 0)$. In fact, the indeterminacy in the calculation of $\frac{dy}{dx}$ is precisely a result of the polar graph as a parametric path having velocity zero.
A: The point $(0,0)$ is a singular point of the cardioid.  Furthermore, it can be classified as a cusp.
In fact, the cardioid itself is a kind of singular “curve“ within the family of limaçons with equation $(x^2+y^2-x)^2 = b^2(x^2+y^2)$.  If $b < 1$, the limaçon has a “dimple,” if $b > 1$, the limaçon has a loop through the origin, and if $b=1$, the curve is a cardioid.
In each case, the curve is the zero level-set of the function
$$
    f(x,y) = b^2 (x^2+y^2) + (x^2+y^2-x)^2
$$
Following the discussion on Wikipedia, we can substitute $y=mx$ to find tangent lines through the origin:
\begin{align*}
    f(x,mx) &= b^2(x^2+m^2x^2) - (x^2 + m^2x^2-x)^2 
    \\&= [b^2m^2+(b^2-1)]x^2 + 2(1+m^2)x^3 + (1+m^2)^2 x^4
\end{align*}
The origin is a double point if the coefficient of $x^2$ is zero.

*

*In the case that $b > 1$, there are two real numbers $m$ which satisfy $b^2 m^2 + (b^2-1) = 0$.  This means that the origin is a crunode.

*In the case that $b < 1$, there are no real solutions for $m$, but two imaginary solutions.  The origin is called an acnode.

*In the case that $b=1$, then $m=0$ is a double solution.  The origin is a cusp.

