We start by finding \(x'(t)\) and \(y'(t)\text{:}\)

\begin{equation*}
x'(t) = -3\sin(t) \cos^2(t) , \qquad y'(t) = 3\cos(t) \sin^2(t)\text{.}
\end{equation*}

Note that both of these are 0 at \(t=0\text{;}\) the curve is not smooth at \(t=0\) forming a cusp on the graph. Evaluating \(\frac{dy}{dx}\) at this point returns the indeterminate form of “0/0”.

We can, however, examine the slopes of tangent lines near \(t=0\text{,}\) and take the limit as \(t\to 0\text{.}\)

\begin{align*}
\lim_{t\to0} \frac{y'(t)}{x'(t)} \amp =\lim_{t\to0} \frac{3\cos(t) \sin^2(t) }{-3\sin(t) \cos^2(t) } \text{ (We can cancel as \(t\neq 0\).) }\\
\amp = \lim_{t\to0} -\frac{\sin(t) }{\cos(t) }\\
\amp = 0\text{.}
\end{align*}

We have accomplished something significant. When the derivative \(\frac{dy}{dx}\) returns an indeterminate form at \(t=t_0\text{,}\) we can define its value by setting it to be \(\lim\limits_{t\to t_0}\)\(\frac{dy}{dx}\text{,}\) if that limit exists. This allows us to find slopes of tangent lines at cusps, which can be very beneficial.

We found the slope of the tangent line at \(t=0\) to be 0; therefore the tangent line is \(y=0\text{,}\) the \(x\)-axis.