Horizontal And Vertical Tangents For Parametric Curves

Recall that if we have a parametric curve in the form $x = f(t)$ and $y = g(t)$ and we want find a derivative for this curve, dy/dx, then it follows that:

(1)
\begin{align} \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \end{align}

# Horizontal Tangents

Recall that with a function f(x), horizontal tangents arise when f'(x) = 0, that is, the numerator of f'(x) is 0. This is similar for parametric curves.

For a parametric curve with a derivative defined by:

(2)
\begin{align} \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \end{align}

If $\frac{dy}{dt} = 0 \quad \mathbf{and} \quad \frac{dx}{dt} ≠ 0$, then a horizontal tangent exists for the specific value of t that satisfies this condition.

## Example 1

Find any horizontal tangents for the parametric curve defined by $x = 2t$ and $y = t^2$.

This is a rather simple example, but let's first differentiate to obtain dx/dt, and dy/dt:

(3)

Now let's set dy/dt = 0. We thus get $2t = 0$, so t = 0. When t = 0, dx/dt ≠ 0 since dx/dt is just a constant. Hence there exists a horizontal tangent at t = 0. We can plug this into our parametric equations to get the coordinates (0, 0) where this occurs.

# Vertical Tangents

When we have a function f(x), there were no such things as vertical tangents (although there was vertical asymptotes). Nevertheless, parametric curves are not necessarily functions, and vertical tangents can exist. If $\frac{dy}{dx} ≠ 0 \quad \mathbf{and} \quad \frac{dx}{dt} = 0$, then a vertical tangent exists for the specific value of t that satisfies this condition.

## Example 2

Find any vertical tangents for the parametric curve defined by $x = \cos \theta$ and $y = \sin \theta$.

Finding dx/dθ and dy/dθ, we obtain:

(4)

Hence dx/dθ = 0 when θ = 0 and π. Also note that dy/dθ ≠ 0 when θ = 0 and when θ = π, hence there exists two vertical tangents when θ = 0 and when θ = π. We can plug these into our parametric equations to get the xy-coordinates (1, 0) and (-1, 0)

# Exceptions

So what are we supposed to do if when dy/dt = 0, dx/dt = 0, or if when dx/dt = 0, dy/dt = 0? Well we have to use limits to determine what happens at these points.

## Example 3

Find any vertical tangents for the parametric curve defined by $x = r(\theta - \sin \theta)$ and $y = r(1 - \cos \theta)$.

First let's differentiate dx/dθ and dy/dθ to obtain dy/dx.

(5)

Therefore:

(6)
\begin{align} \frac{dy}{dx} = \frac{r \sin \theta}{r - r \cos \theta} \\ = \frac{r \sin \theta}{r(1 - \cos \theta)} \\ = \frac{\sin \theta}{1 - \cos \theta} \end{align}

So dx/dθ = 0 when θ = 0, π, 2π, … But notice that when θ = 0, π, 2π, …, then dy/dθ = 0, since sin(0) = 0, sin(π) = 0, …. So we need to use our knowledge of limits in order to figure out what is happening here. Let's just use 2π as an example,

(7)
\begin{align} \lim_{\theta \to 2\pi} \frac{dy}{dx} = \lim_{\theta \to 2\pi} \frac{\sin x}{1 - \cos x} \end{align}

We can use L'Hospital's Rule to evaluate this limit further, first from the right side:

(8)
\begin{align} \lim_{\theta \to 2\pi^+} \frac{\sin x}{1 - \cos x} \\ = \lim_{\theta \to 2\pi^+} \frac{\cos x}{\sin x} \\ = \infty \end{align}

Now let's look at the limit from the left:

(9)
\begin{align} \lim_{\theta \to 2\pi^-} \frac{\sin x}{1 - \cos x} \\ = \lim_{\theta \to 2\pi^+} \frac{\cos x}{\sin x} \\ = -\infty \end{align}

Since 2π exists as a point on our curve, it thus follows that when theta is 2π, a vertical tangent exists.