# Determining the Derivative of a Parametric Equation

Recall that from the page Derivatives for Parametric Curves, that the derivative of a parametric curve defined by $x = x(t)$ and $y = y(t)$, $\frac{dy}{dx}$ is as follows:

(1)## Example 1

**Determine $\frac{dy}{dx}$ given the parametric curve defined by $x = t^2 - 3$ and $y = 4t^3$.**

We will apply our formula:

(2)First let's determine $\frac{dy}{dt}$ and $\frac{dx}{dt}$:

(3)Thus the derivative is: $\frac{dy}{dx} = \frac{2t}{12t^2} = \frac{1}{6t}$

# Calculating Horizontal and Vertical Tangents with Parametric Curves

Recall that with functions, it was very rare to come across a vertical tangent. With parametric curves, vertical tangents are more prominent. Nevertheless, we will look at the techniques for finding both horizontal and vertical tangents of a parametric curve $C$.

## Horizontal Tangent of Parametric Curves

To calculate horizontal tangents, we let $\frac{dy}{dx} =$, which is only possible when $\frac{dy}{dt} = 1$. Note that we must ensure that $\frac{dx}{dt} ≠ 0$. If the derivative of $x$ with respect to $t$ is $0$, then we must apply limit techniques to evaluate what is happening at that point.

For example, let's determine the coordinates of the horizontal tangents defined by the parametric equations $x = 6t^3$ and $y = \sin t$.

When we use our differentiation property we obtain $\frac{dy}{dx}$ in the following manner:

(4)And let's now set our derivative equal to $0$ to obtain:

(5)We are going to ignore multiples of $\frac{\pi}{2}$ now. Thus we obtain a vertical tangent when $t = \frac{-\pi}{2}, \frac{\pi}{2}$. Note that we must check $t = -\frac{\pi}{2}$ and $t = \frac{\pi}{2}$ to ensure that the denominator does not equal $0$. Clearly it does not:

(6)Thus since $\frac{dx}{dt} ≠ 0$, then we can say that there exists a horizontal tangent when $t = -\frac{\pi}{2}, \frac{pi}{2}$. We can find the specific coordinates by plugging these values of $t$ into our parametric equations. We will only look at the coordinates of the tangent when $t = \frac{\pi}{2}$ for this example:

(7)Thus we have a horizontal tangent at $t = \frac{\pi}{2}$ with coordinates $(23.25, 1)$. The graph below shows the location of this horizontal asymptote.

## Vertical Tangents with Parametric Curves

We will continue the analysis of our parametric curve defined by $x = 6t^3$ and $y = \sin t$. A vertical tangent arises when $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} ≠ 0$, essentially the opposite conditions for horizontal tangents.

Let's look at where $\frac{dx}{dt} = 0$.

(8)Note that when $t = 0$, $\frac{dy}{dt} ≠ 0$, so we have a vertical tangent at $t = 0$. To find the specific coordinates, we can plug back into our parametric equations like before.

(9)Thus there is a vertical tangent at $(0, 0)$, which should be evident from the graph of this parametric curve from earlier.

# Exceptions

So what are we supposed to do if when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} = 0$? 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 $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ to obtain $\frac{dy}{dx}$:

(10)Therefore:

(11)So $\frac{dx}{d\theta} = 0$ when $\theta = 0, \pi, 2\pi, ...$. But notice that when $\theta$ is one of these values, then $\frac{dy}{d\theta} = 0$ since $\sin 0 = \sin \pi = ... = 0$. We need to use our knowledge of limits in order to figure out what is happening here. Let's just use $2 \pi$ as an example,

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

(13)Now let's look at the limit from the left:

(14)Since $2 \pi$ exists as a point on our curve, it thus follows that when $\theta = 2 \pi$, a vertical tangent exists.