how to find horizontal tangent

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)

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

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)

\begin{align} \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \end{align}

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

(3)

\begin{align} \frac{dy}{dt} = 2t \quad , \quad \frac{dx}{dt} = 12t^2 \end{align}

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)

\begin{align} \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos t}{18t^2} \end{align}

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

(5)

\begin{align} \frac{dy}{dx} = 0 = \frac{\cos t}{18t^2} \\ 0 = \cos t \\ -\frac{\pi}{2}, \frac{\pi}{2}= t \end{align}

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)

\begin{align} \ 18 (π/2)^2 ≠ 0 \end{align}

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)

\begin{align} x = 6(\frac{\pi}{2})^3 \approx 23.25 \\ y = \sin(\frac{\pi}{2}) = 1 \end{align}

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.

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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)

\begin{align} 18t^2 = 0 \\ t = 0 \end{align}

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)

\begin{align} x = 6(0)^3 = 0 \\ y = \sin(0) = 0 \end{align}

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)

\begin{align} \quad \frac{dx}{d\theta} = r - r\cos \theta \quad \quad \quad \frac{dy}{d\theta} = r\sin \theta \end{align}

Therefore:

(11)

\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 $\frac{dx}{d\theta} = 0$ 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)

\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:

(13)

\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:

(14)

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