Derivatives of Parametric Equations (College Board AP® Calculus BC): Revision Note

You may be asked to find the equation of the line tangent to a parametric curve at a particular point, which requires calculating the parametric first derivative.

If asked to find any local maxima, minima or points of inflection on a parametric curve, start by finding all the points at which the tangent is horizontal (then investigate further, e.g. with a sketch or second-derivative method).

A curve is given parametrically by

$\begin{array}{rcl}x& =& t^3-12t\\ y& =& t^3+3t\end{array}$

Part of the curve is shown below.

(a) Find the slope of the line that is tangent to the curve at $t=3$.

(b) Find the values of $t$ at which the tangent to the curve is vertical. Find the equations of these tangents.

(c) Show that the curve has no horizontal tangents.

Answer:

(a)

You need to use the parametric first derivative, $\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}$

First, differentiate the parametric equations individually

$\begin{array}{rcl}\frac{dx}{dt}& =& 3t^2-12\\ \frac{dy}{dt}& =& 3t^2+3\end{array}$

Then substitute these derivatives into the formula, with $\frac{dy}{dt}$ on the top and $\frac{dx}{dt}$ on the bottom

$\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{3t^2+3}{3t^2-12}$

Substitute in the parameter $t=3$ and simplify

$\frac{dy}{dx}=\frac{3\left(3^2\right)+3}{3\left(3^2\right)-12}=\frac{30}{15}=2$

The slope of the line that is tangent to the curve at $t=3$ is 2

(b)

Vertical tangents are points when $\frac{dx}{dt}=0$ and $\frac{dy}{dt}\ne 0$

Set $\frac{dx}{dt}=0$ and solve

$\begin{array}{rcl}3t^2-12& =& 0\\ t^2& =& 4\\ t& =& \pm 2\end{array}$

Check that $\frac{dy}{dt}\ne 0$ by substituting $t=\pm 2$ into $\frac{dy}{dt}$

$\frac{dy}{dt}=3{\left(\pm 2\right)}^2+3=15\ne 0$

To find the equations of the vertical tangents, remember that vertical lines have the equations $x=k$ where $k$ is a constant

Substitute $t=\pm 2$ into $x=t^3-12t$

$2^3-12\left(2\right)=-16$ and ${\left(-2\right)}^3-12\left(-2\right)=16$

The point at $t=2$ has a vertical tangent with equation $x=-16$

The point at $t=-2$ has a vertical tangent with equation $x=16$

(c)

Horizontal tangents are points when $\frac{dy}{dt}=0$ and $\frac{dx}{dt}\ne 0$

Set $\frac{dy}{dt}=0$ and try to solve

$\begin{array}{rcl}3t^2+3& =& 0\\ t^2& =& -1\end{array}$

There are no real solutions to this equation

There is no value of $t$ for which $\frac{dy}{dt}=0$ so no horizontal tangents