Objectives
- To understand the geometric interpretation of the derivative in the context of parametric equations
- To compute where horizontal and vertical tangent lines occur on the graphs of parametric equations
Section 1.2 Derivatives of Parametric Equations
Subsection 1.2.1 Derivatives of Parametric Equations
The slope of a tangent line for a parametric equation \(c(t)=(x(t), y(t))\) can be found via
\begin{equation}
\frac{y'(t)}{x'(t)} = \frac{dy/dt}{dx/dt} = \frac{dy}{dt} \frac{dt}{dx} = \frac{dy}{dx}.\tag{1.2.1}
\end{equation}
Activity 1.2.1.
For this activity, we will be examining the parametric equations given below:
\begin{equation*}
c(t) = (t^2-9, 8t-t^3).
\end{equation*}
(a)
Write a sentence or two about why a horizontal line (in 2D) has a slope of zero.
(b)
Write a sentence or two about why a vertical line (in 2D) has an undefined slope.
(c)
Explain why the curve plotted in Figure 1.2.1 can’t be written with \(y\) as a function of \(x\) (and thus our first semester calculus ideas do not apply).
(d)
Use the slider to move the point around on the curve given by \(c(t)\text{.}\) Find the two values of \(t\) where the tangent line to the curve is horizontal.
(e)
Use the slider to move the point around on the curve given by \(c(t)\text{.}\) Find all values of \(t\) where the tangent line to the curve is vertical.
(f)
Find \(\frac{dy}{dx}\) (as a function of \(t\)) using the equation at the beginning of this section. Use Equation (1.2.1) to find the values where the tangent line is vertical and where the tangent line is horizontal. Compare your answers to the values you found in Task 1.2.1.d and Task 1.2.1.e.
(g)
Find the parametric equation of the tangent line when \(t=2\) and graph your parametric equation (for the line) on the plot above to confirm your result.
