Parametric Curves: Derivatives and Tangent Lines

Parametric Derivatives and Tangent Lines

By the end of this lesson, you will be able to find $dy/dx$ and $d^2y/dx^2$ for a parametric curve and find tangent lines.

Slope of a Parametric Curve

For $x = f(t)$ and $y = g(t)$, the slope of the tangent line is:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$$

Valid where $f'(t) \neq 0$.

Example: $x = t^2$, $y = t^3 - 3t$ at $t = 2$.

$\frac{dx}{dt} = 2t = 4$; $\frac{dy}{dt} = 3t^2-3 = 9$

$\frac{dy}{dx} = \frac{9}{4}$

Tangent line at $(4, 2)$: $y - 2 = \frac{9}{4}(x-4)$

Second Derivative

$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right] = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{\frac{dx}{dt}}$$

Example: (continued)

$\frac{dy}{dx} = \frac{3t^2-3}{2t}$

$\frac{d}{dt}\left[\frac{dy}{dx}\right] = \frac{6t(2t)-(3t^2-3)(2)}{(2t)^2} = \frac{12t^2-6t^2+6}{4t^2} = \frac{6t^2+6}{4t^2}$

$\frac{d^2y}{dx^2} = \frac{\frac{6t^2+6}{4t^2}}{2t} = \frac{6t^2+6}{8t^3} = \frac{3(t^2+1)}{4t^3}$

Horizontal and Vertical Tangents

Horizontal: $\frac{dy}{dt} = 0$ (and $\frac{dx}{dt} \neq 0$)

Vertical: $\frac{dx}{dt} = 0$ (and $\frac{dy}{dt} \neq 0$)

Both zero simultaneously: need further analysis (indeterminate form).

Summary

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$; $\frac{d^2y}{dx^2} = \frac{(d/dt)(dy/dx)}{dx/dt}$

Horizontal tangent: $dy/dt = 0$; Vertical: $dx/dt = 0$.

Always evaluate at a specific $t$ value by substituting into the formula.