AP Calculus BC / Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions
9.1 Defining and Differentiating Parametric Equations
Notes
A parametric curve gives $x=x(t)$ and $y=y(t)$.
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\quad(dx/dt\ne0)$$
- Horizontal tangent: $\displaystyle \frac{dy}{dt}=0$, $\displaystyle \frac{dx}{dt}\ne0$. Vertical tangent: $\displaystyle \frac{dx}{dt}=0$, $\displaystyle \frac{dy}{dt}\ne0$.
Practice
$x=t^2$, $y=t^3$. Find $\displaystyle \frac{dy}{dx}$.
$\displaystyle \frac{3t^2}{2t}$.
$x=\cos t$, $y=\sin t$. The slope at $\displaystyle t=\frac{\pi}{4}$ is:
$\displaystyle \frac{\cos t}{-\sin t}=-1$.
A parametric curve has a vertical tangent when:
The slope's denominator is 0.
Ready for more? Take the Unit 9 test.