Parametric Equations, Polar Coordinates, and Vector-Valued Functions
Core ideas
Parametric curves describe $x(t)$ and $y(t)$ separately; $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ when $\frac{dx}{dt}\neq0$. Second derivatives require differentiating the slope with respect to $t$ and dividing by $\frac{dx}{dt}$ again. Arc length is $\int\sqrt{(x')^2+(y')^2},dt$.
Polar curves use $r=f(\theta)$. Convert with $x=r\cos\theta$, $y=r\sin\theta$. Area swept is $\frac{1}{2}\int r^2,d\theta$; the slope of a polar tangent is $\frac{dy/d\theta}{dx/d\theta}$. Vector-valued functions $\vec{r}(t)$ have derivative $\vec{r}'(t)$ giving velocity, with speed $|\vec{r}'(t)|$.
Traps
- Forgetting that $\frac{dy}{dx}$ in parametric form is a ratio of derivatives, not $\frac{d}{dt}(y/x)$.
- Computing the second parametric derivative as $\frac{d^2y/dt^2}{d^2x/dt^2}$ - incorrect; divide the derivative of the slope by $\frac{dx}{dt}$.
- Omitting the $\frac{1}{2}$ in polar area or squaring incorrectly.
- Missing limits of integration for polar regions; find where the curve closes or intersects.
Worked example
For $x=t^2$, $y=t^3$, $\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}$. The second derivative is $\frac{d}{dt}\left(\frac{3t}{2}\right)\div\frac{dx}{dt}=\frac{3}{2}\div2t=\frac{3}{4t}$.