Parametric Equations, Polar Coordinates, and Vector-Valued Functions

Parametric equations 参数方程 give $x$ and $y$ each as functions of a parameter $t$ (often time): $x=x(t)$, $y=y(t)$. They trace a curve that need not be a function of $x$. The slope of the curve is found with the chain rule:

Worked example. For $x=t^2$, $y=t^3-t$, find the slope at $t=2$. Here $\dfrac{dx}{dt}=2t$ and $\dfrac{dy}{dt}=3t^2-1$, so $\dfrac{dy}{dx}=\dfrac{3t^2-1}{2t}$; at $t=2$ this is $\dfrac{11}{4}$.

The second derivative is not $\dfrac{d^2y/dt^2}{d^2x/dt^2}$. Instead, differentiate the first derivative with respect to $t$, then divide by $dx/dt$ again:

Use it to test concavity of a parametric curve.

The length of a parametric curve for $t$ from $a$ to $b$ is

This is the parametric version of arc length – summing tiny hypotenuses $\sqrt{dx^2+dy^2}$ over the parameter.

A vector-valued function 向量值函数 $\vec{r}(t)=\langle x(t),\,y(t)\rangle$ gives a position vector for each $t$. Differentiate component-wise: the velocity is $\vec{v}(t)=\langle x'(t),\,y'(t)\rangle$ and the acceleration is $\vec{a}(t)=\langle x''(t),\,y''(t)\rangle$. The speed is the magnitude $|\vec{v}|=\sqrt{x'^2+y'^2}$.

A vector-valued line: start at a, then slide by t lots of the direction b

Integrate a vector function component by component. Given acceleration or velocity plus an initial condition, integrate each component and use the condition to find the constants – recovering velocity from acceleration, or position from velocity.

For a particle moving in a plane: position is $\langle x(t),y(t)\rangle$, velocity and acceleration are its derivatives, speed is $|\vec v|$, and the distance traveled over $[a,b]$ is

Exam skill: these plane-motion problems appear on the BC free-response nearly every year – be fluent finding speed, the position at a later time (initial point plus the integral of velocity), and total distance.

Worked example. A particle has position $\langle t^2,\ t^3-t\rangle$. Its velocity is $\langle 2t,\ 3t^2-1\rangle$, so at $t=1$ the velocity is $\langle 2,\ 2\rangle$ and the speed is $\sqrt{2^2+2^2}=2\sqrt{2}$. Its position at $t=2$ is $\langle 4,\ 6\rangle$.

Polar coordinates 极坐标 locate a point by its distance $r$ from the origin and angle $\theta$: convert with $x=r\cos\theta$, $y=r\sin\theta$. A polar curve $r=f(\theta)$ is a parametric curve in $\theta$, so its slope is

Polar coordinates give a point by its distance r and angle theta

The area swept out by a polar curve $r=f(\theta)$ from $\alpha$ to $\beta$ is

The region is a "fan" of thin triangular sectors; choosing the correct $\theta$-limits (where the curve starts and finishes tracing the region) is the main challenge.

Worked example. Find the area of one petal of the rose $r=2\sin(2\theta)$ (traced for $\theta$ from $0$ to $\tfrac{\pi}{2}$). Using $\sin^2 u=\tfrac12(1-\cos 2u)$,

The shaded region is a fan of thin sectors swept from $\alpha$ to $\beta$; each has area $\tfrac12 r^2\,d\theta$, so the total is $\tfrac12\int_\alpha^\beta r^2\,d\theta$.

For the area between an outer curve $r_1$ and an inner curve $r_2$, subtract the sectors:

Find the intersection angles first (set $r_1=r_2$), and be careful which curve is outer over each interval – they can swap.

• For parametric curves, $\tfrac{dy}{dx}=\tfrac{dy/dt}{dx/dt}$; speed is $\sqrt{(dx/dt)^2+(dy/dt)^2}$.
• A vector-valued function carries the same information — differentiate/integrate it component by component.
• In polar, convert with $x=r\cos\theta$, $y=r\sin\theta$; area swept is $\tfrac12\int r^2\,d\theta$.
• Watch the direction of tracing (the sign of $dx/dt$) and set correct $\theta$-limits for polar area.
• Eliminate the parameter to recover the ordinary $y$-vs-$x$ shape when it helps.