1.8 Parametric and Polar Curves

1.8 Parametric and Polar Curves

A curve may be easier to describe by a parameter or by distance and angle than by a single Cartesian equation. These representations are central to calculus applications and coordinate geometry.

Parametric curves

A parametric curve is $x=x(t)$, $y=y(t)$ for $t$ in an interval. Different parameter values may trace the same point, and increasing $t$ determines orientation. When $dx/dt\ne0$, $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.$ A horizontal tangent has $dy/dt=0$ and $dx/dt\ne0$; a vertical tangent has $dx/dt=0$ and $dy/dt\ne0$. The second derivative is not merely a quotient of second derivatives: $\frac{d^2y}{dx^2}=\frac{d}{dt}(dy/dx)\bigg/\frac{dx}{dt}.$

For $x=t^2-1$, $y=t^3-3t$, the slope is $(3t^2-3)/(2t)$. At $t=1$, the point is $(0,-2)$ and the tangent is horizontal. At $t=0$, $dx/dt=0$ while $dy/dt=-3$, giving a vertical tangent at $(-1,0)$.

Parametric arc length on $a\le t\le b$ is $L=\int_a^b\sqrt{(x'(t))^2+(y'(t))^2}\,dt.$ Signed area under the oriented curve is $\int y,dx=\int_a^b y(t)x'(t),dt$; geometric area may require reversing sign or splitting the interval. Surface areas follow $2\pi\int y,ds$ about the $x$-axis and $2\pi\int x,ds$ about the $y$-axis when the radius is nonnegative.

Polar coordinates

Polar coordinates represent a point by $(r,\theta)$ with $x=r\cos\theta,\qquad y=r\sin\theta,\qquad r^2=x^2+y^2.$ The representation is not unique: $(r,\theta+2k\pi)$ is the same point, and $(-r,\theta)$ equals $(r,\theta+\pi)$. When converting an angle from Cartesian coordinates, use the quadrant rather than relying only on $\tan\theta=y/x$.

Common curves include circles $r=2a\cos\theta$ centered at $(a,0)$ with radius $|a|$, cardioids $r=a(1\pm\cos\theta)$ or $a(1\pm\sin\theta)$, and roses $r=a\cos(n\theta)$ or $a\sin(n\theta)$. A rose has $n$ petals when $n$ is odd and $2n$ petals when $n$ is even.

Polar calculus

For $r=r(\theta)$, differentiate $x=r\cos\theta$ and $y=r\sin\theta$: $\frac{dy}{dx}=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}.$ The area swept from $\theta=\alpha$ to $\theta=\beta$ is $A=\frac12\int_\alpha^\beta r(\theta)^2\,d\theta,$ and arc length is $L=\int_\alpha^\beta\sqrt{r^2+(r')^2}\,d\theta.$ For area between curves, use $\tfrac12\int(r_{outer}^2-r_{inner}^2)d\theta$ only on intervals where the chosen outer curve truly lies farther from the pole.

Symmetry and efficient bounds

A polar equation is symmetric about the polar axis if replacing $\theta$ by $-\theta$ leaves it unchanged; symmetric about the vertical axis if $\theta$ may be replaced by $\pi-\theta$; and symmetric about the pole if replacing $r$ by $-r$ or $\theta$ by $\theta+\pi$ preserves it. These tests are sufficient, not necessary, because polar representations are nonunique.

Worked example: one petal of a rose

For $r=2\cos(3\theta)$, the petal centered on the positive $x$-axis is traced where $r\ge0$ between consecutive zeros $\theta=-\pi/6$ and $\pi/6$. Its area is $A=\frac12\int_{-\pi/6}^{\pi/6}4\cos^2(3\theta)\,d\theta=2\int_{-\pi/6}^{\pi/6}\frac{1+\cos6\theta}{2}\,d\theta=\frac{\pi}{3}.$ Because 3 is odd, the full rose has three petals and total area $\pi$.

Worked example: cycloid slope

For $x=a(t-\sin t)$ and $y=a(1-\cos t)$, $\frac{dy}{dx}=\frac{a\sin t}{a(1-\cos t)}=\cot(t/2).$ At $t=\pi$, the slope is zero. At $t=0$, both derivatives vanish, so the ordinary quotient test is inconclusive; local expansion reveals a cusp.

Common traps

A zero numerator and denominator in the slope formula requires a limit, not an automatic horizontal or vertical classification. Negative $r$ reverses direction. Squaring $r$ in the area formula removes its sign but does not fix incorrect tracing bounds. For the second parametric derivative, remember the final division by $dx/dt$.