Parametric, Polar and Vector-Valued Functions

Study Sheet

Parametric, Polar and Vector-Valued Functions

Everything you need for AP Calculus BC, Topic 11

Parametric Equations & Their Derivatives

Concept
Parametric curves and slope

A parametric curve is given by x=f(t)x=f(t) and y=g(t)y=g(t), where the parameter tt (often time) traces out the points (x,y)(x,y). To find the slope of the tangent line, differentiate each coordinate with respect to tt and divide:

dydx=dy/dtdx/dt(provided dx/dt≠0).\dfrac{dy}{dx}=\frac{dy/dt}{dx/dt}\qquad(\text{provided }dx/dt\neq 0).

The second derivative is the tt-derivative of the first derivative, again divided by dx/dtdx/dt:

d2ydx2=ddt ⁣(dydx)dx/dt.\dfrac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{dx/dt}.
Example
Graphing by eliminating the parameter

Sketch x=3cos⁡t, y=2sin⁡tx=3\cos t,\ y=2\sin t, 0≤t≤2π0\le t\le 2\pi. Since cos⁡t=x3\cos t=\tfrac{x}{3} and sin⁡t=y2\sin t=\tfrac{y}{2}, use cos⁡2t+sin⁡2t=1\cos^2t+\sin^2t=1:

x29+y24=1,\frac{x^2}{9}+\frac{y^2}{4}=1,

an ellipse with xx-radius 33 and yy-radius 22, traced counterclockwise starting at (3,0)(3,0).

Example
First and second derivatives

Let x=2t+1, y=t2−tx=2t+1,\ y=t^2-t. Then dxdt=2\dfrac{dx}{dt}=2 and dydt=2t−1\dfrac{dy}{dt}=2t-1, so

dydx=2t−12=t−12.\frac{dy}{dx}=\frac{2t-1}{2}=t-\tfrac12 .

For the second derivative, differentiate dydx\tfrac{dy}{dx} in tt (getting 11) and divide by dxdt=2\tfrac{dx}{dt}=2:

d2ydx2=12>0⇒the curve is concave up everywhere.\frac{d^2y}{dx^2}=\frac{1}{2}>0\quad\Rightarrow\quad\text{the curve is concave up everywhere.}
Tip

Tip: Do not divide dydx\tfrac{dy}{dx} by xx when finding d2ydx2\tfrac{d^2y}{dx^2}. Always differentiate the slope with respect to tt, then divide by dxdt\tfrac{dx}{dt}.

Arc Length of a Parametric Curve

Concept
Arc length

For a smooth curve x=f(t), y=g(t)x=f(t),\ y=g(t) traced once for a≤t≤ba\le t\le b, the arc length is

L=∫ab(dxdt)2+(dydt)2 dt.L=\displaystyle\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt .

The square root is the speed; arc length accumulates distance along the curve.

Example
A computable arc length

Find the length of x=t, y=23t3/2x=t,\ y=\tfrac23 t^{3/2} for 0≤t≤30\le t\le 3. Here dxdt=1\tfrac{dx}{dt}=1 and dydt=t1/2\tfrac{dy}{dt}=t^{1/2}, so

L=∫031+t dt=[23(1+t)3/2]03=23(8−1)=143.L=\int_0^3\sqrt{1+t}\,dt=\Big[\tfrac23(1+t)^{3/2}\Big]_0^3=\tfrac23(8-1)=\frac{14}{3}.
Tip

Tip: If the curve is retraced or the parameter reverses, arc length still adds up (it is never negative), but you must integrate over an interval that traces the curve once.

Vector-Valued Functions & Motion

Concept
Position, velocity, acceleration, speed

A moving particle has position r(t)=⟨x(t), y(t)⟩\mathbf r(t)=\langle x(t),\,y(t)\rangle. Then

velocity v(t)=r′(t)=⟨x′(t), y′(t)⟩,acceleration a(t)=r′′(t)=⟨x′′(t), y′′(t)⟩.\text{velocity }\mathbf v(t)=\mathbf r'(t)=\langle x'(t),\,y'(t)\rangle,\qquad \text{acceleration }\mathbf a(t)=\mathbf r''(t)=\langle x''(t),\,y''(t)\rangle.
speed=∣v(t)∣=(x′(t))2+(y′(t))2,total distance=∫ab∣v(t)∣ dt.\text{speed}=|\mathbf v(t)|=\sqrt{\big(x'(t)\big)^2+\big(y'(t)\big)^2},\qquad \text{total distance}=\int_a^b|\mathbf v(t)|\,dt .
Example
Velocity, acceleration, and speed

Let r(t)=⟨t3−3t, t2⟩\mathbf r(t)=\langle t^3-3t,\ t^2\rangle. Then v(t)=⟨3t2−3, 2t⟩\mathbf v(t)=\langle 3t^2-3,\ 2t\rangle and a(t)=⟨6t, 2⟩\mathbf a(t)=\langle 6t,\ 2\rangle. At t=2t=2:

v(2)=⟨9,4⟩,a(2)=⟨12,2⟩,speed=∣v(2)∣=92+42=97.\mathbf v(2)=\langle 9,4\rangle,\quad \mathbf a(2)=\langle 12,2\rangle,\quad \text{speed}=|\mathbf v(2)|=\sqrt{9^2+4^2}=\sqrt{97}.
Example
Total distance traveled

A particle has position r(t)=⟨t33−t, t2⟩\mathbf r(t)=\big\langle \tfrac{t^3}{3}-t,\ t^2\big\rangle. Then x′=t2−1, y′=2tx'=t^2-1,\ y'=2t, and the integrand is a perfect square:

(t2−1)2+(2t)2=t4+2t2+1=(t2+1)2 ⇒ ∣v(t)∣=t2+1.(t^2-1)^2+(2t)^2=t^4+2t^2+1=(t^2+1)^2\ \Rightarrow\ |\mathbf v(t)|=t^2+1 .

Total distance on 0≤t≤30\le t\le 3:

∫03(t2+1) dt=[t33+t]03=9+3=12.\int_0^3 (t^2+1)\,dt=\Big[\tfrac{t^3}{3}+t\Big]_0^3=9+3=12.
Tip

Tip: Total distance =∫∣v∣ dt=\int|\mathbf v|\,dt (always ≥0\ge 0). Displacement =r(b)−r(a)=\mathbf r(b)-\mathbf r(a) is a vector; its length can be much smaller if the particle doubles back.

Polar Coordinates & Graphs

Concept
Polar↔\leftrightarrowrectangular and a curve catalog

A polar point (r,θ)(r,\theta) satisfies

x=rcos⁡θ,y=rsin⁡θ,r2=x2+y2,tan⁡θ=yx.x=r\cos\theta,\qquad y=r\sin\theta,\qquad r^2=x^2+y^2,\qquad \tan\theta=\frac{y}{x}.

Common polar graphs:

  • [leftmargin=*]
  • r=ar=a: circle of radius aa centered at the origin.
  • r=a±bcos⁡θr=a\pm b\cos\theta or a±bsin⁡θa\pm b\sin\theta: cardioid if a=ba=b, otherwise a limaçon.
  • r=acos⁡(nθ)r=a\cos(n\theta) or asin⁡(nθ)a\sin(n\theta): a rose with nn petals if nn is odd, 2n2n petals if nn is even.
Example
Converting a point and an equation

The point (r,θ)=(2,π3)\big(r,\theta\big)=\big(2,\tfrac{\pi}{3}\big) becomes x=2cos⁡π3=1, y=2sin⁡π3=3x=2\cos\tfrac{\pi}{3}=1,\ y=2\sin\tfrac{\pi}{3}=\sqrt3, i.e. (1,3)(1,\sqrt3). Convert r=4cos⁡θr=4\cos\theta: multiply by rr to get r2=4rcos⁡θr^2=4r\cos\theta, so x2+y2=4xx^2+y^2=4x, i.e. (x−2)2+y2=4(x-2)^2+y^2=4 --- a circle of radius 22 centered at (2,0)(2,0).

Cardioid r=1+cos⁡θr=1+\cos\theta: maximum r=2r=2 at θ=0\theta=0, and r=0r=0 (the pole) at θ=π\theta=\pi.

Tip

Tip: To graph by hand, make a table of θ=0,π6,π4,π3,π2,…\theta=0,\tfrac{\pi}{6},\tfrac{\pi}{4},\tfrac{\pi}{3},\tfrac{\pi}{2},\dots and plot (r,θ)(r,\theta). Watch for r<0r<0: that point is plotted in the opposite direction, at angle θ+π\theta+\pi.

Slope & Area in Polar

Concept
Slope of a polar curve

For r=f(θ)r=f(\theta), write x=rcos⁡θ, y=rsin⁡θx=r\cos\theta,\ y=r\sin\theta and use the parametric rule:

dydx=dy/dθdx/dθ=r′sin⁡θ+rcos⁡θ r′cos⁡θ−rsin⁡θ .\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta} =\frac{r'\sin\theta+r\cos\theta}{\,r'\cos\theta-r\sin\theta\,}.
Example
Slope at a point

For r=1+cos⁡θr=1+\cos\theta at θ=π2\theta=\tfrac{\pi}{2}: r=1r=1 and r′=−sin⁡θ=−1r'=-\sin\theta=-1. Then

dydθ=r′sin⁡θ+rcos⁡θ=(−1)(1)+(1)(0)=−1,dxdθ=r′cos⁡θ−rsin⁡θ=(−1)(0)−(1)(1)=−1,\frac{dy}{d\theta}=r'\sin\theta+r\cos\theta=(-1)(1)+(1)(0)=-1,\quad \frac{dx}{d\theta}=r'\cos\theta-r\sin\theta=(-1)(0)-(1)(1)=-1,

so dydx=−1−1=1\dfrac{dy}{dx}=\dfrac{-1}{-1}=1.

Concept
Area in polar

The area swept by r=f(θ)r=f(\theta) from θ=α\theta=\alpha to θ=β\theta=\beta is

A=12∫αβr2 dθ.A=\frac12\displaystyle\int_\alpha^\beta r^2\,d\theta .

The area between an outer curve r=R(θ)r=R(\theta) and an inner curve r=ρ(θ)r=\rho(\theta) is

A=12∫αβ(R2−ρ2) dθ.A=\frac12\displaystyle\int_\alpha^\beta\Big(R^2-\rho^2\Big)\,d\theta .
Example
Area enclosed by a cardioid

Area inside r=1+cos⁡θr=1+\cos\theta:

A=12∫02π(1+cos⁡θ)2 dθ=12∫02π(1+2cos⁡θ+cos⁡2θ) dθ=12(2π+0+π)=3π2.A=\frac12\int_0^{2\pi}(1+\cos\theta)^2\,d\theta =\frac12\int_0^{2\pi}\big(1+2\cos\theta+\cos^2\theta\big)\,d\theta =\frac12\Big(2\pi+0+\pi\Big)=\frac{3\pi}{2}.

(We used ∫02πcos⁡2θ dθ=π\int_0^{2\pi}\cos^2\theta\,d\theta=\pi.)

Four-petaled rose r=2cos⁡2θr=2\cos 2\theta (n=2n=2 is even, so 2n=42n=4 petals).

Tip

Tip: Choosing the limits α,β\alpha,\beta is the hard part. For one petal of a rose, integrate between two consecutive θ\theta-values where r=0r=0.

Arc Length of a Polar Curve

Concept
Polar arc length

For r=f(θ)r=f(\theta) on α≤θ≤β\alpha\le\theta\le\beta,

L=∫αβ r2+(drdθ)2   dθ.L=\displaystyle\int_\alpha^\beta\sqrt{\,r^2+\left(\frac{dr}{d\theta}\right)^2\,}\;d\theta .
Example
Arc length of a circle

For r=3r=3 (a circle of radius 33), drdθ=0\tfrac{dr}{d\theta}=0, so on 0≤θ≤2π0\le\theta\le 2\pi:

L=∫02π32+02 dθ=∫02π3 dθ=6π,L=\int_0^{2\pi}\sqrt{3^2+0^2}\,d\theta=\int_0^{2\pi}3\,d\theta=6\pi,

which is exactly the circumference 2π(3)2\pi(3).

Tip

Tip: Simplify r2+(r′)2r^2+(r')^2 before integrating. Trig identities (especially 1+cos⁡θ=2cos⁡2θ21+\cos\theta=2\cos^2\tfrac{\theta}{2}) often collapse the square root, as in the cardioid below.

Going Deeper: Advanced Topics

Example
Motion along a curve: speed and total distance

A particle moves with position r(t)=⟨t−sin⁡t, 1−cos⁡t⟩\mathbf r(t)=\langle t-\sin t,\ 1-\cos t\rangle (a cycloid). Then v(t)=⟨1−cos⁡t, sin⁡t⟩\mathbf v(t)=\langle 1-\cos t,\ \sin t\rangle and

∣v∣=(1−cos⁡t)2+sin⁡2t=2−2cos⁡t=2∣sin⁡t2∣.|\mathbf v|=\sqrt{(1-\cos t)^2+\sin^2 t}=\sqrt{2-2\cos t}=2\left|\sin\tfrac{t}{2}\right|.

On one arch 0≤t≤2π0\le t\le 2\pi, sin⁡t2≥0\sin\tfrac{t}{2}\ge0, so the total distance is

∫02π2sin⁡t2 dt=[−4cos⁡t2]02π=−4(−1)−(−4)(1)=8.\int_0^{2\pi}2\sin\tfrac{t}{2}\,dt=\Big[-4\cos\tfrac{t}{2}\Big]_0^{2\pi}=-4(-1)-(-4)(1)=8 .
Example
Polar area between two curves (find the intersection first)

Find the area inside r=3sin⁡θr=3\sin\theta and outside r=1+sin⁡θr=1+\sin\theta. Set them equal:

3sin⁡θ=1+sin⁡θ ⇒ sin⁡θ=12 ⇒ θ=π6, 5π6.3\sin\theta=1+\sin\theta\ \Rightarrow\ \sin\theta=\tfrac12\ \Rightarrow\ \theta=\tfrac{\pi}{6},\ \tfrac{5\pi}{6}.

Between these angles 3sin⁡θ3\sin\theta is the outer curve, so

A=12∫π/65π/6[(3sin⁡θ)2−(1+sin⁡θ)2]dθ=12∫π/65π/6(3−4cos⁡2θ−2sin⁡θ) dθ.A=\frac12\int_{\pi/6}^{5\pi/6}\Big[(3\sin\theta)^2-(1+\sin\theta)^2\Big]d\theta =\frac12\int_{\pi/6}^{5\pi/6}\big(3-4\cos 2\theta-2\sin\theta\big)\,d\theta .

An antiderivative is 3θ−2sin⁡2θ+2cos⁡θ3\theta-2\sin 2\theta+2\cos\theta; evaluating gives 12(5π2−π2)=π\tfrac12\big(\tfrac{5\pi}{2}-\tfrac{\pi}{2}\big)=\pi.

Example
Tangent lines at the pole

When a polar curve reaches the pole (r=0r=0) at θ=θ0\theta=\theta_0, the tangent line there is simply the line θ=θ0\theta=\theta_0. For the rose r=cos⁡2θr=\cos 2\theta, r=0r=0 when 2θ=π2+kπ2\theta=\tfrac{\pi}{2}+k\pi, i.e. θ=π4,3π4,5π4,7π4\theta=\tfrac{\pi}{4},\tfrac{3\pi}{4},\tfrac{5\pi}{4},\tfrac{7\pi}{4}. These four lines are the tangents to the petals at the origin.

Example
Converting a harder polar equation

Convert r=6cos⁡θ+4sin⁡θr=6\cos\theta+4\sin\theta. Multiply by rr:

r2=6rcos⁡θ+4rsin⁡θ ⇒ x2+y2=6x+4y.r^2=6r\cos\theta+4r\sin\theta\ \Rightarrow\ x^2+y^2=6x+4y .

Complete the square: (x−3)2+(y−2)2=13(x-3)^2+(y-2)^2=13, a circle centered at (3,2)(3,2) with radius 13\sqrt{13}.

Example
Parametric second derivative & concavity

Let x=t2+1, y=t3−3tx=t^2+1,\ y=t^3-3t. Then dxdt=2t\tfrac{dx}{dt}=2t, dydt=3t2−3\tfrac{dy}{dt}=3t^2-3, and

dydx=3t2−32t=32(t−1t).\frac{dy}{dx}=\frac{3t^2-3}{2t}=\frac32\Big(t-\tfrac1t\Big).

Differentiate in tt: ddt(dydx)=32(1+1t2)\tfrac{d}{dt}\big(\tfrac{dy}{dx}\big)=\tfrac32\big(1+\tfrac1{t^2}\big). Divide by dxdt=2t\tfrac{dx}{dt}=2t:

d2ydx2=32(1+1t2)2t=3(t2+1)4t3.\frac{d^2y}{dx^2}=\frac{\tfrac32\big(1+\tfrac1{t^2}\big)}{2t}=\frac{3(t^2+1)}{4t^3}.

Since 3(t2+1)>03(t^2+1)>0, the sign matches t3t^3: concave up for t>0t>0, concave down for t<0t<0.

Example
Arc length of a cardioid

For r=1+cos⁡θr=1+\cos\theta,  r′=−sin⁡θ\ r'=-\sin\theta, and

r2+(r′)2=(1+cos⁡θ)2+sin⁡2θ=2+2cos⁡θ=4cos⁡2θ2.r^2+(r')^2=(1+\cos\theta)^2+\sin^2\theta=2+2\cos\theta=4\cos^2\tfrac{\theta}{2}.

So r2+(r′)2=2∣cos⁡θ2∣\sqrt{r^2+(r')^2}=2\big|\cos\tfrac{\theta}{2}\big|, and the full perimeter is

L=∫02π2∣cos⁡θ2∣ dθ=8.L=\int_0^{2\pi}2\big|\cos\tfrac{\theta}{2}\big|\,d\theta=8 .
Tip

Big picture: Parametric, vector, and polar problems all reduce to the same toolkit --- differentiate each coordinate in the parameter, then combine. Slopes divide by dx/dtdx/dt (or dx/dθdx/d\theta); lengths and distances integrate a square-root speed; polar areas integrate 12r2\tfrac12 r^2. Set up carefully, keep answers exact, and always ask whether the interval traces the curve once.

Formulas, Proofs & Tips

Tip
Parametric derivatives and arc length
dydx=dy/dtdx/dt,L=∫ab(dxdt)2+(dydt)2 dt\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad L=\int_a^b\sqrt{\left(\tfrac{dx}{dt}\right)^{2}+\left(\tfrac{dy}{dt}\right)^{2}}\,dt

What it means. Slope is the ratio of the two rates; length sums tiny hypotenuses.

Example. x=t2x=t^2, y=t3y=t^3: dydx=3t22t=3t2\tfrac{dy}{dx}=\tfrac{3t^2}{2t}=\tfrac{3t}{2}.

Why it works. By the chain rule dydt=dydx⋅dxdt\tfrac{dy}{dt}=\tfrac{dy}{dx}\cdot\tfrac{dx}{dt}, so dividing isolates dydx\tfrac{dy}{dx}. A short step has horizontal part dxdx and vertical part dydy, so its length is dx2+dy2\sqrt{dx^2+dy^2} by Pythagoras — factor out dtdt and integrate.

Tip. This needs dxdt≠0\tfrac{dx}{dt}\neq 0; where it vanishes the curve has a vertical tangent.