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
xy−11234512t = 1.5
The curve x=2t+1x=2t+1, y=t2−ty=t^2-t; at t=1.5t=1.5 the tangent (dashed) has slope dy/dtdx/dt=22=1\tfrac{dy/dt}{dx/dt}=\tfrac{2}{2}=1.

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}.

Over a short time dtdt the point moves dx=x′(t) dtdx=x'(t)\,dt across and dy=y′(t) dtdy=y'(t)\,dt up, so the direction it moves has slope dydx=y′(t) dtx′(t) dt\dfrac{dy}{dx}=\dfrac{y'(t)\,dt}{x'(t)\,dt}; the dtdt cancels. When x′(t)=0x'(t)=0 and y′(t)≠0y'(t)\neq0 the point is moving straight up or down, which is a vertical tangent.

Example
Graphing by eliminating the parameter
t = 0
Counterclockwise from (3,0)(3,0), once around as tt runs over [0,2π][0,2\pi].

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).

Side note
Where it goes wrong

Computing polar area as ∫r dθ\int r\,d\theta. The sector formula has 12r2\tfrac12r^2: a wedge of angle dθd\theta and radius rr has area 12r2dθ\tfrac12r^2d\theta.

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}.

Try it
Quick check

Eliminate the parameter to write yy as a function of xx:  x=t+2, y=t2\ x=t+2,\ y=t^2.

Work it on paper before revealing the answer.

Show the answer

Answer: y=(x−2)2y=(x-2)^2

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.

Try it
Quick check

Find the length of x=t2x=t^2, y=23t3y=\tfrac23t^3 for 0≤t≤80\le t\le\sqrt8.

Show the answer

Answer: 523\dfrac{52}{3}. x′=2tx'=2t, y′=2t2y'=2t^2, so the speed is 4t2+4t4=2t1+t2\sqrt{4t^2+4t^4}=2t\sqrt{1+t^2}. Then ∫082t1+t2 dt=[23(1+t2)3/2]08=23(27−1)\displaystyle\int_0^{\sqrt8}2t\sqrt{1+t^2}\,dt=\left[\tfrac23(1+t^2)^{3/2}\right]_0^{\sqrt8}=\tfrac23(27-1).

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.

Try it
Quick check

For r(t)=⟨t2, t3⟩\mathbf r(t)=\langle t^2,\ t^3\rangle, find the velocity, speed and acceleration at t=1t=1.

Show the answer

Answer: v(1)=⟨2,3⟩\mathbf v(1)=\langle2,3\rangle, speed 13\sqrt{13}, a(1)=⟨2,6⟩\mathbf a(1)=\langle2,6\rangle. Differentiate each component: v=⟨2t,3t2⟩\mathbf v=\langle2t,3t^2\rangle and a=⟨2,6t⟩\mathbf a=\langle2,6t\rangle.

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:

  • 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.

Try it
Quick check

Convert r=6sin⁡θr=6\sin\theta to rectangular form and describe it.

Show the answer

Answer: x2+(y−3)2=9x^2+(y-3)^2=9, the circle of radius 33 centred at (0,3)(0,3). Multiply by rr: x2+y2=6yx^2+y^2=6y, then complete the square.

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
At θ=π2\theta=\tfrac{\pi}{2} the cardioid passes (0,1)(0,1) with tangent slope 11 (dashed).

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 .

Polar regions are swept out by a turning ray, so the natural slices are thin wedges, not rectangles. A wedge of angle dθd\theta and radius rr is almost a circular sector, with area 12r2 dθ\tfrac12r^2\,d\theta, and adding them gives 12∫r2 dθ\tfrac12\int r^2\,d\theta. The limits are the angles at which the ray starts and stops sweeping the region.

Example
Area enclosed by a cardioid
Thin wedges 12r2 dθ\tfrac12r^2\,d\theta fill the cardioid: total area 3π2\tfrac{3\pi}{2}.

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.

Try it
Quick check

Find the area of one petal of r=sin⁡3θr=\sin3\theta.

Show the answer

Answer: π12\dfrac{\pi}{12}. One petal runs from θ=0\theta=0 to θ=π3\theta=\tfrac{\pi}{3} (where rr returns to 00). 12∫0π/3sin⁡23θ dθ=12⋅12∫0π/3(1−cos⁡6θ) dθ=14⋅π3\tfrac12\displaystyle\int_0^{\pi/3}\sin^23\theta\,d\theta=\tfrac12\cdot\tfrac12\int_0^{\pi/3}(1-\cos6\theta)\,d\theta=\tfrac14\cdot\tfrac{\pi}{3}.

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.

Try it
Quick check

Find the length of r=4sin⁡θr=4\sin\theta for 0≤θ≤π0\le\theta\le\pi.

Show the answer

Answer: 4π4\pi. r2+(r′)2=16sin⁡2θ+16cos⁡2θ=16r^2+(r')^2=16\sin^2\theta+16\cos^2\theta=16, so the integrand is 44 and the length is 4π4\pi: the circumference of the circle of diameter 44 that this equation traces once.

Going Deeper: Advanced Topics

Example
Motion along a curve: speed and total distance
One arch of the cycloid traced by a point on a rolling circle of radius 11; its length is 88.

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)
Inside the circle r=3sin⁡θr=3\sin\theta, outside the cardioid r=1+sin⁡θr=1+\sin\theta, between θ=π6\theta=\tfrac{\pi}{6} and 5π6\tfrac{5\pi}{6}.

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.

Try it
Quick check

Set up but do not evaluate the arc-length integral for x=t2, y=t3x=t^2,\ y=t^3 on 0≤t≤20\le t\le 2.

Work it on paper before revealing the answer.

Show the answer

Answer: ∫02(2t)2+(3t2)2 dt=∫024t2+9t4 dt\displaystyle\int_0^2\sqrt{(2t)^2+(3t^2)^2}\,dt=\int_0^2\sqrt{4t^2+9t^4}\,dt

Why It Works

Parametric slope: by the chain rule dydt=dydx⋅dxdt\tfrac{dy}{dt} = \tfrac{dy}{dx}\cdot\tfrac{dx}{dt}, so dydx=dy/dtdx/dt\tfrac{dy}{dx} = \tfrac{dy/dt}{dx/dt} wherever dx/dt≠0dx/dt \ne 0. Arc length is the integral of speed because a tiny piece of the path is nearly straight, with length dx2+dy2=x′2+y′2 dt\sqrt{dx^2 + dy^2} = \sqrt{x'^2 + y'^2}\,dt. The polar area formula integrates the areas of thin circular sectors, each 12r2dθ\tfrac12r^2d\theta, exactly as the disk method integrates thin cylinders.

Reminder — The differentiation rules:  (xn)′=nxn−1,(fg)′=f′g+fg′,(fg)′=f′g−fg′g2,(f(g(x)))′=f′(g(x))g′(x)(x^{n})'=nx^{n-1},\quad (fg)'=f'g+fg',\quad \left(\tfrac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}},\quad \big(f(g(x))\big)'=f'(g(x))g'(x)
Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

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.