MATH 4B - Lecture, Chapter 11

1.8 [Review] Parametric Equations

Both x and y are functions of a third variable t

Tasks:

Table of values (Ex: x=t²-t, y=t³-2t, t=-2..2)

t	x	y
-2	6	-4
-1	2	1
0	0	0
1	0	-1
2	2	4

Plot points

Indicate orientation

Note domain -- depends on t

Eliminate parameter

Solve one for t and sub (Ex: x=3t, y=t^2+4t-3)

Ex above would require quadratic formula

Solve both for t and equate Ex: x=ln t, y=e^t

Solve both for the same func of t: Ex: x=sin(e^t), y=sin²(e^t)

Use an identity

cos^2 + sin^2

sec^2 - tan^2

cosh^2 - sinh^2

How to graph with calculator Ex: x=3sec t, y=2tan t, std zoom

Overnight: try x=5cos t, y=5sin t, zSqr, t=0,,10p, tStep p/2, then p/4, 2p/5, 4p/5 Why?

How to graph with Maple: plot([3*sec(t),4*tan(t),t=0..2*Pi],x=-10..10, y=-10..10);

(first try it without specifying the ranges)

Finding parametric equations

From y=f(x) or x=f(y) Let t = independent variable, sub and solve for the other var.

Know: circle, ellipse, hyperbola (2 forms), line

x-h=a cos t, y-k=a sin t

x-h=a cos t, y-k=b sin t

x-h=a cosh t, y-k=b sinh t

x-h=a sec t, y-k=b tan t

x-h=a f(t), y-k=b f(t); either x or y may be a constant (vert or horiz line)

Modeling special curves --

Ex: cycloid: curve formed by point on rim of wheel of radius a

Finding the point of intersection when a curve loops:

Ex: x = t²-t, y=t³-2t (first example above):

x-coords are the same for two different values of the parameter (so use different letters for t):

(same thing is true for y)

solve u²-u = v²-v and solve u³-2u = v³-2v to get values where u ¹ v

factor by grouping to get: 1st eqn: v=1-u, 2nd eqn v²+uv+u²-2=0

sub to get u=(1±sqrt(5))/2 ~ 1.618 or -1.618. This is t

sub to get x and y: (1,1) is the intersection

Supplementary topic:

Areas: A = int(y*dx,a_x,b_x) = int(y(t)*x'(t)dt,a_t,b_t) (subscripts indicate endpoints in terms of x or t)

Orientation is still the same as for the original integral

App: area inside a loop: take two areas from left end of loop to right end of loop,

subtract top area minus bottom area

Our ex: A = int (t³-2t)(2t-1),t,1/2,(1-sqrt5)/2) - int ((t³-2t)(2t-1),t,1/2,(1+sqrt5)/2).

11.1 Polar Coordinates

Conversions polar to rectangular and vice versa

x=rcosq, y=rsinq, r²=x²+y², tanq=y/x (but pay attn to quadrant)

Points

Ex: (4p/3,6) to rectangular

(-6,-2sqrt(3)) to polar

Note: polar form is not unique; negative radius -- examples above

Equations

Convert: r=3sinq-2cosq, r=2secq, r=q (Archimedian spiral)

y=5x², y=x/sqrt(3)

Note trick of factoring (in or out) a factor of r

Graphs

Cardioid: Ex: r=3-3sinq

If coeffs differ, limaçon (loops and beans)

Rose curves: Ex: r=cos 3q Relate nr of leaves to coeff of q (even vs odd)

Lemniscate (propeller): r² = a cos 2θ

11.2 Tangent Lines and Arc Length for Parametric and Polar Curves

Parametric curves:

See review (above) for Lesson 1.8

Tangent lines

dy/dx = (dy/dt) / (dx/dt) Note that the result is in terms of t, not x or y.

Ex: find the slope of the tangent line to the cycloid x = at - a sin(t), y = a - a cos(t)

dy/dx = sin(t) / (1 - cos(t)) Note: independent of 'a'
Apps: when would the curve have a horizontal tangent? a vertical tangent?

Caution: if both derivs are 0, the curve has a cusp (not smooth)

So we have to calculate limit of dy/dx as x approaches the problem point

Second derivatives: First find dy/dx = y’

Then second derivative is dy’/dx = (dy’/dt) / (dx/dt) – That is, divide again by dx/dt.

Review [from 6.4]: Arc length and surface area for parametric curves

ds = sqrt(dx² + dy²) = (for parametric eqns) sqrt((dx/dt)² + (dy/dt)²) *dt

So L = int (ds) = int (sqrt((dx/dt)² + (dy/dt)²) dt)

Ex: arc length of the circle x=a cos t, y=a sin t for t=0..2p

Ex (elliptic integral of the second kind E(k,φ) arc length of x=acost, y=bsint, 0£t£φ

can be written as b*int(sqrt (1-k²sin²t),t,0,φ) where k²= (b²-a²)/b²

Surface area: As usual, S = int (2p*radius*ds); new ds for parametric equations.

Ex: Surface area of one arch of a cycloid (with a=1) revolved about y=2

Polar curves:

Tangent Lines:

Derivatives: based on parametric eqns x=rcosq and y=rsinq with r as a function of q:

dy/dx = (r'sinq+r cosq) / (r'cosq - r sinq)

Ex: For the cardioid r=1-sinq, find

a. dy/dx = (r'sinq+r cosq) / r'cosq - r sinq) = …(-sin2q+cosq) / (-cos2q-sinq)

= cosq (-2sinq+1) / ((2sinq+1)(sinq-1))

b. slope(s) of tangent line(s) at the pole: solve r=0, find lim of dy/dx as q appr solution p/2

Note form of (-sin2q+cosq) / (-cos2q-sinq) is 0/0, but L'Hopital fails

Instead, mult factored form by conjugate: (sinq + 1)/itself: lim is ¥

c. points where tan line is horiz: solve num = 0. Reject sol of p/2 (based on limit)

d. points where tan line is vertical: solve denom = 0. Note p/2 shows up

Arc length: L = int(sqrt(r²+r'²),q,a,b)

deriv from squares of dx/dq and dy/dq;

x = r cosq, so x'= r'cosq - r sinq; y = r sinq, so y'=r'sinq + r cosq

Ex: Find length of Archimedian spiral r=q from 0 to 4p

11.3 Areas in Polar Coordinates

Area - based on area of a triangle, with h = r, and b = ds = r dq

A = int((1/2)r(q)²,q,a,b)

Toughest part: finding limits of integr

Solving trig eqns

Use symmetry whenever possible

Often (e.g. rose curves) it's best to see when the trig func is max and when it is 0.

Ex: Find area enclosed by one loop of r = cos(5q)

Ex: Find area of region inside circle r=3sinq and outside cardioid r=1+sinq

11.4 Conic Sections

Parabola:

Locus (set) of points equidistant from a fixed point (focus) and a fixed line (directrix)

x^2=4py [vertex at origin, opens upward, focus (0,p), directrix y=-p]

Proof: d((0,p),(x,y)) = d((x,y),(x,y+p))

x^2 + (y-p)^2 = (y+p)^2

Be able to handle different orientations and translations

Examples:

Find facts (focus, vertex, directrix), given eqn (X7a: x² – 4x +2y =1; complete square)

Find eqn given facts (X21b: vertex (0,0), directrix x=7), (X24b: focus (–1,4), directrix x=5)
Ellipse:

Locus (set) of points for which the sum of distances from two fixed points (foci) is constant.

d((x,y),(-c,0)) + d((x,y),(c,0)) = 2a

Lots of alg gets us to x^2/a^2 + y^2/(a^2-c^2) = 1 Let b^2 = a^2-c^2

Notes: major axis: vertices at (±a,0), ends of minor axis at (0,±b),

right triangle relationship among a,b,c

Be able to handle different orientations and translations

Examples:

Find facts given eqn (X14a: complete square: 9x² + 4y² –18x +24y +9 = 0)

Find eqn given facts (X28b: length of minor axis 8, foci (0,±3))
Hyp:

Same def, except "difference"; same eqn except - (setup and final result, except that b²=c²-a²

Notes: major axis: vertices at (±a,0), ends of "pseudo-axis" at (0,±b)

Be able to handle different orientations and translations

Asymptotes: slopes ±b/a or a/b, depending on orientation. Solve eqn for y²; limit: y = ±b/a*x

Examples:

Find facts given eqn (X20b: complete square: 4y² – x² –40y – 4x = –60)

Find eqn given facts (asymptotes y=1/3*x+4, -1/3*x-2, passes through P(-3,4)

Sol: center (-9,1) (by simult eq: asymptotes cross at the center);

P is above asymp (y(-3)=3<4, so orientation is vertical

eqn so far is (y-1)^2/a^2 - (x+9)^2 = 1

At P, we get 9/a^2 - 36/b^2 = 1

vertical, so slopes are ±a/b=1/3, so b = 3a

subs into eqn, get a^2=5, so b^2=45; rewrite eqn.

eccentricity = c/a. Ellipse: e<1; Hyperbola: e>1

Show that dist from main axis to end of latus rectum (focal chord) is b²/a for ellipse and hyperbola.

11.5 Rotation of Axes; Second Degree Equations

cot(2t)=(A-C)/B

From this, sketch right triangle

From right triangle, find cos(2t)

From cos(2t), find both cos(t) and sin(t) from the power reducing formula cos²t or sin²t= (1±cos(2t)

rotation formulas

x = x' cos t - y' sin t

y = x' sin t + y' cos t

x' = x cos t + y sin t

y' = -x sin t + y cos t

11.6 Conic Sections in Polar Coordinates

Tasks: find e, rect eqn, slope of asym for hyp, sketch

Alt def for conic sections: Let F = focus, L = directrix; e = eccentricity. Conic section is set of all points P for which the ratio dist(P,F) / dist (P,L) = e; parabola, ellipse, or hyperbola if e =, <, or > 1.

Thm 11.7.6: A polar eqn in the form r = ed / (1±e cosq) (or sinq) is a conic section w eccentricity e and a focus at the pole (origin); d = signed dist to directrix [exists for all];

e = eccentricity = c/a for ellipse and hyperbola.

ellipse if e<1, hyperbola if e>1. Parabola if e=1 (c/a is irrelevant for parabola)

Be able to sketch conic from eqn based on 4 quadrantal angles

Ex: r = 6 / (1+2sinq)

How to tell, based on 4 plots, whether we have

For asymptotes, as r appr ¥, denom appr 0, solve for q, m = tanq, y-int=y-coord of center

Be able to find rectangular eqn of conic from clues in the polar eqn

orientation, center, c & a directly, b based on a and c.

Demo: convert from polar form to rectangular form

clear frac, convert rsinq or rcosq, isolate r, square both sides, convert r², result=conic. compl sq