Unit 9 Highlights

Differentiating Vectors

<x(t), y(t)> →<x’(t), y’(t)>

Parametric Equation

x = f(t) and y = g(t); t is parameter; can set equations equal to by substituting for t

Vector Basics

have magnitude & direction; represented as directed line segments; have horizontal (x) and vertical (y) component <x(t), y(t)>

Integrating Vectors

< [initial x position + ∫x’(t)], [initial y position + ∫y’(t)] >

Position Vector

< x(t), y(t) >

velocity vector

< (dx/dt), (dy/dx) >

Slope of tangent

(dy/dt) / (dx/dt)

position at time t

x(t) = x(c) + ∫ x’(t) dt

y(t) = y(c) + ∫ y’(t) dt

Speed

√(dx/dt)² + (dy/dt)²

Total distance traveled (arc length)

∫ √(dx/dt)² + (dy/dt)² dt

Second Derivative

(d[dy/dx]/dt) / (dx/dt)

llvll

√(v₁² + v₂²)

unit vector

has same direction of original vector but with length 1; u = v / (llvll)

v = <x, y> →linear combination

v = xi + yj

unit vector with θ from x-axis to u

u = icosθ + jsinθ

vector with θ from x-axis to v

v = llvll <cosθ, sinθ> = llvll icosθ + llvll jsinθ

Orthogonal vectors

u • v = 0

dot product

u • v = u₁v₁ + u₂v₂

if θ is between two non-zero vectors, u and v

cosθ = (v • u) / (llull • llvll) or u • v = llull • llvll cosθ

vector valued functions

r(t) = f(t)I + g(t)j

Limits of vector-valued functions

limr(t) = [limf(t)]i + [limg(t)]j as t→a; if limit of a component fails to exist, overall fails to exist

Continuity of Vector-Valued Function

limr(t) = r(a) as t→a

Derivative Rules for Vector Valued Functions

same as if they were regular functions

If r(t) = f(t)i + g(t)j where f and g are continuous on [a,b],

∫r(t)dt = [ ∫f(t)dt]i +[ ∫g(t)dt]j + C₁i + C₂j and ∫r(t)dt = [ ∫f(t)dt]i +[ ∫g(t)dt]j as a→b

velocity v(t)

r’(t) = x’(t)i + y’(t)j

acceleration a(t)

r’’(t) = x’’(t)i + y’’(t)j

speed, llv(t)ll

llr’(t)ll = √[x’(t)]² + [y’(t)]²