Question

Calc III Tutorial: Vector-Value Functions in 3 Dimensions

Answer 1

\({\bf{Definitions:}}\)
let x = f(t), y = g(t), z = h(t) for t on some interval

(x,y,z) is the curve/path of a particle as a function of t, where f g and h are component functions

the vector r(t) defines the vector from the origin to the point P, vector OP
r(t) = f(t) i + g(t) j + h(t) k

in general, a vector-valued function is a rule that assigns a vector to each domain element D. this is in contrast to a scalar function which assigns a scalar to each domain element (so more like the functions in lower-level math)

Answer 2

\({\bf{Limits~and~Continuity:}}\)

\(\lim_{t \rightarrow t_{0}}r(t)=L\)
means that the function r has limit of vector L as t approaches t0

and for every ε > 0 there is δ> 0 such that |r(t) - L| < ε for 0 < |t - t0| < δ (review delta-epsilon proofs if you don't remember this)

r(t) is continuous at t if r(t) approaches r(t0) as t approaches t0; if every point on the function satisfies this condition the whole function is continuous

Answer 3

\({\bf{Derivatives:}}\)

r'(t) = (df/dt) i + (dg/dt)j + (dh/dt)k (basically just take the derivative of each component)

curve is "smooth" if the derivative is is continuous and nonzero at all points (note: for the derivative to be 0, all components must be 0, not just one of them)

the scalar multiple of the derivative ( basically (1/Δt)*(r(t+Δt) - r(t)) ) points in the same direction as the position vector between f(t) and f(t+Δt) for positive t. for infinitely small t (Δt --> 0) this vector approaches the tangent at f(t)

r'(t) points in the direction of motion

Answer 4

Source material is section 13.1 of Thomas' Calculus, Early Transcendentals, 14th edition by Hass, Heil, Weir, et. al.