College Trigonometry, 2011a

1056 Applications of Trigonometry
2. When parameterizing line segments, we think: ‘starting point + (displacement)t’. For the
first part of the path, we get {x =3t, y =4t for 0 ≤ t ≤ 1, and for the second part
we get {x =3+2t, y =4− 4t for 0 ≤ t ≤ 1. Since the first parametrization leaves off
at t = 1, we shift the parameter in the second part so it starts at t = 1. Our current
description of the second part starts at t =0,soweintroducea‘timedelay’of1unit
to the second set of parametric equations. Replacing t with (t − 1) in the second set of
parametric equations gives {x =3+2(t − 1), y=4− 4(t − 1) for 0 ≤ t − 1 ≤ 1. Simplifying
yields {x =1+2t, y =8− 4t for 1 ≤ t ≤ 2. Hence, we may parametrize the path as
{x = f(t), y= g(t) for 0 ≤ t ≤ 2where
{
f(t) =
3t, for 0 ≤ t ≤ 1
1+2t, for 1 ≤ t ≤ 2
{
and g(t) =
4t, for 0 ≤ t ≤ 1
8 − 4t, for 1 ≤ t ≤ 2
3. We know that {x = cos(t), y=sin(t) for 0 ≤ t