34 Curves in Space
or
., d^2 s _.. Ids\^2 du(s) .„ „„ ,
°<«>=d* "<'>+(*) -11. (1.3.19)Equation (1.3.19) shows that the acceleration a(t) has two components: the
tangential acceleration (d^2 s/dt^2 ) u{t) and the normal acceleration
(ds/dt)^2 (du(s)/ds). By Lemma 1.1, page 26, the derivative of a unit
vector is always orthogonal to the vector itself, and so the tangential and
normal accelerations are mutually orthogonal. The derivation also shows
that the decomposition (1.3.19) of the acceleration into the tangential and
normal components does not depend on the coordinate system.
EXERCISE 1.3.12.c In (1.3.20) below, r — r(t) represents the position of
point mass m at time t in the Cartesian coordinate system:
r(t) = t^2 i + 2t^2 j + t^2 k; r(t) = 2cos7rfz + 2sin7r£ j;
r{t) = 2cost^2 i + 2smt^2 j\ r(t) = cos t^2 i + 2 sin t^2 j.For each function r — r(t),(1.3.20)- Sketch the corresponding trajectory;
- Compute the velocity and acceleration vectors as functions of t;
- Draw the trajectory for 0 < t < 1 and draw the vectors r '(1), r "(1);
- Compute the normal and tangential components of the acceleration and
draw the corresponding vectors when t = 1; - Verify your results using a computer algebra system.
We will now write the decomposition (1.3.19) for the CIRCULAR MOTION
IN A PLANE. Let C be a circle with radius R and center at the point O.
Assume a point mass moves along C. Choose the cartesian coordinates
i, j, k with origin at O and i, j in the plane of the circle. Denote by 9(t)
the angle between i and the position vector r(t) of the point mass. Suppose
that the function 6 = 6(t) has two continuous derivatives in t, |#'(t)| > 0,
t > 0, and 0(0) = 0. Then
r(t) = Rcos9(t) i + Rsin6(t)j,
v = r'(t) = -0'(t)Rsm6(t)i + 0'(t)Rcose(t)j,
\\v\\ = (r'.r'^=R\e'(t)\,