34.73
Derive the following quotient rule:
PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION
283
By the product rule,
34.74
Assume that an object moves on a circle of radius r with constant speed v > 0. Show that the acceleration
vector is directed toward the center of the circle and has length v
2 /r.
The position vector R(r) satisfies |R(f)l = f and I
R '(')I
= v - Let 0 be the angle from the positive x-axis to
R(f) and let 5 be the corresponding arc length on the circle. Then s = r6, and, since the object moves
with constant speed v, s = vt. Hence, 6 = vtlr. We can write R(r) = r(cos 6, sin 6). Bv the chain rule.
Again by the chain rule,
R"(0 =
Hence, the acceleration vector
R"(<) points in the opposite direction to R(/)>
tnat is, toward the center of the circle, and
34.75
Let R(t) = (cos(7re'/2),sin(-7re72)). Determine R(0), v(0), and a(0), and show these vectors in a diagram.
Let
thus,
0 = -ne'12;
R(t) = (cos 0, sine). By the chain rule, v(f) = R'(t) = (-sin 0, cos 6)
0(-sin0, cos 9), and
When
and
f = 0,
0=77/2,
R(0) = (0,l),
See Fig. 34-15.
34.76
Let R(t) = (10cos27rf, 10sin27rf). Find the acceleration vector.
Fig. 34-15
Let 0 = 2irt. Then dO/dt = 2-n- and R(f) = 10(cos 0, sin 9). By the chain rule, v(f) = R'(0 =
10(-sin0, cos0)
= 20Tr(-sin 6, cosO). By the chain rule again, the acceleration vector a(t) = d\/dt =
20ir(-cos 6, -sin 0)
= -407r
2
(cos 0, sin 0) = -47r
2 RO). Note that this is a special case of Problem 34.74.
34.77
Let R(t) = (t,l/t). Find v(f), the speed |v(f)|, and a((). Describe what happens as f-*•+<».
v(/) = (l,-l/r
2 ),
and a(r) = (0,2/r
3 ). As f-»+», the direction of R(r) approaches that of the positive *-axis and its length approaches +°°. 1 he velocity vector approaches the unit vector
i = (1,0), and the speed approaches 1. The acceleration vector always points along the positive y-axis, and its
length approaches 0.
34.78
Let R(t) = (t + cos t, t-sin t). Show that the acceleration vector has constant length.
R'(0 = (l -sinf, 1 -cosf), and R"(f) = (-cos t, sin t). Hence, |R"(')| =
34.79
Prove that, if the acceleration vector is always perpendicular to the velocity vector, then the speed is constant.
This follows from Problem 34.66, substituting R'(0 for R(t).