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CHAPTER 34
34.93 Generalize the result of Problem 34.83 to any motion at constant speed.
If dv/dt = 0, Problem 34.92 gives T' = (l/y)aor N = (l/u)|T'|a.
34.94
Define the curvature K and radius of curvature p of a curve R(f)As in Problem 34.89, let $ denote the angle between the velocity vector R'(') and the positive jr-axis. The
curvature K is defined as d/ds, where s is the arc length. The curvature measures how fast the tangent vector
turns as a point moves along the curve. The radius of curvature is defined as p = |l//c|.
34.95
For a circle of radius a, traced out in the counterclockwise direction, show that the curvature is 1 la, and the radius
of curvature is «, the radius of the circle.
If (x a , y 0 ) is the center of the circle, then R(t) = (x 0 + a cos t, y g + a sin t) traces out the circle, where t is
the angle from the positive jc-axis to R(f) — (x a , y 0 ). Then R'(0
= a (~sin t, cos t), and dsldt — a. The
angle made by the positive *-axis with R'(f) is tan~
l
Hence,
= tan ' (-cot t), or that angle + ir.
So, K = l/a, and by definition, the radius of curvature is a.
34.96
Find the curvature of a straight line R(f) = A + rB.
R'(f) = B. Since R'(0 is constant, $ is constant, and, therefore, K = d(f>/ds = 0.
34.97
Show that, for a curve y =/(*), the curvature is given by the formula K = y"/[l + (y')
2 ]
3 '
2 . (We assume
that ds/dx>0, that is, the arc length increases with x.)
Since y' is the slope of the tangent line,
tan -y'.
Hence, differentiating with respect to s,
sec
2 (d<(>/ds) = y"/(ds/dx).
But, sec
2 = 1 + tan
2 4> = 1 + (y')
2 , and dsldx = [1 + (y')
2 ]"
2 . Thus,
34.98
Find the curvature of the parabola y = x
2
at the point (0,0).
y' = 2x and y" = 2. Hence, by Problem 34.97, K = 2/(l + 4x
2 )
3 '
2 . When x = 0, x = 2.
34.99
Find the curvature of the hyperbola xy = 1 at (1,1).
y' = -l/*
2
and y" = 2/*
3
. By Problem 34.97,
When x = l, K =2/2V5 = V5/2.
34.100 Given a curve in parametric form R(t) = (x(t), y(t)),
show that K =
(Here, the dots
indicate differentiation with respect to /, and we assume that x > 0 and
Substitute the results of Problem 34.19 into the formula of Problem 34.97.
34.101 Find the curvature at the point 6 = IT of the cycloid R(0) = a(0 - sin 8,1 - cos 6).
Use the formula of Problem 34.100, taking t to be 0. Then x = a(l -cos0), y = asin0, x = asin0,
y = a cos 6, (x)
2 + (y )
2 = a
2 (l - cos 0)
2 + a
2 sin
2 0 = a\2 - 2 cos 0). Then
When 0 = 77, « = -l/4a.