PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION
287
34.102 Find the radius of curvature of y = In x at x = e.
and
By Problem 34.97,
When
and
34.103 Find the curvature of R(0 = (cos
3 t, sin
3 t) at f=ir/4.
Use Problem 34.100.
x = -3 cos
2 1 sin t,
y = 3 sin
2 1 cos t,
x = -3(cos
3
1 - 2 sin
2 1 cos t) =
-3 cos r(3 cos
2 1 - 2). y = 3(-sin
3 / + 2 cos
2 1 sin 0 = 3 sin r(3 cos
2 f - 1). [(x)
2 + (.y)
2 ]
3 '
2 = (9 cos
4 1 sin
2 1 +
9 sin
4 t cos
2 O
3 '
2 = (9 cos
2 1 sin
2 r)
3 '
2 = 27 cos
3 1 sin
3 1. Now, xy - yx = (-3 cos
2 1 sin 0(3 sin 0(3 cos
2 t - 1) -
(3 sin
2 1 cos 0(~3 cos 0(3 cos
2 r - 2) = -9 cos
2 1 sin
2 f. *c = -9 cos
2 1 sin
2 f/27 cos
3 1 sin
3 t = - 1 /(3 cos f sin 0When /=7r/4, K = -§.
34.104 For what value of x is the radius of curvature of y = e* smallest?
y'=y" = e*. By Problem 34.97, K = e*l(\ + e
2
*)
3
'
2
, and the radius of curvature p is (1 + e
2 *)
3 '
2 /e*.
Then
Setting dpldx = Q, we find 2e
2
* = l, 2x = In | = -In 2, x=-(ln2)/2. The first-derivative test shows
that this yields a relative (and, therefore, an absolute) minimum.
34.105 For a given curve R(0, show that T"(0 = (v/p)N(t).
By definition of N(0, T'(0 = |T'(0|N(0- We must show that \T(t)\ = v/p. By the chain rule, T'(0 =
By Problem 34.89,
Hence, |T'(Ol = \d/dt\. But,
and, therefore,
Since
Thus,
34.106 Assume that, for a curve R(0, v = dsldt > 0. Then the acceleration vector a can be represented as the
following linear combination of the perpendicular vectors T and N: a = (d
2 s/dt
2 )T + (u
2 /p)N. The coefficients
of T and N are called, respectively, the tangential and normal components of the acceleration vector. (Since
the normal component v
2 /p is positive, the acceleration vector points "inside" the curve, just as N does.)
By definition of T, T = R'(0/"- So, R'(0 = vT. By the product rule, R"(0 = vT + (dv/dt)T. By
Problem 34.105, T' = (u/p)N. Hence, a = R"(0 = (d
2 s/dt
2 )T + (v
2 /p)N.
34.107 Find the tangential and normal components of the acceleration vector for the curve R(0 = (e', e
2 ') (a motion
along the parabola y = x
2 ).
Hence,
Since
R" = (dv /dt)T + (u
2 /p)N,
\R"\
2 = (dvldt)
2 + (v
2 /p)
2 . Hence,
So,
and T and N are perpendicular, the Pythagorean theorem yields
287
34.102 Find the radius of curvature of y = In x at x = e.
and
By Problem 34.97,
When
and
34.103 Find the curvature of R(0 = (cos
3 t, sin
3 t) at f=ir/4.
Use Problem 34.100.
x = -3 cos
2 1 sin t,
y = 3 sin
2 1 cos t,
x = -3(cos
3
1 - 2 sin
2 1 cos t) =
-3 cos r(3 cos
2 1 - 2). y = 3(-sin
3 / + 2 cos
2 1 sin 0 = 3 sin r(3 cos
2 f - 1). [(x)
2 + (.y)
2 ]
3 '
2 = (9 cos
4 1 sin
2 1 +
9 sin
4 t cos
2 O
3 '
2 = (9 cos
2 1 sin
2 r)
3 '
2 = 27 cos
3 1 sin
3 1. Now, xy - yx = (-3 cos
2 1 sin 0(3 sin 0(3 cos
2 t - 1) -
(3 sin
2 1 cos 0(~3 cos 0(3 cos
2 r - 2) = -9 cos
2 1 sin
2 f. *c = -9 cos
2 1 sin
2 f/27 cos
3 1 sin
3 t = - 1 /(3 cos f sin 0When /=7r/4, K = -§.
34.104 For what value of x is the radius of curvature of y = e* smallest?
y'=y" = e*. By Problem 34.97, K = e*l(\ + e
2
*)
3
'
2
, and the radius of curvature p is (1 + e
2 *)
3 '
2 /e*.
Then
Setting dpldx = Q, we find 2e
2
* = l, 2x = In | = -In 2, x=-(ln2)/2. The first-derivative test shows
that this yields a relative (and, therefore, an absolute) minimum.
34.105 For a given curve R(0, show that T"(0 = (v/p)N(t).
By definition of N(0, T'(0 = |T'(0|N(0- We must show that \T(t)\ = v/p. By the chain rule, T'(0 =
By Problem 34.89,
Hence, |T'(Ol = \d
and, therefore,
Since
Thus,
34.106 Assume that, for a curve R(0, v = dsldt > 0. Then the acceleration vector a can be represented as the
following linear combination of the perpendicular vectors T and N: a = (d
2 s/dt
2 )T + (u
2 /p)N. The coefficients
of T and N are called, respectively, the tangential and normal components of the acceleration vector. (Since
the normal component v
2 /p is positive, the acceleration vector points "inside" the curve, just as N does.)
By definition of T, T = R'(0/"- So, R'(0 = vT. By the product rule, R"(0 = vT + (dv/dt)T. By
Problem 34.105, T' = (u/p)N. Hence, a = R"(0 = (d
2 s/dt
2 )T + (v
2 /p)N.
34.107 Find the tangential and normal components of the acceleration vector for the curve R(0 = (e', e
2 ') (a motion
along the parabola y = x
2 ).
Hence,
Since
R" = (dv /dt)T + (u
2 /p)N,
\R"\
2 = (dvldt)
2 + (v
2 /p)
2 . Hence,
So,
and T and N are perpendicular, the Pythagorean theorem yields