PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION
279
34.36
Find the arc length of x=^
2 , y = £(6f + 9)
3
'
2
, from f = 0 to f = 4.
dxldt=t,
dy/dt = (6t + 9)
1 '
2 ,
(dxldt? = t\
(dy/dt)
2 = 6t + 9.
So,
s =
(t + 5)dt = (^t" + 30 ]o = 8+12 = 20.
34.37 Find the arc length of x - a cos
3 6, y = a sin
3 6, from 6 = 0 to 0 = Tr/2.
Thus,
34.38
Find the arc length of x = cos t + (sin t, y = sin t - tcos t, from t=irl6 to r=ir/4.
dxldt = -sin t + sin t + t cos t = t cos t, dyldt = cos t - cos f + t sin f = t sin f. Hence,
34.39 Find the length of one arch of the cycloid x = a(0 -sin 6), y = a(l-cosO), 0<0<27r.
So,
<£c/d0 = a(l-cos0), dy/d6 = asin0.
34.40 Find the arc length of x = u, y = w
3 '
2 , 0
12 . Hence,
34.41
Find the arc length of * = lnsin0, y = 6, 77/6 <0 < 7r/2.
dx/d0 = col6, dyldO = l. Hence, 5 =
34.42
Rework Problem 34.33 in polar coordinates (r, 6), where r —
tanfl =y/x.
In
On the given curve,
and y/x = tan t. Thus, replacing r by 6. we
have as the equation of the curve in polar coordinates: r=e or 0 = In r (a logarithmic spiral). Using the
arc-length formula
we retrieve s =
VECTOR-VALUED FUNCTIONS
34.43
If F(«) = (/(«), g(")) is a two-dimensional vector function, lim F(w) = (lim/(«), lim g(w)), where the
limit on the left exists if and only if the limits on the right exist. Taking this as the definition of vector convergence, show that F'(w) = (f'(u), g'(u)).
This last limit is, by the definition, equal to
34.44 If R(0) = (r cos 0, r sin 0), with fixed r > 0, show that R'(0) X R(0).
By Problem 34.43, R'(0) = (-rsin 0, rcos 0). Then R(0)-R'(0) = (rcos 0, r sin 0)- (-rsin 0, rcos 0) =
-r
2 cos 0 sin 0 + r
2 sin 0 cos 0 = 0.
34.45
Show that, if R(w) traces out a curve, then R'(«) is a tangent vector pointing in the direction of motion along the
curve.
Refer to Fig. 34-13. Let OP=R(u) and OQ = R(u + &u). Then PQ = R(u + AM) - R(M) and
As Aw-»0, Q approaches P, and the direction of PQ (which is the direction of Pg/A«) approaches the direction of R'(u), which is thus a tangent vector at P.
279
34.36
Find the arc length of x=^
2 , y = £(6f + 9)
3
'
2
, from f = 0 to f = 4.
dxldt=t,
dy/dt = (6t + 9)
1 '
2 ,
(dxldt? = t\
(dy/dt)
2 = 6t + 9.
So,
s =
(t + 5)dt = (^t" + 30 ]o = 8+12 = 20.
34.37 Find the arc length of x - a cos
3 6, y = a sin
3 6, from 6 = 0 to 0 = Tr/2.
Thus,
34.38
Find the arc length of x = cos t + (sin t, y = sin t - tcos t, from t=irl6 to r=ir/4.
dxldt = -sin t + sin t + t cos t = t cos t, dyldt = cos t - cos f + t sin f = t sin f. Hence,
34.39 Find the length of one arch of the cycloid x = a(0 -sin 6), y = a(l-cosO), 0<0<27r.
So,
<£c/d0 = a(l-cos0), dy/d6 = asin0.
34.40 Find the arc length of x = u, y = w
3 '
2 , 0
12 . Hence,
34.41
Find the arc length of * = lnsin0, y = 6, 77/6 <0 < 7r/2.
dx/d0 = col6, dyldO = l. Hence, 5 =
34.42
Rework Problem 34.33 in polar coordinates (r, 6), where r —
tanfl =y/x.
In
On the given curve,
and y/x = tan t. Thus, replacing r by 6. we
have as the equation of the curve in polar coordinates: r=e or 0 = In r (a logarithmic spiral). Using the
arc-length formula
we retrieve s =
VECTOR-VALUED FUNCTIONS
34.43
If F(«) = (/(«), g(")) is a two-dimensional vector function, lim F(w) = (lim/(«), lim g(w)), where the
limit on the left exists if and only if the limits on the right exist. Taking this as the definition of vector convergence, show that F'(w) = (f'(u), g'(u)).
This last limit is, by the definition, equal to
34.44 If R(0) = (r cos 0, r sin 0), with fixed r > 0, show that R'(0) X R(0).
By Problem 34.43, R'(0) = (-rsin 0, rcos 0). Then R(0)-R'(0) = (rcos 0, r sin 0)- (-rsin 0, rcos 0) =
-r
2 cos 0 sin 0 + r
2 sin 0 cos 0 = 0.
34.45
Show that, if R(w) traces out a curve, then R'(«) is a tangent vector pointing in the direction of motion along the
curve.
Refer to Fig. 34-13. Let OP=R(u) and OQ = R(u + &u). Then PQ = R(u + AM) - R(M) and
As Aw-»0, Q approaches P, and the direction of PQ (which is the direction of Pg/A«) approaches the direction of R'(u), which is thus a tangent vector at P.