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Vector Functions in Space. Divergence and Curl. Line Integrals solved questions
Vector Functions in Space. Divergence and Curl. Line Integrals solved questions
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CHAPTER 45Vector Functions in Space.
Divergence and Curl. Line Integrals
45.1 For the space curve R(t) = (t, t, t ), find a tangent vector and the tangent line at the point (1,1, 1).
A tangent vector is given by the derivative R'(t) = (l,2t,3t ). At (1,1,1), < = 1 . Hence, a tangent
vector is (1, 2, 3). Parametric equations for the tangent line are x = 1 + w, y = 1 + 2a, z = 1 + 3w. As a
vector function, the tangent line can be represented by (1,1,1) + u(l, 2, 3).
45.2 Find the speed of a particle tracing the curve of Problem 45.1 at time t = 1. (The parameter/is usually, but not
necessarily, interpreted as the time.)
If 5 is the arc length, is the speed. In general, if
and For this particular case,
When
45.3 Find the normal plane to the curve of Problem 45.1 at t = \.
The tangent vector at t = l is (1,2,3). That vector is perpendicular to the normal plane. Since the
normal plane contains the point (1,1,1), its equation is \(x - 1) + 2(y - 1) + 3(z - 1) = 0, or equivalently,
x + 2y + 3z = 6.
45.4 Find a tangent vector, the tangent line, the speed, and the normal plane to the helical curve R(t) =
(a cos 2irt, a sin 2irt, bt) at f = l.
A tangent vector is R'(Q = (-2ira sin 2irt, 2-rra cos 2irt, b) = (0, 2-rra, b). The tangent line is (a,0, b) +
u(0, 2TTd, b). The speed is R'(0l = V4?r V + b2. The normal plane has the equation (0)(;e - a) +
(2trd)(y - 0) + (b)(z - b) = 0, or equivalently, 2-rray + bz = b2.
45.5 Prove that the angle 6 between a tangent vector and the positive z-axis is the same for all points of the helix of
Problem 45.4.
R'(0 = (-2TTO sin 2Trt, 27ra cos 2irt, b) has a constant z-component; thus, 0 is constant.
45.6 Find a tangent vector, the tangent line, the speed, and the normal plane to the curve R(t) - (t cos t, t sin t, t) at
t=TT/2.
R'(t) = (cos t- tsin t, sin t+ teas t, !) = (- it 12,1,1) is a tangent vector when t=TT/2, The tangent
line is traced out by (0, it I I , it 12) + u(-ir/2,1, 1), that is, x = (-ir/2)u, y = -rrl2+u, z = it 12 + u.
The speed at t = ir/2 is An equation of the normal plane is
(-7T/2)(jc-0) + ( y - 7 r / 2 ) + ( z - 7 r / 2 ) = 0 , or equivalently, (-ir/2)x + y + z = ir.
45.7 If G(0 = (M 3 ,lnf) and ¥(t) = t2G(t), find F'(0-
In general, (The proof in Problem 34.53 is valid for arbitrary vector
functions.) Hence, F'(0 = r(l, 3f 2 ,1 It) + 2t(t, t\ In t) = (3/2, 5f4, t + 2t In t).
45.8 If G(t) = (e',cost,t), find
By the formula of Problem 45.7,
(cos t ) ( e , cos t, t) = (e'(sin t + cos t), cos 2 1 - sin 2 1, sin r + t cos t).
425
, 426 CHAPTER 45
45.9 If F(t) = (sin /, cos t, t) and G(t) = (t, 1, In t), find [F(0-G(0].
The product formula [F(0-G(0] = F(0'G'(0 + F'(0-G(0 of Problem 34.56 holds for arbitrary vector
functions. So [F(r)-G(r)] = (sin t,cost, O ' O . 0 , + (cos/, -sin t, !)•(*, 1, In /) = sin t + 1 + t cos t -
sin / + In t = 1 + / cos / + In t.
45.10 If G(3) =•(!,!, 2) and G'(3) = (3,-2,5), find [rG(f)] at / = 3.
[r 2 G(/)J = t2G'(t) + 2tG(t). Hence, at t = 3, [t2G(t)] = 9(3, -2, 5) + 6(1,1,2) = (33, -12, 57).
45.11 Let F(0 = G(0 x H(/). Prove F'= (G X H') + (G1 x H).
45.12 Prove that [R(/> X R'(/)] = R(/) X «•(/).
By Problem 45.11, [R(0 x R'(01 - [R(f) x R"(t)l + [R'(f) x R'(f)l. But Ax A= 0 for all A.
Hence, the term R'(/) x R'(/) vanishes.
45.13 If G'(0 is perpendicular to G(f) for all t, show that |G(f)| is constant, that is, the point G(t) lies on the surface of £
sphere with center at the origin.
Since G'(0 is perpendicular to G(r), G'(f) • G(t) = 0. Then [G(0-G(r)] = G(0-G'(0 +
G'(0 ' G(f) = 0 + 0 = 0. Therefore, |G(r)|2 is constant, and hence |G(f)| is constant.
45.14 Derive the converse of Problem 45.13: If |G(f)| is constant, then G(f) • G'(0 = 0.
Let |G(/)| = c for all t. Then G(t)-G(t) = |G(f)| 2 = c2 So [G(0-G(0] = 0. Hence,
G(0-G'(0 + G'(0'G(r) = 0. Thus, 2G(r)-G'(0 =0, and, therefore, G(f)-G'(f) = 0.
45.15 Assume that R(f) ^ 0 and R'(f)-^0 for all t. If the endpoints of R(f) is closest to the origin at t = t0,
show that the position vector R(t) is perpendicular to the velocity vector R'(0 at f = t0. [Recall that, when t is
the time, R'(') is called the velocity vector.)
has a relative minimum at f = 0. Hence,
[R(0 • R(r)] = R(f) • R'(/) + R'(0' R(0 = 2R(0 • R'(0- Therefore, R(r)-R'(0=0 at t = t0. Since
R(0^0 and R'(r)^0, R(t) 1 R'(0 at t = ta.
45.16 Find the principal unit normal vector N to the curve R(f) = (t, t2, t3) when f=l.
The unit tangent vector is The principal unit normal vector is
Now,
When
Therefore, Thus,
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