Exam (elaborations)
Vector Functions in Space. Divergence and Curl. Line Integrals solved questions
- Course
- Institution
Vector Functions in Space. Divergence and Curl. Line Integrals solved questions
[Show more]
Seller
Follow
jureloqoo
Content preview
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,
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller jureloqoo. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $8.63. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
76449 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now