190 method ii fundamental theorem of line integral by

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Unformatted text preview: ent) is 30 meters when there are three complete revolutions (t = 6π ), and so 6πλ = 30 Therefore, r(t) = 6 cos ti + 6 sin tj + and so r (t) = −6 sin ti + 6 cos tj + Now, apply the formula for line integral: F • dr = = 0 6π 0 z =⇒ λ= 5 π 5 tk 0 ≤ t ≤ 6π, π 5 k 0 ≤ t ≤ 6π. π C F(r(t)) • r (t)dt 0i + 0j − (90 − 6π 0 6π z )gk • 15 dt − 6 sin ti + 6 cos tj + 5 k dt π 5 =−g π 90 − t 3π 6π 0 t2 5 = − g 90t − π 6π = −2670g So the work done is 2670g kg-m2 s−2 (against the gravity). 190 Method II. (Fundamental Theorem of Line Integral) By the analysis above, we know that the force is given by z F(x, y, z ) = 0i + 0j − (90 − )gk, 15 and by Component Test, it is easy to see that F is conservative, and the corresponding potential function f is f (x, y, z ) = −(90 − z2 z )gdz = −90z + . 15 30 Then we can apply the Fundamental Theorem, and get W= C ∇f • dr = f (0, 0, 30) − f (0, 0, 0) = −2670g Method III. (Path Independent) In the official solution file, the solution is using the direct computation, i.e., using the formula: C b a F • dr = F (r (t)) • r (t)dt. Actually, the function F is conservative, so we have another more straightforward solution. Notice that F is given by z − 90)gk. 15 (Using component test, we can easily check F here is conservative.) F (x, y, z ) = 0i + 0j + ( One of the most important property of conservative fields is that the integration is not depending on the path of integration. Then, we choose the path to be a line segment from the starting point to the ending point. Luckily, F is along this direction, so the integration here is an integration of one variable. C F • dr = 30 ( 0 z2 z − 90)gdz = [( − 90z )g]30 = 2670g. 0 15 30 Method IV. (Physical Solution, Not Acceptable in the Exam) We can also use potential energy to solve the problem. Since the water leaks steadily during the man’s ascent, then you can do the problem like this: Suppose the man is just carrying 9kg of water from the bottom to the top, where 9kg is the average of 10kg and 8kg. The work done is equal to the increment of potential energy. Thus, in the question, the man carried 9kg of water to the place of 30m height, then the potential energy is increased by (Wm + Wp ) ∗ g ∗ 30 = (80 + 9) ∗ g ∗ 30 = 2670g. We have the same answer here. 2 191 4. Evaluate the line integral C xy 2 dx + x3 dy , where C is the rectangle with vertices (0, 0), (2, 0), (2, 3) and (0, 3) oriented anticlockwise by (i) direct computation, and (ii) using Green’s Theorem. Preliminaries Green’s Theorem Let D be a bounded region in the xy -plane and ∂D the boundary of D. Suppose P (x, y ) and Q(x, y ) has continuous partial derivatives on D . Then P dx + Qdy = ∂D D ∂ Q ∂P − ∂x ∂y...
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