Now our goal is to verify the Gradient Theorem – twice! The Gradient Theorem claims that: of - dr = vf •dr = f(B) – [(A) d) First evaluate the leftmost expression directly, the line integral of the gradient of f along the line L. e) Next evaluate the middle expression directly, the line integral of the gradient of f along the helix H. ) Finally evaluate the rightmost expression directly, which is f(0,1,1) – f(1,0,0). This is by far the easiest part of this problem, so you're not missing something if you think it's simple. All three answers from d, e, and f should match, verifying that the Gradient Theorem is true in this case.
Now our goal is to verify the Gradient Theorem – twice! The Gradient Theorem claims that: of - dr = vf •dr = f(B) – [(A) d) First evaluate the leftmost expression directly, the line integral of the gradient of f along the line L. e) Next evaluate the middle expression directly, the line integral of the gradient of f along the helix H. ) Finally evaluate the rightmost expression directly, which is f(0,1,1) – f(1,0,0). This is by far the easiest part of this problem, so you're not missing something if you think it's simple. All three answers from d, e, and f should match, verifying that the Gradient Theorem is true in this case.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Please I need help with d) e) f) g) h) i)

Transcribed Image Text:The Fundamental Theorem for Line Integrals / The Gradient Theorem
1. Consider the scalar field
f(x,y, z) = 3xy + z?
and the two points A = (1,0,0) and B = (0,1,1). Define two different curves that start at A and end at B:
L: r(t) = (1 - t, t,t) 0sts1
H: r(t) = (cos t, sint, 2t/m) 0stsn/2
Note that L is a line segment, while H is (part of) a helix.
a) First verify that both curves truly do begin at A and end at B. That is, plug in to prove that r(0) = A and r(1) = B for
L, and r(0) = A and r(1/2) = B for H.
b) Graph L, H, A and B using Geogebra, some other program, or by hand, to get a feel for what's going on.
c) Find Vf, the gradient of f. You will use this below.
Now our goal is to verify the Gradient Theorem – twice! The Gradient Theorem claims that:
(vf - dr = vf •dr = f(B) – f(A)
d) First evaluate the leftmost expression directly, the line integral of the gradient of f along the line L.
e) Next evaluate the middle expression directly, the line integral of the gradient of f along the helix H.
f) Finally evaluate the rightmost expression directly, which is f(0,1,1) – f(1,0,0). This is by far the easiest part of this
problem, so you're not missing something if you think it's simple.
All three answers from d, e, and f should match, verifying that the Gradient Theorem is true in this case.
But wait, there's more! Let F(x, y,z) = (-y,x,xy).
8) Compute S, F · dr.
h) Computef, F · dr.
i) The two values from parts (g) and (h) are not the same. Why not? What property of the vector field F prevents these
two line integrals from being equal?
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