5. Find the work done by F = (x² + y)i + (y² + x)j + ze²k over the following paths from (1, 0, 0) to (1, 0, 1). a. The line segment x = 1, y = 0,0 ≤z ≤ 1 The helix r(t) = (cost)i + (sin t)j + (-) k; 0 ≤ t ≤ 2π C. The x-axis from (1, 0, 0) to (0, 0, 0) followed by the parabola z = y², y = 0 from (0, 0, 0)to (1, 0, 1)
5. Find the work done by F = (x² + y)i + (y² + x)j + ze²k over the following paths from (1, 0, 0) to (1, 0, 1). a. The line segment x = 1, y = 0,0 ≤z ≤ 1 The helix r(t) = (cost)i + (sin t)j + (-) k; 0 ≤ t ≤ 2π C. The x-axis from (1, 0, 0) to (0, 0, 0) followed by the parabola z = y², y = 0 from (0, 0, 0)to (1, 0, 1)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 12T
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![5. Find the work done by F = (x² + y)i + (y² + x)j + ze²k over the following paths from (1, 0, 0) to (1, 0, 1).
a.
The line segment x = 1, y = 0,0 ≤ z ≤ 1
b.
The helix r(t) = (cost)i + (sin t)j + (1) k; 0 ≤ t ≤ 2n
C.
The x-axis from (1, 0, 0) to (0, 0, 0) followed by the parabola z = y², y = 0 from (0, 0, 0)to (1, 0, 1)
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5. F = (x²+4)i + (x²+x)j + 2e²1
(uy|F= (3 (20²) - (x²+x)); +
(32 (x²+4) -3
(28²)]) j
(3 (x²+x) - (x²+x)) k
= 0
=) (Curl F = 0 =) F i conservative
vector field.
·Let Jf(x11,2)= F
af (*²/47₁2) =
x² +7
dx
"F(x₁4₁2) =
√(x²+4)dx+G(1₁2)
af (x₁4/2) =
+ xy + g(1₁2)
x+39(4₁2) = y²+x
ag(412) =
y² = 1 g (1,2)=√√4² cly+h(2)
Jy
Jv12) =
+7 (2)
393
=h²(²) = 2 c²
f(x17²) = x³ + x + xy +(²-1) c²
3
f (1,0,0) = — 2/3 it (1,0,1)= 1/3/201
a.
√₁ F. dr = f(1,0,1) - |(1,0,0) = 1
b. Sc F.dr=f(1,0,1) -+(1,0,0)=1
J₁ F. dr = f(1,0,1) - † (1,0,0) = 1
2ل
+
n(2)=√ze²dz
=(2-1)6²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd920315e-5ad2-4fdd-a656-6468e885a914%2F15eab01f-1ea4-4028-a19b-1c1fc2d7bcde%2Fs7c3cj_processed.png&w=3840&q=75)
Transcribed Image Text:5. Find the work done by F = (x² + y)i + (y² + x)j + ze²k over the following paths from (1, 0, 0) to (1, 0, 1).
a.
The line segment x = 1, y = 0,0 ≤ z ≤ 1
b.
The helix r(t) = (cost)i + (sin t)j + (1) k; 0 ≤ t ≤ 2n
C.
The x-axis from (1, 0, 0) to (0, 0, 0) followed by the parabola z = y², y = 0 from (0, 0, 0)to (1, 0, 1)
Show transcribed data
Expert Answer
Anonymous answered this
14,501 answers
5. F = (x²+4)i + (x²+x)j + 2e²1
(uy|F= (3 (20²) - (x²+x)); +
(32 (x²+4) -3
(28²)]) j
(3 (x²+x) - (x²+x)) k
= 0
=) (Curl F = 0 =) F i conservative
vector field.
·Let Jf(x11,2)= F
af (*²/47₁2) =
x² +7
dx
"F(x₁4₁2) =
√(x²+4)dx+G(1₁2)
af (x₁4/2) =
+ xy + g(1₁2)
x+39(4₁2) = y²+x
ag(412) =
y² = 1 g (1,2)=√√4² cly+h(2)
Jy
Jv12) =
+7 (2)
393
=h²(²) = 2 c²
f(x17²) = x³ + x + xy +(²-1) c²
3
f (1,0,0) = — 2/3 it (1,0,1)= 1/3/201
a.
√₁ F. dr = f(1,0,1) - |(1,0,0) = 1
b. Sc F.dr=f(1,0,1) -+(1,0,0)=1
J₁ F. dr = f(1,0,1) - † (1,0,0) = 1
2ل
+
n(2)=√ze²dz
=(2-1)6²
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