Pearson eText Calculus and Its Applications, Brief Edition -- Instant Access (Pearson+)
Pearson eText Calculus and Its Applications, Brief Edition -- Instant Access (Pearson+)
12th Edition
ISBN: 9780136880257
Author: Marvin Bittinger, David Ellenbogen
Publisher: PEARSON+
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Chapter E, Problem 39E
To determine

To Show: a=5 yield to an equivalent result as a=5 , since a2=25 

In the integral

  1x225dx

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• • Let > be a potential for the vector field F = (−2 y³, −6 xy² − 4 z³, −12 yz² + 4 2). Then the value of sin((-1.63, 2.06, 0.57) – (0,0,0)) is - 0.336 -0.931 -0.587 0.440 0.902 0.607 -0.609 0.146
The value of cos(4M) where M is the magnitude of the vector field with potential ƒ = e² sin(лy) cos(π²) at x = 1, y = 1/4, z = 1/3 is 0.602 -0.323 0.712 -0.816 0.781 0.102 0.075 0.013
There is exactly number a and one number b such that the vector field F = conservative. For those values of a and b, the value of cos(a) + sin(b) is (3ay + z, 3ayz + 3x, −by² + x) is -0.961 -0.772 -1.645 0.057 -0.961 1.764 -0.457 0.201
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