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Sliva [168]
3 years ago
13

For each of the following vector fields F , decide whether it is conservative or not by computing the appropriate first order pa

rtial derivatives. Type in a potential function f (that is, \nabla f = \mathbf{F} ). If it is not conservative, type N.
A. \mathbf{F} \left( x, y \right) = \left( -10 x + 7 y \right) \mathbf{i} + \left( 7 x + 6 y \right) \mathbf{j}
f \left( x, y \right) =

B. \mathbf{F} \left( x, y \right) = -5 y \mathbf{i} - 4 x \mathbf{j}
f \left( x, y \right) =

C. \mathbf{F} \left( x, y, z \right) = -5 x \mathbf{i} - 4 y \mathbf{j} + \mathbf{k}
f \left( x, y, z \right) =

D. \mathbf{F} \left( x, y \right) = \left( -5 \sin y \right) \mathbf{i} + \left( 14 y - 5 x \cos y \right) \mathbf{j}
f \left( x, y \right) =

E. \mathbf{F} \left( x, y, z \right) = -5 x^{2} \mathbf{i} + 7 y^{2} \mathbf{j} + 3 z^{2} \mathbf{k}
f \left( x, y, z \right) =

Note: Your answers should be either expressions of x, y and z (e.g. "3xy + 2yz"), or the letter "N"
Mathematics
1 answer:
Mazyrski [523]3 years ago
6 0

Answer:

(a)

Conservative

(b)

Not conservative

(c)

Conservative.

Step-by-step explanation:

(a)

\mathbf{F}(x,y) = (-10x+7y,7x+6y)

Notice that

\frac{\partial\mathbf{F}_y}{\partial x} = 7

and

\frac{\partial\mathbf{F}_x}{\partial y} = 7

Therefore the field is conservative.

(b)

Notice that

\mathbf{F}(x,y) = (-5y,-4x)

and

\frac{\partial\mathbf{F}_y}{\partial x} = -4

but

\frac{\partial\mathbf{F}_x}{\partial y} = -5

Therefore is not conservative.

(c)

Notice that

To prove that the vector field is conservative you have to compute the curl of the vector field and you would get that.

\mathbf{F}(x,y,z) = (-5x,-4y,1)

\nabla \times \mathbf{F} =  (0,0,0)

Therefore your vector field is conservative.

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Answer:

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Step-by-step explanation:

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Answer:

i). x³ + 9x² + yz - 15

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Step-by-step explanation:

Question (38)

i). Two expressions are -5x² - 4yz + 15 and x³+ 4x²- 3yz

  By subtracting expression (1) from expression (2) we can the expression by addition which we can get expression (1).

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