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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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4 years ago
A little help please?
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Answer:

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Step-by-step explanation:

7 0
3 years ago
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1/3(3+2x)−1=10 Drag and drop the statement that shows the solution to the equation into the box.
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Given
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Therefore, x = 15
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I think I know this much so far for A (but I could be wrong):
Eduardwww [97]
Part A. You have the correct first and second derivative.

---------------------------------------------------------------------

Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

-------------------------------------------------------------

Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h  ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0. 
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well. 
6 0
3 years ago
Can you help me please
user100 [1]

The answer to this question is 10. The working out is shown in the image attached.

*** Sorry instead of writing 170 I wrote 270 for the big angle.. other than that it's correct

or another way to solve this is....

we know that a straight line is 180 degrees and the angle APD is 170 all we need to do is subtract 170 from 180 to get the answer of DPB

Hope this helped

3 0
3 years ago
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