Query133Mth

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course Mth 277

Question: Determine if the vector field F = (y- x^2)i + (2x + y^2)j is conservative, and if it is find a scalar potentialYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

F = (y- x^2)i + (2x + y^2)j , with F of the form

F(x,y)=u(x,y)i + v(x,y)j, where du/dy = 1 and dv/dx = 2

Based on this, the field is not conservative.

confidence rating #$&*:

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Given Solution:

The vector field is conservative if, and only if, its curl is zero.

The curl of the field M i + N j + P k is (P_y - N_z) i + (M_z - P_x) j + (N_x - M_y) k.

For the current function there is no z dependence and P = 0, so the expression becomes just (N_x - M_y) k.

M_y = 1 and N_x = 2, so the field is not conservative.

If we did try to find the scalar potential function, we would try to find f(x, y) such that del f = F. If this was so then we would have f_x = M and f_y = N.

If f_x = (y - x^2), then integrating with respect to x we conclude that f = y x - x^3 / 3 + g(y), where g(y) is any function of y.

If f_y = (2 x + y^2), then integrating with respect to y we conclude that f = 2 x y + y^3 / 3 + h(x), where h(x) is any function of x.

We could let g(y) = y^3 / 3, and h(x) = - x^3 / 3. However the first integration yields y x, and the second yields 2 x y, both functions of x as well as y. That is, any f function such that del f = F must contain the term x y as well as the term 2 x y. We can't accomplish this by adjusting either g(y) or h(x), and we conclude that no scalar potential function f exists for this F.

Had the original function been F(x, y) = (2 y - x^2) i + (2 x + y^2) j then :

Our integrals would have been 2 y x - x^3 / 3 + g(y) and 2 x y + y^3 / 3 + h(x). Letting g(y) = y^3 / 3 and h(x) = -x^3 / 3, our function would be the same using either the x or the y integral; either way we would get 2 x y - x^3 / 3 + y^3 / 3.

M_y would have been 2 and N_x would have been 2, so that our curl would have been (N_x - M_y) k = 0.

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Self-critique (if necessary):OK

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Question: Show that F = e^-y i - xe^-y j is conservative and find a scalar potential f for F. Evaluate the line integral Int[ F dot dR, C], where C is any smooth path connecting (0,0) and (1,1).

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Your solution:

F = e^-y i - xe^-y j

Int[ F dot dR, C]

(0,0) (1,1)

du/dy = -e^-y

dv/dx = -e^-y

Based on this, F is conservative.

confidence rating #$&*:

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Given Solution:

The curl of this function, which is of the form M i + N j + P k with M = e^-y, N = - x e^(-y) and P = 0, is easily found to be

(P_y - N_z) i + (M_z - P_x) j + (N_x - M_y) k = (N_x - M_y) k = ( -e^-y - (-e^(-y)) ) k = 0.

So the function is conservative and a scalar potential function exists.

We therefore integrate:

Integrating e^-y with respect to x we get x e^-y + g(y).

Integrating - x e^(-y) with respect to y we get x e^-y + h(x).

The general form of a scalar potential for our given vector field F is therefore f(x, y) = x e^-y + g(y) + h(x), where g(y) and h(x) are arbitrary functions of y and x, respectively.

The simplest such function is obtained by letting g(y) and h(x) both be zero, giving us

f(x, y) = x e^-y

We can easily verify that del f = F.

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Self-critique (if necessary):

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Question: OK

Consider the integral Int[ 2x^2y dx + x^3 dy, C]. Evaluate it on each of the following curves C:

The semicircle x = sqrt(1-y) from -1 <= y <= 1.

The line segment from (0,-1) to (1,1) and then the line segment from (1,1) to (0,1).

The square with vertices (1,1), (-1,1), (-1,-1), (1,-1) traversed once clockwise.

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Your solution:

Without a solution Im not entirely sure if I am taking the right approach, but here is my attempt.

Int[ 2x^2y dx + x^3 dy, C]

x = sqrt(1-y) from -1 <= y <= 1

line from (0,-1) to (1,1) and another from (1,1) to (0,1)

square with vertices (1,1), (-1,1), (-1,-1), (1,-1)

Int[ 2x^2y dx + x^3 dy, C]

Int(2(cos(t)^2y *(-sin(t)) + (cos(t))^3 (cos(t)) dt,-`pi/2,`pi/2)

Int(2(cos(t)^2y *(-sin(t)) + (cos(t))^3 (cos(t)) dt,-`pi/2,`pi/2)

Int(2(cos(t)^2*(sin(t))*(-sin(t))dt,-`pi/2,`pi/2) + Int(cos(t))^4 dt,-`pi/2,`pi/2)

Int(2(cos(t)^2*(sin(t))*(-sin(t))dt,-`pi/2,`pi/2)

-2*Int((sin(t)cos(t))^2 dt)

-2Int((sin(t)cos(t))^2 dt)

-2Int((sin(2t)/2)^2 dt)

-2Int(sin(2t)^2/4 dt)

-2Int(sin(2t)^2/4 dt)

-2Int((1 - cos(4t)/2)/4 dt)

-2Int((1 - cos(4t))/8 dt)

Int((1/4)( 1 + 2cos(2x) + cos^2(2x)) dt)

(3/8)t +(1/4)sin(2t)+(1/32)sin(4t)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Question: Verify that the integral Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] is independent of path and evaluate it where C is any path from (0, pi/18) to (1, pi/6).

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Your solution:

Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C]

(0, pi/18) to (1, pi/6)

I started to get confused on this problem.

confidence rating #$&*:

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Given Solution:

This integral is of the form F dot ds for F = (x y cos(x y) + sin(x y) ) i + x^2 cos(x y) j, of the form M i + N j.

We easily verify that curl F = (N_x - M_y) k = ( (2 x cos(xy) - x^2 y sin(x y)) - (x cos(x y) - x^2 y sin(x y) + x cos(x y) ) ) k = 0.

We conclude that a scalar potential function exists, and we integrate to find it:

Integrating x y cos(x y) + sin(x y) with respect to x we get cos( x y ) / y + x sin(x y) - cos(x y) / y = x sin(x y), to which we can add the arbitrary integration 'constant' g(y).

Integrating x^2 cos(x y) with respect to y we get x^2 ( sin(x y) / x) = x sin(x y) + h(x).

Our general scalar potential is therefore f(x, y) = x sin(x y) + g(y) + h(x), for arbitrary g(y) and h(x). The simplest scalar potential function is just f(x, y) = x sin(x y), and we will use this to evaluate our integral.

For conservative field F our fundamental theorem says that

integral ( F dot ds , between points (x0, y0) and (xf, yf) ) = f(xf, yf) - f(x0, y0)

Our current integral is conservative. Its path originates at (0, pi/18) and terminates at (1, pi/6), so

Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C]

= f(1, pi/6) - f(0, pi/18)

= 1 sin(1 * pi / 6) - 0 sin(0 * pi/18)

= 1/2.

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Self-critique (if necessary):

After looking at the solution I can better piece together the problem and how to go about solving it. "

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Good.

Let me know if you have questions.

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