#$&* 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
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* 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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Verify that the integral Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] is independent of path and evaluate it where C is any path from (0, pi/18) to (1, pi/6). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] (0, pi/18) to (1, pi/6) I started to get confused on this problem. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After looking at the solution I can better piece together the problem and how to go about solving it. "