#$&* course Mth 277 5/2 2 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): ********************************************* 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: du/dy = -e^-y dv/dx = -e^-y So because partial derivatives are equal F is a conservative. Int( u(x,y) dx) = Int( e^(-y) dx) = x*e^(-y) + k(y), where k(y) is constant that is fn. of y alone. f_y(x,y) = d/dy[ x*e^(-y) + k(y)] = -x(e^(-y)) + (dk/dy), setting equal to v(x,y) to solve for dk/dy v(x,y) = - xe^-y = -x(e^(-y)) + (dk/dy), so we see that dk/dy = 0. So k(y) = 0 + C So that any such function has a scalar potential of F of the form f(x,y) = -xe^(-y), choosing c =0 for simplicity. 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): I’m not following the same procedure, well I probably am just not seeing the connections. Remind me again how the curl plays a part in this problem?????? ********************************************* Question: 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: We first need a parameterization of the circle. x = 1*cos(t), y = (1)sin(t) With these parameter’s we will have to find the range of t -(`pi/2)<=t<=(`pi/2) Computing `ds dx/dt = -sin(t), dy/dt= cos(t) ds = `sqrt( (-sin(t))^2 + (cos(t)^2) dt = 1 dt The line integral is then, 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) Finally…-2Int((1 - cos(4t))/8 dt) = -2*(t/8 - 1/32sin(4t)) =-t/4 + sin(4t)/16 , -`pi/2,`pi/2 = -`pi/4 Int(cos(t))^4 dt,-`pi/2,`pi/2) = Int((cos(t)^2)^2 dt) = Int((1 +cos(2t)/2))^2 dt) Int((1 +cos(2t)/2))^2 dt) = Int((1/4)(1 +cos(2t)))^2 dt) =Int((1/4)( 1 + 2cos(2x) + cos^2(2x)) dt) = Int((1/4)( 1 + 2cos(2x) + (1 +cos(4x))/2) dt) finally we have Int((1/4)( 1 + 2cos(2x) + cos^2(2x)) dt) = (3/8)t +(1/4)sin(2t)+(1/32)sin(4t), evaluated = 3`pi/8 So that 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)^2*(sin(t))*(-sin(t))dt,-`pi/2,`pi/2) + Int(cos(t))^4 dt,-`pi/2,`pi/2) = -`pi/4 + 3`pi/8 = `pi/8 = approx. .3927 For 2nd line integral having trouble setting it it For 3rd line integral would equal 0 right, because it’s a closed path????? ?????I don’t think this is correct so next approach is….????? ????I’m getting really confused, I have spent almost 6 hrs over the last 2 days trying to fully understand so here what I’ve got??????? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Verify that the integral Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] is independent of path and evaluate it where C is any path from (0, pi/18) to (1, pi/6). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I’m still a little confused, even after looking at solution. So we don’t separate function up into i components and j components, we find curl just wrt entire integral? Also Can you break down how you calculated the Curl because I think I missing something???? Also I’m I right with my line of thinking that there are a few different logical processes going on in this section. 1) when given vector field we are to find whether it’s conservative or not, once it has been determined to be conservative or not is when we are to find the potential fn which we can integrate because it is easier or the only way to perform the integration? 2) When dealing with vector fields we can see integration as differences in init and end points of vector so that if we can parameterize we could use i component as init position and j component as end point? 3) Scalar potential? It seems there is so much info given within this section that it’s hard to separate it all out and figure out what applies to what and everything that is being said. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Mth 277 5/2 2 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): ********************************************* 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: du/dy = -e^-y dv/dx = -e^-y So because partial derivatives are equal F is a conservative. Int( u(x,y) dx) = Int( e^(-y) dx) = x*e^(-y) + k(y), where k(y) is constant that is fn. of y alone. f_y(x,y) = d/dy[ x*e^(-y) + k(y)] = -x(e^(-y)) + (dk/dy), setting equal to v(x,y) to solve for dk/dy v(x,y) = - xe^-y = -x(e^(-y)) + (dk/dy), so we see that dk/dy = 0. So k(y) = 0 + C So that any such function has a scalar potential of F of the form f(x,y) = -xe^(-y), choosing c =0 for simplicity. 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): I’m not following the same procedure, well I probably am just not seeing the connections. Remind me again how the curl plays a part in this problem??????
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* 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: I’m still a little confused, even after looking at solution. So we don’t separate function up into i components and j components, we find curl just wrt entire integral? Also Can you break down how you calculated the Curl because I think I missing something???? Also I’m I right with my line of thinking that there are a few different logical processes going on in this section. 1) when given vector field we are to find whether it’s conservative or not, once it has been determined to be conservative or not is when we are to find the potential fn which we can integrate because it is easier or the only way to perform the integration?
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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.