#$&* course Mth 277 Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
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Given Solution: del f = (x^3 y^2)_x i - (x sqrt(y) )_y j = 3 x^2 y^2 i - x / (2 sqrt(y)) j. If we know that the field F( x, y ) = F_1(x, y) i + F_2(x, y) j is conservative then the potential function f(x, y) has the following characteristics: F_1(x, y) = f_x (x, y) F_2(x, y) = f_y (x, y). Thus if we integrate f_x (x, y) = F_1 (x, y) with respect to x (treating y as constant) we should get our function f(x, y). The integration constant can be a function of y, since y is treated as a constant. If we integrate f_y (x, y) = F_2 (x, y) with respect to y (treating x as constant) we should also get our function f(x, y). This integration also has an integration constant which, since x is treated as constant, can be a function of x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. If F(x, y) = F_1(x, y) i + F_2(x, y) j = (3 x^2 y^2 - sqrt(y)) i + (2 x^3 y - x / (2 sqrt(y) + 2 y) j, then What is the general integral of F_1(x, y) with respect to x? What is the general integral of F_2(x, y) with respect to y? Is it possible to reconcile the two general integrals? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: What is the general integral of F_1(x, y) with respect to x? What is the general integral of F_2(x, y) with respect to y? Is it possible to reconcile the two general integrals? F(x, y)= F_1(x, y) i + F_2(x, y) j = (3 x^2 y^2 - sqrt(y)) i + (2 x^3 y - x / (2 sqrt(y) + 2 y) j f(x, y)=x^3y^2-xsqrt(y)+g(y)=x^3y^2-xsqrt(y)+y^2 f(x, y)=x^3y^2-xsqrt(y)+y^2+h(x)=x^3y^2-xsqrt(y)+y^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Integrating F_1 ( x, y) = 3 x^2 y^2 - sqrt(y) with respect to x, we treat y as a constant. Our general antiderivative is x^3 y^2 - x sqrt(y) + c, where c is an integration constant. Since y is treated as a constant, c can be any function of y only. So we can write our general antiderivative as f(x, y) = x^3 y^2 - x sqrt(y) + g(y), where g(y) is any function of y. Integrating F_2 (x, y) = 2 x^3 y - x / (2 sqrt(y)) + 2 y with respect to y, we treat x as a constant. Our general antiderivative is x^3 y^2 - x sqrt(y) + y^2 + c, where c is an integration constant. Since x is treated as a constant, c can be any function of x only. So we can write our general antiderivative as f(x, y) = x^3 y^2 - x sqrt(y) + y^2 + h(x), where h(x) is any function of x. Thus we have f(x, y) = x^3 y^2 - x sqrt(y) + g(y) and also f(x, y) = x^3 y^2 - x sqrt(y) + y^2 + h(x). These two expressions for f(x, y) appear to be different, but they do agree on the first two terms x^3 y^2 - x sqrt(y). If the remaining expressions g(y) from the first and y^2 + h(x) in the second can be reconciled, then the two expressions can be the same. Thus if we can find functions g(y) and h(x) with the property that g(y) = y^2 + h(x), we will have a solution. In fact this works out easily, if we just let g(y) = y^2 and h(x) = 0. Our first expression for the function becomes f(x, y) = x^3 y^2 - x sqrt(y) + g(y) = x^3 y^2 - x sqrt(y) + y^2 and our second becomes f(x, y) = x^3 y^2 - x sqrt(y) + y^2 + h(x) = x^3 y^2 - x sqrt(y) + y^2 It is easy to verify that for this function, del (f(x, y) ) is equal to our function F ( x, y ). Our function x^3 y^2 - x sqrt(y) + y^2 is called a scalar potential for F ( x, y ). It is worth noting that adding a constant number k doesn't affect this result. The function x^3 y^2 - x sqrt(y) + y^2 + k, where k is a constant number, is also a scalar potential for our function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK #$&* If a vector field F(x, y) has a scalar potential f(x, y), then if A = (x1, y1) and B = (x2, y2) are any two points of the x-y plane, then the integral F (x, y) dot `ds is path-independent for all paths connecting A and B, in the following sense: If C is any path starting at A and ending at B, then the integral of F (x, y) dot `ds over the path is equal to f(x2, y2) - f(x1, y1). It follows that if C1 and C2 are both paths both starting at A and ending at B, the integral of F (x, y) dot `ds over the path C2 is equal to integral of F (x, y) dot `ds over the path C1.