If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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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.
If our F function is the divergence del f of some function f(x, y), our work integral over a path C will be just a function of the initial point (x(t1), y(t1)) and the final point (x(t2), y(t2)). We call f a potential function and our line integral is the potential difference of the function f between the points.
It is easy to prove that the curl of a conservative field is 0. Since a conservative field F is the gradient of a potential function f, we need only prove that del X del f = 0. Using determinant notation and the fact that del f = f_x i + f_y j + f_z k, this is only a matter of writing out the determinant, evaluating it and showing that the result is zero.
It takes a little more work to prove that if the curl of a vector field F is always zero, then the field is the gradient of some function f, and is therefore conservative.
Question:
`q001. Find F ( x, y ) = del f, if f(x, y) = x^3 y^2 - x sqrt(y). Then find del X F(x, y).
Your solution:
Confidence rating:
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:
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):
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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
Your solution:
Confidence rating:
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.
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#$&*
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.
Question: `q003. For the function F(x, y) = (3 x^2 y^2 - sqrt(y)) i + (2 x^3 y - x / (2 sqrt(y) + 2 y) j of the preceding problem, find the line integral of F dot `ds for the straight line connecting the points (0, 0) and (2, 1), and also the line integral of F dot ds for the path defined by x = 2 t, y = t^2, 0 <= t <= 1.
Your solution:
Confidence rating:
Given Solution:
The 'cheap' solution is as follows:
- The second path connects the t = 0 point (2 * 0, 0^2) = (0, 0) to the point (2 * 1, 1^2) = (2, 1).
- So does the first path.
The scalar potential f(x, y) corresponding to the given function F (x, y) was found in the preceding problem to be f(x, y) = x^3 y^2 - x sqrt(y) + y^2.
The integral along any path connecting (0, 0) and (2, 1) is therefore
- integral = f(2, 1) - f(0, 0) = 2^3 * 1^2 - 2 sqrt(1) + 1^2 = 8 - 2 + 1 = 7.
We can detail the solution for either path:
For the second path, dx/dt = 2 and dy/dt = 2 t, so dx = 2 dt and dy = 2 t dt so `ds = ( 2 i + 2 t j ) `dt.
It is straightforward to calculate F dot `ds and verify that the integral does indeed come out to be 7.
For the first path we can use the parameterization x = 2 t, y = t and obtain the same result.
Self-critique (if necessary):
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