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. 

 

 

 

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

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:
 

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).

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:
 

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: 

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

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:
 

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).

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:
 

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

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):