Final answer:
By computing the necessary partial derivatives, it was determined that the first and third vector fields are conservative because the partial derivatives matched; the second vector field is not conservative as they were different. Furthermore, the magnitude of a specific conservative force at x=y=1 m is 4√2 N/m.
Step-by-step explanation:
To determine whether a vector field f is conservative, one must compute the first order partial derivatives and check if the partial derivative of the x-component of the force with respect to y (dFx/dy) is equal to the partial derivative of the y-component of the force with respect to x (dFy/dx). This equality is a necessary condition for a force to be conservative. A conservative force is related to a potential energy function, and therefore the force can be expressed as the negative gradient of that potential energy function.
Let's examine each given vector field:
- f(x,y)=(6x-4y)i+(-4x+14y)j
Let's calculate the partial derivatives:
(dFx/dy) = -4
(dFy/dx) = -4
Since these partial derivatives are equal, the force is conservative. - f(x,y)=3yi+4xj
For this vector field, we have:
(dFx/dy) = 3
(dFy/dx) = 4
The partial derivatives are different, which means this force is not conservative. - f(x,y)=(3siny)i+(-8y+3xcosy)j
Here, the partial derivatives are:
(dFx/dy) = 3cosy
(dFy/dx) = 3cosy
As the derivatives are equal, the force field is conservative.
Lastly, to find the magnitude of the force at the point x = y = 1 m for a conservative force satisfying (dFx/dy) = (dFy/dx) = (4 N/m3)xy, we use the given condition:
Fx = 4xy = 4(1)(1) = 4 N/m
Fy = 4xy = 4(1)(1) = 4 N/m
Thus, by applying the Pythagorean theorem, the magnitude of the force is √(Fx2 + Fy2) = √(16 + 16) = √32 = 4√2 N/m.