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Thus we must have: We can use this condition to find the potential function. But how do we know there is an f that will satisfy this
So a potential function exists only if: . Check this first, then move on to finding the potential function. Example:
Let First check So our potential function is:
Thus Conservative Vector Field There is more than one way to find the potential function. After we found: We could then take the derivative of this with respect to y to get: Set this equal to N(x,y) since Thus
The advantage of this way is that you only have to integrate once.
Let First check Thus Another perspective of the Potential Function:
Let’s work backward: Let y(x) be a solution to a first-order differential equation where y is a function of x and is
implicitly defined by:
f Differentiate each side with respect to x: Thus the differential equation is:
And we know the solution to this differential equation is: Let’s show our potential function from our last example satisfies the differential equation: Should satisfy the differential equation: x y x Plug and into the equation: If you graph the family of potential functions over the vector field you will find that the curves are perpendicular to the
vectors on the vector field. We’ll look at this example in Mathematica. We will use this idea in applications later.
It would also be nice to know the curves which are parallel to the vectors on the vector field. These curves would then
be the solutions to the differential equation: Instead of using arrows to represent vector functions we sometimes use families of curves called field lines, flow lines,
streamlines, or integral curves.
A curve y = f(x) is a field line of a vector function if at each point on the curve, is tangent to the curve. So we must solve the above differential equation to find these curves. This cannot
always be done symbolically and we would be stuck with graphical and numerical methods for the solutions.
We look at the...
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