Unformatted text preview: Mathematica solution of this differential equation……
The only differential equations we will solve are the Exact Differential Equation to find the Potential function and a
Differential Equation that can be solved by Separation of Variables.
Find equations for the flow lines for the vector field defined by:
Thus the differential equation is:
So we have a family of curves: We will look at the vector field and flow lines in Mathematica…. Summary of 14.1:
Use Mathematica to graph vector fields, streamlines, and potential curves
Know how to calculate a gradient field
Determine if a vector field is conservative and if so find its potential function
Find the equations for flow lines Section 14.2 Line Integrals
A line integral is a generalization of a definite integral. A definite integral chops up the x-axis which is a straight line (1D).
Why don’t we take some arbitrary curve C in 2D space (xy-plane) instead and chop that up! We had
length of one of the subintervals on the x-axis. We still want to maintain this idea but since C is not a straight line we
have to measure the length of the subinterval with arc length
. An application of the definite integral was the area
under the curve y=f(x). Since our curve C is in 2D our function above it must be a surface
. The area then
represents a curtain that extends from the xy-plane to the surface. Why not extend this idea to 3D and have the curve
floating in space. Then the function
could be considered the height at each point on the curve C . So our
curtain area would be floating in 3D space. This leads to the definition for the line integral with respect to Arc Length: The curve C is called the path of integration. Let a denote the start of the path and b the end of the path.
If the curve C is parameterized by arc length then the line integral is simple to evaluate: Example:
Curve C: where Find:
We must first parameterize the curve with respect to arc length.
Thus our new parameterization:
There should be an easier way and there is…..
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