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Unformatted text preview: Vector Calculus 1 The first rule in understanding vector calculus is draw lots of pictures. This subject can become rather abstract if you let it, but try to visualize all the manipulations. Try a lot of special cases and explore them. Keep relating the manipulations to the underlying pictures and don’t get lost in the forest of infinite series. Along with the pictures, there are three types of derivatives, a couple of types of integrals, and some theorems relating them. 9.1 Fluid Flow When water or any fluid moves through a pipe, what is the relationship between the motion of the fluid and the total rate of flow through the pipe (volume per time)? Take a rectangular pipe of sides a and b with fluid moving at constant speed through it and with the velocity of the fluid being the same throughout the pipe. It’s a simple calculation: In time Δ t the fluid moves a distance v Δ t down the pipe. The cross-section of the pipe has area A = ab , so the volume that move past a given flat surface is Δ V = Av Δ t . The flow rate is the volume per time, Δ V/ Δ t = Av . (The usual limit as Δ t → isn’t needed here.) A v Δ t (a) A v Δ t (b) Just to make the problem look a little more involved, what happens to the result if I ask for the flow through a surface that is tilted at an angle to the velocity. Do the calculation the same way as before, but use the drawing (b) instead of (a). The fluid still moves a distance v Δ t , but the volume that moves past this flat but tilted surface is not its new (bigger) area A times v Δ t . The area of a parallelogram is not the product of its sides and the volume of a parallelepiped is not the area of a base times the length of another side. A h v Δ t α ˆ n α The area of a parallelogram is the length of one side times the perpendicular distance from that side to its opposite side. Similarly the volume of a parallelepiped is the area of one side times the perpendicular distance from that side to the side opposite. The perpendicular distance is not the distance that the fluid moved ( v Δ t ). This perpendicular distance is smaller by a factor cos α , where α is the angle that the plane is tilted. It is most easily described by the angle that the normal to the plane makes with the direction of the fluid velocity. Δ V = Ah = A ( v Δ t )cos α The flow rate is then Δ V/ Δ t = Av cos α . Introduce the unit normal vector ˆ n , then this expression can be rewritten in terms of a dot product, Av cos α = A~v . ˆ n = ~ A . ~v (9 . 1) James Nearing, University of Miami 1 9—Vector Calculus 1 2 where α is the angle between the direction of the fluid velocity and the normal to the area. ~ C ~ D θ ~ C × ~ D This invites the definition of the area itself as a vector, and that’s what I wrote in the final expression. The vector ~ A is a notation for A ˆ n , and defines the area vector. If it looks a little odd to have an area be a vector, do you recall the geometric interpretation of a cross product? That’s the vector perpendicular togeometric interpretation of a cross product?...
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This note was uploaded on 01/30/2012 for the course PHYS 315 taught by Professor Nearing during the Fall '08 term at University of Miami.
- Fall '08