<p> Calc 3 Lecture Notes Section 14.1 Page 1 of 10</p><p>Section 14.1: Vector Fields Big idea: A vector field is used to represent situations where there is a vector associated with every point in space.</p><p>Big skill: You should be able to graph vector fields, compute flow lines for velocity fields, and compute gradient fields and potential functions.</p><p>Definition 1.1: Vector Field A vector field (in the plane) is a function F(x, y) that maps points in R2 into the set of two- dimensional vectors V2. We write this as:</p><p>F( x, y) = f1( x , y) , f 2( x , y) = f 1( x , y) i + f 2 ( x , y) j</p><p> for scalar functions f1(x, y) and f2(x, y). </p><p>In 3D space, a vector field is a function F(x, y, z) that maps points in R3 into the set of three- dimensional vectors V3. We write this as:</p><p>F( x, y , z) = f1( x , y , z) , f 2( x , y , z) , f 3( x , y , z) = f 1( x , y , z) i + f 2( x , y , z) j + f 3 ( x , y , z ) k</p><p> for scalar functions f1(x, y, z) and f2(x, y, z), and f3(x, y, z).</p><p>We can visualize vector fields by drawing a collection of vectors F(x, y), each with their initial point at (x, y).</p><p>Some important vector fields you should know: (There is a decent vector field graphing tool online at http://www.pierce.ctc.edu/dlippman/g1/GrapherLaunch.html)</p><p>Uniform vector fields Example: F(x, y) = 1, 0. Calc 3 Lecture Notes Section 14.1 Page 2 of 10</p><p>Radial Vector Fields Example: F(x, y) = x, y.</p><p>Example: F(x, y) = -x, -y.</p><p> x, y Example: F( x, y) = x2+ y 2</p><p>-x, - y Example: F( x, y) = x2+ y 2 Calc 3 Lecture Notes Section 14.1 Page 3 of 10</p><p> x, y , z Example: F( x, y , z) = x2+ y 2 + z 2</p><p> x, y , z F x, y , z = Example: ( ) 3/ 2 ( x2+ y 2 + z 2 )</p><p>Rotational (Circular) Vector Fields Example: F(x, y) = -y, x.</p><p>Example: F(x, y) = y, -x. Calc 3 Lecture Notes Section 14.1 Page 4 of 10</p><p>-y, x Example: F( x, y) = x2+ y 2</p><p>Velocity Fields Many times vector fields are used to represent the velocity of a particle as a function of its position in space, like the air flow over a car in a wind tunnel test. When this is the case, we can find the path of a particle that starts at any given point (x0, y0) by solving the differential equations: v= F( x, y)</p><p> xⅱ( t), y( t) = f1( x , y) , f 2 ( x , y)</p><p>� x( t) f1 ( x, y)</p><p>� y( t) f2 ( x, y)</p><p> with initial conditions x(t0) = x0, and y(t0) = y0. The path followed by a particle for a given initial condition is called a flow line.</p><p>Practice: 1. Find the flow lines for the uniform vector field F(x, y) = 1, 0. Calc 3 Lecture Notes Section 14.1 Page 5 of 10</p><p>2. Use winplot to graph the vector field F(x, y) = 2, 1 + 2xy and numerically compute some flow lines.</p><p>In the second problem above, we have not learned how to solve that type of couple differential equation. However, there is a technique to obtain the equation of a flow line, even if we can not find x(t) and y(t) explicitly. This technique involves using the chain rule to obtain a differential equation for y = f(x). </p><p>Let’s say x = g(t) and y = h(t) for some functions g and h that are too hard to compute from the velocity field. Note that x = g(t) = f1(x, y) and y = h(t) = f2(x, y). By the chain rule, dt 1 dy dy dt = = . We can show that dx dx using theorem 5.2 from section 2.5 on page 192: dx dt dx dt Calc 3 Lecture Notes Section 14.1 Page 6 of 10</p><p> x= g( t) t= g-1 ( x) dt d = g-1 ( x) dx dx 1 = g( g-1 ( x)) 1 = g( t) 1 = dx dt</p><p>So, dy dy dt = dx dt dx dy dt = dx dt y( t) = x( t) dy f( x, y) = 2 dx f1 ( x, y)</p><p>When the right hand side is separable, we know how to solve this differential equation…</p><p>Practice: 1. Find the flow lines for the circular vector field F(x, y) = y, -x. Calc 3 Lecture Notes Section 14.1 Page 7 of 10</p><p>2. Find the flow lines for the radial vector field F(x, y) = x, y.</p><p>Definition 1.2: Gradient Fields and Potential Functions The vector field F = f is called gradient field for any scalar function f(x, y). f is also called the potential function for F. Whenever F = f for some scalar function f, we refer to F as a conservative vector field.</p><p>Practice: 1. Compute the gradient field for f (x, y)  x 2 y  y 3 . Calc 3 Lecture Notes Section 14.1 Page 8 of 10</p><p>1 2 2 2 f( x, y , z) = 2. Compute the gradient field for f( x, y , z) = x + y + z and 2 2 2 . x+ y + z Compare them to the unit radial vector field.</p><p>Clearly, given any differentiable scalar function f (x, y) , one can always define a corresponding vector field by computing f .</p><p>However, the converse of this statement is not true. That is, given a vector field F(x, y), it is not always possible to find a corresponding potential function f(x, y) such that F = f (but sometimes it is…).</p><p> f f To find a potential function, either integrate or , making sure to leave your constant of x y integration as a function of the other variable, then take the partial derivative of your answer with respect to the other variable to find the “constant function” of integration.</p><p>Practice: 1. Find a potential function for the vector field F(x , y )= 6 x + 5 y , 5 x + 4 y . Calc 3 Lecture Notes Section 14.1 Page 9 of 10</p><p>2. Find a potential function for the vector field F(x , y , z )= y cos( xy ), x cos( xy ), - sin z .</p><p>3. Show that the vector field F(x , y )= y , - x has no potential function. (This example illustrates a non-conservative vector field; i.e., not every vector field is a gradient field.) Calc 3 Lecture Notes Section 14.1 Page 10 of 10</p><p>4. Find the electric fields generated by a single positive charge, by a single negative charge, and by an electric dipole, which consists of equal magnitude positive and negative charges separated by a distance d.</p>