A Little Review: Calc 3, Div, Grad and Curl

A Little Review: Calc 3, Div, Grad and Curl The Gradient Given a scalar valued function, z = f (x, y) or w = f (x, y, z), then the gradient of f is a vector: z = f (x, y) ⇒ ∇f = hfx , fy i w = f (x, y, z) ⇒ ∇f = hfx , fy , fz i Two important facts about the gradient: • The directional derivative is how we measure the rate of change of f at x = a in the direction of the unit vector w: Dw (f ) = ∇f (a) · w • The maximum rate of change of f is in the direction of the gradient (and the magnitude of that change is the magnitude of the gradient). • The gradient vector at a point is orthogonal to the level curve (or level surface) at that point. For example, let f (x, y) = x2 + y 2 . The gradient is ∇f = hfx , fy i. Consider the level curve f (x, y) = k. Then the slope of the tangent line to a point on that curve is: fx + fy dy =0 dx ⇒ dy −fx = dx fy Thus, a vector moving in that direction: h1, −fx /fy i, or hfy , −fx i, and here its easy to see that this is orthogonal to the gradient. ~ is the heat flow, then the heat flow is in the opposite • If u(x, y) is the temperature and φ direction of the gradient of u. This is Fourier’s law extended to multiple dimensions: ~ = −K0 ∇u φ  • Often we use operator notation for the gradient: ∇ = ∂ ∂ ∂ , , ∂x ∂y ∂z  where ∇f is understood as hfx , fy , fz i. The Laplacian For shorthand, ∇2 = ∇ · ∇. For example, if u is a function of x and y, then:   ∂ ∂ 2 , · hux , uy i = uxx + uyy ∇ u = ∇ · (∇u) = ∂x ∂y This operation comes up so often, we give it a name: The Laplacian. The notation for it looks like ∆ (but it should be clear from the context what we’re talking about): ∆u = ∇2 u = uxx + uyy 1 Divergence and Curl The divergence of a vector field F = hP, Q, Ri is a scalar quantity: div(F) = Px + Qy + Rz or div(F) = ∇ · F In a fluid flow, the divergence is the flux per unit volume, or the rate of change of the density of the fluid. The curl of a vector field is a vector field. The definition is easiest to recall as the cross product: curl(F) = ∇ × F = i j k ∂/∂x ∂/∂y ∂/∂z P Q R = (Ry − Qz )i − (Rx − Pz )j + (Qx − Py )k (Remember the negative sign for the j term!) The curl measures the tendency of the fluid to rotate at that point, the curl points to the axis of rotation, and the magnitude is speed of rotation. Surface Integrals R We saw that evaluating line integrals C f (x, y) ds directly requires us to have a parametric form for the curve, r(t) = hx(t), y(t)i so that ds = |r 0 (t)| dt, and the integral is a standard integral in t. Similarly, to evaluate a surface integral directly also requires a parametric form for the surface (you might note that a “surface” is a mapping of the plane into three dimensions). The Stewart calculus text uses r(u, v) = hx(u, v), y(u, v), z(u, v)i where (u, v) ∈ D (domain is D). For example, if z = f (x, y) is the surface of some function, then an obvious way to parametrize it is: r(x, y) = hx, y, f (x, y)i Finally, a couple of facts about surface integrals: • The area of surface S parametrized by r over domain D is: ZZ ZZ dS = |ru × rv | dA S D Quick example: Set up a double integral representing the surface area of S, which is the graph of z = y + x2 , with domain D: 0 ≤ x, y ≤ 1. Then r = hx, y, y + x2 i ⇒ rx = h1, 0, 2xi ry = h0, 1, 1i rx × ry = h−2x, −1, 1i 2 Therefore, |rx × ry | = √ 4x2 + 2, and the surface area is: Z 1 Z 0 1 √ 4x2 + 2 dy dx 0 • The flux of a vector field F over a surface S with unit normal n: ZZ ZZ F · (ru × rv ) dA F · n dS = D S (Note: n = (ru × rv )/|(ru × rv )|) • Fourier’s Law: If the temperature at (x, y, z) is given by u(x, y, z), then the flux at that point is φ = −K∇u = −Khux , uy , uz i ⇒ ∇φ = −K∇2 u = −Khuxx , uyy , uzz i • The Divergence Theorem: Let E be a simple solid region with boundary surface S, with positive orientation. Then: ZZ ZZZ F · n dS = div(F) dV S E • Example: Find the flux of the vector field F = hz, y, xi over the unit sphere. SOLUTION: The divergence of F: div(F) = ∇ · F = Px + Qy + Rz = 0 + 1 + 0 = 1 Therefore, the flux of F across the unit sphere can be computed as a triple integral: ZZ ZZZ 4π F · n dS = 1 dV = Volume of unit sphere = 3 S E Conversion to Polar and Spherical Coordinates Finally, a quick review of coordinates: • Polar coordinates: (x, y) ↔ (r, θ) where r 2 = x2 + y 2 tan(θ) = y/x x = r cos(θ) y = r sin(θ) When integrating, dxdy becomes r dr dθ 3 • Cylindrical Coordinates: This coordinate system is polar in the xy plane, and then z is the third coordinate. That is, cylindrical coordinates are: (x, y, z) ↔ (r, θ, z) where r, θ are as in the previous section. When integrating, the volume element is dV = r dz dr dθ. • Spherical Coordinates: A sphere can be defined by angle φ (measures from the north pole down), the angle θ is in the xy−plane as normal, and its radius ρ. The conversion formulas are: x = r cos(θ) r = ρ sin(φ) ρ2 = x2 + y 2 + z 2 y = r sin(θ) cos(φ) = z/ρ z = ρ cos(φ) cos(θ) = x/r When integrating, dx dy dz is replaced by ρ2 sin(φ) dρ dθ dφ Exercise Set for Section 1.5 1. Show that u(x, y, t) = 4kt + x2 + y 2 solves ut = k∇2 u 2. Show that, if u(x, t) and v(y, t) each solve the one dimensional heat equation (with the same constant k), then w(x, y, t) = u(x, t)v(y, t) will solve the two dimensional heat equation, ut = k∇2 u 3. (*) Verify the divergence theorem directly, by computing the flux integral directly, then comparing it to the volume integral of the divergence, if the vector field is F = h0, 0, zi and S is the planar surface with corners (0, 0, 0), (1, 0, 0), (0, 1, 3), and (1, 1, 3). NOTE: The purpose of this question is to review, so open a calc book as necessary. 4. Use the divergence theorem to compute the flux of the vector field: hx2 + y 2 , y 2 + z 2 , x2 + z 2 i through the cube: 0 ≤ x, y, z ≤ 1. NOTE: In this case, use the divergence theorem and don’t compute the flux integral directly (you would have to compute it for each side of the cube). 5. (*)(1.5.11 in the text) 6. (1.5.13 in the text- You may assume that we’ve already shown 1.5.12(c) 4