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Your book defines the first and second order Taylor polynomials on page 196. Using more compact notation, we can write the Taylor Forumla for the function f about the point x0 as p2 (x) = f (x0 ) + Df (x0 )(x − x0 ) + (x − x0 )t Hf (x0 )(x − x0 ) 1. Write out Taylor’s Forumla explicitly for scalar valued functions of one, two, and three variables. In dimension 1 this is the same Taylor polynomial that we worked with in single variable calculus. We will let our point a = a. p1 (x) = f (a) + f 0 (a)(x − a) p2 (x) = f (a) + f 0 (a)(x − a) + f 00 (a) (x − a)2 2 Now for n = 2 the two dimensional case, rather than using x1 and x2 as variables we will use x and y, and we will write a = (a, b) p1 (x, y) = f (a) + fx (a)(x − a) + fy (a)(y − b) p2 (x, y) = f (a) + fx (a)(x − a) + fy (a)(x − a) + fxx (a) fyy (a) (x − a)2 + (y − b)2 + fxy (a)(x − a)(y − b) 2 2 Finally if n = 3, the 3 dimensional case, we will use x, y, z for our variables, and let the point a = (a, b, c) we have p1 (x, y, z) = f (a) + fx (a)(x − a) + fy (a)(y − b) + fz (a)(z − c) p2 (x, y, z) = f (a) + fx (a)(x − a) + fy (a)(y − b) + fz (a)(z − c) + 2. Consider the function f (x, y) = sin(xy) and • Compute the Hessian matrix of f at the point a = (0, 0). he Hessian matrix in this situation will be given as H= ∂2f ∂x2 ∂2f ∂2f ∂y∂x ∂y 2 ∂2f ∂x∂y ! To begin we will compute ∂f = fx (x, y) = y cos(xy) ∂x ∂f = fy (x, y) = x cos(xy) ∂y fxx (a) fyy (a) fyy (a) (x − a)2 + (y − b)2 + (z − c)2 + f 2 2 2 Building off of this result, we can easily compute the whole Hessian Which we can easily compute explicitly H= −y 2 sin(xy) cos(xy) − xy sin(xy) cos(xy) − xy sin(xy) −x2 sin(xy) • Compute the degree 2 Taylor Polynomial of f (x, y) at the point a = (0, 0) Using the general form we compute above, and the Hessian matrix (which we also computed) p2 (x, y) = sin(0 · 0) + 0 · cos(0 · 0)(x − 0) + 0 · cos(0 · 0)(y − 0) − 02 sin(0 · 0) − 02 sin(0 · 0) + cos(0 · 0) − 0 · 0 sin(x 3. Find the second order Taylor expansion about the point (0, 0) of the function f (x, y) = exy We begin by computing the matrix of partial derivatives of f . Df (x, y) = (exy y, exy x) From this we compute the Hessian matrix Hf (x, y) = exy y 2 xy e + exy xy exy + exy xy exy x2 Then we evaluate at the point (0, 0) and find Df (0, 0) = (0, 0) 0 1 Dg(0, 0) = 1 0 Now we put these together to compute the degree 2 Taylor polynomial 1 x T2 (x, y) = f (0, 0) + Df (0, 0) · (x, y) + (x, y)Hf (0, 0) y 2 Expanding out the linear algebra we obtain T2 (x, y) = 1 + xy Which is the degree 2 Taylor polynomial about (0, 0) of the function f (x, y) = exy 4. Find and classify all critical points of the function f (x, y) = x3 + x2 y + y 2 + xy + x + 1 As in the last problem we will begin by computing both the matrix of partial derivatives and the Hessian matrix. Df (x, y) = (3x2 + 2xy + y, x2 + 2y + x) 6x + 2y 2x + 1 Hf (x, y) = 2x + 1 2 Now to find the critical values we compute where Df (x, y) = (0, 0) We easily find the following points (0, 0), (1, −1), and (1/2, −3/8). Now using our criteria on each of these points we find that the points (0, 0) and (1, −1) are both saddle points, while (1/2, −3/8) is a local minimum. 5. Let F(x, y, z) = (xy 2 , yx2 , −z 2 ), use the divergence theorem to calcualte the total flux out of a canister W , which can be expressed by the equations x2 + y 2 ≤ 9 and 1 ≤ z ≤ 3. We will solve this problem using the divergence theorem. As the problem is stated we want compute a flux (surface) integral over the boundary of W . That is we want to compute ZZ FdS ∂W Using the divergence theorem this becomes ZZZ ÷W dA W It is not difficult to compute the div of F ∂ ∂ ∂ xy 2 + yx2 + −z 2 = ∂x ∂y ∂z 2 2 = y + x − 2z ÷F = So we setup the integral according to the divergence theorem, and writing the region in cylindrical coordinates ZZZ W = y 2 + z 2 − 2z = Z3 Z 1 0 2π Z 0 3 (r2 − 2z)rdrdθdz = 9π 6. Use the Divergence Theorem to compute the value of the flux integral ZZ FdS S Where F(x, y, z) = (y 3 + 3x, xz + y, z + x4 cos(x2 y)) and S is the boundary of the region bounded by x2 + y 2 = 1, x ≥ 0, y ≥ 0 and 0 ≤ z ≤ 1 Using the divergence theorem we will compute this rather difficult integral into a (hopefully) simpler triple integral over the divergence of F. To begin let us compute ÷F ÷(F) = ∂ ∂ ∂ , , ∂x ∂y ∂z · y 3 + 3x, xz + y, z + x4 cos(x2 y) = 5 We then can setup the triple integral as ZZZ 5dV W Looking at the region we want to compute in cylindrical coordinates as π Z1 Z2 Z1 5rdrdθdz = 0 0 5π 4 0 Which is the value of the surface integral, by the Divergence theorem.