MAT 211 Final Exam. Spring Jennings. Show your work!

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1 MAT 211 Final Exam. pring 215. Jennings. how your work! Hessian D = f xx f yy (f xy ) 2 (for optimization). Polar coordinates x = r cos(θ), y = r sin(θ), da = r dr dθ. ylindrical coordinates x = r cos(θ), y = r sin(θ), z = z, dv = r dr dθ dz. pherical coordinates x = ρ cos(θ) sin(φ), y = ρ sin(θ) sin(φ), z = ρ cos(φ), dv = ρ 2 sin(φ) dρ dφ dθ. Q Green s Theorem P, Q dr = R x P da if R is a closed, bounded region in the plane and y is its positively oriented boundary curve. Increment of area on surface parametrized by r(u, v) = (x(u, v), y(u, v), z(u, v)): d = r u r v du dv. nd = d = r u r v du dv. tokes Theorem F dr = curl F n d if is a surface in 3D space and is its oriented boundary curve. Divergence Theorem F nd = div F dv if B is a bounded 3D solid with outward-oriented boundary surface. B 1. (1 points) Find the angle between the vectors v = 1, 2, 3 and w = 1,, 1. If θ is the angle between the vectors then so v w = v w cos(θ) v w = 4 v = = 14 w = = 2 ( ) 4 θ = arccos Let find A = 3, 2, 1, B =, 1, 1

2 (a) (1 points) A vector that is perpendicular to both A and B. A B = 1, 3, 3 (b) (1 points) The area of the parallelogram whose edges are A and B. A B = = 19 sq. units (c) (1 points) Find equation(s) for the plane through the points P = (1, 2, 3), Q = P + A = (4, 4, 4), R = P + B = (1, 3, 4). The vector A B = 1, 3, 3 is perpendicular to the plane so is an equation for the plane. = 1(x 1) 3(y 2) + 3(z 3) (d) (1 points) Find equation(s) for the line that passes through the points P and Q. The vector A points from P to Q so (x(t), y(t), z(t)) = P + ta = (1, 2, 3) + t 3, 2, 1 = (1 + 3t, 2 + 2t, 3 + t) parametrizes the line. 3. A curve is parametrized by Find (a) (5 points) its velocity at time t = 1, r(t) = t, t 2, t 3. r (t) = 1, 2t, 3t 2 r (1) = 1, 2, 3 (b) (5 points) its acceleration at time t = 1. r (t) =, 2, 6t r (1) =, 2, 6 4. (1 points) ketch a contour plot for the function s(x, y) = x 2 + y 2 Page 2

3 showing at least four level curves. On the same plot sketch at least four different gradient vectors s. Don t worry too much about the length of s but be sure they point in the right direction. Below is a plot that I created with age. tarting from the center the contour liness are s(x, y) = 1, 2, 3, 4, 5, 6, 7. (s = is the pont at the center, it is a an absolute minimum of s). The arrows are gradients of s, rescaled so they all fit. The gradient vectors are longer near the edges because s increases fastest near the edges, and the gradients are perpendicular to the level courves. Of course if you plot this by hand then you won t have all this detail, but at least the shape of the level curves and the directions of the gradients should be right, so they should be perpendicular to the level curves and point away from the center in the direction where s increases the fastest (1 points) Let f(x, y, z) = x 2 + 2xy + yz 2 Find the directional derivative of f at the point P = (1, 2, 3) in the direction of the unit vector (u) = 1/9, 4/9, 8/9. f u = f P u P f = 2x + 2y, 2x + z 2, 2yz f (1,2,3) = 6, 1, 12 f u = 6, 1, 12 1/9, 4/9, 8/9 = 146 P 9 6. (1 points) Find all the local maxima, minima, and saddle points of the function p(x, y) = x 3 3x + y 3 12y. Page 3

4 The critical points (x, y) occur where and so Apply the second derivative test. = p x = 3x 2 3 = 3(x + 1)(x 1) = p y = 3y 2 12 = 3(y + 2)(y 2) critical points are ( 1, 2), ( 1, 2), (1, 2), (1, 2). p xx = 6x, p xy =, p yy = 6y D = p xx p yy p 2 xy = 36xy (x, y) p xx p yy p 2 xy p xx type ( 1, 2) 72 6 local maximum ( 1, 2) 72 doesn t matter saddle (1, 2) 72 doesn t matter saddle (1, 2) 72 6 local minimum 7. (1 points) Find the maximum and minimum of the function f(x, y) = xy on the ellipse x 2 + 4y 2 = 4. either x = y = or x, y, λ and objective: f(x, y) = xy constraint: g(x, y) = x 2 + 4y 2 = 4 Lagrange multiplier condition: f = λ g y, x = λ 2x, 8y y = λ2x, and x = λ8y y 2x = λ = x 8y y 2x = x 8y 8y 2 = 2x 2 4y 2 = x 2 Page 4

5 ince x 2 + 4y 2 = 4 it follows that x = y = is impossible so x 2 + 4y 2 = 4 and 4y 2 = x 2 4y 2 + 4y 2 = 4 y = ± 1 2 and x = ± 2 The ellipse is a closed, bounded curve and the objective function f(x, y) is continuous so there must be an absolute max and min on the ellipse. They must occur at critical points. Evaluating f at at the critical points, one obtains (x, y) f(x, y) = xy type ( 2, 1/2) 1 absolute max ( 2, 1/2) 1 absolute min ( 2, 1/2) 1 absolute min ( 2, 1/2) 1 absolute max 8. (1 points) Let T be the solid tetrahedron with vertices (,, ), (1,, ), (, 1, ) and (,, 1). At each point (x, y, z) in the tetrahedron the mass density is mass density: µ(x, y, z) = x grams per cubic centimeter. Find the mass of the tetrahedron. (Please work this integral all the way out until you get a number, and show your work!) The tetrahedron has a triangular base bounded by the x-axis, the y-axis, and the line x + y = 1 in the x, y-plane. The plane x + y + z = 1 forms the roof of the tetrahedron. Thus (x, y, z) is in the tetrahedron if and only if x 1 and y 1 x and z 1 x y Page 5

6 so the mass is 1 1 x 1 x y 1 1 x = = 1 x dz dy dx = x x 2 xy dy dx = 1 1 x 1 xz z=1 x y z= x x 2 x 2 + x (x 2x2 + x 3 ) dx = = = 1 24 dy dx xy x 2 y 1 2 xy2 y=1 x 1 y= dx 1 2 x x x3 dx 9. (1 points) hange the order of integration but do not integrate. Just set up any integral that you d use, you do not need to work it out x xy 2 dy dx 6 y/3 xy 2 dx dy 1. (1 points) onsider the 2-dimensional vector field F = 2x + 3y, 3x + 8y. Is F the gradient F = f of some function f? If it is a gradient find the function f. If it is not a gradient, show why it is not a gradient. F is (locally) a gradient because it satisfies the mixed partials are equal condition: (2x + 3y) y = 3 = It remains to find the function f where f = F. We need 3x + 8y x f = 2x + 3y, x (1) f = 3x + 8y, y (2) Page 6

7 Equation (1) says f(x, y, z) = 2x + 3y dx = x 2 + 3xy + (y) (3) for some function (y) that depends only on y. Then equation (2) says hence 3x + 8y = f y = 3x + (y) 8y = (y) so (y) = 8y dy = 4y 2 + K (4) for some genuine constant K. Plug equation (4) into equation (3) to obtain f(x, y) = x 2 + 3xy + 4y 2 + K (In this context it s OK to assume K = ). 11. (1 points) Find the line integral G dr where G(x, y) = yî + xĵ and is the upper half of the unit circle, oriented counterclockwise : x 2 + y 2 = 1 and y. Parametrize by r(t) = cos(t), sin(t), t π dr = sin(t), cos(t) dt Along the vector field is G = y, x = sin(t), cos(t) so G dr = π sin(t), cos(t) sin(t), cos(t) dt = π sin 2 (t) + cos 2 (t) dt = π 12. (1 points) Let be an oriented curve that starts at (1, 2, 3) and ends at (7, 8, 9). Let Find the line integral of the gradient of g over : g(x, y, z) = x(z y). g dr Page 7

8 (Hint: if this looks hard, stop and think again!) By the fundamental theorem of calculus g dr = g(7, 8, 9) g(1, 2, 3) = 7 1 = (1 points) Let H be the vector field H(x, y) = x 4, y 3 and let be a circle of radius R centered at the origin, oriented in the counterclockwise direction. Use Green s theorem to calculate the line integral H dr. Let D be the interior of the circle. y 3 H dr = x x4 y da = D D da = 14. (1 points) A surface in 3 dimensional space is parametrized by the function Find the area of the surface. r(u, w) = u + 1, 2u + 3w + 5, 4w + 1, u 1, w 2. area = da = 2 1 r u r du dw w r = 1, 2, u r =, 3, 4 w r u r = 8, 4, 3 w so the area is du dw = (1 points) Use the divergence theorem to calculate the flux F n dr where F = x y, y z, z x Page 8

9 and is the tin-can-shaped surface that forms the boundary of the solid cylinder oriented with outward pointing normal vector. B : x 2 + y 2 1 and z 1 Let B be the interior of the cylinder. The divergence theorem says F n dr = div F dv also so div F = (x y) x F n dr = (y z) + y B + B (z x) z = = 3 3 dv = 3(volume of B) = 3π Page 9