Solution of final examination

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1 of final examination Math 20, pring 201 December 9, 201 Problem 1 Let v(t) (2t e t ) i j + π cos(πt) k be the velocity of a particle with initial position r(0) ( 1, 0, 2). Find the accelaration at the point (e 1, 1, 2). We have r(t) r(0) t 0 v(s) ds r(t) (e 1, 1, 2) + t 0 (2s e s ) i j + π cos(πs) k ds (e 1, 1, 2) + (t e t ) i t j + sin(πt) k (t 2 + e e t, 1 t, 2 + sin(πt)) Thus the point (e 1, 1, 2) corresponds to parameter t 0. For getting acceleration, we simply differentiate the expression of v(t) to obtain a(t) (2 + e t ) i π 2 sin(πt) k Hence accelaration at point (e 1, 1, 2) is a(0) (, 0, 0). Problem 2 (a) Find the equation of the plane that passes through the points (2, 2, 0), (,, 1) and (0,, ). (b) Find the parametric equation of the line that passes through the point (1, -2, 1) and is perpendicular to the plane in part (a). 1

2 (a) To get the equation of plane we first get two vectors along the plane. o here we can choose them as v 1 (,, 1) (2, 2, 0) < 1, 1, 1 > v 2 (0,, ) (2, 2, 0) < 2, 5, > o perpendicular vector to the plane will be i j k n v 1 v < 2, 1, > 2 5 Thus the equation of plane is given by (choosing point (2, 2, 0) as base point) < x 2, y 2, z > < 2, 1, > 0 2x 4 + y 2 z 0 2x + y z 6 (b) Perpendicular vector to the plane is already calculated in previous part as n < 2, 1, >. ince the sought-after line is perpendicular to plane so this will be a vector along the plane. Thus the parametric equation of line is going to be where t R. Problem Let x(t) 1 + 2t y(t) 2 + t z(t) 1 t f(x, y) x2 z y (a) Find the direction in which f is increasing fastest at the point (2, 2, 1). (b) Find the rate of change of f at the point (2, 2, 1) in the direction of the point (, 1, 0). (a) f increases fastest in the direction given by f, where f ( 2xz y, x2 z y, x2 2 y ). o the direction is f(2, 2, 1) < 2, 1, 2 > 2

3 (b) The vector along the direction of change is given by (, 1, 0) (2, 2, 1) < 1, 1, 1 >. o the unit vector along this direction is u 1 < 1, 1, 1 > Hence the rate of change along the given direction is f 1 < 1, 1, 1 >< 2, 1, 2 > 1 < 1, 1, 1 > 1 ( ) 1 Problem 4 Let f(x, y) x 2 y 2 4x + 2y + 2 (a) Find the point on the graph of f where the tangent plane is parallel to the xy plane. (b) Find the equation of the tangent plane to the graph of f at the point found in part (a) (a) We recall that the normal vector to tangent plane of graph of function z f(x, y) is given by o in the above example we have n < f, f, 1 > n(x, y, z) ( 2x + 4, 2y 2, 1) Thus the set of points on graph of f where tangent plane is parallel to XY plane i.e. the set of points on graph of f where normal vector to tangent plane is along z axis is { (x, y, z) R 2x + 4 0, 2y 2 0, z f(x, y) } which is (2, 1, f(2, 1)) or (2, 1, 1). (b) The normal vector at the point (2, 1, -1) is n(2, 1, 1) < 0, 0, 1 > Thus the equation of tangent plane is given by < x 2, y 1, z + 1 > < 0, 0, 1 > 0 z Note that we could have directly written the answer as z 1 since (by part (a)) wqe know that tangent plane is parallel to XY plane and the z coordinate of point lying on tangent plane, i.e. z coordinate of point (2, 1, 1) is -1.

4 Problem 5 Find the extreme values of the function on the region x 2 + y 2 16 f(x, y) 2x 2 + y 2 4x 5 We first look at the interior of the domain, which is an open disk of radius 4. Here the critical points of the function is given by f 0, f 0 4x 4 0, 6y 0 (x, y) (1, 0) ince the point (1, 0) falls inside the open domain so we record it in the table (printed at the bottom of answer sheet). Now, along the boundary we have x 2 + y 2 16, so to extremize with this constraint we will use Lagrange multiplier method. Let us define the auxillary function g(x, y, λ) as g(x, y, λ) f(x, y) λ(16 x 2 y 2 ) 2x 2 + λx 2 + y 2 + λy 2 4x 16λ 5 Thus the conditions g g 0, 0, as well as constraint x2 + y 2 16 give 4x + 2λx 4 0 6y + 2λy 0 x 2 + y 2 16 In above system of equations the second equation yields y 0 or λ. From there the solutions, we get are: { (x, y, z) (4, 0), ( 4, 0), ( 2, 2 ), ( 2, 2 } ) (x, y) f(x, y) Behavior (1, 0) -7 Minimum (4, 0) 11 (-4, 0) 4 ( 2, 2 ) 47 Maximum ( 2, 2 ) 47 Maximum Problem 6 Evaluate the iterated integral I π 0 π y sin(x 2 ) dx dy 4

5 ince we don t know the anti-derivative of function sin(x 2 ) so we switch the order of integral to get I y π y0 x π xy sin(x 2 ) dx dy x π x0 yx y0 sin(x 2 ) dy dx x π x0 x sin(x 2 ) dx 1 2 cos(x2 ) x π x0 1 ( 1 1) 1 2 Problem 7 Let the curve be positively oriented triangle with vertices (0, 0), (0, 1), and (1, 1). Evaluate (2x 5 + 2xy)dx + (xy y 4 )dy We can parametrize the curve and calculate the integral directly but Green s theorem will give the same result in shorter amount of time. o denoting the interior region of curve by the triangle T we get (2x 5 + 2xy)dx + (xy y 4 )dy ( (xy y4 ) (2x5 + 2xy) )da Problem 8 T x x0 x0 y0 (y 2x)dydx x0 T (y 2x)dA 2 (1 x)2 2x(1 x)dx 7 2 x2 5x + 2 dx 7 6 x 5 2 x2 + 2 x 1 x0 1 6 For each of the following, determine if the vector field F is conservative and (if conservative) find a function f such that F f. Furthermore, evaluate the line integral F dr along the given curve F(x, y) (2x 2 + y )i + (y x 2 )j. is the line segment from (1, 0) to (2, 2). F(x, y) 6x y 2 i + x 4 y j. consists of the curve y xe x2 1 + sin(πx) from (0, 0) to (1, 1) followed by the curve x 1 + (y 1) from (1,1) to (0,0). F(x, y, z) e y i + xe y j + 2e 2z k. is the curve given by r(t) t 4 i + t j + t k, where 1 t 1. 5

6 For F(x, y) (2x 2 + y )i + (y x 2 )j, we have (y x 2 ) 2x (2x2 + y ) 6y 2 so that the vector field F is not conservative. Moreover the curve is parametrized as {x(t) 1 + t, y(t) 2t 0 t 1} o we have dx dy dt 1, dt 2. Thus we get line integral as F dr (2x 2 + y )dx + (y x 2 )dy t0 t0 For F(x, y) 6x y 2 i + x 4 y j, we have ( (2(1 + t) 2 + (2t) )1 + ((2t) (1 + t) 2 )2 ) dt (24t + 12t)dt 6t 4 + 6t 2 1 t (x 4 y) 12x y (6x y 2 ) Moreover the domain of F is whole plane R 2, which is simply connected. Hence the vector field F is conservative vector field. Regarding finding f a function f such that F f i.e. 6x y 2 f, x4 y we see (by insepction) that f(x, y) 2 x4 y 2. ince the given curve is closed so F dr 0 i j k For F(x, y, z) e y i + xe y j + 2e 2z k, curl(f) z e y xe y 2e 2z 0. Furthermore, domain of F is simply connected and so vector field F is simply connected. Regarding finding a function f such that F f f i.e. ey f, xey f, z 2e2z we see again by inspection that f(x, y, z) xe y + e 2z. (Alternatively we can integrate f z 2e2z to get f(x, y, z) e 2z + h(x, y), then f h ey f, h xey. o that h(x, y) xe y f(x, y, z) xe y + e 2z.) Regarding the integration along curve, we have F dr f dr f(r(1)) f(r( 1)) f(1, 1, 1) f(1, 1, 1) e + e 2 e 1 e 2 6

7 Problem 9 Let F(x, y, z) y 2 i + x j + z 2 k Evaluate F dr over each of the following curves (with positive orientation) is the boundary of the part of the plane 2x + y + z 6 that lies in the first octant. is the curve of intersection of the cylinder x 2 + y 2 9 and the plane x + z 0. ince the vector field F has complicated expression in it, it beehooves us to use toke s theorem (provided that curl(f) has simple expression). Here we have i j k curl(f) z y 2 x z 2 (1 + 2y) k. o that we will calculate F dr as curl(f) d, where is a surface (with compatible orientation of ), whose boundary is curve. is the boundary of the part of the plane 2x + y + z 6 that lies in the first octant. o we can choose surface as triangle with vertices (, 0, 0), (0, 2, 0), (0, 0, 6) with a normal pointing towards the positive z axis. Thus rewriting the surface as z f(x, y) 6 2x y, we see that its projection along XY (say xy ) plane is again a traingle with vertices (0, 0), (, 0), and (0, 2). o using the surface integral formula we have (noting that f 2, f ) F dr curl(f) d < 0, 0, 1 + 2y > < 2,, 1 > da x0 6 2x y0 4x x 5x xy (1 + 2y) dydx x x0 x ( 4x x ) dx is the curve of intersection of the cylinder x 2 + y 2 9 and the plane x + z 0. o here we can choose surface to be the ellipse formed by taking and all its inside points together. Its projection along XY plane is clearly the disk x 2 + y 2 9, and the surface is described by equation z f(x, y) x so we have f, f 0. Also normal to the 7

8 surface will be towards the positive z axis. Thus we have F dr curl(f) d < 0, 0, 1 + 2y > <, 0, 1 > da Problem 10 x 2 +y 9 2π r0 θ0 9π + 0 9π xy (1 + 2y) dydx 2π rdrdθ + r0 θ0 r0 2π θ0 (1 + 2r sin(θ)) rdrdθ 2r sin(θ) rdrdθ 2π r2 2 Let be the boundary of the solid region enclosed by the paraboloid z 5 x 2 y 2 that lies above the plane z 4. Find the flux of vector field across by F(x, y, z) y i + x j + z k Without using divergence theorem wih divergence theorem We have two surfaces here namely part of the plane z 4 (say 1 ), with projection x 2 + y 2 1 along XY plane. The second surface (say 2 ) is portion of the paraboloid z f(x, y) 5 x 2 y 2 with same projection x 2 + y 2 1 along XY plane. we note that surface 2 is up, while surface 1 sits down 2. o that natural choice of orientation of 1 will be downward while the natural choice of orientation for surface 2 will be upward. Keeping this thing in mind, we have F d F d + F d 1 2 omputing each of the integral separately, we have for 1 the surface is described by z f(x, y) 4 with normal vector < 0, 0, 1 >. Hence F d < y, x, 4 > < 0, 0, 1 > da 4dA 1 x 2 +y 2 1 x 2 +y (area of unit disk) 4π r0 2π + 2r 2 dr sin(θ)dθ r0 θ0 8

9 For surface 2, we have z f(x, y) 5 x 2 y 2, so that f 2x, f 2y. Thus we get F d < y, x, 5 x 2 y 2 > < 2x, 2y, 1 > da 2 x 2 +y 2 1 2π r0 r0 (5 + 4r 2 sin(θ) cos(θ) r 2 )rdrdθ θ0 2π 5rdr θ0 10π π 2 9π 2 Hence we finally get our flux as x 2 +y 2 1 2π 2π dθ + 4 2r dr sin(2θ)dθ r dr dθ r0 θ0 r0 θ0 F d 4π + 9π 2 π 2 Using divergence theorem, we can choose our solid region V to be everything contained inside closed surface. Moreover div(f) + + z z 1. o we have F d div(f)dv 1dV Using cylindrical coordinate, we get V 1dV r0 V 2π 5 r 2 θ0 z4 1 2π r2 2 r4 4 2π r0 4 π 2 V r drdθdz 2π (r r )dr r0 (5 + 4xy x 2 y 2 )da 9