1. (a) (5 points) Find the unit tangent and unit normal vectors T and N to the curve. r(t) = 3cost, 4t, 3sint

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1 1. a) 5 points) Find the unit tangent and unit normal vectors T and N to the curve at the point P, π, rt) cost, t, sint ). b) 5 points) Find curvature of the curve at the point P. Solution: a) r t) sint,, cost, r t) point) Tt) r t) r t) 5 sint, 5, 5 cost +1 point) At the point P: t π, sin π 1, cos π point) ) π Then T 5, 5, 5 1, 5, +1 point) 1 T t) 5 cost,, 5 sint, T t) 5. Nt) T t) T t) + points) cost,, sint, N ) π,, 1,, 1. b) r π ),,, r t) cost,, sint, r π ) ) π π r r ) i j k,,, 6, 9, 6, + points) π r ) π r ) 15, π r ) 15, κ points) 5 1

2 . 1 points) Use LINEAR approximation to approximate the number.+e.8. Solution: enote fx,y) x+e y. Then f x x,y) 1 f x+ey, y x,y) e y x+ey. + points) We are looking for a linear approximation near the point x,y ),). +1 point) f,), f x,) 1, f y,) point) Let Lx,y) be the linearization of fx,y) near,). Then Therefore, fx,y) Lx,y) + 1 x )+ 1 y. + points).+e.8 f.,.8) L.,.8) + 1. )+ 1.8) or.+e points)

3 . 1 points) Find all critical points of the function fx,y) x x xy. For each critical point determine if it is a local maximum, local minimum or a saddle point. Solution: f x x,y) 9x y, f y x,y) xy. +1 point) The equation f y x,y) gives two solutions x or y. +1 point) Case x : f x,y) y gives solutions y or y and points, ) and, ). +1 point) Case y : f x x,) 9x gives solutions x or x and points ), and, ). +1 point) Critical points are, ),, ), ) ),, and,. +1 point) f xx x,y) 18x, f xy x,y) y, f yy x,y) x. +1 point) x,y) f xx x,y)f yy x,y) f xyx,y) 7x 16y 89x y ). +1 point), ), ) <. Therefore,, ) and, ) are saddle points. +1 point) ), >, f xx ), 1 >. Therefore, ), is a point of a local minimum. +1 point) ) ), >, f xx, 1 <. Therefore, ), is a point of a local maximum. +1 point)

4 . 1 points) Find the volume of the solid E bounded by y x, x y, z x+y + 5, and z. Solution: E {x,y,z) x 1, x y x, z x+y+5}. + points) The volume V of the solid E is V dv x x+y+5 dzdydx + points) E x x x x+y +5)dydx ] x+5)y + y x x dx x / +5x 1/ + 1 x x 5x 1 ) x dx + points) 5 x5/ + 1 x/ + 1 x 1 x 5 x 1 ] 1 1 x points)

5 5. 1 points) Find the y coordinate of the center of mass of a lamina that occupies the region bounded by y x +, x, and y and has density ρx,y) y. Simplify your answer as much as possible. Solution: The region is R {x,y) y, y x }. + points) The mass of the lamina is m ρx,y)da +1 point) R ydxdy y +y)dy y 1 y +y ] points) The y coordinate of the center of mass of the lamina is ȳ 1 m R yρx,y)da +1 point) 1 y dxdy 1 y +y )dy + points) 1 1 y 1 5 y5 + ] y 5 + ] points) 15 5

6 6. 1 points) Evaluate the integral e x y da R where R is the parallelogram ABC with vertices A,), B,1), C 7,), and,) using the transformation x u+v and y u+v. Simplify your answer as much as possible. Solution: T : x u+v, y u+v. The inverse transformation is T 1 : u 1 x y), v 1 x+y). + points) 9 In the uv-plane the region that corresponds to the parallelogram ABC can be found if we apply T 1 to its vertices: A 1 T 1 A),), B 1 T 1 B) 1,), C 1 T 1 C) 1,1), 1 T 1 ),1), which is the square R 1 A 1 B 1 C 1 1 {u,v) u 1, v 1}. + pts) The Jacobian of the transformation is J x u x v y u y v points) e x y da e u v 9 da +1 point) R 9 R 1 e u v dudv 9 e u v ] 1 dv 9 e 1 ) e v dv e 1 ) e 1 ) e 1)e v] 1 e 1 e 1 +e ). + points) 6

7 7. 1 points) Evaluate the line integral e x+y dx+e y dy C along the negatively oriented closed curve C, where C is the boundary of the triangle with the vertices,),,1), and 1,). Solution: Px,y) e x+y, Qx,y) e y. Hence P y x,y) ex+y, Q x,y). + points) x The triangle bounded by C is {x,y) x 1, y x+1}. + points) Using Green s Theorem +1 point) and negative orientation of C we get e x+y dx+e y dy ) e x+y da + points) C x+1 e x+y dydx e x+y ] x+1 dx e x+1 e x) dx e x+1 1 ex ] 1 e 1 e e+ 1 1 e e points) 7

8 8. 1 points) Evaluate the integral S 1 z)ds, where S is the part of the surface z 5 x y inside the cylinder x +y 1. Solution: z gx,y) 5 x y, g x x,y) x, g x,y) y, +1 point) y ) ) g g ds 1+ + da 1+x x y +y da. + points) The domain in the xy-plane is the disk {x,y) x +y 1}. In polar coordinates x rcosθ, y rsinθ, and {r,θ) r 1, θ < π}. +1 point) Hence 1 z)ds 1 1 x y ) ) 1+x +y da S x +y ) 1+x +y da π r 1+r r drdθ π r 1+r r dr + points) Substitution: u 1+r gives r u 1, du rdr, rdr 1 du. Then π r 1+r r dr π u 1)u 1/ 1 du +1 point) 1 Hence S π π 1 z)ds 15 1 u / u 1/ )du π 5 u5/ ] u/ 1 8 ) ) +1 π ) π. + points) 8

9 9. 1 points) Evaluate the line integral C F dr for the vector field Fx,y,z) yi+xj zk, where the closed curve C is the boundary of the triangle with vertices,,5),,,1), and,,) traced in this order. Solution: Use Stokes Theorem F dr curlf ds. +1 point) C S i j k curlf x y z k,,. + points) y x z S is the triangle with vertices P,,5), Q,,1), R,,). To find the equation of S we consider vectors PQ,, and PR,,. A normal vector n to the surface S is i j k n PQ PR 1, 6, point) S lies in the plane with the equation 1x+6y+6z 5) or z 5 x y. +1 pt) Using x and y as parameters we define S by rx,y) x, y, 5 x y, +1 pt) where x,y) and is the triangle with vertices,),,), and,) in the xy-plane. r x x,y) 1,,, r y x,y), 1, 1. i j k r x r y 1, 1, 1. + points) 1 1 Then F dr curlf ds,,, 1, 1 da da A) 6, C S where A) is the area of the triangle. Therefore, C F dr 6. + points) 9

10 1. 1 points) Evaluate the flux of Fx,y,z) z yi+x yj+x+y)k over S, where S is the closedsurfaceconsisting ofthecoordinateplanesandthepartofthespherex +y +z in the first octant x, y, z, with the normal pointing outward. Solution: By the ivergence Theorem the flux is FdS div F dv, +1 point) S E where divf x +1 point) and E is the region bounded by S. In the spherical coordinates x ρsinφcosθ, y ρsinφsinθ, z ρcosφ the region is E {ρ,θ,φ) ρ, θ π/, φ π/}. + points) Then divfdv π/ π/ ρ sin φcos θ ρsin φ dρdφdθ E π/ cos θdθ π/ sin φdφ ρ dρ. +1 point) π/ π/ sin φ dφ cos θdθ π/ 1 φ sinφ]π/ + 1 π/ cosθ+1 1 cosφ π/ dθ 1 ) dφ 1 cosφ+1 ] π/ 1 sinθ+θ π, +1 point) π/ dφ π ρ dρ. +1 point) 1 cosφ+cos φ ) dφ ] π/ 1 sinφ+φ π, + points) 16 Hence π/ cos θdθ π/ sin φdφ ρ dρ π 16 π 16 π. +1 point) Therefore, the flux is 16 π. 1