1. (a) (5 points) Find the unit tangent and unit normal vectors T and N to the curve. r (t) = 3 cos t, 0, 3 sin t, r ( 3π

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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 3, 3π, r t) 3 cos t, 4t, 3 sin t 3 ). b) 5 points) Find curvature of the curve at the point P. olution: a) r t) 3 sin t, 4, 3 cos t, r t) T t) r t) r t) 35 sin t, 45, 35 cos t At the point P: t 3π 4, sin 3π 4 1, cos 3π 4 1. ) 3π Then T 3 4 5, 4 5, , 4 5, 3 1 T t) N t) T t) T t) b) r 3π 4 35 cos t,, 35 sin t, T t) 3 5. ) ) 3π cos t,, sin t, N 4,, 3, 4, 3, r t) 3 cos t,, 3 sin t, r 3π 4 ) 3,, 3, 1,, 1. r 3π 4 ) 3π r 4 ) i j k , 9, 6, 3π r 4 ) 3π r 4 ) 15, 3π r 4 ) 3 15, κ

2 . 1 points) Use LINEAR approximation to approximate the number e.8. olution: enote fx, y) x + e y. Then f x x, y) 1 x + e, f y y x, y) e y x + e y. We are looking for a linear approximation near the point x, y ) 3, ). f3, ), f x 3, ) 1 4, f y 3, ) 1 4. Let Lx, y) be the linearization of fx, y) near 3, ). Then Therefore, fx, y) Lx, y) x 3) y e.8 f3.4,.8) L3.4,.8) ) + 1.8) or e

3 3. 1 points) Find all critical points of the function fx, y) 4x 3x 3 xy. For each critical point determine if it is a local maximum, local minimum or a saddle point. olution: f x x, y) 4 9x y, f y x, y) 4xy. The equation f y x, y) gives two solutions x or y. ase x : f x, y) 4 y gives solutions y or y and points, ) and, ). ase y : f x x, ) 4 9x gives solutions x 3 or x 3 and points ) 3, and 3, ). ritical points are, ),, ), 3 ) ),, and 3,. f xx x, y) 18x, f xy x, y) 4y, f yy x, y) 4x. x, y) f xx x, y) f yy x, y) f xyx, y) 7x 16y 89x y )., ), ) 3 <. Therefore,, ) and, ) are saddle points. 3 ), 3 >, f xx 3 ), 1 >. Therefore, 3 ), is a point of a local minimum. ) ) 3 3, 3 >, f xx, 1 <. Therefore, 3 ), is a point of a local maximum.

4 4. 1 points) Find the volume of the solid E bounded by y x, x y, z x + y + 5, and z. olution: E {x, y, z) x 1, x y x, z x + y + 5}. The volume V of the solid E is x x+y+5 V dv dz dy dx E x x x x + y + 5) dy dx ] x + 5)y + y x x dx x 3/ + 5x 1/ + 1 x x3 5x 1 ) x4 dx 5 x5/ x3/ x 1 4 x4 5 3 x3 1 ] 1 1 x

5 5. 1 points) Find the y coordinate of the center of mass of a lamina that occupies the region bounded by y x + 4, x, and y and has density ρx, y) y. implify your answer as much as possible. olution: The region is R {x, y) y, y 4 x }. The mass of the lamina is m ρx, y) da R y dx dy y 3 + 4y) dy y y4 + y ] The y coordinate of the center of mass of the lamina is ȳ 1 m R y ρx, y) da 1 4 y dx dy 1 4 y 4 + 4y ) dy y y5 + 4 ] 3 y ]

6 6. 1 points) Evaluate the integral e x y da R where R is the parallelogram AB with vertices A, ), B 4, 1), 7, 4), and 3, 3) using the transformation x 4u + 3v and y u + 3v. implify your answer as much as possible. olution: T : x 4u + 3v, y u + 3v. The inverse transformation is T 1 : u 1 3 x y), v 1 x + 4y). 9 In the uv-plane the region that corresponds to the parallelogram AB can be found if we apply T 1 to its vertices: A 1 T 1 A), ), B 1 T 1 B) 1, ), 1 T 1 ) 1, 1), 1 T 1 ), 1), which is the square R 1 A 1 B {u, v) u 1, v 1}. The Jacobian of the transformation is J x u x v y u y v e x y da e u 3v 9 da R 9 R 1 e u 3v du dv 9 e u 3v ] 1 dv 9 e 1 ) e 3v dv 3 3 e 1 ) e 3 1 ) e 1) e 3v] 1 3 e 1 e 1 + e 3).

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

8 8. 1 points) Evaluate the integral 1 z) d, where is the part of the surface z 5 x y inside the cylinder x + y 1. olution: z gx, y) 5 x y, g g x, y) x, x ) ) g g d da 1 + x x y + y da. x, y) y, y The domain in the xy-plane is the disk {x, y) x + y 1}. In polar coordinates x r cos θ, y r sin θ, and {r, θ) r 1, θ < π}. Hence 1 z) d 1 1 x y ) ) 1 + x + y da x + y ) 1 + x + y da π r 1 + r r dr dθ π r 1 + r r dr ubstitution: u 1 + r gives r u 1, du r dr, r dr 1 du. Then π r 1 + r r dr π u 1)u 1/ 1 du 1 Hence π π 1 u 3/ u 1/ ) du π 5 u5/ ] 3 u3/ 1 8 ) ) + 1 π z) d ) π.

9 9. 1 points) Evaluate the line integral F dr for the vector field Fx, y, z) y i + x j z k, where the closed curve is the boundary of the triangle with vertices,, 5),,, 1), and, 3, ) traced in this order. olution: Use tokes Theorem F dr curlf d. i j k curlf x y z k,,. y x z is the triangle with vertices P,, 5), Q,, 1), R, 3, ). To find the equation of we consider vectors P Q,, 4 and P R, 3, 3. A normal vector n to the surface is i j k n P Q P R 4 1, 6, lies in the plane with the equation 1x + 6y + 6z 5) or z 5 x y. Using x and y as parameters we define by rx, y) x, y, 5 x y, where x, y) and is the triangle with vertices, ),, ), and, 3) in the xy-plane. r x x, y) 1,,, r y x, y), 1, 1. i j k r x r y 1, 1, Then F dr curlf d,,, 1, 1 da da A) 3 6, where A) is the area of the triangle. Therefore, F dr 6.

10 1. 1 points) Evaluate the flux of Fx, y, z) z y i + x y j + x + y) k over, where is the closed surface consisting of the coordinate planes and the part of the sphere x +y +z 4 in the first octant x, y, z, with the normal pointing outward. olution: By the ivergence Theorem the flux is F d div F dv, E where div F x and E is the region bounded by. In the spherical coordinates x ρ sin φ cos θ, y ρ sin φ sin θ, z ρ cos φ the region is E {ρ, θ, φ) ρ, θ π/, φ π/}. Then div F dv π/ π/ ρ sin φ cos θ ρ sin φ dρ dφ dθ E π/ cos θ dθ π/ sin 3 φ dφ ρ 4 dρ. π/ π/ sin 3 φ dφ cos θ dθ π/ π/ cos θ + 1 dθ 1 1 cos φ ) sin φ dφ u cos φ] ] π/ 1 sin θ + θ π 4, 1 u ) du 3, Hence ρ 4 dρ 3 5. π/ cos θ dθ π/ sin 4 φ dφ ρ 3 dρ π π. Therefore, the flux is π.

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