Solutions for Review Problems

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1 olutions for Review Problems 1. Let be the triangle with vertices A (,, ), B (4,, 1) and C (,, 1). (a) Find the cosine of the angle BAC at vertex A. (b) Find the area of the triangle ABC. (c) Find a vector that is perpendicular to the plane that contains the points A, B and C. (d) Find the equation of the plane through A, B and C. (e) Find the distance between (, 1, 1) and the plane through A, B and C. (f) Find the volume of the parallelepiped formed by AB, AC and A. (a) Using A (,, ), B (4,, 1) and C (,, 1), We have AB <,, 1 >, AC <, 1, 1 > and BC <, 1, >. Let θ be the angle BAC at vertex A. AB We have cos(θ) AC AB AC <,, 1> <,1, 1> (b) First we find AB AC i j k i 1 1 j + 1 k i + j + k < 1,, > Thus the area of the triangle ABC 1 AB AC (c) The vector AB AC < 1,, > is perpendicular to the plane that contains the points A, B and C. (d) The normal vector of the plane is AB AC < 1,, >. o the equation of the plane thru A (,, ) with normal vector < 1,, > is < x, y, z > < 1,, >, i.e. x + y 4 + z 4. o the plane thru A, B and C is x + y + z 1. (e) From previous problem, we know that the equation of the plane through A, B and C is x + y + z 1. o the distance between (, 1, 1) and the plane x + y + z 1 is (f) The volume of the parallelepiped formed by ( AB AC) A. AB, AC and A. Find the distance between the planes x y + z 1 and 4x y + 4z 7. The plane 4x y + 4z 7 can be rewritten as x y + z 7. Using the distance formula between planes, the distance between P 1 : x y + z 1 Math 85 ec 4: page 1 of 14

2 olutions for Review Problems Math 85 ec 4: page of 14 and P : x y + z 7 is ( 1) 1 6. A plane P is drawn through the points A (1, 1, ), B (, 1, 1) and C (1,, 1). Find the distance between the plane P and the point (1, 1, 1).. (a) Find a vector equation of the line through (, 4, 1) and (4, 5, ) (b) Find a vector equation of the line through (1, 1, 1) that is parallel to the line through (, 4, 1) and (4, 5, ). (c) Find a vector equation of the line through (1, 1, 1) that is parallel to the line x y z. 1 (a) Let P (, 4, 1) and Q (4, 5, ). The direction of the line is P Q < 4, 5 4, 1 ><, 1, >. The vector equation is < x, y, z ><, 4, 1 > +t <, 1, > or x + t, y 4 + t and z 1 + t. (b) The direction of the line is P Q < 4, 5 4, 1 ><, 1, >. The starting point of the line is (1, 1, 1). o the equation of the line is x 1+t, y 1 + t and z 1 + t. (c) The direction of the line x y z is <, 1, >. The starting 1 point of the line is (1, 1, 1). o the equation of the line is x 1 + t, y 1 t and z 1 + t. 4. (a) Find the equation of a plane perpendicular to the vector i j + k and passing through the point (1, 1, 1). (b) Find the equation of a plane perpendicular to the planes x + y z 7 and x 4y + z and passing through the point (1, 1, 1). (a) The equation of the plane with normal vector i j + k and passing through the point (1, 1, 1) is 1, 1, 1 x 1, y 1, z 1 (x 1) (y 1)+(z 1) or x y+z 1. (b) The plane that is perpendicular to the planes x + y z 7 and x 4y + z has normal vector,, 1 1, 4,, 7, 14. Thus the equation of the plane is 7y 14z 1, that is y + z. 5. Find the arc-length of the curve r(t) t, e t, e t when t ln().

3 Math 85 ec 4: page of 14 olutions for Review Problems Given r(t) t, e t, e t, we have r (t), e t, e t and r (t) + e t + e t (e t + e t ) e t + e t. Hence the arc-length of the curve r(t) t, e t, e t between t ln() is ln() r (t) dt ln() (e t + e t )dt e t + e t ln() e ln() + e ln() ( 1 + 1) 1 +. Note that e ln() 1 1. e ln() 6. Find parametric equations for the tangent line to the curve r(t) t, t, t at the point ( 1, 1, 1). Note that r(t) t, t, t. We have r( 1) 1, 1, 1. Taking the derivative of r(t), we get r (t) t, 1, t. Thus the tangent vector at t 1 is r ( 1), 1,. Therefore parametric equations for the tangent line is x 1 + t, y 1 + t and z 1 + t. 7. Find the linear approximation of the function f(x, y, z) x + y + z at (1,, ) and use it to estimate (1.1) + (.1) + (1.9). The partial derivatives are f x (x, y, z) f y (x, y, z) x x, f y(x, y, z) +y +z z x, f x(1,, ) 1 and f +y +z y(1,, ) and f z(1,, ). The linear approximation of f(x, y, z) at (1,, ) is L(x, y, z) f(1,, ) + f x (1,, )(x 1) + f y (1,, )(y ) + f z (1,, )(z ) y x +y +z, + 1 (x 1) + (y ) + (z ). Thus L(1.1,.1, 1.9) + 1(1.1 1) + (.1 ) (1.9 ) Hence (1.1) + (.1) + (1.9) is about.. 8. (a) Find the equation for the plane tangent to the surface z x y + x at (1,, 1). (b) Find the equation for the plane tangent to the surface x + xy + xyz 4 at (1, 1, ). (a) Let f(x, y) x y + x. We have f x 6x +, f y y, f x (1, ) 8 and f y (1, ) 4. The equation of the tangent plane through the point (1,, 1) is z f(1, ) + f x (1, )(x 1) + f y (1, )(y + ) 1 + 8(x 1) + 4(y + ) 8x + 4y + 1.

4 olutions for Review Problems Math 85 ec 4: page 4 of 14 (b) In general, the normal vector for the tangent plane to the level surface of F (x, y, z) k at the point (a, b, c) is F (a, b, c). The surface x +xy +xyz 4 can be rewritten as F (x, y, z) x +xy +xyz 4, F (x, y, z) x + y + yz, xy + xz, xy and F (1, 1, ) 5, 4, 1 Thus the equation of the tangent plane to the surface x + xy + xyz 4 at the point (1, 1, ) is 5, 4, 1 x 1, y 1, z which yields 5x 5 + 4y 4 + z. It can be simplified as 5x + 4y + z uppose that over a certain region of plane the electrical potential is given by V (x, y) x xy + y. (a) Find the direction of the greatest decrease in the electrical potential at the point (1, 1). What is the magnitude of the greatest decrease? (b) Find the rate of change of V at (1, 1) in the direction, 4. (a) We have V (x, y) V x (x, y), V y (x, y) (x xy + y ) x, (x xy + y ) y x y, x + y ince V (x, y) x y, x + y the direction of the greatest decrease in electrical potential is V (1, 1) 1, 1 and the magnitude is V (1, 1). (b) The unit vector in the direction, 4 is u 1, 4. Thus the rate of 5 change of V at (1, 1) in the direction, 4 is V (1, 1) u 1, 1 1 5, Find the local maxima, local minima and saddle points of the following functions. ecide if the local maxima or minima is global maxima or minima. Explain. (a) f(x, y) x y + y x y (b) f(x, y) x + y xy

5 Math 85 ec 4: page 5 of 14 olutions for Review Problems (a) To find critical points, set f x (x, y) 1 6x and f y (x, y) 6 y. Hence, (, ) is the only critical point. We also have f xx 6, f xy f yx and f yy. ( f(x, y)) ( 6 ) ince det( f(, )) 1 > and f xx (, ) <, the second derivative test implies that f has a local maximum at (, ). Because f is a quadratic function, it follows the graph of f is an elliptical paraboloid and (, ) is a global maximum. We can also see that f has a global maximum at (, ) be completing the square: f(x, y) 1 (x ) (y ). (b) The system of equations f x (x, y) x y f y (x, y) y x implies that x y and ( y y) y ( y ). Thus, (, ) and (9/4, /) are the critical points. We also have f xx, f xy f yx and f yy 6y. ( ) ( f(x, y)) 6y ince et( f(x, y)) f xx f yy f xy ()(6y) ( ) 1y 9, et( f(, )) 9 <, et( f(9/4, /)) 18 >, the second derivative test establishes that f has a saddle point at (, ) and a local minimum at (9/4, /). Because lim y f(, y) lim y y, we see that (9/4, /) is not a global minimum. 11. Use Lagrange multipliers to find the maximum or minimum values of f subject to the given constraint. (a) f(x, y, z) x y, x + y (..1) (..) (..) Let f(x, y) x y and g ( x, y) x + y. The necessary conditions for the optimizer (x, y) are f(x, y) λ g ( x, y) and the constraint equations x + y which are: ince f(x, y) (x, y) and g(x, y) (x, y), thus (x, y) must satisfy x λx y λy x + y

6 olutions for Review Problems Math 85 ec 4: page 6 of 14 From (4), (5), we get 4x + 4y 4λ (x + y ). ince x + y m we have λ 1. o λ ±1. If lambda 1, then eq(4) is always true and we get y by eq(5). Using x +y, we get x ±. If lambda 1, then eq(5) is always true and we get x by eq(4). Using x + y, we get y ±. o the candidates are (, ), (, ), (,, ) and (,, ). o f((, )) f((, )) and f((,, )) f((,, )). Thus the maximum is, the minimum is, the maximizers are (, ), (, ), and the minimizers are (,, ) and (,, ). (b) f(x, y, z) x + y + z, x + y + z 1. Let f(x, y, z) x + y + z and g(x, y, z) x + y + z 1. We have f(x, y, z) (1, 1, 1) and g(x, y, z) (x, y, z). The necessary conditions for the optimizer (x, y, z) are f(x, y, z) λ g(x, y, z) and the constraint equations which are: (..4) (..5) (..6) (..7) 1 λx 1 λy 1 λz x + y + z 1 From (7),(8) (9) and (1), we know that λ, x 1, y 1 and z 1. λ λ λ 1 Plugging into (1), we get , 1 and λ ±. o 4λ 4λ 4λ 4λ (x, y, z) ( 1, 1, 1 ) ( 1 1 λ λ λ, 1, ) or (x, y, z) (( 1, 1, 1 ). We have f(( 1 1, 1, )) and f(( 1, 1, 1 )). Thus the maximizers are ( 1, 1, 1 ) with maximum. The minimizers are ( 1, 1, 1 ) with minimum. 1. Compute the following iterated integrals. (a) 1 1 y ye x dxdy x Let {(x, y) y x 1, y 1} Then y x and x 1. o is the same as {(x, y) x 1, y x }. We have 1 1 y ye x dxdy 1 x ye x dydx 1 y e x x dx 1 xe x dx x x x e x 4 1 e (b) e x y dydx 4 x

7 Math 85 ec 4: page 7 of 14 olutions for Review Problems The region of integration is {(x, y) x, 4 x y { }. The is the region in fourth quadrant. In polar coordinates, it is R (r, θ) : r, π θ }. We also have x + y r and e x y dydx 4 x π π e r e r rdrdθ dθ ( e 4 1 ) π. (c) 1 1 x 1 x y 1 x x +y (x + y ) dzdydx The region of integration is {(x, y, z) 1 x 1, 1 x y 1 x, x + y z x y }. In cylindrical coordinates, it is R { (r, θ, z) : r 1, θ π, r z r }. Recall that x r cos(θ), x r sin(θ) We have (x + y ) r and 1 1 x 1 x y (x + y ) 1 x x +y dzdydx π 1 r r rdzdrdθ r π 1 r4 z r drdθ r π 1 r4 ( r )drdθ π 1 ( r5 r7 ) 1 dθ 5 7 π 8π 4 dθ 5 5. (d) 4 y 4 x y 4 x y x + y + z dzdxdy y In spherical coordinates, the region E {(x, y, z) x 4 y, y, 4 x y z 4 x y }

8 olutions for Review Problems Math 85 ec 4: page 8 of 14 is described by the inequalities ρ, θ ππ and φ π. Note that y ρ sin(φ) cos(θ) Hence, the integral is 4 y 4 x y π π π π y x 4 x + y + z dzdxdy y ρ sin (φ) cos (θ)(ρ) ρ sin(φ) dρ dθ dφ ρ 5 sin (φ) cos (θ) dρ dθ dφ ( π ) ( π ) ( ) cos (θ)dθ sin (φ) dφ ρ 5 dρ ( π ) ( 1 + cos(θ) π ) ( dθ (1 cos (φ)) sin(φ) dφ ( ( θ + sin(θ) ) ( ) π ( cos(φ) + cos (φ) ) ( ) π ρ 6 ) 4 6 π π 9 ) ρ 5 dρ 1. Find the volume of the following regions: (a) The solid bounded by the surface z x x + y and the planes x, x 1, y, y 1 and z. The volume is 1 1 xdx, xdx du and x x + ydx u 1/ o x x + ydxdy 1 (1+y) / (y)/ dx (1+y)5/ x x + ydxdy Let u x + y. Then du u/ du + C (x +y) / + C. (x +y) / 1 dy (y)5/ ()5/ ( 15 ) ()5/ (b) The solid bounded by the plane x + y + z, x, y and z. The region E bounded by the xy, yz, xz planes and the plane x + y + z is the set { (x, y, z) R : x, y x, z

9 Math 85 ec 4: page 9 of 14 olutions for Review Problems E x x y }. The volume of E is dv ) x x y x y dy dx ( x) x( x) ( x) dz dy dx ) x y xy y dx x 9 x x + dx 9x x z x y dy dx dx (by substitution u4-x-y) 9 x + x + x (x 6x + 9) dx + x (c) The region bounded by the cylinder x + y 4 and the plane z and y + z. The region is bounded above by the plane z y and below by z. In polar coordinates, this region x + y 4 is R { (r, θ) : r, θ π }. Note that z y r cos(θ) Hence, we can compute the volume of the region by Volume π π π ( r cos(θ) ) rdr dθ r r cos(θ) dr dθ π [ r 1 r cos(θ) ] dθ [ 6 8 cos(θ)] dθ 1π. 14. Let C be the oriented path which is a straight line segment running from (1, 1, 1) to (, 1, ). Calculate fds where f (x + y + z). C is parametrized by x(t) 1 t, y(t) 1 t and z(t) 1+t where t 1. We have f(x(t), y(t), z(t)) x(t) + y(t) + z(t) 1 t + 1 t t t and ds x (t) + y (t) + z (t) dt ( 1) + ( ) + dt 9dt dt. o C fds 1 ( t)dt t t Calculate the following line integrals C F d r: (a) F y sin(xy) i + x sin(xy) j and C is the parabola y x from (1, ) to (, 18). (b) F x i 4y j + (z ) k and C is the line from (1, 1, 1) to (,, 1).

10 olutions for Review Problems Math 85 ec 4: page 1 of 14 (a) If f(x, y) cos(xy) then f y sin(xy) i + x sin(xy) j F. Hence, the Fundamental Theorem for line integrals implies that F d r f(, 18) f(1, ) cos() cos(54). C (b) If f(x, y, z) x y + z z then f x i 4y j + (z ) k F. Hence, the Fundamental Theorem for line integrals implies that F d r f(,, 1) f(1, 1, 1) C 16. Calculate the circulation of F around the given paths. (a) F xy j around the square x 1, y 1 oriented counterclockwise. (b) F (x +y) i+(x+y ) j around the triangle with vertices (, ), (, ), (, ) oriented counterclockwise. (c) F y i + xy j around the unit circle oriented counterclockwise. (d) F xz i + (x + yz) j + x k and C is the circle x + y 1, z oriented counterclockwise when viewed from above. R T (a) If R {(x, y) : x 1, y 1} then Green s Theorem implies that ( F F d r x F ) 1 1 ( 1 ) ( 1 ) 1 da y dy dx dx y dy 1 y. R (b) If T is the triangle with vertices (, ), (, ), (, ) then Green s Theorem gives ( F F d r x F ) 1 da da Area(T ) 1 (4)() 6. y T (c) If is the unit disk then Green s Theorem yields ( F F d r x F ) 1 da y da y da y T dy π. Indeed, the function f(x, y) y is symmetry about the origin [which means f( x, y) f(x, y)] so the integral of f(x, y) over is zero. (d) If is the disk given by x + y 1 and ] z oriented upwards then C. ince F det [ i j k x y z xz x+yz x y i+x j + k, tokes Theorem implies that F d r ( F ) da ( y i + x j + k) k da Area() π.

11 Math 85 ec 4: page 11 of 14 olutions for Review Problems 17. Calculate the area of the region within the ellipse x /a + y /b 1 parameterized by x a cos(t), y b sin(t) for t π. If R is region in the plane and F x j then Green s Theorem implies that Area(R) da ( ) F F 1 da F d r. Using this idea, we R R x y R can calculate the area of the region enclosed by the given parameterized curves. For the ellipse, we have Area ab R F d r π [ 1 cos(t) sin(t) + t π a cos(t)b cos(t) dt ab ] π cos (t) dt abπ. 18. Compute the flux of the vector field F through the surface. (a) F x i + y j and is the part of the surface z 5 (x + y ) above the disk of radius 5 centered at the origin oriented upward. (b) F y i + z k and is the part of the surface z y + 5 over the rectangle x 1, y 1 oriented upward. (c) F y i + j xz k and is the surface y x + z with x + z 1 oriented in the positive y-direction. (d) F x i + (y xy) j + 1z k and is the sphere of radius 5 centered at the origin oriented outward. (e) F z i + x k and is a square pyramid with height and base on the xy-plane of side length 1. (f) F y j and is a closed vertical cylinder of height with its base a circle of radius 1 on the xy-plane centered at the origin. (a) The surface is given by r(s, t) s cos(t) i + s sin(t) j + (5 s ) k for s 5, t π. ince r s r i j k t det cos(t) sin(t) s s cos(t) i + s sin(t) j + s k, s sin(t) s cos(t)

12 olutions for Review Problems Math 85 ec 4: page 1 of 14 F d A we have π 5 π 5 π 5 π 5 F ( r(s, t) ) ( r s r ) t ds dt ( s cos(t) i + s sin(t) j ) (s cos(t) i + s sin(t) j + s k ) ds dt s cos (t) + s sin (t) ds dt [ s s 4 ds dt π ] 5 75π. (b) The surface is given by r(s, [ t) s i + t j + (t + 5) k for s 1, t 1. ince r r det ] i j k s t 1 t j + k, we have 1 t 1 1 F da F ( r(s, t) ) ( r s r ) ds dt t 1 1 ( t i + (t + 5) k ) ( t j + k) dt ds 1 1 ( 1 t + 5 dt ds ) ( 1 ) ds t + 5 dt [ 1 t + 5t ] (c) The surface is given by r(s, t) s cos(t) i + s j + s sin(t) k for s 1, t π. ince r t r i j k s det s sin(t) s cos(t) s cos(t) i + s j s sin(t) k cos(t) s sin(t) F d A we have 1 π 1 π 1 π 1 F ( r(s, t) ) ( r t r ) s ds dt ( s i + j s sin(t) cos(t) k ) ( s cos(t) i + s j s sin(t) k ) dt ds s 4 cos(t) + s + s 4 sin (t) cos(t) dt ds [ s 4 sin(t) + st + s4 sin (t) ] π ds 1 πs ds π.

13 Math 85 ec 4: page 1 of 14 olutions for Review Problems (d) If is the ball of radius 5 centered at the origin then. The ivergence Theorem implies that F da ( F ) dv x + (1 x) + 1 dv 11 Volume(B) 55π. (e) If P is the solid square pyramid with height and base on the xy-plane of side length 1 then P. Hence, the ivergence Theorem gives: F da ( F ) dv dv. P P (f) If W is the solid vertical cylinder of height with its base a circle of radius 1 on the xy-plane centered at the origin then W. Applying the ivergence Theorem yields F da ( F ) dv 1 dv Volume(W ) π. W W W P 19. Let F (8yz z) j + ( 4z ) k. (a) how that G 4yz i + x j + xz k is a vector potential for F. (b) Evaluate F d A where is the upper hemisphere of radius 5 centered at the origin oriented upwards. (a) ince G det [ i j k x y z 4yz x xz ] (8yz z) j + ( 4z ) k F, we see that G is a vector potential for F. (b) If C is the circle of radius 5 in the xy-plane centered at the origin oriented counterclockwise when viewed from above, then C. Hence, toke s Theorem implies that F da ( G) da G d r. ince C can C be parameterized by r(t) 5 cos(t) i + 5 sin(t) j for t π, we have F da C π 75 G d r π π G ( r(t) ) r (t) dt ( 15 cos(t) j ) ( 5 sin(t) i + 5 cos(t) j ) dt [ 1 cos (t) dt 75 cos(t) sin(t) + t ] π 75π. Alternatively, let be the disk of radius 5 in the xy-plane centered at the origin oriented upwards. Hence, C and toke s Theorem

14 olutions for Review Problems Math 85 ec 4: page 14 of 14 implies that F da ( G) da G d r ( G) da C F da F k da ( 4() ) da Area() 75π.. For constants a, b, c and m consider the vector field F (ax + by + 5z) i + (x + cz) j + (y + mx) k. (a) uppose that the flux of F through any closed surface is. What does this tell you about the values of the constants a, b, c, m? (b) uppose instead that the circulation of F around any closed curve is. What does this tell you about the values of the constants a, b, c, m? (a) Let W be any solid region in -space. ince the flux of F through any closed surface is the ivergence Theorem implies that F da F dv. W By choosing W to be a sequence of balls centered at (x, y, z) that contract down to (x, y, z) in the limit, we deduce that F everywhere. However, F a which means a. (b) Let be any smooth oriented surface with piecewise smooth boundary. ince the circulation of F around any closed curve is, tokes Theorem implies that F d r ( F ) da. By choosing be to a small disk perpendicular to the curl of F at (x, y, z), we deduce that F. However, we have i j k F det x y z ( c) i + (5 m) j + (1 b) k ax + by + 5z x + cz y + mx which means that b 1, c and m 5. W

This makes sense. t 2 1 + 1/t 2 dt = 1. t t 2 + 1dt = 2 du = 1 3 u3/2 u=5

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