# Thus z x f x f z 3 x 2 3 yz 3 z 2 3 xy x 2 yz z 2 xy

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Thus, z x = - F x F z = - 3 x 2 + 3 yz 3 z 2 - 3 xy = - x 2 + yz z 2 - xy , z y = - F y F z = - 3 y 2 + 3 xz 3 z 2 - 3 xy = - y 2 + xz z 2 - xy .
Problem 13(b) - Spring 2008 (b) Is the tangent plane to the surface x 3 + y 3 + z 3 - 3 xyz = 0 at the point (1 , 0 , - 1) perpendicular to the plane 2 x + y - 3 z = 2? Justify your answer with an appropriate calculation.
Problem 13(b) - Spring 2008 (b) Is the tangent plane to the surface x 3 + y 3 + z 3 - 3 xyz = 0 at the point (1 , 0 , - 1) perpendicular to the plane 2 x + y - 3 z = 2? Justify your answer with an appropriate calculation. Solution: Since F ( x , y , z ) = x 3 + y 3 + z 3 - 3 xyz is constant along the surface F ( x , y , z ) = 0, F is normal (orthogonal) to the surface. Calculating, we obtain: F = h 3 x 2 - 3 yz , 3 y 2 - 3 xz , 3 z 2 - 3 xy i F (1 , 0 , - 1) = h 3 , 3 , 3 i , which is the normal vector to the tangent plane of the surface. Since the normal to the plane 2 x + y - 3 z = 2 is h 2 , 1 , - 3 i and h 3 , 3 , 3 i · h 2 , 1 , - 3 i = 0, it is perpendicular .
Problem 14(a) - Spring 2008 (a) Consider the vector field G ( x , y ) = h 4 x 3 + 2 xy , x 2 i . Show that G is conservative (i.e. G is a potential or a gradient vector field), and use the fundamental theorem for line integrals to determine the value of R C G · d r , where C is the contour consisting of the line starting at (2 , - 2) and ending at ( - 1 , 1).
Problem 14(a) - Spring 2008 (a) Consider the vector field G ( x , y ) = h 4 x 3 + 2 xy , x 2 i . Show that G is conservative (i.e. G is a potential or a gradient vector field), and use the fundamental theorem for line integrals to determine the value of R C G · d r , where C is the contour consisting of the line starting at (2 , - 2) and ending at ( - 1 , 1). Solution: Since y (4 x 3 + 2 xy ) = 2 x = x ( x 2 ) , there exists a potential function F ( x , y ), where F = G . Note that: F y = x 2 = F ( x , y ) = x 2 y + g ( x ) , where g ( x ) is some function of x . Since F x = 4 x 3 + 2 xy , F x = ( x 2 y + g ( x )) x = 2 xy + g 0 ( x ) = 4 x 3 + 2 xy = g 0 ( x ) = 4 x 3 = g ( x ) = x 4 + constant . Hence, F ( x , y ) = x 4 + x 2 y is a potential function . By the fundamental theorem of calculus for line integrals, Z C G · d r = Z C F · d r = F ( - 1 , 1) - F (2 , - 2) = 2 - 8 = - 6 .
Problem 14(b) - Spring 2008 (b) Now let T denote the closed contour consisting of the triangle with vertices at (0,0),(1,0), and (1,1) with the counterclockwise orientation, and let F ( x , y ) = h 1 2 y 2 - y , xy i . Compute R T F · d r directly (from the definition of line integral).
Problem 14(b) - Spring 2008 (b) Now let T denote the closed contour consisting of the triangle with vertices at (0,0),(1,0), and (1,1) with the counterclockwise orientation, and let F ( x , y ) = h 1 2 y 2 - y , xy i . Compute R T F · d r directly (from the definition of line integral). Solution: The curve T is the union of the segment C 1 from (0 , 0) to (1 , 0), the segment C 2 from (1 , 0) to (1 , 1) and the segment C 3 from (1 , 1) to (0 , 0) . Parameterize these segments as follows: C 1 ( t ) = h t , 0 i 0 t 1 C 2 ( t ) = h 1 , t i 0 t 1 C 2 ( t ) = h 1 - t , 1 - t i 0 t 1 Z T F · d r = Z C 1 F · d r + Z C 2 F · d r + Z C 3 F · d r = Z 1 0 h 0 , 0 i·h 1 , 0 i dt + Z 1 0 h 1 2 t 2 - t , t i·h 0 , 1 i dt + Z 1 0 1 2 (1 - t ) 2 - (1 - t ) , (1 - t ) 2 ·h- 1 , - 1 i dt = 0 + Z 1 0 t dt + Z 1 0 - 3 2 t 2 + 2 t - 1 2 dt = Z 1 0 - 3 2 t 2 + 3 t - 1 2 dt = - 1 2 t 3 + 3 2 t 2 - 1 2 t 1 0 = - 1 2 + 3 2 - 1 2 = 1 2 .
Problem 14(c) - Spring 2008 Let F ( x , y ) = h 1 2 y 2 - y , xy i .