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Quiz4_sol

# Quiz4_sol - ∇ V = h 38 6 12 i it changes most rapidly And...

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NAME: Spring 2011, MAC 2313, Quiz 4 UFID: Section: 3124 1. Find ∂z/∂x and ∂z/∂y for x - z = arctan( y - z ). (2 points) Solution . Let F ( x, y, z ) = x - z - arctan( y - z ) = 0, then F x ( x, y, z ) = 1 , F y ( x, y, z ) = - 1 1 + ( y - z ) 2 , F z ( x, y, z ) = - ( y - z ) 2 1 + ( y - z ) 2 (1) Therefore, we have ∂z ∂x = - F x F z = 1 + ( y - z ) 2 ( y - z ) 2 ∂z ∂y = - F y F z = 1 + ( y - z ) 2 (2) 2. Suppose that over a certain region of space the electronical potential V is given by V ( x, y, z ) = 5 x 2 - 3 xy + xyz . (a) Find the rate of change of the potential at point P (3 , 4 , 5) in the direction of the vector v = h 1 , 1 , - 1 i . (b) What is the maximum rate of change at P ? (2 points) Solution . First we have V x ( x, y, z ) = 10 x - 3 y + yz, V y ( x, y, z ) = - 3 x + xz, V z ( x, y, z ) = xy, (3) and therefore V x (3 , 4 , 5) = 38 , V y (3 , 4 , 5) = 6 , V z (3 , 4 , 5) = 12). Furthermore, the unit vector of direction v is v | v | = 1 3 h 1 , 1 , - 1 i . Hence, the directional derivative is D v V = V · v | v | = 38 + 6 - 12 3
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Unformatted text preview: ∇ V = h 38 , 6 , 12 i , it changes most rapidly. And the directional derivative is D ∇ V V = |∇ V | = √ 38 2 + 6 2 + 12 2 = 40 . 3 . (5) 3. Find the equation of (a) the tangent plane and (b) the normal line to z +1 = xe y cos z at the point (1 , , 0). (2 points) Solution . We let F ( x,y,z ) = z + 1-xe y cos z = 0, then F x ( x,y,z ) =-e y cos z, F y ( x,y,z ) =-xe y cos z, F z ( x,y,z ) = 1 + xe y sin z, (6) So we have F x (1 , , 0) =-1 ,F y (1 , , 0) =-1 ,F z (1 , , 0) = 1. Therefore the tangent plane is-( x-1)-y + z =-x-y + z + 1 = 0 , (7) and the normal line is x-1-1 = y-1 = z 1 . (8)...
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