# quiz4_sol - P =(0 1 2 to the surface z-y e x y = 1 is grad...

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Date: Feb 09, 2011 MA3160-08 Quiz 4 No Calculators! Justify all answers! Name (print): Solutions 1. (5pts.) Find the directional derivative of f ( x , y ) = xy - y 2 at the point P = (1 , 2) in the direction making an angle of 3 π/ 4 with the positive x-axis. Solution The direction vector is ~ u = cos(3 π/ 4) ~ i + sin(3 π/ 4) ~ j = - 1 2 ~ i + 1 2 ~ j Note that this direction vector is already a unit vector. The gradient of f ( x , y ) at P = (1 , 2) is grad f (1 , 2) = f x (1 , 2) ~ i + f y (1 , 2) ~ j = [ y ~ i + ( x - 2 y ) ~ j ] | (1 , 2) = 2 ~ i - 3 ~ j . Therefore, the derivative in the direction ~ u at the point P = (1 , 2) is f ~ u (1 , 2) = grad f (1 , 2) · ~ u = - 2 2 - 3 2 = - 5 2 2. (5pts.) Find the equation of the tangent plane at P = (0 , 1 , 2) to the surface z - y e x / y = 1. Solution Note that the point P = (0 , 1 , 2) actually lies on the surface z - y e x / y = 1. The surface z - y e x / y = 1 is a level surface of the function w = f ( x , y , z ) = z - y e x / y . Therefore, the normal vector at
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Unformatted text preview: P = (0 , 1 , 2) to the surface z-y e x / y = 1 is grad f (0 , 1 , 2) = f x (0 , 1 , 2) ~ i + f y (0 , 1 , 2) ~ j + f z (0 , 1 , 2) ~ k = [-e x / y ~ i-(e x / y-xy y 2 e x / y ) ~ j + ~ k ] (0 , 1 , 2) =-~ i-~ j + ~ k . Let Q = ( x , y , z ) be a general point on the tangent plane. Then the vector ~ PQ = ( x-0) ~ i + ( y-1) ~ j + ( z-2) ~ k lies on the tangent plane and is perpendicular to grad f (0 , 1 , 2). Therefore, grad f (0 , 1 , 2) · ~ PQ = 0 gives the equation of the tangent plane, i.e.,-x-( y-1) + ( z-2) = 0 or, equivalently, z = x + y + 1...
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## This note was uploaded on 04/04/2011 for the course MA 3160 taught by Professor Staff during the Spring '08 term at Michigan Technological University.

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