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math226fall10fes

math226fall10fes - 1(a Find the equation of the tangent...

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1 . (a) Find the equation of the tangent plane to the surface S : yz = ln( x + z ) at A (0 , 0 , 1) . What is the distance from the tangent plane to the point B (3 , 1 , 2)? Answer. S : f ( x, y, z ) = ln( x + z ) - yz = 0. We find f ( x, y, z ) = 1 x + z , - z, 1 x + z - y . The normal vector at (0 , 0 , 1) is ~n = h 1 , - 1 , 1 i . The equation of the tangent plane is x - y + z - 1 = 0 . The distance D = -→ AB · ~n | ~n | = h 3 , 1 , 1 i · h 1 , - 1 , 1 i 1 3 = 3 . (b) Write a parametric equation for the line of intersection of the tangent plane in (a) and the plane z = x + 2 y - 1 . Write the equation of the line in symmetric form. What is the distance from the line to the point B (3 , 1 , 2)? Answer. Normal vector to the second plane is ~n 2 = ( z - x - 2 y ) = h- 1 , - 2 , 1 i . We choose a point in the intersection of two planes z = x +2 y - 1 = y - x + 1 , 2 x + y = 2: for example, C (1 , 0 , 0) . The line passes through it in the direction of ~ τ = ~n × ~n 2 = h 1 , - 2 , - 3 i . The equation in parametric form: x = 1 + t, y = - 2 t, z = - 3 t . Symmetric form (eliminating t ): x - 1 = - y/ 2 = - z/ 3 . Since ~a = C ~ B = h 2 , 1 , 2 i , | ~ τ | = 14 ,~a · ~ τ = - 6, the distance to B (1 , 2 , 3) is D = v u u t | ~a | 2 - ~a · ~ τ | ~ τ | ! 2 = s 9 - 36 14 = s 45 7 . 2. (a) Determine whether f ( x, y, z ) = x 2 + 3 yz is increasing or de- creasing at ( - 1 , 1 , 1) in the direction of ~a = h 1 , - 1 , 1 i . At what rate? Answer. We find the gradient f ( x, y, z ) = * x x 2 + 3 yz , 3 z 2 x 2 + 3 yz , 3 y 2 x 2 + 3 yz + and f ( - 1 , 1 , 1) = D - 1 2 , 3 2 , 3 2 E . The rate of change in the direction of ~a is D ~u f ( - 1 , 1 , 1) = f ( - 1 , 1 , 1) · ~a | ~a | = D - 1 2 , 3 2 , 3 2 E · h 1 , - 1 , 1 i 1 3 = - 1 2 3 < 0. So, f is decreasing at the rate - 1 2 3 .

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