advertisement

LAST NAME : FIRST NAME : QUIZ 2, Version A : MATH 251, Section 505 last name : . . . . . . . . first name : . . . . . . . . GRADE : . . . . . . . . ”An Aggie does not lie, cheat or steal, or tolerate those who do” signature : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Write up your result, detail your calculations if necessary and BOX your final answer 1. [25pts] Find the differential dz of the function z = ex + x cos y. 2. [25pts] Use the Chain Rule to find Gx where G = ln(u2 + v 2 ) and u = x + y, v = x + y 2 . y . x (a) [25pts] Find the direction in which f increases most rapidly at the point (1, 1). 3. Let f (x, y) = x3 + (b) [25pts] Find the directional derivative of f at the point (1, 1) in the direction < 4, 3 >. 1. We have dz = fx (x, y)dx + fy (x, y)dy, so dz = (ex + cos y)dx − x sin ydy . 2. Denote G = ln(u2 + v 2 ), where DG = R2 \ {(0, 0)}. So, ∂u ∂G ∂v ∂ G + , ∂x ∂u ∂x ∂v 2u 2v = 2 + , u + v 2 u2 + v 2 2(x + y) 2(x + y 2 ) + , = (x + y)2 + (x + y 2 )2 (x + y)2 + (x + y 2 )2 Gx = = 4x + 2y + 2y 2 . (x + y)2 + (x + y 2 )2 3. f increases most rapidly in the direction of the gradient, so the gradient at (1, 1) is y 1 ∇f (1, 1) = hfx (1, 1), fy (1, 1)i. We have fx (x, y) = 3x2 − 2 and fy (x, y) = on Df = {(x, y) ∈ R2 | x 6= 0}. x x So, fx (1, 1) = 2 and fy (1, 1) = 1. ∇f (2, 1) = h2, 1i. 4. The directional derivative of f at (1, 1) is Du f (1, 1) = ∇f (1, 1)• u < 4, 3 > 8+3 11 . = h2, 1i• = = |u| 5 5 5