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LAST NAME : FIRST NAME : QUIZ 2, Version A : MATH 251, Section 506 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 p 1. [25pts] Find the domain and the range of z = f (x, y) = 1 − x2 − y 2 , z ≥ 0. 2. [25pts] Identify the level curves of the surface defined by z = x2 + 4y. 3. [25pts] Find all of the first partial derivatives of f (x, y, z) = yexyz . 4. [25pts] Find an equation of the tangent plane z = x2 − y 2 at the point (3, −2, 5). 1. The domain Df = {(x, y) ∈ R2 | 1 − x2 − y 2 ≥ 0} = {(x, y) ∈ R2 | x2 + y 2 ≤ 1}. It’s the disc of center (0, 0) and radius 1. Since z ≥ 0, f (x, y) ≥ 0 and f (x, y) = 0 for all points on the unit circle. Moreover, f (x, y) ≥ f (0, 0) = 1, so the values taken by f is between 0 and 1, so [0, 1] . 2. We look for curves with equation k = x2 + 4y, where k is a real number. So, rewrite the equation as k 4y − k = −x2 and we recognize the equation of parabolas with y-axis and vertex (0, ) . 4 3. f is a product of standard functions. For all (x, y, z) in R3 , fx (x, y, z) = y 2 zexyz , fy (x, y, z) = exyz + xyzexyz , fz (x, y, z) = xy 2 exyz . 4. The function f (x, y) = x2 − y 2 is polynomial. We have f (3, −2) = 5, so (3, −2, 5) is well in the surface described by f . We have fx (x, y) = 2x and fy (x, y) = −2y. The equation of the tangent plane at this point is : z − f (3, −2) = fx (3, −2)(x − 3) + fy (3, −2)(y + 2), z − 5 = 6(x − 3) + 4(y + 2), z = 6x + 4y − 5. The equation is : z = 6x + 4y − 5 .