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LAST NAME : FIRST NAME : QUIZ 2, Version B : 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) = 4 − x2 − y 2 , z ≥ 0. 2. [25pts] Identify the level curves of the surface defined by z = y 2 + x. 3. [25pts] Find all of the first partial derivatives of f (x, y, z) = x2 eyz . 4. [25pts] Find an equation of the tangent plane to the surface z = x2 + 4y 2 at the point (2, 1, 8). 1. The domain Df = {(x, y) ∈ R2 | 4 − x2 − y 2 ≥ 0} = {(x, y) ∈ R2 | x2 + y 2 ≤ 4}. It’s the disc of center (0, 0) and radius 2. Since z ≥ 0, f (x, y) ≥ 0 and f (x, y) = 0 for all points on the circle with center (0, 0) and radius 2. Moreover, f (x, y) ≥ f (0, 0) = 2, so the values taken by f is between 0 and 1, so [0, 2] . 2. We look for curves with equation k = y 2 + x, where k is a real number. So, rewrite the equation as x = k − y 2 and we recognize the equation of parabolas with x-axis and vertex (k, 0) . 3. f is a product of standard functions. For all (x, y, z) in R3 , fx (x, y, z) = 2xeyz , fy (x, y, z) = x2 zeyz , fz (x, y, z) = x2 yeyz . 4. The function f (x, y) = x2 + 4y 2 is polynomial. We have f (2, 1) = 8, so (2, 1, 8) is well in the surface described by f . We have fx (x, y) = 2x and fy (x, y) = 8y. The equation of the tangent plane at this point is : z − f (2, 1) = fx (2, 1)(x − 2) + fy (2, 1)(y − 1), z − 8 = 4(x − 2) + 8(y − 1), z = 4x + 8y − 8. The equation is : z = 4x + 8y − 8 .