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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 an equation of the tangent plane of z = (x + 1)2 + 4y 2 at (−1, 1, 4). 2. [25pts] Find the differential of the function z = √ x ln(1 + y). 3. [25pts] Find the gradient of f (x, y) = x2 y + xey at (1, 0). 4. [25pts] Use the chain rule to find gs with g(s, t) = f (x(s, t), y(s, t)) where f (x, y) = ln(x + y 2 ) and x(s, t) = 2s + t and y(s, t) = s2 + t2 . 1. Let f (x, y) = (x + 1)2 + 4y 2 . f is a polynomial function ( defined on R2 ). The surface S is given as a graph of function of 2 variables, so we apply the formula seen in class : an equation of the tangent plane is of the form z − 4 = fx (−1, 1)(x + 1) + fy (−1, 1)(y − 1). fx (x, y) = 2(x + 1) so fx (−1, 1) = 0 and fy (x, y) = 8y so fy (−1, 1) = 8. Then an equation of the tangent plane to the surface S = {(x, y, z) | f (x, y) = z} is z − 4 = 8(y − 1) 2. ⇔ 8y − z = 4 . √ 1 x dz(x, y) = fx (x, y)dx + fy (x, y)dy = ln(1 + y) √ dx + dy . 1+y 2 x 3. f est a function defined on R2 . She has also continuous partial derivatives. We have fx (x, y) = 2xy+ey and fy (x, y) = x2 + xey . Its gradient at (1, 0) is ∇f (1, 0) =< fx (1, 0), fy (1, 0) >=< 1, 2 > . 4. g, f, x, y are all of them differentiable. By the Chain Rule, ∂y ∂f ∂x ∂f + , ∂s ∂x ∂s ∂y 1 2y(s, t) = 2. + 2s. , 2 x(s, t) + y (s, t) x(s, t) + y 2 (s, t) gs (s, t) = = 2 + 4s(s2 + t2 ) . 2s + t + (s2 + t2 )2