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LAST NAME : FIRST NAME : QUIZ 4, 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 = 2x2 + 4y 2 at (1, 1, 6). 2. [25pts] Find the differential of the function z = 3. [25pts] Find √ yex . dy for xy = sin(x + y 2 ). dx 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) = 2x2 + 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 − 6 = fx (1, 1)(x − 1) + fy (1, 1)(y − 1). fx (x, y) = 4x so fx (1, 1) = 4 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 − 6 = 4(x − 1) + 8(y − 1) ⇔ 4x + 8y − z = 6 . 2. dz(x, y) = fx (x, y)dx + fy (x, y)dy = √ 1 yex dx + √ ex dy . 2 y 3. We use here the implicit Theorem on f (x, y) = 0 where f (x, y) = xy − sin(x + y 2 ). So, we assume that fy (x, y) = x − 2y cos(x + y 2 ) 6= 0. Then the Theorem says that y is an implicit function of x in a neighborhood of these points and dy −fx (x, y) y − cos(x + y 2 ) . (x) = = − dx fy (x, y) x − 2y cos(x + y 2 ) 4. g, f, x, y are all of them differentiable. By the Chain Rule, ∂x ∂f ∂y ∂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 + 2s(s2 + t2 ) . 2s + t + (s2 + t2 )2