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QUIZ 9 : MATH 251, Section 516 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. [30pts] Does there exist a vector field G such that Curl G = h−xz 2 , 2x, 3yzi (support your answer) ? 2. Let S the surface given by ~r(u, v) = hu3 − 5v, v 2 , u2 − 3i. (a) [35pts]Find the normal vector to S at the point corresponding to (u, v) = (2, 1). (b) [35pts]Find an equation of the tangent plane to S passing through the point (3, 1, 1). Correction. 1. Thanks to a theorem seen in class, if such a vector field G exists then div(curl G) = 0 for all (x, y, z). So, compute divh−xz 2 , 2x, 3yzi = ∂(−xz 2 ) ∂(2x) ∂(3yz) + + ∂x ∂y ∂z = −z 2 + 3y 6= 0 for some (x, y, z) ∈ R3 . So, by contraposition, there doesn’t exist such a G . 2. (a) A normal vector to a parametric surface S is ~ru × ~rv (u, v). ~ru (u, v) = h3u2 , 0, 2ui, so for (u, v) = (2, 1), ~ru (2, 1) = h12, 0, 4i. ~rv (u, v) = h−5, 2v, 0i, so for (u, v) = (2, 1), ~rv (2, 1) = h−5, 2, 0i ~i ~j ~k ~ru × ~rv (1, 2) = 12 0 4 −5 2 0 = h−8, −20, 24i. (b) The point (3, 1, 1) corresponds to r(2, 1). Then an equation of the tangent plane to S at (3, 1, 1) is ~ru × ~rv (2, 1) • hx − 3, y − 1, z − 1i = −8(x − 3) − 20(y − 1) + 24(z − 1) = 0 ⇔ −8x − 20y + 24z = −20 ⇔ 2x + 5y − 6z = 5 .