LAST NAME :
FIRST NAME :
QUIZ 5, 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. [30pts] Does there exist a vector G such that CurlG =< xy 2 z, x, z + x2 > ? (Explain Why)
2. Let S the surface given by ~r(u, v) = hv 2 − 5u, u2 , v − 3i, 0 ≤ u ≤ 1, 0 ≤ v ≤ 1.
(a) [35pts]Find the normal vector to S at the point corresponding to (u, v) = (1, 1).
(b) [35pts]Find an equation of the tangent plane to S passing through the point (−4, 1, −2).
1. If such a vector G exists then div(CurlG) = 0 but
div(CurlG) = y 2 z + 1 6= 0, for some (x, y, z).
Then there is no such G.
2. (a) A normal vector to a parametric surface S is ~ru × ~rv (u, v) and r(1, 1) = (−4, 1, −2).
~ru (u, v) = h−5, 2u, 0i, so for (u, v) = (1, 1), ~ru (1, 1) = h−5, 2, 0i.
~rv (u, v) = h2v, 0, 1i, so for (u, v) = (1, 1), ~rv (1, 1) = h2, 0, 1i
~i
~ru × ~rv (1, 1) = −5
2
~j
2
0
~k 0
1
= h2, 5, −4i.
(b) The point (−4, 1, −2) corresponds to r(1, 1). Then an equation of the tangent plane to S at
(−4, 1, −2) is
~ru ×~rv (1, 1)•hx − (−4), y − 1, z − (−2)i = 2(x+4)+5(y −1)−4(z +2) = 0 ⇔ 2x + 5y − 4z = 5 .