QUIZ 4 : 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
y
1. Let f (x, y, z) = x + .
z
(a) [25pts] Find the direction in which f increases most rapidly at the point (2, 1, 1).
(b) [25pts] Find the directional derivative of f at the point (2, 1, 1) in the direction < 4, 3, −1 >.
2. Let f (x, y) = x2 + y 2 + 4x − 6y.
(a) [25pts] Find the critical points of this function.
(b) [25pts] Classify these points ( say if these are maximum, minimum points or saddle points by
justifying) .
1. (a) f is defined on Df = {(x, y, z) ∈ R3 | z 6= 0}. f is differentiable on Df . f increases most rapidly
in the direction of gradient :
y
1
∇f (x, y, z) = hfx (x, y, z), fy (x, y, z), fz (x, y, z)i = h1, , − 2 i.
z z
So, at (2, 1, 1), we have ∇f (2, 1, 1) = h1, 1, −1i .
(b) Since f is differentiable on Df , its differential and its directional derivative coincide at (2, 1, 1) ∈
u
4
3
−1
Df . So, denote u = h4, 3, −1i. The unit vector in the direction of u is :
=< √ , √ , √ >.
|u|
26 26 26
4
3
−1
Du f (2, 1, 1) = ∇f (2, 1, 1)• < √ , √ , √ >
26 26 26
4
3
−1
= h1, 1, −1i• < √ , √ , √ >
26 26 26
4+3+1
8
= √
= √ .
26
26
2. f is a polynomial function. It admits second partial derivatives at any points in R2 .
(a) Determine the critical points of f i.e the sets of points (x, y) such that fx (x, y) = 2x + 4 = 0 =
fy (x, y) = 2y − 6. Solve the following system
2x + 4 = 0
x = −2
⇔
2y − 6 = 0,
y = 3,
We obtain only one critical point : (−2, 3) .
(b) Now, classify this point. We have fxx (x, y) = 2, fyy (x, y) = 2, fxy (x, y) = 0 = fyx (x, y) (by the
Clairaut’s Theorem since the function is polynomial). So, fxx (−2, 3)fyy (−2, 3) − (fxy (−2, 3))2 =
4 > 0 and fxx (−2, 3) = 2 > 0, so (−2, 3) is a local minimum point.