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Ma 116 / Prof. Miller
Name:
Exam 3
April 18, 2002
Solutions
ID:
x2
+ 3y 3 .
2
Determine which are local maxima, which are local minima, and which are saddle points.
1. [9 pts] Find all critical points of f (x, y) = 2 + 3x y +
Solution:
f x (x, y) = 3y + x = 0
⇒ x = −3y
2
f y (x, y) = 3x + 9y = 0 ⇒ −9y + 9y 2 = 0
⇒ y(y − 1) = 0
y = 0 ⇒ x = 0,
y = 1 ⇒ x = −3
Two critical points:
(0, 0) and (−3, 1)
f x x (x, y) = 1, f yy (x, y) = 18y, f x y (x, y) = 3
D(x, y) = f x x f yy − [ f x y ]2 = 18y − 9
D(0, 0) = −9 < 0
⇒ (0, 0) is a saddle point
D(−3, 1) = 18 − 9 > 0, f x x (−3, 1) > 0 ⇒ (−3, 1) is a local minimum
2. [8 pts]
For the function f (x, y) =
p
4 − 4x 2 − y 2 ,
(a) determine the domain and range of f (x, y).
(b) calculate the partial derivatives
Solution:
Part (a): Domain and Range
Domain: 4 − 4x 2 − y 2 ≥ 0
∂f
∂2 f
and
.
∂x
∂ y∂ x
or
f min = 0 (when 4x 2 + y 2 = 4),
Range: 0 ≤ f (x, y) ≤ 2.
Part (b):
x 2 + (y/2)2 ≤ 1.
f max = 2 (when x = y = 0)
Partial derivatives
∂f
∂x
=
−1/2
2
2
1/2 4 − 4x − y
−8x
∂2 f
∂ y∂ x
=
−3/2
−4x −1/2 4 − 4x 2 − y 2
−2y =
= p
−4x
4 − 4x 2 − y 2
−4x y
4 − 4x 2 − y 2
3/2
3. [8 pts]
For the the function f (x, y) = x 2 + 2y 2 + x y,
(a) find the directional derivative of f at the point (1, 1) in the direction h −1, −1 i.
(b) in which direction does the function f decrease most rapidly at the point (1, 1).
Solution:
Part (a):
Directional derivative
∇ f (x, y) = h2x + y, 4y + xi −→ ∇ f (1, 1) = h3, 5i
h−1, −1i
(unit vector in direction of h−1, −1i)
uE =
√
2
√
h3, 5i ◦ h−1, −1i
−3 − 5
Du f (1, 1) = ∇ f (1, 1) ◦ uE =
= √
= −4 2
√
2
2
Part (b): The gradient vector points in the direction of steepest ascent.
Therefore the function is decreasing most rapidly in the direction opposite the gradient.
Direction of steepest descent = −∇ f (1, 1) = h−3, −5i.
4. [8 pts] For the function f (x, y) =
√
x y, determine the equation of the tangent plane
√
at P = (2, 8). Use this plane to estimate (2.02) · (7.96) .
Solution:
r
1 y
f x (x, y) = (1/2)(x y)
y=
−→ f x (2, 8) = 1
2 x
r
1 x
f y (x, y) = (1/2)(x y)−1/2 x =
−→ f x (2, 8) = 1/4
2 y
−1/2
Tangent Plane:
z = f (2, 8) + f x (2, 8)(x − 2) + f y (2, 8)(y − 8)
z = 4 + (x − 2) − (y − 8)/4
f (2.02, 7.96) ≈ 4 + (2.02 − 2) − (7.96 − 8)/4 = 4 + 0.02 − 0.01 = 4.01
√
5. [8 pts] Consider the surface obtained by rotating the curve y = xe−x about the x-axis
on the interval 0 ≤ x ≤ 3. Determine a parametric representation for this surface. [ Be sure
to include the parameter domain. ]
Solution: Cross-sections at x = xo are circles in the yz-plane with radius
√
r (xo ) = xo e−xo . Let u parametrize the axis of rotation (x = u) and v be the angle of
rotation about the x-axis, the polar angle in the yz-plane.
√
√
x = u,
y = u e−u cos v,
z = u e−u sin v
0 ≤ u ≤ 3,
0 ≤ v ≤ 2π
6. [9 pts] For the surface defined by the equation F(x, y, z) = x 4 + z 2 + x yz = 3,
(a) determine the equation of the tangent plane to this surface at the point P = (1, −1, 2).
(b) assuming z to be a function of x and y, determine the value of the partial derivative
∂z
at the point P = (1, −1, 2).
∂x
Solution:
Part (a):
Tangent Plane
∇ F(x, y, z) = 4x 3 + yz, x z, 2z + x y
∇ F(1, −1, 2) = h4 − 2, 2, 4 − 1i = h2, 2, 3i
Tangent Plane:
∇ F(1, −1, 2) ◦ hx − 1, y + 1, z − 2i = 0
2(x − 1) + 2(y + 1) + 3(z − 2) = 0 −→ 2x + 2y + 3z = 6
Part (b):
Assume z = z(x,y)
− ∂∂ Fx P
∂z 2
= −
= ∂F ∂x P
3
∂z
P