MAT 211 Test 2. Spring 2015
SHOW YOUR WORK!!
1. (10 points) Below is a contour plot of the function f (x, y) = x2 + y 2 − x4 − y 4 . The numbers on the
contour lines indicate values of f . On the contour plot, draw the gradient vectors at the points (0.6, 0.6),
(0.3, 0.3), and (0, 0.5).
Solution: The gradient vectors are perpendicular to the contour lines and point in the direction in
which f increases fastest. (The actual gradient vectors are twice as long as the ones depicted here.)
0
40
0.
0.1
0
0.05
00
.2.15
0
5
0.35
0 0.4
.4
5
0
0
.1
0
5
.2
0
0
.2
5
0.
35
0.
10 0.05
0
.2
0
5
.3
0
25
0.
0.45
50
.1
00
.2
-0.5
0
0.105.20
0.05
0
0
.2
.3
0.05
0.30
0.
45
0.5
0.10
5
0
0.3
0.1
1
0.4
5
0
.4
0
0.30
-1
-1
0 5
0.20.1
0.05
-0.5
0
0.3
0.10
5
0.5
40
0.
1
2. (15 points) Find all the local maxima, minima, and saddle points of the function
f (x, y) = x2 + y 2 − x4 − y 4 . There are nearly ten of them! Be sure to say which are maxima, which are
minima, and which are saddle points.
Solution:
fx = 2x − 4x3
fy = 2y − 4y 3
fxx = 2 − 12x2
fxy = 0
fyy = 2 − 12y 2
Critical points occur where
0 = fx and 0 = fy
0 = 2x − 4x3 and 0 = 2y − 4y 3
0 = x(1 − 2x2 ) and 0 = y(1 − 2y 2 )
p
p
(x = 0 or x = ± 1/2) and (y = 0 or y = ± 1/2)
So there are nine critical points:
p
p
(− 1/2,p− 1/2)
p(0, − 1/2)
p
( 1/2, − 1/2)
p
(− 1/2, 0)
p(0, 0)
( 1/2, 0)
p
p
(− 1/2,
p 1/2)
(0,
1/2)
p
p
( 1/2, 1/2)
The Hessian is
p
p
2
D = fxx fyy − fxy
= (2 − 12x2 )(2 − 12y 2 )D < 0 at (0, ± 1/2) and (± 1/2, 0)
D > 0 at the other critical points
so
(0, ±
p
1/2), (±
p
1/2, 0) are saddle points,
and the other critical points are maxima and minima.
fxx > 0 at (0, 0)
p
p
fxx < 0 at (± 1/2, ± 1/2)
so
(±
p
(0, 0) is a local minimum
p
1/2, ± 1/2) are local maxima
3. (a) (10 points) Find the linear approximation (that is, the tangent plane) for the function
f (x, y) = 5 − x2 y 3
at the point (x, y) = (1, 1)
Solution:
f (1, 1) = 4
fx = −2xy 3
2 2
fy = −3x y
fx (1, 1) = −2
fy (1, 1) = −3
so the equation of the tangent plane is
z = 4 − 2(x − 1) − 3(y − 1)
This is the best linear approximation to z = f (x, y) around (x, y) = (1, 1).
(b) (5 points) Use your answer from part (a) to approximate f (1.02, 0.97).
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Solution:
f (1.02, 0.97) ≈ 4 − 2(1.02 − 1) − 3(0.97 − 1) = 4 − 0.04 + 0.09 = 4.05
4. (10 points) Let
W (s, t) = F (x(s, t), y(s, t))
where F , x, and y are differentiable and
x(1, 0) = 2
y(1, 0) = 3
xs (1, 0) = −2
ys (1, 0) = 5
xt (1, 0) = 6
yt (1, 0) = 4
Fx (2, 3) = −1
Fy (2, 3) = 10.
Find Ws (1, 0) and Wt (1, 0).
Solution: The chain rule says
Ws = Fx xs + Fy ys
W t = Fx x t + Fy y t
Assume (s, t) = (1, 0) so x, y, Ws , Wt , xs , ys , xt , yt are all evaluated at (s, t) = (1, 0). Also x(1, 0) = 2
and y(1, 0) = 3 so Fx , Fy are evaluated at (x, y) = (2, 3). Thus
Ws (1, 0) = Fx (2, 3)xs (1, 0) + Fy (2, 3)ys (1, 0)
= (−1)(−2) + (10)(5)
= 52
Wt (1, 0) = Fx (2, 3)xt (1, 0) + Fy (2, 3)yt (1, 0)
= (−1)(6) + (10)(4)
= 34
5. (10 points) The radius of a right circular cone is increasing at a rate of 1.8 m/s while its height is
decreasing at a rate of 2.5 m/s. At what rate is the volume of the cone changing when the radius is
120 m and the height is 140 m? (Hint: the volume of a cone is one-third the area of its base times its
height.)
Solution: Let V represent the volume of the cone, r its radius, h its height, and t time. We are
given
dh
dr
= 1.8 and
= −2.5
dt
dt
meters per second when
r = 120 and h = 140
meters. Also
V =
1 2
πr h
3
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cubic meters. The chain rule says the rate of change of volume is
dV
dV dr dV dh
=
+
dt
drdt
dh dt dr
dh
π
2rh + r2
=
3
dt
dt
π
=
2(120)(140)(1.8) + (1202 )(−2.5) .
3
This works out to be approximately 25635 cubic meters per second.
6. (a) (8 points) Find the directional derivative of g(x, y) = x2 + 2xy + 3y 2 at the point (x, y) = (2, 3) in
the direction < 3/5, 4/5 >.
Solution:
∇g =< 2x + 2y, 2x + 6y >
so at the point (2, 3) the gradient is
∇g|(2,3) =< 10, 22 >
and the directional derivative in the direction < 3/5, 4/5 >is
∇g|(2,3) · < 3/5, 4/5 >=< 10, 22 > · < 3/5, 4/5 >= 118/5
(b) (7 points) In what direction is the function g(x, y) = x2 + 2xy + 3y 2 increasing fastest at the point
(2, 3)?
Solution: It increases fastest in the direction of the gradient
√
1
< 10, 22 >
+ 222
102
7. (15 points) Use Lagrange multipliers to find the points where k(x, y) = x − y has extreme values (maximum or minimum) on the circle x2 + y 2 = 1. Be sure to say which point(s) are maxima and which are
minima.
Solution:
objective: k(x, y) = x − y
constraint: g(x, y) = x2 + y 2 = 1
We want to find maxima and minima of the objective, along the circle where the constraint is
satisfied. The Lagrange multiplier approach suggests solving the equation
∇k = λ∇g
< 1, −1 >= λ < 2x, 2y >
1 = 2λx
and
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− 1 = 2λy
The last pair of equations show that x 6= 0 and y 6= 0 so we can divide by them and optain
1
−1
=λ=
2x
2y
y = −x
Plugging that back into the constraint equation one has
x2 + (−x)2 = 1
2x2 = 1
p
x = ± 1/2.
y = −x so the candidates are
(x, y) = (
p
1/2, −
p
1/2) or (x, y) = (−
p
1/2,
p
1/2)
The circle where the constraint is satisfied is a closed, bounded set, and k is continuous so it must
have an absolute maximum and an absolute minimum on the set where the constraint is satisfied.
Evaluating at the two candidate points we find that
p (x, y)
p
( p
1/2, −p1/2)
(− 1/2, 1/2)
fp
(x, y)
2 p1/2
−2 1/2
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type
absolute max
absolute min