Exam 2, Fall 2010—Solutions
Problem 1. (10 points)
Let f (x, y) = 2e2x−y + 2.
(a) (4 pts) Find the linearization of the function f at the point (5, 10).
Solution: The linearization L(x, y) of f (x, y) at P = (x0 , y0 ) is defined by
L(x, y) = f (P ) + fx (P ) · (x − x0 ) + fy (P ) · (y − y0 ).
In our case,
f (5, 10) = 4,
fx (x, y) = 4e2x−y ,
fx (5, 10) = 4,
fy (x, y) = −2e2x−y ,
fy (5, 10) = −2.
Thus, the linearization L(x, y) of f (x, y) = 2e2x−y + 2 at the point P = (5, 10) is equal to
L(x, y) = 4 + 4(x − 5) − 2(y − 10) = 4 + 4x − 2y.
(b) (3 pts) Using the linear approximation, estimate the value f (5.1, 10.1).
Solution: Recall that the linear approximation to f (x, y) near the point (x0 , y0 ) is
f (x, y) ≈ L(x, y) = f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ).
In our case, (x, y) = (5.1, 10.1) is close to the point (5, 10) and using (a) we get
f (5.1, 10.1) ≈ L(5.1, 10.1) = 4 + 4 · (5.1 − 5) − 2 · (10.1 − 10) = 4.2.
(c) (3 pts) Write an equation of the tangent line to the level curve f (x, y) = 4 at the point (5, 10).
Solution: The tangent line to a level curve f (x, y) = c at the (x0 , y0 ) is perpendicular to the
gradient vector ∇f (x0 , y0 ). Thus, the point (x, y) lies on the tangent line through (x0 , y0 ) exactly
when the vector $x − x0 , y − y0 % is perpendicular to ∇f (x0 , y0 ). So, an equation of the tangent line
is
∇f (x0 , y0 )•$x − x0 , y − y0 % = 0.
In our case, ∇f (5, 10) = $4, −2% and an equation of the tangent line through (5, 10) is
4(x − 5) − 2(y − 10) = 0,
or 2x − y = 0.
Problem 2. (14 points)
(a) (9 pts) Let g(u, v) be a differentiable function of two variables satisfying
∂g
∂g
(1, 2) = 3 and
(1, 2) = 4.
∂u
∂v
!
"
∂G
∂G
and
when x = 1 and y = 1.
Define G(x, y) = g x/y, x2 + y 2 . Find
∂x
∂y
Solution: To calculate the partial derivatives of G with respect to x and y we use the chain rule.
We have
∂G
∂g
∂u
∂g
∂v
(x, y) =
(u, v) ·
(x, y) +
(u, v) ·
(x, y)
∂x
∂u
∂x
∂v
∂x
∂G
∂g
∂u
∂g
∂v
(x, y) =
(u, v) ·
(x, y) +
(u, v) ·
(x, y)
∂y
∂u
∂y
∂v
∂y
1
In our case,
∂G
∂g 1 ∂g
=
· +
· (2x)
∂x
∂u y ∂v
#
$
∂G
∂g
x
∂g
=
· − 2 +
· (2y)
∂y
∂u
y
∂v
and at the point (x, y) = (1, 1) we have u = (1/1, 12 + 12 ) = (1, 2) so that
∂G
(1, 1) = 3 · 1 + 4 · 2 = 11
∂x
∂G
(1, 1) = 3 · (−1) + 4 · 2 = 5.
∂y
(b) (5 pts) The equation xyz + z 3 = 6 implicitly defines z = z(x, y) as a function of x and y. Find
∂z
at the point (−1, 1, 2).
the value of
∂x
Solution: We use the chain rule to perform the implicit differentiation. Differentiating with
respect to x we get
∂z
∂z
yz + xy
+ 3z 2
= 0.
∂x
∂x
Evaluating at the point (−1, 1, 2),
∂z
∂z
∂z
(−1, 1) + 12 (−1, 1) = 2 − 11 (−1, 1)
∂x
∂x
∂x
∂z
and solving for ∂x (−1, 1), we get
∂z
2
(−1, 1) = .
∂x
11
2−
Problem 3. (11 points )
√
Let F (x, y, z) = 2xy 2 + z 3 and P = (0, 2, −1).
(a) (6 pts) In what direction does F have the maximum rate of change at P ? (Write the result as
a unit vector.) What is this maximum rate of change?
Solution: A function F (x, y, z) has a maximum rate of change at P in the direction of the gradient
∇F (P ) and this maximum rate of change is equal to the magnitude |∇F (P )|. The unit vector u
∇F (P )
in the direction of ∇F (P ) is equal to u = |∇F
(P )| . In our case,
√
∇F (x, y, z) = $2y 2 , 4xy, 3z 2 %,
∇F (0, 2, −1) = $4, 0, 3%.
√
Thus, the maximum rate of change at the point P = (0, 2, −1) is in the direction of the vector
$4, 0, 3%. The maximum
rate of change at P is equal to |$4, 0, 3%| = 5, and the unit vector in the
√
direction of ∇F (0, 2, −1) is equal to 51 $4, 0, 3%.
(b) (5 pts) Write an equation of the tangent plane to the level surface F (x, y, z) = −1 at the point
P.
Solution: Recall that the tangent plane to the level surface F (x, y, z) = c at the point P =
(x0 , y0 , z0 ) is given by
fx (P )(x − x0 ) + fy (P )(y − y0 ) + fz (P )(z − z0 ) = 0.
In our case, the tangent plane has an equation
√
4 · (x − 0) + 0 · (y − 2) + 3 · (z + 1),
2
i.e.,
4x + 3z = −3.
Problem 4. (14 points)
(a) (11 pts) Let f (x, y) = 3y − y 3 − 3x2 y. Find all the critical points of f and classify each as a
local minimum, a local maximum, or a saddle point. Put your final answers in the table below.
Solution: Recall that the critical points (x, y) are where ∇f (x, y) = $0, 0%. In our case, ∇f (x, y) =
$−6xy, 3 − 3y 2 − 3x2 %. So we get two equation
3 − 3y 2 − 3x2 = 0.
−6xy = 0,
From the first equation, we have either x = 0 or y = 0. If x = 0, then the second equation gives
y 2 = 1, that is, y = −1 or y = 1, and we have points (0, −1) and (0, 1). If y = 0, then the second
equation gives x2 = 1 and the points (−1, 0) and (−1, 0). We classify these points by the second
derivative test. Since
fxx = −6y,
fxy = fyx = −6x,
2
and D = fxx fyy − fxy
= 36(y 2 − x2 ),
fyy = −6y,
we have the following classification of the critical points.
Critical Point
D
fxx
Classification
(0,-1)
36
6
local minimum
(0,1)
36
-6
local maximum
(-1,0)
-36
–
saddle point
(1,0)
-36
–
saddle point
R2 .
(b) (3 pts) Let g(x, y) be a continuous functions on
On which of the following domains the
function g is guaranteed to have a global maximum value and a global minimum value? Circle your
answer(s).
Solution: Recall that a continuous function g(x, y) on a closed and bounded set D has a global
maximum and a global minimum. The only sets which are closed and bounded are C = {(x, y)| x2 +
y 2 ! 1} and D = {(x, y)| 0 ! x ! 1, 0 ! y ! 1}.
(a)
A = {(x, y)| x ≥ 0, y ≥ 0}
(b) B = {(x, y)| x2 + y 2 < 1}
• (c)
C = {(x, y)| x2 + y 2 ! 1}
• (d) D = {(x, y)| 0 ! x ! 1, 0 ! y ! 1}
(e)
E = {(x, y)| 0 ! x ! 1, 0 < y < 1}.
Problem 5. (12 points)
Let f (x, y) = x2 + y 2 − 2x.
(a) (8 pts) Use the method of Lagrange multipliers to find the extreme values of f subject to
x2 + y 2 = 9.
Solution: First note the circle x2 + y 2 = 9 is bounded and closed and the function f is continuous.
Hence f attains a global maximum and a global minimum. To find the points where f attains
extreme values, we solve the equation ∇f (x, y) = λ∇g(x, y) and g(x, y) = 9. In our case these
3
equations take the form
(1)
2x − 2 = 2λx
(2)
2y = 2λy
2
x + y 2 = 9.
(3)
(2) is equivalent to
(4)
y(1 − λ) = 0
which implies that either y = 0 or λ = 1. If λ = 1, then (1) implies that 2x − 2 = 2x which is
impossible. So, λ (= 1 and y = 0. Then (3) gives , x2 = 9, that is x = −3 or x = 3. So we get two
points (−3, 0) and (3, 0) at which f (−3, 0) = 15 and f (3, 0) = 3. Thus, the global maximum of f
subject to x2 + y 2 = 9 is equal to 15 and the global minimum is equal to 3. (b) (4 pts) Find the
extreme values of f on the disk D = {(x, y)| x2 + y 2 ! 9}.
Solution: The disk D is closed and bounded and f is continuous on D. So, f attains its extreme
values on D. The extreme values of f occur either at the critical points of f in the interior of
D. i.e., in the set {(x, y)| x2 + y 2 < 9}, or at points found by the Lagrange’s multipliers on the
boundary of D, i.e., on the circle x2 + y 2 = 9. We already know the extreme values of f on the
boundary of the disk D. So we look for critical points in the interior of D.
The critical points of f are the solution of ∇f (x, y) = $0, 0%. In our case, $2x − 2, 2y% = $0, 0%.
Thus, the point (1, 0) is a critical point of f and it belongs to the open disk x2 + y 2 < 9. The value
of f at (1, 0) is equal to f (1, 0) = −1. Combining this fact with part (a), the global maximum of
f on D is equal to 15 and the global minimum is equal to −1.
Problem 6. (12 points)
Consider the integral
I=
%
0
4% 2
√
x
y3
1
dy dx.
+1
(a) (4 pts) Use the axes below to sketch the region of integration for the integral I.
(b) (8pts) Write the integral I using the order dx dy and evaluate the resulting integral.
4
%
0
4% 2
√
x
1
dy dx =
3
y +1
%
0
2 % y2
0
% 2&
=
0
y3
1
dx dy
+1
'y2
%
dy =
2
y2
dy
3
0 y +1
0
(2
(
(y 3 + 1)$
1
3
dy = ln(y + 1)((
3
y +1
3
0
x
3
y +1
%
1 2
=
3 0
1
= ln 9.
3
Problem 7. (12 points)
Consider the double integral
I=
%
2
−2
%
√
4−x2
e−x
2 −y 2
dy dx.
0
(a) (4 pts) Sketch the region R over which the integration is performed.
Solution: The region of the integration is R = {(x, y)| − 2 ! x ! 2, 0 ! y !
√
4 − x2 }.
(b) (8 pts) Write the integral I using polar coordinates and evaluate the resulting integral.
Solution: In polar coordinates the region R can be described as R = {(r, θ)| 0 ! θ ! π, 0 ! r ! 2}.
Thus, the integral is equal to
% 2 % √4−x2
% π
2
−x2 −y 2
e
dy dx =
e−r r dr dθ
−2
0
0
% π
2$
2(
e−r ((2
e−r $
−
=
−
dr dθ =
dθ
2
2 (0
0
0
0
%
" π!
"
1 π!
=
1 − e−4 =
1 − e−4 .
2 0
2
%
π
% 2#
Problem 8. (15 points)
The region D in R3 is bounded by the planes
x = 0,
y = 0, z = 0, and 2x + 2y + z = 4.
))) z
(a) (9 pts) Express the the integral
D e dV as an iterated integral with the order of integration
dz dy dx and then evaluate the resulting integral.
Solution: The tetrahedron D and its projection R on the xy-plane are sketched below.
5
z
4
y
2
x+y =2
2
y
x
0
2
2
x
Figure 1. The tetrahedron and its projection onto the xy-plane.
The region D can be described as D = {(x, y, z)| (x, y) ∈ R, 0 ! z ! 4 − 2x − 2y}, where R is the
triangle R = {(x, y)|0 ! x ! 2, 0 ! y ! 2 − x}. Thus,
'
% % &% 4−2x−2y
% 2 % 2−x % 4−2x−2y
%%%
z
z
e dV =
e dz dA =
ez dz dy dx.
D
R
0
The inner integral is equal to
% 4−2x−2y
0
0
0
0
(4−2x−2y
(
e dz = e (
= e4−2x−2y − 1.
z
z(
0
The middle integral is equal to
(2−x
$
% 2−x #
(
1
5
1
1
1 4−2x−2y
4−2x−2y
− y ((
= − e0 + e4−2x + (x − 2) = − + x + e4−2x .
e
− 1 dy = − e
2
2
2
2
2
0
0
Finally, the outer integral is equal to
(
$
% 2#
5
1 4−2x
x2 1 4−2x ((2
5
− +x− e
dx = − x +
− e
(
2
2
2
2
4
0
0
1 1 4
13 1
= −5 + 2 − + e = − + e4 .
4 4
4
4
))) z
(b) (3 pts) Rewrite the triple integral
D e dV as an iterated integral with the order of integration
dy dz dx. (Do NOT evaluate it integral.)
))) z
(c) (3 pts) Rewrite the triple integral
D e dV as an iterated integral with the order of integration
dx dy dz. (Do NOT evaluate the integral.)
Solution:
(b) The region D can be described as
D = {(x, y, z)| (x, z) ∈ R1 , 0 ! y ! 2 − x − z/2}
= {(x, y, z)| 0 ! x ! 2, 0 ! z ! 4 − 2x, 0 ! y ! 2 − x − z/2}
6
z
(b)
z
4
(c)
4
2x + z = 4
2y + z = 4
x
y
2
0
2
0
Figure 2. (a) The projection R1 of the tetrahedron onto the xz-plane and (b) the
projection R2 of the tetrahedron onto the zy-plane.
so that
%%%
z
e dV =
% % &%
R1
D
2−x−z/2
z
e
0
'
dA =
%
0
2 % 4−2x % 2−x−z/2
0
ez dy dz dx.
0
(c) The region D can be described as
D = {(x, y, z)| (y, z) ∈ R2 , 0 ! x ! 2 − y − z/2}
= {(x, y, z)| 0 ! z ! 4, 0 ! y ! 2 − z/2, 0 ! x ! 2 − y − z/2}
so that
%%%
D
z
e dV =
% % &%
R2
2−y−z/2
z
e
0
'
dA =
7
%
0
4 % 2−z/2 % 2−y− z2
0
0
ez dx dy dz.