Second Midterm Exam
Math 212 Fall 2010
Instructions: This is a 90 minute exam. You should work alone, without
access to any book or notes. No calculators are allowed. Do not discuss this
exam with anyone other than your instructor. When you have completed the
exam, write out and sign the Honor Code pledge on the front.
The exam consists of 6 questions. You must show all of your work on each
problem to receive full credit, and be sure to clearly indicate your final
answer to each question.
1. [15 Points] Let f (x, y, z) = (xz, sin(xy)) and g(u, v) = (u2 , euv , ln v)
(a) Compute the derivatives Df and Dg.
Solution:
"
Df =
∂f1
∂x
∂f2
∂x
∂f1
∂y
∂f2
∂y
 ∂g1
Dg =
∂u
 ∂g2
∂u
∂g3
∂u
∂f1
∂z
∂f2
∂z
#
=
∂g1 
∂v
∂g2 
∂v
∂g3
∂v
z
0
x
,
y cos xy x cos xy 0


2u
0
= veuv ueuv  .
0
1/v
(b) Using the multivariable chain rule, compute the derivative D(f ◦ g). (Your final answer should be a matrix whose entries are functions of u and v.)
Solution:
D(f ◦ g) = (Df )(Dg)


2u
0
z
0
x  uv
ve
ueuv 
=
y cos xy x cos xy 0
0
1/v
z(2u)
x/v
=
(y cos xy)(2u) + (x cos xy)(veuv ) (x cos xy)(ueuv )
2u ln v
u2 /v
=
2ueuv cos(u2 euv ) + u2 veuv cos(u2 euv ) u3 euv cos(u2 euv )
1
(c) Using the multivariable chain rule, compute the derivative D(g ◦ f ). (Your final answer should be a matrix whose entries are functions of x, y, and z.)
Solution:
D(g ◦ f ) = (Dg)(Df )


2u
0
z
0
x
uv
uv 

ue
= ve
y cos xy x cos xy 0
0
1/v


2uz
0
2ux
= veuv z + ueuv y cos cy ueuv x cos xy veuv x
x cos xy
y cos xy
0
v
v


2xz 2
0
2x2 z
= (sin xy)exz sin xy z + xzexz sin xy y cos xy xzexz sin xy x cos xy (sin xy)exz sin xy x
x cos xy
y cos xy
0
sin xy
sin xy
2. [20 Points] Let f (x, y) = 2x2 + y 4 − 4xy.
(a) Find the critical points of f .
Solution:
We set
0 = ∇f = (4x − 4y, 4y 3 − 4x),
so x = y and 4y 3 − 4y = 4y(y − 1)(y + 1) = 0, and we get three critical points,
(−1, −1), (0, 0), (1, 1) .
(b) Compute the Hessian matrix of f .
" 2
Hf =
∂2f
∂x∂y
∂2f
∂y 2
∂ f
∂x2
∂2f
∂y∂x
#
=
4
−4
−4 12y 2
(c) Using the multivariable second derivative test, classify the critical points of f .
2
Solution: We have ∂∂xf2 = 4 > 0 everywhere, and the determinant of Hf is 16(3y 2 − 1), so by
the second derivative test, (1, 1) and (−1, −1) are local minima and (0, 0) is a saddle point.
(d) Starting from the point (3, 2), in which direction does the function f decrease most
rapidly?
Solution:
The negative of the gradient points in the direction of greatest decrease:
1
−∇f (3, 2) = (−4, −20), so the direction is √
(−4, −20) .
42 + 202
2
Z 4Z
3. [15 Points] Let I =
0
2
√
x
y5
x
dy dx
+1
(a) Write I as the double integral over a region D in the plane. Sketch the region D.
(b) Reverse the order of integration, i.e. rewrite I as an iterated integral “dx dy”.
Solution:
2
Z
0
Z
y2
y5
0
x
dx dy
+1
(c) Compute I.
Solution:
Z
0
2
Z
0
y2
y5
x
dx dy =
+1
Z
0
2
2
1 2 y
x 2
y 5 + 1 x=0
!
dy
Z
1 2 y4
=
dy
2 0 y5 + 1
Z 33
1
1
=
du where u = y 5 + 1, du = 5ydy
10 1 u
1
=
ln 33
10
4. [15 Points] Find the dimensions of a right circular cylinder of the maximum possible
volume that can be inscribed in a sphere of radius 16. (Make sure to clearly explain what
your variables mean.)
3
Solution: Let the sphere by given by x2 +y 2 +z 2 = 162 . By symmetry, we may as well assume
that the axis of the cylinder is the z-axis. Let (x, y, z) be a point on both the surface of the
sphere and p
the surface of the cylinder. The volume of the inscribed cylinder is then (πr2 )(2z),
where r = x2 + y 2 is the radius and 2z is the height. Thus, we are trying to maximize the
function f (r, z) = r2 z, subject to the constraint g(r, z) = r2 + z 2 = 162 .
To do so, we solve the Lagrange multiplier equation ∇f = λ∇g, where ∇f = (2rz, r2 )
r2
= 2z
, so r2 = 2z 2 . √Substituting this into the constraint
and ∇g = (2r, 2z). This gives 2rz
2r
√
equation yields 3z 2 = 162 , so z = √163 , and r = 2z = 16√32 . Thus the largest volume cylinder
√
32
16 2
has height √ and radius √ .
3
3
5. [20 Points] Let W be the region in R3 defined by the inequalities y ≥ x2 , z ≤ 3 − y, and
z ≥ 0. Let f (x, y, z) be a continuous function on W .
(a) Sketch the region W .
4
ZZZ
(b) Write
f (x, y, z) dV as an iterated integral in the order “dz dy dx”.
W
Solution:
√
ZZZ
Z
f (x, y, z) dV =
W
3
√
− 3
Z
3
Z
3−y
f (x, y, z)dz dy dx .
x2
0
ZZZ
(c) Write
f (x, y, z) dV as an iterated integral in the order “dx dy dz”.
W
5
Solution:
ZZZ
Z
3
Z
3−z
f (x, y, z) dV =
W
0
√
Z
y
√
− y
0
f (x, y, z)dx dy dz .
ZZZ
y dV
(d) Evaluate
W
Solution:
Z
0
3
Z
0
3−z
Z
√
y
√
− y
Z
3
3−z
Z
y
y dx dy dz =
0
Z
0
3
3−z
Z
=
=
Z
Z0 3 0
y
√
− y
1 dx dy dz
√
2y y dy dz
4 5/2 3−z
y
5
y=0
0
Z
√
dz
3
− z)5/2 dz
0
3
8
= − 35
(3 − z)7/2 z=0
√
8 7/2
= 35
3 = 216
3.
35
=
6
4
(3
5
6. [15 Points] Let f (x, y) = x2 + y 2 − 2x − 4y. Find the absolute maximum and minimum
of f , subject to the constraints x2 + y 2 ≤ 2 and y ≥ 0.
Solution:
First, we set 0 = ∇f = (2x − 2, 2y − 4), which tells us that (1, 2) is the only
critical point of f , but it is outside the region defined by the constraints. Thus the minimum
and maximum of f will be on the boundary. The √
curve bounding
√ the region has two smooth
components, meeting non-smoothly at the points ( 2, 0) and (− 2, 0).
The first component is given implicitly by the equation g1 (x, y) = x2 +y 2 = 2. Thus we
set ∇f = λ∇g1 , giving (2x − 2, 2y − 4) = λ(2x, 2y). After checking that there are no solutions
with x = 0 or y = 0, we get 2x−2
= 2y−4 , so 1 = y2 , and y = 2x. The line y = 2x meets the
2x q 2y q x
q q 2
2
2
circle x2 + y 2 = 2 in two points, ±
,
but
only
,
2
, 2 25 satisfies y ≥ 0.
5
5
5
The second boundary component is given implicitly by the equation g2 (x, y) = y = 0.
We set ∇f = λ∇g2 , giving (2x − 2, 2y − 4) = λ(0, 1), so that x = 1, giving us the point (1,
√ 0).
We now have four candidates for extrema of f on our region, the two points (± 2, 0)
where the boundary components
q qmeet,
and the two constrained critical points from the Lagrange
2
multiplier calculations,
, 2 25 and (1, 0). To find the absolute maximum and minimum,
5
we just evaluate f at these four points:
√ √
f
2, 0 = 2 − 2 2
√ √
f − 2, 0 = 2 + 2 2 absolute maximum
q q √
2
f
, 2 25 = 2 − 2 10 absolute minimum
5
f (1, 0) = −1
7