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Practice Midterm 2 – Math 2153
1. Decide if the following statements are TRUE or FALSE and circle your answer. You do NOT need to justify
your answers.
(a) (1 point) If both partial derivatives fx and fy exist at (a, b) then f is differentiable at (a, b).
Solution: F (fx and fy must exist and be continuous on an open set containing the point (a, b) to
ensure that f is differentiable at (a, b).)
(b) (1 point) If f has a local maximum at the point (a, b, c) then ∇f = 0.
Solution: F (f can also have a local maximum at (a, b, c) when ∇f does not exist.)
(c) (1 point) If f has a saddle point at (a, b) then f cannot have a local minimum at (a, b).
Solution: T (The definition of “saddle point” precludes it from being a local maximum or minimum)
(d) (1 point) If f is differentiable at (a, b, c) then magnitude of the gradient vector ∇f (a, b, c) is the
maximal directional derivative
Du (a, b, c)
where u ranges over all unit vectors in R3 .
Solution: T
2. Give examples of the following. Be as explicit as possible. You do NOT need to justify your answers.
(a) (2 points) Give an example of a function f (x, y) continuous on R2 such that there are infinitely
many points (a, b) ∈ R2 such that f has a local maximum at (a, b).
Solution:
f (x, y) = 0
(b) (2 points) Give an example of a function F (x, y, z) for which the graph of z = sin(xy) is a level
surface.
Solution:
F (x, y, z) = z − sin(xy)
(c) (2 points) Give an example of a function f (x, y) with domain R2 for which fx (0, 0) exists but
fy (0, 0) does not exist.
Solution:
f (x, y) = |y|
(d) (2 points) Sketch level curves for a function f (x, y) with four local maximuma and no local minima.
Make sure to include enough level curves to illustrate these properties.
Solution:
3. Compute the following:
∂ ln(xy)
(a) (2 points)
∂y
xy
Solution:
∂
∂y
ln(xy)
xy
∂
=
∂y
=
=
(b) (2 points) Du (3x3 y − y) where u =
D
−2 √
√
, −15
5
ln(xy)
ey ln x
yey ln x
yx
− ln(x)ey ln x ln(xy)
e2y ln x
xy−1 −ln(x)xy ln(xy)
x2y
E
.
Solution: Let f (x, y) = 3x3 y − y.
Du (3x3 y − y) = ∇f · u
−2 −1
= 9x2 y, 3x3 − 1 · √ , √
5
5
=
−18
√ x2 y
5
−
−1
√
(3x3
5
− 1)
(c) (2 points) Find the unit vector u = hu1 , u2 , u3 i pointing in the direction of maximum increase for
the function f (x, y, z) = xyz 2 at the point (1, 1, 1).
Solution:
∇f = yz 2 , xz 2 , 2xyz
∇f (1, 1, 1) = h1, 1, 2i
∇f (1, 1, 1)
|∇f (1, 1, 1)|
h1, 1, 2i
=
|h1, 1, 2i|
D
E
= √16 , √16 , √26
u=
Page 2
Z
2
Z
π
sin(x + y) dx dy.
(d) (2 points)
0
1
Solution:
Z
2
Z
π
2
Z
π
− cos(x + y) sin(x + y) dx dy =
dy
x=0
1
0
1
2
Z
− cos(π + y) + cos y dy
=
1
2
Z
=
cos y + cos y dy
1
2
= 2 sin y x=1
= 2 sin 2 − 2 sin 1
Z
2
Z
2
Z
(e) (2 points)
1
0
0
1
xz
dz dx dy.
y
Solution:
Z
1
2
Z
0
2
1
Z
0
xz
dz dx dy =
y
Z
2
Z
2
Z
1
Z
0
=
1
Z
2
=
1
Z
2
1
xz 2 dx dy
2y z=0
2
x
dx dy
0 2y
2
x2 dy
4y x=0
2
1
dy
y
1
2
= ln y =
y=1
= ln 2 − ln 1
= ln 2
(f) (2 points) Compute
∂z
∂x
in terms of x, y and z if z satisfies the implicit equation
xy + yz + xz = 7
Solution:
∂
∂
(xy + yz + xz) =
(0)
∂x
∂x
∂z
∂z
y+y
+z+x
=0
∂x
∂x
∂z
∂z
y
+x
= −y − z
∂x
∂x
∂z
(y + x) = −y − z
∂x
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∂z
=
∂x
−y−z
y+x
4. Change the order of integration for the following double integrals. You may have to express the new integral
as a sum of double integrals. You do not need to evaluate the integrals.
Z 2 Z ln x
xy 2 dy dx.
(a) (2 points)
x−1
1
Solution:
2
Z
Z
ln x
2
Z
2
x−1
Z
−xy 2 dy dx
xy dy dx =
x−1
1
1
Z
ln x
ln 2 Z ey
=
ln 2
Z
xy 2 dx dy +
ey
0
π/4
Z
1
−xy 2 dx dy
y+1
Z
ln 2
2
xy 2 dx dy
y+1
cos y
xy 2 dx dy.
(b) (2 points)
0
Z
2
Z
ln 2
y+1
=
1
Z
y+1
0
Z
Z
−xy 2 dx dy +
sin y
Solution:
Z
π/4
Z
√
cos y
2
Z
2/2
arcsin x
Z
xy dx dy =
0
Z
2
xy dy dx +
sin y
0
0
1
√
2/2
Z
arccos x
xy 2 dy dx
0
5. (5 points) Give a triple integral which computes the volume of the bounded region in R3 enclosed by the
surfaces
x2 + y 2 − z 2 = 9, z = 4, z = −4
Solution: Let D be the region in R3 described above. Then
ZZZ
Volume of D =
dV
D
Z
√
4
Z
=
−4
9+z 2
√
− 9+z 2
√
Z
9+z 2 −x2
√
− 9+z 2 −x2
dy dx dz
6. (5 points) Compute the maximum and minimum values for the function f (x, y) = x − y 2 + xy on the region
in the plane bounded by the ellipse y 2 + 9x2 = 9.
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Solution: First we find the critical points of f .
∇f = h1 + y, −2y + xi
Thus 0 = 1 + y and 0 = −2y + x. Hence y = −1 and 0 = (−2)(−1) + x. Thus the critical point of f is
(−2, −1). This point is not in the region on which we are maximizing since (−1)2 + 9(−2)2 > 9. Thus
there are no critical points in the interior of the ellipse.
Now let g(x, y) = y 2 + 9x2 − 9. We wish to maximize f subject to the constraint g(x, y) = 0. Applying
the method of Lagrange multipliers we compute
∇g = h18x, 2yi
Setting ∇f = λ∇g gives the two equations
1 + y = λ18x
−2y + x = λ2y
Solving the second equation for λ we get λ =
−2y+x
2y .
Substituting this into the first equation we get
1+y =
−2y + x
2y
18x
2y + 2y 2 = −36xy + 18x2
2y + 2y 2 + 36xy − 18x2 = 0
From the constraint equation we get y 2 = 9 − 9x2 so
2y + 2(9 − 9x2 ) + 36xy − 18x2 = 0
2y + 18 + 36xy − 36x2 = 0
(2 + 36x)y = 36x2 − 18
9(2x2 − 1)
1 + 18x
4
81(4x
+ 4x2 − 1)
y2 =
1 + 36x + 324x2
81(4x4 + 4x2 − 1)
9 − 9x2 =
1 + 36x + 324x2
y=
You might have noticed by now that computing x here is not a reasonable task ...
7. (5 points) Use the method of Lagrange multipliers to find the maximum volume for a lidless can with surface
area 1.
Solution: The volume of a can with radius r and height h is
V (r, h) = πr2 h.
The surface area for a lidless can with radius r and height h is
S(r, h) = πr2 + 2πrh.
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We wish to maximize V subject to the constraint that S(r, h) = 1
∇V =h2πrh, πr2 i
∇S =h2πr + 2πh, 2πri
The vector equation ∇V = λ∇S gives the two equations
2πrh = λ(2πr + 2πh)
πr2 = λ2πr
Solving for λ in the second equation we get
r
2
λ=
Substituting λ =
r
2
into the first equation and cancelling π we get
2rh =
r
(2r + 2h)
2
and hence
2rh = r2 + rh
rh = r2
h=r
Substituting h = r back into our constraint S(r, h) = 1 gives
1 = πr2 + 2πr2
1 = 3πr2
1
=r
±√
3π
Radii must be positive so only thepositive root makes sense. We have h = r so the maximum volume
must occur at (r, h) = √13π , √13π . Thus the maximum volume is
V
1
1
√ ,√
3π
3π
=π
=
1
√
3π
2 1
√
3π
1
√
3 3π
8. (5 points) Estimate the change in z = xy 2 − x2 + y when (x, y) changes from (1, 2) to (1.1, 1.9).
Solution: Let f (x, y) = xy 2 − x2 + y. Then
fx = y 2 − 2x
fy = 2xy + 1
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so
fx (1, 2) = 22 − 2(1) = 2
fy (1, 2) = 2(1)(2) + 1 = 5.
Thus we can estimate the change ∆z in z using the differential dz
∆z ≈ dz
= fx (1, 2) dx + fy (1, 2) dy
= fx (1, 2) · (1.1 − 1) + fy (1, 2) · (1.9 − 2)
= 2 · (0.1) + 5 · (−0.1)
= −0.3
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