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MATH 51 FINAL EXAM SOLUTIONS (AUTUMN 2001)
- Compute the following.
(a)
− 1
Solution.
(b) The angle between
(^) and
Solution.
cos θ =
v · w
‖v‖‖w‖
=⇒ θ =
3 π
(c) The area of the triangle with vertices (0, 0 , 0), (− 1 , 4 , 1) and (2, − 2 , 1).
Solution. The area of this triangle is half the area of the parallelogram gen-
erated by v =
(^) and w =
. Since v × w =
, the area of the
triangle is
1 2
‖v × w‖ =
9 2
. Equivalently, using the result from part (b), the
triangle has a base of ‖w‖ = 3 and a height of ‖v‖ sin θ = 3, so the area is 1 2
9 2
- Let
A =
(a) For which vectors b =
b 1
b 2
b 3
(^) does the equation Ax = b have a solution? Express
your answer as one or more equations of the form ?b 1 +?b 2 +?b 3 =?.
Solution. Reducing the augmented matrix for the system Ax = b gives
b 1
b 2
b 3
b 1
b 2 − b 1
b 3 − 7 b 1
b 1
b 2 − b 1
b 3 − 7 b 1 − 4(b 2 − b 1 )
The system is therefore consistent (i.e. b is in C(A)) if and only if − 3 b 1 −
4 b 2 + b 3 = 0.
(b) Find a basis for the null space of A.
Solution. Continuing with the elimination from part (a) gives
rref(A) =
so a basis for N (A) is
(c) Find a basis for the column space of A.
Solution. Since the pivots of rref(A) are in the first two columns, the first
two columns of A (^)
form a basis for C(A).
(d) What is the rank of A?
Solution. 2
- (a) Let
b =
v 1 =
v 2 =
v 3 =
Solution 1. Verify the three subspace properties directly.
(i) T( 0 ) = 0 = 5 0 , so 0 is in V.
(ii) Suppose x and y are in V. Then T(x + y) = T(x) + T(y) = 5x + 5y =
5(x + y), so x + y is in V.
(iii) Suppose x is in V and c ∈ R. Then T(cx) = cT(x) = c(5x) = 5(cx), so
cx is in V.
Solution 2. Let A denote the matrix for T. Then
V = {x ∈ R
n | Ax = 5x} = {x ∈ R
n | (A − 5 In)x = 0 } = N (A − 5 In),
and the null space of any n × n matrix is a subspace of R
n .
- (a) Suppose T : R
3 → R
5 is a linear transformation such that
T(e 1 ) =
T(e 1 + e 2 ) =
T(e 1 + e 2 + e 3 ) =
Find the matrix A such that T(x) = Ax for all x ∈ R
3 .
Solution. The columns of A are T(e 1 ), T(e 2 ) and T(e 3 ). We are given T(e 1 ),
T(e 2 ) = T(e 1 + e 2 ) − T(e 1 ) =
and
T(e 3 ) = T(e 1 + e 2 + e 3 ) − T(e 1 + e 2 ) =
so
A =
(b) The matrix for rotation by 45
◦ about the x-axis in R
3 is
A =
√^1 2
√^1 2
0 √^1 2
√^1 2
and the matrix for rotation by 45
◦ about the z-axis in R
3 is
B =
√^1 2
√^1 2
√^1 2
√^1 2
(You need not verify these results.) Let T be the linear transformation obtained
by first rotating by 45
◦ about the x-axis and then rotating by 45
◦ about the z-axis.
Find the matrix for T.
Solution. BA =
1 √ 2
1 2
1 2
√^1 2
1 2
1 2
0 √^1 2
√^1 2
- Consider the ellipse 2x
2
2 = 1, and let T : R
2 → R
2 be the linear transfor-
mation with matrix A =
[
]
(a) Show that points (u, v) = T(x, y) in the image of the ellipse under T lie on the
circle u
2
2 = 5.
Solution. u = x + 2y and v = 3x + y, so
u
2
2 = (x + 2y)
2
2
= x
2
2
2
2
= 10x
2
2
= 5(2x
2
2 )
(b) Use the result of part (a) to find the area enclosed by the ellipse.
Solution. Since det(A) = −5, the area of the circle is 5 times the area of the
ellipse. Since the area of the circle is 5π, the area of the ellipse is π.
(c) Parametrize the ellipse. Hint: Parametrize the circle first and use A
− 1 .
Figure 1 Figure 2 Figure 3
−1 0 1
−
0
1
−1 0 1
−
0
1
−1 0 1
−
0
1
Figure 4 Figure 5 Figure 6
−1 0 1
−
0
1
−1 0 1
−
0
1
−1 0 1
−
0
1
Figure 7 Figure 8 Figure 9
−1 0 1
−
0
1
−1 0 1
−
0
1
−1 0 1
−
0
1
- Answer each question True or False. No explanation is necessary. Each correct answer
is worth 1 point.
(a) There exists a number c for which the function g(x, y) =
xy x^2 +y^2
(x, y) 6 = (0, 0)
c (x, y) = (0, 0)
is continuous at (0, 0).
Solution. False.
(b) There exists a number c for which the function g(x, y) =
xy √ x^2 +y^2
(x, y) 6 = (0, 0)
c (x, y) = (0, 0)
is continuous at (0, 0).
Solution. True.
(c) On the domain D = {(x, y) | x
2
2 ≤ 1 } the function f (x, y) = e
x^2 − 2 xy cos(xy)
attains a maximum value.
Solution. True.
(d) On the domain D = {(x, y) | x
2
2 < 1 } the function f (x, y) = x + y attains a
maximum value.
Solution. False.
(e) On the domain D = {(x, y) | x
2
2 < 1 } the function f (x, y) = 5 attains a
maximum value.
Solution. True.
(f) Suppose f (x, y) is differentiable and ∇f (1, 2) = (3, −7). Then there exists a
direction u in which Duf (1, 2) = 8.
Solution. False.
(g) If f is differentiable at a, then D−uf (a) = −Duf (a) for every unit vector u.
Solution. True.
(h) If f (x, y) has a local minimum at (0, 0) along every line through (0, 0), then f has
a local minimum at (0, 0).
Solution. False.
(i) There exists a function f (x, y) such that ∇f (x, y) = (2xy, x
2 ).
Solution. True.
(j) There exists a function f (x, y) such that ∇f (x, y) = (x
2 , 2 xy).
Solution. False.
- Find the maximum and minimum values of f (x, y) = x
3
2 − 9 x + y
2 − 2 y on the
square domain D = {(x, y) | 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 } and all points at which they are
attained.
Solution. Let f (x, y, z) = x
3 +xyz+z
3
. Then ∇f (x, y, z) = (3x
2 +yz, xz, xy+
3 z
2 ), so ∇(f (1, 1 , 1) = (4, 1 , 4) is a vector normal to the tangent plane to the
level surface f (x, y, z) = 3 at (1, 1 , 1). Thus the equation of the tangent plane
is 4(x − 1) + 1(y − 1) + 4(z − 1) = 0.
(b) Regarding z = z(x, y) as a function of x and y near the point (1, 1 , 1), compute
∂z
∂x
Solution 1. Differentiate with respect to x to get
3 x
2
∂z
∂x
2 ∂z
∂x
At (x, y, z) = (1, 1 , 1) this gives
∂z ∂x
Solution 2. Rewrite the equation of the tangent plane from part (a) as
z = 1 − 1(x − 1) −
1 4
(y − 1) and recall that the equation of the tangent plane
to the graph of z = f (x, y) at (a, b, f (a, b)) is given by
z = f (a, b) +
∂f
∂x
(a, b)(x − a) +
∂f
∂y
(y − b)
so
∂z ∂x
is just the coefficient if the (x − 1) term, −1.
- Let f : R
3 → R be a differentiable function and suppose that
∂f
∂x
(x 0 , y 0 , z 0 ) = 4
∂f
∂y
(x 0 , y 0 , z 0 ) = 5
∂f
∂z
(x 0 , y 0 , z 0 ) = 8
(a) Let u be the unit vector
. Compute D uf^ (x 0 , y 0 , z 0 ).
Solution.
Duf (x 0 , y 0 , z 0 ) = ∇f (x 0 , y 0 , z 0 ) · u = (4, 5 , 8) · (1/ 3 , 2 / 3 , 2 /3) = 10.
(b) Find a vector which points in the direction in which f is decreasing most rapidly
at (x 0 , y 0 , z 0 ).
Solution. Any positive scalar multiple of −∇f (x 0 , y 0 , z 0 ) = (− 4 , − 5 , −8).
(c) Suppose we know that f (x 0 , y 0 , z 0 ) = 5. Determine the gradient of the function
g(x, y, z) = (f (x, y, z))
2 at (x 0 , y 0 , z 0 ).
Solution. ∇g(x 0 , y 0 , z 0 ) = 2f (x 0 , y 0 , z 0 )∇f (x 0 , y 0 , z 0 ) = (40, 50 , 80).
- Let f (x, y) = x
2 − x ln y.
(a) Find Jf (2, 1).
Solution. Jf (x, y) =
[
2 x − ln y −
x y
]
, so Jf (2, 1) =
[
]
(b) Find the linear approximation of f at (2, 1) and use it to approximate f (1. 99 , 1 .02).
Solution. f (2, 1) = 4, so f (x, y) ≈ 4 + 4(x − 2) − 2(y − 1), and thus
f (1. 99 , 1 .02) ≈ 3 .92.
(c) Find Hf (2, 1).
Solution. Hf (x, y) =
[
1 y
−
1 y
x y^2
]
, so Hf (2, 1) =
[
]
(d) Find the second degree Taylor Polynomial of f at (2, 1).
Solution. 4 + 4(x − 2) − 2(y − 1) +
1 2
[2(x − 2)
2 − 2(x − 2)(y − 1) + 2(y − 1)
2 ]
(e) Near (2, 1) does the graph of f lie above its tangent plane, below its tangent plane,
or neither? Explain.
Solution. Hf (2, 1) is positive definite since AC − B
2 > 0 and A > 0. Thus
the graph of f lies above its tangent plane near (2, 1).
- (a) Find all the critical points of the function f (x, y) = 12xy − 2 x
2 − 9 y
4 .
Solution. ∇f (x, y) = (12y − 4 x, 12 x − 36 y
3 ). At a critical point therefore
x = 3y, and thus 36y(1 − y
2 ) = 0. The critical points are therefore (0, 0),
(3, 1) and (− 3 , −1).
(b) At each critical point, determine whether f has a local maximum, local minimum,
or saddle point.
Solution. Hf (x, y) =
[
12 − 108 y
2
]
At the first critical point Hf (0, 0) =
[
]
is indefinite since AC − B
2
− 144 < 0 and therefore f has a saddle at (0, 0).
At the other two critical points Hf (3, 1) = Hf (− 3 , −1) =
[
]
is
negative definite since AC − B
2 = 432 − 144 > 0 and A < 0. Thus f has local
maxima at (3, 1) and (− 3 , −1).
