Download Solutions to Math 51 Second Exam — February 18, 2016

Solutions to Math 51 Second Exam — February 18, 2016
1. (9 points) (a) Write down a 2×2 matrix A such that, for any vector v ∈ R2 , Av is the vector obtained
by first rotating v ∈ R2 counterclockwise by 45 degrees and then doubling the x-coordinate.
√
√
cos(π/4) − sin(π/4)
2/2 −√ 2/2
√
The rotation matrix is
=
. The scaling matrix is
sin(π/4) cos(π/4)
2/2
2/2
2 0
. One first rotates and then scales, so the computation is:
0 1
√
√ √
√
2 0
− 2
√2/2 −√ 2/2 = √ 2 √
.
0 1
2/2
2/2
2/2
2/2
(b) Describe, in words, what multiplication by A−1 does to a vector v; that is, describe A−1 v in terms
of v. Your description should be similar to the description of A given in part (a).
It first divides the x-coordinate of v by 2, and then rotates the resulting vector by 45 degrees
clockwise.
(c) Let S : R2 → R2 be a linear transformation that rotates vectors counterclockwise by 17.5 degrees.
If B is the matrix of S, what is det(B)? Give a brief justification.
cos(17.5◦ ) − sin(17.5◦ )
is cos2 (17.5◦ ) + sin2 (17.5◦ ), which is equal to 1.
The determinant of
sin(17.5◦ ) cos(17.5◦ )
h 1 i
(d) Let T : R3 → R3 be orthogonal projection onto the line span 2 . If C is the matrix of T ,
3
what is det(C)? Give a brief justification.
C is not invertible, and therefore has determinant 0. There are several ways to see 
this,
 but
1
here is, in my mind, the simplest one. The column space of C is exactly the span of 2. So
3
therefore it is not all of R3 . Since invertible matrices have full rank, C is not invertible.
Math 51, Winter 2016
Solutions to Second Exam — February 18, 2016
Page 2 of 11
2. (10 points) Let f (x, y, z) = x2 y 2 + yz + z 2 x.
(a) Write down the linear approximation of f near (1, 1, 1).
Linear approximation of f near a point a is given by f (x) ≈ Lf (x) = f (a) + Df (a)(x − a).
In our case Df (x, y, z) = (2xy 2 + z 2 , 2x2 y + z, y + 2zx), evaluating at a = (1, 1, 1) gives
Df (1, 1, 1) = (3, 3, 3). We also have f (1, 1, 1) = 3, so the linear approximation near (1, 1, 1)
is


x−1
f (x, y, z) ≈ 3 + (3, 3, 3) y − 1 = 3 + 3(x − 1) + 3(y − 1) + 3(z − 1)
z−1
(b) Compute an equation for the tangent plane at (1, 1, 1) to the level set f (x, y, z) = 3.
 
3
We already computed the entries of ∇f (1, 1, 1) = 3, so the equation of the tangent plane
3
becomes
  

3
x−1



0 = ∇f (a) · (x − a) = 3 · y − 1 = 3(x − 1) + 3(y − 1) + 3(z − 1)
3
z−1
Notes: Since all the necessary computation was done in (a), credit was only given for the fully
correct equation (for example 3(x-1)+3(y-1)+3(z-1) is not an equation).
(c) Starting from (1, 1, 1), in which direction does f decrease the most rapidly? Give your answer as
a unit vector in R3 .
The function f decreases most rapidly in the direction of −∇f (1, 1, 1), so the unit vector in that
direction is
 
 
3
1
∇f (1, 1, 1)
1  
1  
3 = −√ 1
−
= −√
k∇f (1, 1, 1)k
27 3
3 1
Math 51, Winter 2016
Solutions to Second Exam — February 18, 2016
Page 3 of 11
3. (11 points) A frog is jumping around on a sheet of paper. It is found to be extremely predictable:
when it jumps from position (x, y), it jumps, without fail, to the position
J(x, y) = (−x − y 2 + 1, y + x2 y).
For example, starting from position (1, 0) it jumps to position (0, 0).
(a) Compute DJ(0, 0) and DJ(1, 0); show your steps.
−1
−2y
2xy 1 + x2
−1 0
DJ(0, 0) =
0 1
−1 0
DJ(1, 0) =
0 2
DJ(x, y) =
(b) The frog starts at (0, 0), and then jumps four times. Where does it end up? Show your steps.
We have J(0, 0) = (1, 0) and J(1, 0) = (0, 0), so J(J(J(J(0, 0)))) = (0, 0), that is:
J
J
J
J
(0, 0) −
→ (1, 0) −
→ (0, 0) −
→ (1, 0) −
→ (0, 0)
(c) Now the frog starts at position (0.01, 0.01). Estimate its position after four jumps; show your
reasoning.
Linear approximation says:
0.01
0
0
0.01
0
(J ◦ J ◦ J ◦ J)
≈ (J ◦ J ◦ J ◦ J)
+ D(J ◦ J ◦ J ◦ J)
·
−
0.01
0
0
0.01
0
The chain rule says:
D(J ◦ J ◦ J ◦ J)(0, 0) = DJ(J(J(J(0, 0))) · DJ(J(J(0, 0)) · DJ(J(0, 0)) · DJ(0, 0)
−1 0 −1 0 −1 0 −1 0
=
0 2
0 1
0 2
0 1
1 0
=
0 4
Combining these two equations gives:
(J ◦ J ◦ J ◦ J)
0.01
0
1 0 0.01
≈
+
0.01
0
0 4 0.01
0.01
=
0.04
Math 51, Winter 2016
Solutions to Second Exam — February 18, 2016
Page 4 of 11
4. (8 points) Below is a collection of level sets of a function f : R2 → R. You may assume that f and its
first and second derivatives are continuous, and the length scales in the x- and y-directions are equal.
(a) Sketch, on the plot, the direction of the gradient vector at A.
(b) Sketch, on the plot, the direction of the gradient vector at B.
See picture for (a). For parts (a) and (b), the gradient points in the direction in which f increases
most rapidly. This also means the gradient is perpendicular to the level sets. For drawings where
no good will could make the vector look perpendicular to level sets, no credit was given.
For the remaining parts, circle “pos.” if the quantity is positive; “neg.” if the quantity is negative.
(c) fxx at D.
pos.
neg.
Math 51, Winter 2016
Solutions to Second Exam — February 18, 2016
Page 5 of 11
(d) fyy at E.
pos.
neg.
(e) fxx at F .
pos.
neg.
(f) fyy at F .
pos.
neg.
Math 51, Winter 2016
Solutions to Second Exam — February 18, 2016
Page 6 of 11
5. (9 points) In this question, no justification is needed for any part.
(a) Which of the following statements about matrix multiplication is/are valid? Circle the one or
more statement(s) that is/are true for every choice of three square matrices A, B, C of the same
size.
AB = BA
A(BC) = (AB)C
A(B + C) = AB + AC
det(AB) = det(BA)
The grading scheme for part (a) was that you got .75 points for a correct answer and -.75 points
for an incorrect answer. You could not get fewer than 0 points.
• AB = BA is not always valid since matrix multiplication does not always commute.
• A(BC) = (AB)C is always true because multiplication of matrices is associative.
• A(B + C) = AB + AC is always true because multiplicaiton and addition of matrices is
distributive.
• det(AB) = det(BA) is always true because both the left-hand side and the right-hand side
are equal to det(A) det(B).
(b) Let A be a 5 × 5 matrix. Among the statements listed below, circle each which, considered by
itself, is a guarantee that A will not be invertible. (There may be more than one statement to
circle, but each statement should be considered on its own — not combined with other ones.)
• det(A) = 0.
• The first and second columns of A are the same.
• Every column of A has only one non-zero entry in it.
• All the diagonal entries of A are zero: a11 = a22 = a33 = a44 = a55 = 0.
The grading scheme for part (b) was that you got .75 points for a correct answer and -.75 points
for an incorrect answer. You could not get fewer than 0 points.
• True. det(A) = 0 if and only if the matrix A is not invertible so certainly this implies that
A is not invertible.
• True. The first and second columns of A being the same implies that det(A) = 0 so A is
not invertible.
• False. Every column of A has only one non-zero entry in it is not enough to imply that
A is not invertible. One counterexample isA the 5 × 5 identity matrix. It has only one
non-zero entry in each column but it is invertible.
• False. For a counterexample, take the matrix A where the first column of A is e2 , the
second column of A is e3 , the third column of A is e4 , the fourth column of A is e5 , and the
fifth column of A is e1 . The column vectors form a basis of R5 so rref (A) is the identity
and A must be invertible. However, the diagonal entries of A are all zero.
Math 51, Winter 2016
Solutions to Second Exam — February 18, 2016
Page 7 of 11
(c) A population of birds lives on two islands. On January 1, 2016, there are N birds on the first
island, and M birds on the second island.
Each spring, ten percent of the birds from the first island move to the second one. Each fall, ten
percent of the birds from the second island move back to the first island. For the purpose of this
question, ignore all births and deaths.
Let N 0 be the number of birds on the first island as of December 31, 2017. Let M 0 be the number
of birds on the second island as of December 31, 2017. Define the matrices
0.9 0.1
0.9 0
1.0 0
1.0 0.1
A=
, B=
, C=
, D=
.
0 1.0
0.1 1.0
0.1 0.9
0 0.9
Which of the following expressions equals
Do not circle more than one.)
• (AC)2
N M
N • (CA)2 M
N • A2 C 2 M
N • C 2 A2 M
N0 M0
? Circle the correct answer. (There is only one.
• D2 B 2
• B2D
N M
2 N
M
• (DB)
2 N
• (BD)
2 N
M
M
The grading scheme for part (c) was all or nothing.
During spring, birds from island one will move. The associated matrix for that linear transformation is
0.9 0
.
B=
0.1 1.0
During the fall, birds from island two will move. The associated matrix for that linear transformation is
1.0 0.1
.
D=
0 0.9
Between January 1st, 2016 and December 31st, 2017, two springs and two falls will pass. The
exact order is one spring, one fall, one spring, and one fall will pass. Therefore
the number of
N = (DB)2 N .
birds on each island at the end of 2017 will be DBDB M
M
Math 51, Winter 2016
Solutions to Second Exam — February 18, 2016
Page 8 of 11
6. (11 points) A certain hill has height (in meters) given by the function
h(x, y) = 20e
((1/2) − (x/100)2 − (y/200)2 )
.
Here (x, y) measures position relative to the summit of the hill; thus (0, 0) is the hill summit, x measures
the distance east of the summit, and y measures distance north of the summit. x, y, h are all measured
in meters.
(a) Compute a linear approximation to h near (50, 100).
(a) Let (x, y) be a point close to (50, 100), then the linear approximation formula gives
x − 50
h(x, y) ≈ h(50, 100) + ∇h(50, 100) ·
.
y − 100
We compute
2
2
100
1
50
1
1
1
h(50, 100) = 20e 2 −( 100 ) −( 200 ) = 20e 2 − 4 − 4 = 20e0 = 20.
Separately, we compute the partial derivatives of h with respect to x and y at (50, 100):
2
y
1
x 2
∂h
−2x
∂h
−2 · 50
1
−
−
)
(
)
(
200
(x, y) = 20 · e 2 100
·
so that
(50, 100) = 20 · 1 ·
=−
2
2
∂x
100
∂x
100
5
and
2
y
1
x 2
∂h
(x, y) = 20 · e 2 −( 100 ) −( 200 ) ·
∂y
−2y
2002
∂h
so that
(50, 100) = 20 · 1 ·
∂y
−2 · 100
2002
We have then
h(x, y) ≈ 20 + − 51
1
− 10
x − 50 y − 100
x − 50
= 20 −
·
−
.
y − 100
5
10
=−
1
.
10
Math 51, Winter 2016
Solutions to Second Exam — February 18, 2016
Page 9 of 11
(b) You start at the point (x, y) = (50, 100) and set out walking. You want to walk so that your path
is as level as possible (i.e. your altitude doesn’t change). In which direction should you begin
walking? (Your answer should be a unit vector in R2 describing the direction.)
We look for a direction v where the slope of the hill at (x, y) = (50, 100) is as close to zero
(that is, as level) as possible. Therefore, we want the directional derivative Dv h(50, 100) of h at
(50, 100) to be as close to zero as possible. Recalling that
Dv h(50, 100) = ∇h(50, 100) · v,
it turns out that we are looking for a unit vector v orthogonal to ∇h(50, 100).
We computed the gradient in the previous part, and we obtained
1
∇h(50, 100) = − 51 − 10
1 1
10
so that a vector w orthogonal to it is (for example)
, which is parallel to w =
. It
−2
− 15
remains to turn w into a unit vector having the same direction, and to do this we divide w by
its length:
" 1 #
√
w
w
5
v=
=√
=
.
− √25
||w||
1+4
" −1 #
√
The vector −v =
5
√2
5
is also a correct answer.
(c) You start at the point (50, 100) and would like to climb up the mountain as fast as possible.
However, because of a leg injury, your path absolutely cannot rise with a slope greater than 14 at
any point. (In other words, for each four meters of horizontal motion, your altitude should not
rise by more than one meter.) In which direction should you begin walking? Your answer should
be a unit vector in R2 describing the direction. Please justify your answer.
We are looking for a direction v where the slope of the hill is as large as possible but still less
than 14 . As in the previous part, the slope of the hill in the direction v is given by Dv h(50, 100) =
∇h(50, 100) · v. Moreover, we know from the general theory that the direction of highest ascent
is precisely given by the gradient, so the largest possible slope of the hill at (50, 100) is obtained
∇h(50,100)
by taking v = ||∇h(50,100)||
which yields
Dv h(50, 100) = ∇h(50, 100) ·
s
=
√
−1
5
2
||∇h(50, 100)||2
∇h(50, 100)
=
= ||∇h(50, 100)|| =
||∇h(50, 100)||
||∇h(50, 100)||
+
−1
10
2
r
=
1
1
+
=
25 100
r
√
4+1
5
=
.
100
10
5
1
Finally we notice that 105 < 14 , because squaring both sides yields 100
< 16
. Hence the direction
∇h(50,100)
1
of highest ascent already has slope less than the prescribed 4 , so we can take v = ||∇h(50,100)||
=
"
#
−1 −2
√
10
5
5
√
=
−1
−1 .
5
√
10
5
Math 51, Winter 2016
Solutions to Second Exam — February 18, 2016
7. (10 points) Suppose that A is a 4 × 4 matrix such that A−1

3
4
=
0
2
2
2
3
1
1
2
1
4
Page 10 of 11

2
1
.
0
3
Please note that you have not been given the matrix A, only its inverse A−1 . Also, you can solve all
the questions below without inverting any 4 × 4 matrices.
1
4
(a) Find all vectors v ∈ R satisfying Av = 00 .
0
Apply A−1
3
to the equation: we get v = 40 .
2
(b) What is the reduced row echelon form of A?
A is invertible, so rref is the identity 4 × 4 matrix.
0
(c) Let v1 , v2 , v3 , v4 be the columns of A. Let w =
1
1
0
and let B = {v1 , v2 , v3 , v4 }. What is [w]B ?
In other words, what are the coordinates of w with respect to B?
0
−1
This is A · 11 .
0
(d) The determinant of A−1 is −65. You do not need to prove this. What is the determinant of A?
det(A−1 ) =
1
det(A)
so det(A) =
1
−65
1
= − 65
.
Math 51, Winter 2016
Solutions to Second Exam — February 18, 2016
Page 11 of 11
8. (8 points) Let f : R2 → R and g : R2 → R be two functions. Suppose that, at (0, 0), the unit
0 . (You may
direction of fastest increase of f is [ 10 ] and the unit direction of fastest increase of g is −1
assume that the gradients of both f and g are nonzero at (0, 0). )
Each of the statements below is either always true (“T”), or always false (“F”), or sometimes true and
sometimes false, depending on the situation (“MAYBE”). For each part, decide which and circle the
appropriate choice; you do not need to justify your answers.
1
(a) ∇f (0, 0) =
.
T
F
MAYBE
0
(a) Maybe: ∇f is proportional to [ 10 ], but need not be equal to it.
(b)
∂g
∂y (0, 0)
< 0.
T
F
MAYBE
T
F
MAYBE
True: the gradient of g is in the dirction of (0, −1), so gy < 0.
f (x, y)
(c) The derivative DF (0, 0), where F =
, is a diagonal matrix.
g(x, y)
True.
T
F
MAYBE
(d) The level set of g that passes through (0, 0) is tangent to the x-axis.
0 True: the level set is normal to the vector −1 which makes it tangent to the x-axis.
(e) At (0, 0), the
of fastest increase of f + g is a positive scalar
direction
1
.
multiple of
−1
T
F
MAYBE
(e) Maybe! The direction of fastest increase isthesum of the gradients of f, g which is of the
a
1
form [ −b
] for a, b both positive. This could be −1
but might not be.
(f) At (0, 0), the
direction of fastest increase of f + g is a positive scalar
1
multiple of
.
1
T
F
MAYBE
T
F
MAYBE
(f) This is false, see discussion from (e).
1
(g) At (0, 0), f + g is increasing in the direction
.
1
a
Maybe. If the gradient of f + g is [ −b
], its directional derivative in [ 11 ] direction is a − b. Could
be positive, could be negative!
(h) At (0, 0), f + g is decreasing in the direction
1
.
−1
T
F
MAYBE
True. (h) Similar to the previous part – the directional derivative is a+b, so is definitely positive.
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