Solutions For Calculus Midterm Exam #1
1. (a) To find the eigenvalues of L, we must solve det(L − λI) = det
·0.1 − λ 1.2 0.4 0.3 − λ
¸
= λ2− 0.4λ − 0.45
= (λ − 0.9)(λ + 0.5) = 0.
We find λ1 = 0.9 and λ2 = −0.5.
To find an eigenvector associate with the eigenvalue λ1 = 0.9, we must determine x and y (not both equal to 0), so that
·0.1 1.2 0.4 0.3
¸ ·x y
¸
= 0.9
·x y
¸ . Writing this as a system of linear equations, we find
½ −0.8x + 1.2y = 0 0.4x − 0.6y = 0.
We see that both equations are the same. Simplifying yields 2x = 3y.
Setting x = 3, we find y = 2, that is v1 =
·3 2
¸
is an eigenvector associated with the eigenvalue λ1 = 0.9.
To find an eigenvector associate with the eigenvalue λ2 = −0.5, we must determine x and y (not both equal to 0), so that
·0.1 1.2 0.4 0.3
¸ ·x y
¸
= −0.5
·x y
¸ . Writing this as a system of linear equations, we find
½ 0.6x + 1.2y = 0 0.4x + 0.8y = 0.
We see that both equations are the same. Simplifying yields x = −2y.
Setting x = 2, we find y = −1, that is v2 =
· 2
−1
¸
is an eigenvector associated with the eigenvalue λ2 = −0.5.
(b) In the case of the Leslie matrix,
the associated eigenvector to the largest eigenvalue is a stable age distribution.
Since v1 =
·3 2
¸
, we find that
10000 × 3 3 + 2 10000 × 2
3 + 2
=
·6000 4000
¸
is a population of size 10,000 for which the proportion of the population in each age group stays the same from one year to the next.
(c) In the case of the Leslie matrix,
the largest eigenvalue is interpreted as the growth parameter;
that is, it determines the rate at which the population changes.
Since λ1 = 0.9 is the largest eigenvalue,
we find that the population declines each year by the factor 0.9.
(d) We first represent X(0) =
·3000 1000
¸
as a linear combination of v1 and v2. For this, we must find a and b so that
·3000 1000
¸
= a
·3 2
¸ + b
· 2
−1
¸ .
Writing this as a system of linear equations yields
½ 3a + 2b = 3000 2a − b = 1000.
Solving the system, we obtain a = 5000
7 and b = 3000 7 . Therefore
X(n) = Ln
·3000 1000
¸
= Ln[5000 7
·3 2
¸
+3000 7
· 2
−1
¸ ]
= 5000 7 Ln
·3 2
¸
+3000 7 Ln
· 2
−1
¸
= 5000 7 (0.9)n
·3 2
¸
+ 3000
7 (−0.5)n
· 2
−1
¸ .
2. (a) graph: Figure 3; level curves: Figure 6.
(b) graph: Figure 2; level curves: Figure 4.
(c) graph: Figure 1; level curves: Figure 5.
3. (a) If we substitute y = mx in the preceding limit, then (x, y) → (0, 0) reduces to x → 0 and we find
x→0lim 2xy
x3+ yx = lim
x→0
2mx2
x3+ mx2 = lim
x→0
2m x + m = 2.
(b) If we substitute y = x2 in the preceding limit, then (x, y) → (0, 0) reduces to x → 0 and we find
x→0lim 2xy
x3+ yx = lim
x→0
2x3
x3+ x3 = 1.
(c) From (a) and (b) we observe that the limits along two respective paths are different.
4. (a) The Jacobi matrix is
∂u
∂x
∂u
∂v ∂y
∂x
∂v
∂y
=
exsin y excos y 2
2x − y − 1 2x − y
.
(b) The linear approximation of f(x,y) at (1,1) is
L(x, y) =
·u(1, 1) v(1, 1)
¸
+ (Df)(1, 1)
·x − 1 y − 1
¸
=
·e sin 1 0
¸ +
·e sin 1 e cos 1
2 −1
¸ ·x − 1 y − 1
¸
=
·e sin 1 0
¸ +
·e sin 1(x − 1) + e cos 1(y − 1) 2(x − 1) − (y − 1)
¸ .
(c) The linear approximation of f(x,y) at (1.1,0.9) is L(1.1, 0.9) =
·2.3692 0.3
¸ .
And
f(1.1, 0.9) =
·2.3532 0.2624
¸ .
We see that the linear approximation is close to the actual value.
5. (a) The gradient vector of f (x, y) is
Of (x, y) =
∂f (x, y)
∂f (x, y)∂x
∂y
=
2x x2+ y2
2y x2+ y2
.
(b) At (1,1), f (x, y) increases most rapidly in the direction of the gradient vector Of (1, 1) and decreases most rapidly in the direction of the gradient vector −Of (1, 1).
Evaluating −Of at (1,1), we find
−Of (1, 1) =
·−1
−1
¸ . This vector has length p
(−1)2+ (−1)2 =√ 2.
Normalizing this vector yields
u =
"
−√12
−√12
# .
(c) The vector pointing from (1,1) to (-3,4) is
·−3 − 1 4 − 1
¸
=
·−4 3
¸
with the length p
(−4)2+ 32 = 5.
Therefore, the unit vector of this direction is v =
·−4/5 3/5
¸
along which the directional derivative is
Dvf (1, 1) = (Of (1, 1)) · v =
·1 1
¸
·
·−4/5 3/5
¸
= −1 5.
(d) The gradient vector of f at (-3,4) is perpendicular to the level curve through (-3,4).
Evaluating Of at (-3,4), we find
Of (−3, 4) =
·−256
8 25
¸ .
Therefore, the line perpendicular to the level curve passing through (-3,4) has the form
·−3 4
¸ + t
− 6 825 25
, for t ∈ R.
6. Using the chain rule, we find
dω
dt = ∂ω
∂x
∂x
∂t + ∂ω
∂y
∂y
∂t
= eyet+ xey˙2t
= et2et+ etet2˙2t
= (1 + 2t)et+t2. 7. (a) To compute ∂f∂x at (0,0),
we set y = 0, then f (x, 0) = 1, and
∂f (0, 0)
∂x = 0.
Likewise, setting x = 0, we find f (0, y) = 1, and
∂f (0, 0)
∂y = 0.
Therefore
Of (0, 0) =
·0 0
¸ . To compute ∂g∂x at (0,0), by definition
∂g(0, 0) g(0 + h, 0) − g(0, 0) g(h, 0)
Likewise,
∂g(0, 0)
∂y = lim
h→0
g(0, 0 + h) − g(0, 0)
h = lim
h→0
g(h, 0) h = 0.
Therefore
Og(0, 0) =
·0 0
¸ .
(b) f (x, y) is not continuous at (0,0), hence not differentiable at (0,0).
Therefore there is no tangent plane at (0,0).
g(x, y) is differentiable at (0,0) since the partial derivatives ∂g∂x and ∂y∂g are continuous on an open disk centered at (0,0).
Therefore, z = 0 defines the tangent plane to the graph of g at (0,0).
8. (a) The curve C1 is the graph of the given function restricted on v = 0. The slope of the tangent line to C1 at the point A is, by definition, the partial derivative with respective to u. That is,
∂f
∂u(2, 1) = 2e2ln 3 + 1 3e2. Therefore, the tangent line equation to C1 at the point A is
z − e2ln 3 = (2e2ln 3 + 1
3e2)(x − 2).
(b) To obtain the parametric representation for the curve C2, we have to determine a straight line
L : (2, 1) + t
·ω1 ω2
¸
, t ∈ R.
which is the projection of C2 onto the u-v plane. Without loss of generality, we assume
·ω1 ω2
¸
is a unit vector.
To obtain ω1 and ω2, we solve Of (2, 1) ·
·ω1 ω2
¸
= (2e2ln 3 +1
3e2)ω1+ 1
3e2ω2 = 0.
and find
·ω1 ω2
¸
= t
· −1
6 ln 3 + 1
¸
, t ∈ R.
The curve C2 is the mapping of L by f (u, v),
therefore the parametric representation for the curve C2 is z(t) = f (L)
= f ((2, 1) + t(−1, 6 ln 3 + 1))
= f (2 − t, 1 + (6 ln 3 + 1)t)
= e(2−t)22 ln(3 + (6 ln 3)t), t ∈ R.
