Solutions for Calculus Quiz♯1
1. y = (1/2)x = eln(1/2)x
= e−(ln 2)x = e−µx.Thus µ = ln 2.
2. We compare these functions and obtain thatfor 0 < x < 1, x−1 > x−
1
2 > x1
3 > x1
2 ;for x > 1, x
1
2 > x1
3 > x−1
2 > x−1.Then we graph them in the following figure.
n=–1
n=–1/2
n=1/3
n=1/2
0
2
4
6
8
10
y
2 4 6 8 10
x
3. Let E5 be the energy released by an earthquake of magnitude 5.Let E2 be the energy released by an earthquake of magnitude 2.Then
log10 E5 = 11.8 + 1.5 × 5 = 19.3
log10 E2 = 11.8 + 1.5 × 2 = 14.8.
It follows that
log10
E5
E2= log10 E5 − log10 E2 = 19.3 − 14.8 = 4.5.
⇒E5
E2= 104.5.
1
4. Denote the size of population at time t by N(t). And we know that
N(t) = N0 · 3t, t = 0, 1, 2, ...
describes a population with N0 that triples in size every unit of time.Hence N(t) = 20 · 3t, t = 0, 1, 2, ....
5. (a) Suppose a is a fixed point of {an}. Then
a =1
2(a +
4
a).
Hence a = ±2.That is the fixed points of {an} are 2,−2.
(b) Since a0 = 1 > 0, then
an+1 =1
2(an +
4
an) > 0 ∀n ∈ N.
Thus limn→∞an = 2.
6. (a) Referring to Figure 2.14, we can see that the straight line in Figure 2.14 has slope
(1 −1
R)/K.
Thus
Nt
Nt+1
=1
R+
1 − 1R
K· Nt =
1
3+
2
45Nt.
(b) From (a), Nt+1 =RNt
1 + R−1K
N.
Solving the fixed point is to find N such that
N =RN
1 + R−1K
N.
If N 6= 0, we can see that N = K is the nontrivial fixed point.Thus in this problem, K = 15 is the nontrivial fixed point.
(c) We know that limt→∞Nt = K = 15. And we use the formula
Nt+1 =RNt
1 + R−1K
Nto compute as the following.
N0 = 1
N1 =RN0
1 + R−1k
N0
=3 · 1
1 + 215
· 1=
45
17.
2
Similarly,
N2 =135
23
N3 =405
41
N4 =243
19
N5 =3645
257≈ 14.1829
N6 =10935
743≈ 14.7173
where 15 − N6 < 0.5.
3
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