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.

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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.

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