Solutions for Calculus Quiz♯1 1.

Solutions for Calculus Quiz ♯1

1. y = (1/2)^x= e^ln(1/2)x = e^{−(ln 2)x} = e^−µx. Thus µ = ln 2.

2. We compare these functions and obtain that for 0 < x < 1, x⁻¹ > x⁻¹² > x¹³ > x¹²;

for x > 1, x¹² > x¹³ > x⁻¹² > 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

log₁₀E5 = 11.8 + 1.5 × 5 = 19.3 log₁₀E2 = 11.8 + 1.5 × 2 = 14.8.

It follows that

log₁₀ E5

E₂ = log₁₀E5− log₁₀E2 = 19.3 − 14.8 = 4.5.

⇒ E5

E2

= 10^4.5.

1

4. Denote the size of population at time t by N(t). And we know that N(t) = N0· 3^t, t = 0, 1, 2, ...

describes a population with N0 that triples in size every unit of time.

Hence N(t) = 20 · 3^t, 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

N_t Nt+1

= 1

R +1 − _R¹

K · N_t= 1 3+ 2

45N_t. (b) From (a), Nt+1 = RNt

1 + ^R−1_K N.

Solving the fixed point is to find N such that

N = RN

1 + ^R−1_K 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 lim_t→∞N_t = K = 15. And we use the formula Nt+1 = RNt

1 + ^R−1_K N to compute as the following.

N0 = 1 N1 = RN₀

1 + ^R−1_k N0

= 3 · 1

1 + ₁₅² · 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