Answer to Question #350903 in Calculus for Kueska

Question

1. A certain radioactive element has a half-life of 5 hours. If its initial mass is at

6,470 grams, how many grams are left after 2 days?

2. Carlo’s online investment is worth Php870,000 on its 3rd year and Php650, 000

on its 7th year. What is the worth of his initial investment?

3. The pressure exerted on us by the atmosphere decreases exponentially as you

go up. The pressure at ground level is 1,013 hPa and decreases to 965 hPa at

381 meters. What is the pressure at the summit of Mt. Apo which is at 2,954

meters?

4. A strain of bacteria growing on the palm of your hands is 9 bacteria after 6

minutes. If you start with only one bacterium and follows an exponential growth

model, how many bacteria could be present at the end of 1 hour?

5. A certain breed of rats doubles its population every 2 months. Assuming there

were only 6 rats initially at a certain area, how many months will it take for the

population to grow to 68 rats?

Answer 1

1.

$$m(t)=m_0(2)^{-t/t_{1/2}}$$

Given $m_0=6470\ g, t_{1/2}=5\ h.$

$$2\ days=48\ hours$$

$$m(48)=6470(2)^{-48/5}=8.337\ g$$

$8.337$ grams

2.

$$A(t)=P(1+r)^t$$

We see that if $t_2\ge t_1,$ then $A(t_2)\ge A(t_1).$

$$A(3)=870000, A(7)=650000$$

Since $A(3)>A(7),$ there is no solution.

Carlo’s online investment is worth Php $650,000$ on its 3rd year and Php$870, 000$ on its 7th year. What is the worth of his initial investment?

$$A(t)=P(1+r)^t$$

$$650000=P(1+r)^3$$

$$870000=P(1+r)^7$$

$$(1+r)^4=\dfrac{870000}{650000}$$

$$P=650000(\dfrac{87}{65})^{-3/4}$$

$$P=522346.80$$

3.

$$p(h)=p(0)e^{-kh}$$

Given $p(0)=1013\ hPa, p(381)=965\ hPa.$

$$965=1013e^{-k(381)}$$

$$k=-\dfrac{1}{381}\ln(\dfrac{965}{1013})$$

$$p(2954)=1013(\dfrac{965}{1013})^{2954/381}$$

$$p(2954)=695.27\ hPa$$

4.

$$N(t)=N(0)e^{kt}$$

Given $N(0)=1, N(6)=9$

$$9=1\cdot e^{k(6)}$$

$$e^{6k}=9$$

$$N(60)=1\cdot e^{k(60)}=9^{10}=3486784401$$

5.

$$n(t)=n(0)\cdot (2)^{t/t_{2}}$$

Given $n(0)=6, t_{2}=2.$

$$68=6\cdot (2)^{t/2}$$

$$t=2\cdot\dfrac{\ln(68/6)}{\ln(2)}$$

$$t=7$$