Answer to Question #211896 in Calculus for paul

Question

Solve the following real-life problems involving growth and decay.

1. There are 10.14 million people in Region III in 2010. It rose to11.22 million in

the year 2015, respectively. If it follows an exponential growth, what will be

our population in the year 2020?

2. After 15 years, what will be the mass of a 648-gram radioactive material

that has a half-life of 9 years?

3. Mr. Alfredo’s investment on a certain product is valued at P578,000

after 5 years and rose to Php 789,000 after 7 years. How much did he

invest initially?

4. (from question no. 3.) What will be the value of his investment after 10

years?

5. The population of pigs in a certain province is 6,500. But because of

African Swine flu, it went down to 4200 after 6 months. How many

months will it take to reduce the whole population to 400, if intervention

measures to save the pigs wer e not applied?

Answer 1

1.

$$N=N_0e^{kt}$$

$N_0=10.14\ million$

$$N=10.14e^{kt}$$

$N(5)=11.22\ million$

$$11.22=10.14e^{5t}=>e^t=(\dfrac{11.22}{10.14})^{1/5}$$

$$N(10)=10.14e^{10t}=10.14(\dfrac{11.22}{10.14})^{2}$$

$$N(10)=12.415\ million$$

2.

$$N=N_0e^{-kt}$$

$$\dfrac{N_0}{2}=N_0e^{-kt_h}$$

$$e^{kt_h}=2$$

$$k=\dfrac{\ln2}{t_h}$$

$$N=N_0(2)^{-t/t_h}$$

$$N(15)=648(2)^{-15/9}$$

$$N(15)=204\ g$$

3.

$$A=P(1+i)^n$$

$$A(5)=578000=P(1+i)^5$$

$$A(7)=789000=P(1+i)^7$$

$$1+i=(\dfrac{789000}{578000})^{1/2}$$

$$P=\dfrac{578000}{(\dfrac{789}{578})^{5/2}}$$

$$P=265494.13$$

4.

$$A(10)=P(1+i)^{10}$$

$$A(10)=578000(\dfrac{789}{578})^{5/2}$$

$$A(10)=1258347.96$$

5.

$$N=6500e^{-kt}$$

$$N(6)=6500e^{-6k}=4200$$

$$N(t_1)=6500e^{-kt_1}=400$$

$$e^k=(\dfrac{65}{42})^{1/6}$$

$$k=\dfrac{1}{6}\ln(\dfrac{65}{42})$$

$$\dfrac{1}{6}\ln(\dfrac{65}{42})t_1=\ln(\dfrac{65}{4})$$

$$t_1=6\cdot\dfrac{\ln(\dfrac{65}{4})}{\ln(\dfrac{65}{42})}$$

$$t_1=38.305\ months$$