Respuesta :
Answer:
a) - r=5%: [tex]S=$ 5,136.10[/tex]
- r=4%: [tex]S=$ 4,885.61 [/tex]
- r=3%: [tex]S=$ 4,647.34 [/tex]
b) - r=5%: t=14 years
- r=4%: t=17 years [/tex]
- r=3%: t=23 years [/tex]
c) The amount obtained is
- Compuonded quarterly: $5,191.83
- Compuonded continously: $5,200.71
The latter is always greater, since the more often it is capitalized, the greater the effect of compound interest and the greater the capital that ends up accumulating.
Explanation:
The rate of accumulation of money is
[tex]dS/dt=rS[/tex]
To calculate the amount of money accumulted in a period, we have to rearrange and integrate:
[tex]\int dS/S=\int rdt=r \int dt\\\\ln(S)=C*r*t\\\\S=C*e^{rt}[/tex]
When t=0, S=S₀ (the initial capital).
[tex]S=S_0=Ce^{r*0}=Ce^0=C\\\\C=S_0[/tex]
Now we have the equation for the capital in function of time:
[tex]S=S_0e^{rt}[/tex]
a) For an initial capital of $4000 and for a period of five years, the amount of capital accumulated for this interest rates is:
- r=5%: [tex]S=4000e^{0.05*5}=4000*e^{0.25}= 5,136.10[/tex]
- r=4%: [tex]S=4000e^{0.04*5}=4000*e^{0.20}= 4,885.61 [/tex]
- r=3%: [tex]S=4000e^{0.03*5}=4000*e^{0.15}= 4,647.34 [/tex]
b) We can express this as
[tex]S=S_0e^{rt}\\\\2S_0=S_0e^{rt}\\\\2=e^{rt}\\\\ln(2)=rt\\\\t=ln(2)/r[/tex]
- r=5%: [tex]t=ln(2)/0.05=14[/tex]
- r=4%: [tex]t=ln(2)/0.04=17 [/tex]
- r=3%: [tex]t=ln(2)/0.03= 23 [/tex]
c) When the interest is compuonded quarterly, the anual period is divided by 4. In 5 years, there are 4*5=20 periods of capitalization. The annual rate r=0.0525 to calculate the interest is also divided by 4:
[tex]S = 4000 (1+(1/4)(0.0525))^{5*4}=4000(1.013125)^{20}\\\\S=4000*1.297958= 5,191.83[/tex]
If compuonded continously, we have:
[tex]S=S_0e^{rt}=4000*e^{0.0525*5}=4000*1.3= 5,200.71[/tex]
The amount obtained is
- Compuonded quarterly: $5,191.83
- Compuonded continously: $5,200.71
The latter is always greater, since the more often it is capitalized, the greater the effect of compound interest and the greater the capital that ends up accumulating.
