Notes: `r = 5%` Find the time necessary for $1000 to double when it is invested at a rate of r compounded (a) anually, (b) monthly, (c) daily, and...

Friday, 16 January 2015

$\displaystyle{r}={5}%$ Find the time necessary for $1000 to double when it is invested at a rate of r compounded (a) anually, (b) monthly, (c) daily, and...

Formula for compounding n times per year: $\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

Formula for compounding continuously: $\displaystyle{A}={P}{{e}}^{{{r}{t}}}$

A=Final Amount

P=Initial Amount

r=rate of investment expressed as a percent

n=number of compoundings per year

t=time in years

a) r=5% n=1 (annually)

$\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

$\displaystyle{2000}={1000}{{\left({1}+\frac{{.05}}{{1}}\right)}}^{{{1}\cdot{t}}}$

$\displaystyle{2}={{1.05}}^{{t}}$

$\displaystyle{\ln{{\left({2}\right)}}}={t}{\ln{{\left({1.05}\right)}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{\ln{{\left({1.05}\right)}}}}={t}$

$\displaystyle{14.21}={t}$

Final Answer: 14.21 years

b) r=5% n=12 (monthly)

$\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

$\displaystyle{2000}={1000}{{\left({1}+\frac{{.05}}{{12}}\right)}}^{{{12}\cdot{t}}}$

$\displaystyle{2}={{\left({1.00416}\right)}}^{{{12}{t}}}$

$\displaystyle{\ln{{\left({2}\right)}}}={12}{t}{\ln{{\left({1.00416}\right)}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{{12}{\ln{{\left({1.00416}\right)}}}}}={t}$

$\displaystyle{13.89}={t}$

Final Answer: 13.89 years

c) r=5% n=365 (daily)

$\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

$\displaystyle{2000}={1000}{{\left({1}+\frac{{.05}}{{365}}\right)}}^{{{365}\cdot{t}}}$

$\displaystyle{2}={{\left({1.000136}\right)}}^{{{365}{t}}}$

$\displaystyle{\ln{{\left({2}\right)}}}={365}{t}{\ln{{\left({1.00136}\right)}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{{365}{\ln{{\left({1.00136}\right)}}}}}={t}$

$\displaystyle{13.86}={t}$

Final Answer:...

Formula for compounding n times per year: $\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

Formula for compounding continuously: $\displaystyle{A}={P}{{e}}^{{{r}{t}}}$

A=Final Amount

P=Initial Amount

r=rate of investment expressed as a percent

n=number of compoundings per year

t=time in years

a) r=5% n=1 (annually)

$\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

$\displaystyle{2000}={1000}{{\left({1}+\frac{{.05}}{{1}}\right)}}^{{{1}\cdot{t}}}$

$\displaystyle{2}={{1.05}}^{{t}}$

$\displaystyle{\ln{{\left({2}\right)}}}={t}{\ln{{\left({1.05}\right)}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{\ln{{\left({1.05}\right)}}}}={t}$

$\displaystyle{14.21}={t}$

Final Answer: 14.21 years

b) r=5% n=12 (monthly)

$\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

$\displaystyle{2000}={1000}{{\left({1}+\frac{{.05}}{{12}}\right)}}^{{{12}\cdot{t}}}$

$\displaystyle{2}={{\left({1.00416}\right)}}^{{{12}{t}}}$

$\displaystyle{\ln{{\left({2}\right)}}}={12}{t}{\ln{{\left({1.00416}\right)}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{{12}{\ln{{\left({1.00416}\right)}}}}}={t}$

$\displaystyle{13.89}={t}$

Final Answer: 13.89 years

c) r=5% n=365 (daily)

$\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

$\displaystyle{2000}={1000}{{\left({1}+\frac{{.05}}{{365}}\right)}}^{{{365}\cdot{t}}}$

$\displaystyle{2}={{\left({1.000136}\right)}}^{{{365}{t}}}$

$\displaystyle{\ln{{\left({2}\right)}}}={365}{t}{\ln{{\left({1.00136}\right)}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{{365}{\ln{{\left({1.00136}\right)}}}}}={t}$

$\displaystyle{13.86}={t}$

Final Answer: 13.86 years

d) $\displaystyle{A}={P}{{e}}^{{{r}{t}}}$

$\displaystyle{2000}={1000}{{e}}^{{{.05}\cdot{t}}}$

$\displaystyle{2}={{e}}^{{{.05}{t}}}$

$\displaystyle{\ln{{\left({2}\right)}}}={.05}{t}{\ln{{e}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{{.05}{\ln{{e}}}}}={t}$

$\displaystyle{13.86}={t}$

Final Answer: 13.86 years

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