Question

When interest is compounded continuously, the amount of moneyincreases at a rate proportional to the amount Spresent at time t,that is,$\mathbf{dS}\mathbf{/}\mathbf{dt}\mathbf{=}\mathbf{rS}$, where ris the annual rate of interest.

(a) Find the amount of money accrued at the end of 5yearswhen $500is deposited in a savings account drawing$\mathbf{5}\frac{3}{4}$% annual interest compounded continuously.

(b) In how many years will the initial sum deposited have doubled?

(c) Use a calculator to compare the amount obtained in part (a)with the amount$\mathit{S}\mathbf{=}\mathbf{5000}{\left(1+\frac{1}{4}(0.0575)\right)}^{\mathbf{5}\mathbf{\left(}\mathbf{4}\mathbf{\right)}}$that is accruedwhen interest is compounded quarterly.

Short answer

(a) The amount of money accrued is 6665.5 dollars.

(b) The time at which the initial amount is doubled is 12.05 years.

(c) The amount is lower than the amount in part (a) by 13.7 dollars when the comparison happens between this amount and the amount in part (a).

Step-by-step solution

Define growth and decay

The initial-value problem, $\frac{\text{dx}}{\text{dt}}{\text{= kx, x(t}}_{\text{0}}{\text{) =x}}_{\text{0}}$where k is a constant of proportionality, serves as a model for diverse phenomena involving either growth or decay. This is in the form of a first-order reaction (i.e.) a reaction whose rate, or velocity, $\text{dx/dt}$is directly proportional to the amount x of a substance that is unconverted or remaining at time t.

Solve for first order growth and decay equation.

(a)

Let the linear equation with the amount of money S as be,

$\frac{\text{dS}}{\text{dt}}\text{= rS}$… (1)

And with the conditions, $\text{S(t = 0) = 5000 dollars}$and $\text{r =}\frac{\text{5}\frac{\text{3}}{\text{4}}}{\text{100}}\text{= 0.0575}$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

role="math" localid="1663840585610" $\begin{array}{rcl}\frac{\mathrm{dS}}{\mathrm{S}}& =& \mathrm{rdt}\\ \int \frac{\mathrm{1}}{\mathrm{S}}\mathrm{dS}& =& \mathrm{r}\int \mathrm{dt}\\ \mathrm{lnS}& =& \mathrm{rt}+{\mathrm{c}}_{\mathrm{1}}\\ & & {\mathrm{e}}^{\ln \left(\mathrm{S}\right)}={\mathrm{e}}^{\left(\mathrm{rt}+{\mathrm{c}}_{\mathrm{1}}\right)}\end{array}$

Then, the equation becomes,

$$\begin{gathered} {\text{S =e}}^{\text{rt}}{\text{e}}^{{\text{c}}_{\text{1}}} \\ {\text{= ce}}^{\text{0.0575t}} \end{gathered}$$… (2)

Obtain the values of constants.

To find the values of constants, apply the point$\text{(S,t) = (5000,0)}$ in the equation (2), then

$$\begin{gathered} {\text{5000 = ce}}^{\text{0}} \\ \text{c = 5000} \end{gathered}$$

Substitute the value of c in the equation (2).

$S=5000e^{0.0575t}$… (3)

Obtain theamount of accrued money.

Substitute the value into the equation (3).

$\begin{array}{rcl}S& =& 5,000e^{0.0575\times 5}\\ & =& 5,000\times e^{0.2875}\\ & =& 5,000\times 1.333\\ & =& 6,665.5dollars\end{array}$

Obtain the time at which the initial amount is doubled.

(b)

Substitute the value $\text{S = 10,000}$into the equation (3).

role="math" localid="1663840799530" $$\begin{gathered} {\text{10,000 = 5,000e}}^{0.0575\mathrm{t}} \\ 2={\mathrm{e}}^{0.0575\mathrm{t}} \\ \ln 2=\; 0.0575\mathrm{t} \\ \mathrm{t}=\frac{\ln 2}{0.0575} \\ =\frac{0.693}{0.0575} \\ =\; 12.05\mathrm{years} \end{gathered}$$

Obtain the amount of accrued money using calculator.

(c)

$$\begin{gathered} \text{S = 5,000}{\left(\text{1 +}\frac{\text{1}}{\text{4}}\text{(0.0575)}\right)}^{\text{5(4)}} \\ \text{S = 6,651.8} \end{gathered}$$

Hence, the amount is lower than the amount in part (a) by$\text{13.7}$ dollars when the comparison happens between this amount and the amount in part (a).