Math / Calculus

Questiondeposited in it a particular savings account is compounded continuously. The account initially had $\$ 2,200$ $A(t)=2200 e^{05}$ (a) By what percent will the worth of the account increase per year? Round to the nearest hundredth of a percent. $\begin{array}{l} \text { e.0s(1) } \\ =1.05127-1 \\ =0.05127(100) \\ =5.127 \\ =5.13 \% \end{array}$ (b) To the nearest tenth of a year, how long will it take for the worth of the account to triple? (c) If another investment began with a principal of $\$ 2,500$ and earned simplest interest of $3.8 \%$ applied once per year, which investment would be worth more after 10 years? Justify.

Studdy Solution

STEP 1

1. The savings account is compounded continuously, which means we use the formula $A(t) = P e^{rt}$ , where $P$ is the principal, $r$ is the rate, and $t$ is time.
2. We are given the function $A(t) = 2200 e^{0.05t}$ for the continuously compounded account.
3. Simple interest is calculated using the formula $A = P(1 + rt)$ .

STEP 2

1. Calculate the annual percentage increase for the continuously compounded account.
2. Determine the time required for the account to triple in value.
3. Compare the value of the continuously compounded account and the simple interest account after 10 years.

STEP 3

To find the annual percentage increase, we need to evaluate the expression $e^{0.05}$ and convert it to a percentage increase:
$e^{0.05} \approx 1.05127$
Subtract 1 to find the increase:
$1.05127 - 1 = 0.05127$
Convert to a percentage:
$0.05127 \times 100 = 5.127\%$
Round to the nearest hundredth:
$5.13\%$

STEP 4

To find how long it will take for the account to triple, set up the equation:
$3 \times 2200 = 2200 e^{0.05t}$
Simplify:
$3 = e^{0.05t}$
Take the natural logarithm of both sides:
$\ln(3) = 0.05t$
Solve for $t$ :
$t = \frac{\ln(3)}{0.05}$
Calculate $t$ :
$t \approx \frac{1.0986}{0.05} \approx 21.972$
Round to the nearest tenth:
$t \approx 22.0 \text{ years}$

STEP 5

Calculate the value of the continuously compounded account after 10 years:
$A(10) = 2200 e^{0.05 \times 10} = 2200 e^{0.5}$
Calculate $e^{0.5} \approx 1.64872$ :
$A(10) \approx 2200 \times 1.64872 \approx 3627.18$
Calculate the value of the simple interest account after 10 years:
$A = 2500(1 + 0.038 \times 10) = 2500(1 + 0.38)$
$A = 2500 \times 1.38 = 3450$
Compare the two amounts:
The continuously compounded account is worth approximately $\$3627.18$ , while the simple interest account is worth $\$3450$ . Thus, the continuously compounded account is worth more after 10 years.
The continuously compounded account increases by $5.13\%$ per year, takes approximately $22.0$ years to triple, and is worth more than the simple interest account after 10 years.

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