In a bank, principal increases continuously …
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Preeti Dabral 1 year, 2 months ago
Let P be the principal at any time t. According to the given problem, {tex}\frac{d p}{d t}=\left(\frac{5}{100}\right) \times \mathrm{P}{/tex}
or {tex}\frac{d p}{d t}=\frac{\mathrm{P}}{20}{/tex} ...(i)
separating the variables in equation (i), we get
{tex}\frac{d p}{\mathrm{P}}=\frac{d t}{20}{/tex} ...(ii)
Integrating both sides of equation (ii), we get
log P = {tex}\frac{t}{20}{/tex} + C1
or P = {tex}e^{\frac{t}{20}} \cdot e^{\mathrm{C}_{1}}{/tex}
or P = {tex}\mathrm{C} e^{\frac{t}{20}}{/tex} (where eC1 = C) ...(iii)
Now, P = 1000, when t = 0
Substituting the values of P and t in (iii), we get C = 1000. Therefore, equation (iii), gives
P = 1000 et/20
Let t years be the time required to double the principal. Then 2000 = 1000 et/20 {tex}\Rightarrow{/tex} t = 20 loge2.
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