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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS4 Marks
Question
It is known that, if the interest is compounded continuously, the principal changes at the rate equal to the product of the rate of bank interest per annum and the principal. Let $P$ denotes the principal at any time $t$ and rate of interest be $r\%$ per annum.

Based on the above information, answer the following question.
  1. Find the value of $\frac{\text{dP}}{\text{dt}}.$
  1. $\frac{\text{Pr}}{1000}$
  2. $\frac{\text{Pr}}{100}$
  3. $\frac{\text{Pr}}{10}$
  4. $\text{Pr}$
  1. If $P_0$ be the initial principal, then find the solution of differential equation formed in given situation.
  1. $\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{100}$
  2. $\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{10}$
  3. $\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\text{rt}$
  4. $\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=100\text{rt}$
  1. If the interest is compounded continuously at $5\%$ per annum, in how many years will $₹\ 100$ double itself?
  1. $12.728$ years
  2. $14.789$ years
  3. $13.862$ years
  4. $15.872$ years
  1. At what interest rate will $₹\ 100$ double itself in $10$ years? $(\log_\text{e}2 = 0.6931 ).$
  1. $9.66\%$
  2. $8.239\%$
  3. $7.341\%$
  4. $6.931\%$
  1. How much will $₹\ 1000$ be worth at $5\%$ interest after $10$ years? $(e^{0.5} = 1.648)$.
  1. $₹\ 1648$
  2. $₹\ 1500$
  3. $₹\ 1664$
  4. $₹\ 1572$
✓

Answer

  1. $(b) \frac{\text{Pr}}{100}$
Here, $P$ denotes the principal at any time $t$ and the rate of interest be $r\%$ per annum compounded continuously, then according to the law given in the problem, we get
$\frac{\text{dP}}{\text{dt}}=\frac{\text{Pr}}{100}$
  1. $(a) \log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{100}$
We have, $\frac{\text{dP}}{\text{dt}}=\frac{\text{Pr}}{100}$
$\Rightarrow\frac{\text{dP}}{\text{P}}=\frac{\text{r}}{100}\text{dt}$
$\Rightarrow\int\frac{\text{1}}{\text{P}}\text{dP}=\frac{\text{r}}{100}\int\text{dt}$
$\Rightarrow\log\text{P}=\frac{\text{rt}}{100}+\text{C}...\text{(i)}$
$\text{At t}=0,\ \ \text{P}=\text{P}_0.$
$\therefore\ \ \text{C}=\log\text{P}_0$
So, $\log\text{P}=\frac{\text{rt}}{100}+\log\text{p}_0$
$\Rightarrow\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{100}...\text{(ii)}$
  1. $(c) 13.862$ years
We have, $r = 5, P_{0 }= ₹\ 100$ and $P = ₹\ 200 = 2P_0$ Substituting these values in $(2),$ we get
$\log2=\frac{5}{100}\text{t}$
$\Rightarrow\text{t}=20\log_\text{e}$
$2 = 20 \times 0.6931\ \text{years} = 13.862\ \text{years}$
  1. $(d) 6.931\%$
We have,
$P_0 = ₹\ 100, P = ₹\ 200 = 2P_{0 }$ and $t = 10$ years Substituting these values in $(2),$ we get
$\log2\frac{10\text{r}}{100}\Rightarrow\text{r}=10\log2=10\times0.6931=6.931$
  1. $(a)\ ₹ 1648$
We have $P_0 = ₹\ 1000, r = 5$ and $t = 10$ Substituting these values in $(2),$ we get
$\log\Big(\frac{\text{P}}{1000}\Big)=\frac{5\times10}{100}=\frac{1}{2}=0.5$
$\Rightarrow\frac{\text{P}}{1000}=\text{e}^{0.5}$
$\Rightarrow\text{ P} = 1000 × 1.648 = ₹ \ 1648$

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