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Differential Equations — Maths STD 12 Science — Question

Gujarat 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 P0 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? (e0.5 = 1.648).
  1. ₹ 1648
  2. ₹ 1500
  3. ₹ 1664
  4. ₹ 1572
✓

Answer

  1. (b) $\frac{\text{Pr}}{100}$

Solution:

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) $\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{100}$

Solution:

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

Solution:

We have, r = 5, P0 = ₹ 100 and P = ₹ 200 = 2P0 Substituting these values in (2), we get

$\log2=\frac{5}{100}\text{t}$

$\Rightarrow\text{t}=20\log_\text{e}$

$2 = 20 × 0.6931\ \text{years} = 13.862\ \text{years}$

  1. (d) 6.931%

Solution:

We have,

P0 = ₹ 100, P = ₹ 200 = 2P0 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

Solution:

We have P0 = ₹ 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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