56. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).

9 Views|Posted 8 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

8 months ago

Let P, r and t be the principal rate and time respectively.

Then, increase in principal $\frac{d P}{d t} = P * r %$

$\begin{array}{l} \Rightarrow \frac{d P}{d t} = P . \frac{r}{100} \\ \Rightarrow \frac{d P}{P} = \frac{r}{100} d t \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{d P}{P} = \int \frac{r}{100} d t \\ \Rightarrow l o g P = \frac{r t}{100} + c \\ \Rightarrow P = e^{\frac{r t}{100} + c} \end{array}$

Given at t=0,P=100

So, $100 = e^{\frac{r * 0}{100} + c}$

$\begin{array}{l} \Rightarrow 100 = e^{0} * e^{c} \\ \Rightarrow e^{c} = 100 (? e^{0} = 1) \end{array}$

And at $t = 10, P = 2 * 100 = 200$

So, $200 = e^{\frac{r 10}{100} + c}$

$\begin{array}{l} \Rightarrow 200 = e^{\frac{r}{10}} . e^{e} \\ \Rightarrow e^{\frac{r}{10}} = \frac{200}{100} = 2 \end{array}$

$\begin{array}{l} \Rightarrow \frac{r}{10} = l o g 2 \\ \Rightarrow \frac{r}{10} = 0.6931 \\ \Rightarrow r = 6.931 \end{array}$

Hence, the rate is 6.931%

Upvote

Similar Questions for you

Let α be the angle between the lines whose direction cosines satisfy the equations I + m – n = 0 and I² + m² – n² = 0. Then the value of sin⁴α + cos⁴α is:

Vishal Baghel

l + m – n = 0

l + m = n . (i)

l² + m² = n²

Now from (i)

l² + m² = (l + m)²

=> 2lm = 0

=>lm = 0

l = 0 or m = 0

=> m = n Þ l = n

if we take direction consine of line

$0, \frac{1}{\sqrt[]{2}}, \frac{1}{\sqrt[]{2}} a n d \frac{1}{\sqrt[]{2}}, 0, \frac{1}{\sqrt[]{2}}$

cos a = $\frac{1}{2}$

$s i n^{4} α + c o s^{4} α = {(\frac{\sqrt[]{3}}{2})}^{4} = {(\frac{1}{2})}^{4} = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8}$

Vishal Baghel

$\Rightarrow \int_{}^{} \frac{d y}{1 + y^{2}} = - 2 \int_{}^{} \frac{e^{x} d x}{1 + {(e^{x})}^{2}} + C$

$\Rightarrow t a n^{- 1} y = - 2 \cdot t a n^{- 1} e^{x} + C$

x = 0, y = 0

$\Rightarrow 0 = 2 t a n^{- 1} + C$

$C = + \frac{π}{2}$

now at x = $l n \sqrt[]{3}$

$t a n^{- 1} y = - 2 t a n^{- 1} (e^{l n \sqrt[]{3}}) + \frac{π}{2}$

$6 (y' (0) + {(y (l n \sqrt[]{3}))}^{2}) = 6 (- 1 + \frac{1}{3}) = - 4$

If x₁ $\underset{x \to \infty}{l i m} \frac{c o t^{- 1} (\sqrt[]{x + 1} - \sqrt[]{x})}{s e c^{- 1} ({(\frac{2 x + 1}{x - 1})}^{x})}$ then $\frac{(3 e^{x_{1}} + x_{2})}{4}$ equals

Vishal Baghel

$x_{1} = \underset{x \to \infty}{l i m} \frac{2^{\frac{x^{n}}{e^{x}}} - 3^{\frac{x^{n}}{e^{x}}}}{\frac{x^{n}}{e^{x}}}, p u t \frac{x^{n}}{e^{x}} = t$

$x_{2} = \underset{x \to \infty}{l i m} \frac{c o t^{- 1} (\sqrt[]{x + 1} - \sqrt[]{x})}{s e c^{- 1} ({(\frac{2 x + 1}{x - 1})}^{x})}$

$x_{2} = \frac{2}{π} \underset{x \to \infty}{l i m} t a n^{- 1} (\sqrt[]{x + 1} + \sqrt[]{x})$

$x_{2} = \frac{2}{π} . \frac{π}{2} = 1$

Let $y = y (x)$ be a function of $x$ satisfying $y \sqrt[]{1 - x^{2}} = k - x \sqrt[]{1 - y^{2}}$ where $k$ is a constant and $y (\frac{1}{2}) = - \frac{1}{4}$ . Then $\frac{d y}{d x}$ at $x = \frac{1}{2}$ , is equal to:


Alok Kumar Singh

Differentiating

y. $\frac{- 2 x}{2 \sqrt[]{1 - x^{2}}} + \sqrt[]{1 - x^{2}} \cdot y^{'} = \frac{x \cdot 2 y y^{'}}{2 \sqrt[]{1 - y^{2}}} - \sqrt[]{1 - y^{2}}$

Put $x = \frac{1}{2}, y = - \frac{1}{4}$ and $x, y = - \frac{1}{8}$

$y^{'} = \frac{- \sqrt[]{5}}{2}$

Alok Kumar Singh

dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.