Advertisements
Advertisements
प्रश्न
The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?
Advertisements
उत्तर
Let the population at any time t be P.
Given:- \[\frac{dP}{dt} \alpha P\]
\[\Rightarrow \frac{dP}{dt} = \beta P\]
\[ \Rightarrow \frac{dP}{P} = \beta dt\]
\[ \Rightarrow \log\left| P \right| = \beta t + \log C . . . . . . . . \left( 1 \right)\]
Now,
\[\text{ At }t = 1990, P = 200000\text{ and at }t = 2000, P = 250000\]
\[ \therefore \log 200000 = 1990\beta + \log C . . . . . . . . \left( 2 \right) \]
\[ \log 250000 = 2000\beta + \log C . . . . . . . . . \left( 3 \right)\]
\[\text{ Subtracting }\left( 3 \right)\text{ from }\left( 2 \right), \text{ we get }\]
\[\log 200000 - \log 250000 = 10\beta\]
\[ \Rightarrow \beta = \frac{1}{10}\log\left( \frac{5}{4} \right)\]
\[\text{ Putting }\beta = \frac{1}{10}\log \left( \frac{5}{4} \right) \text{ in }\left( 2 \right),\text{ we get }\]
\[\log 200000 = 1990 \times \frac{1}{10}\log\left( \frac{5}{4} \right) + \log C\]
\[ \Rightarrow \log 200000 = 199\log\left( \frac{5}{4} \right) + \log C \]
\[ \Rightarrow \log C = \log 200000 - 199\log\left( \frac{5}{4} \right) \]
\[\text{ Putting }\beta = \frac{1}{10}\log \left( \frac{5}{4} \right), \log C = \log 200000 - 199 \log\left( \frac{5}{4} \right) \text{ and }t = 2010\text{ in }\left( 1 \right),\text{ we get }\]
\[\log\left| P \right| = \frac{1}{10} \times 2010\log \left( \frac{5}{4} \right) + \log 200000 - 199 \log\left( \frac{5}{4} \right)\]
\[ \Rightarrow \log\left| P \right| = 201 \log \left( \frac{5}{4} \right) + \log 200000 - 199\log\left( \frac{5}{4} \right)\]
\[ \Rightarrow \log\left| P \right| = \log \left( \frac{5}{4} \right)^{201} - \log \left( \frac{5}{4} \right)^{199} + \log 200000\]
\[ \Rightarrow \log\left| P \right| = \log\left\{ \left( \frac{5}{4} \right)^{201} \left( \frac{4}{5} \right)^{199} \right\} + \log 200000\]
\[ \Rightarrow \log\left| P \right| = \log\left\{ \left( \frac{5}{4} \right)^2 \right\} + \log 200000\]
\[ \Rightarrow \log\left| P \right| = \log\left( \frac{25}{16} \times 200000 \right)\]
\[ \Rightarrow \log\left| P \right| = \log 312500\]
\[ \Rightarrow P = 312500\]
APPEARS IN
संबंधित प्रश्न
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]
C' (x) = 2 + 0.15 x ; C(0) = 100
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
(y + xy) dx + (x − xy2) dy = 0
Solve the following differential equation:
(xy2 + 2x) dx + (x2 y + 2y) dy = 0
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]
Solve the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\], given that y = 1, when x = 0.
(y2 − 2xy) dx = (x2 − 2xy) dy
\[\frac{dy}{dx} = \frac{y}{x} + \sin\left( \frac{y}{x} \right)\]
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x} , y\left( 1 \right) = 0\]
The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.
If the interest is compounded continuously at 6% per annum, how much worth Rs 1000 will be after 10 years? How long will it take to double Rs 1000?
Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e2x.
Find the equation to the curve satisfying x (x + 1) \[\frac{dy}{dx} - y\] = x (x + 1) and passing through (1, 0).
Find the equation of the curve which passes through the origin and has the slope x + 3y− 1 at any point (x, y) on it.
The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).
The differential equation of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = C\] is
Find the equation of the plane passing through the point (1, -2, 1) and perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2).
Form the differential equation from the relation x2 + 4y2 = 4b2
Solve the differential equation sec2y tan x dy + sec2x tan y dx = 0
Solve the differential equation (x2 – yx2)dy + (y2 + xy2)dx = 0
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
Choose the correct alternative:
General solution of `y - x ("d"y)/("d"x)` = 0 is
A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is
