# How to do this application question 3?

Feb 9, 2018

$L = 120 \left(45 - t\right)$.

#### Explanation:

Let $L \mathmr{and} t$ be as defined in the Problem.

We are given that, the amt. of water decreases at a constant rate.

This means that, $\frac{\mathrm{dL}}{\mathrm{dt}} = k , k \text{ const.}$

$\therefore \mathrm{dL} = k \mathrm{dt} ,$ which is a separable variable type Diff. Eqn.

To get its General Solution (GS), integrating term-wise,

$\int \mathrm{dL} = \int k \mathrm{dt} + c , k \text{ const.}$

$\therefore L = k t + c \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(\square\right)$.

To determine the consts. $k \mathmr{and} c$, we use the given conditions.

$\left(1\right) : t = 20 , L = 3000 , \mathmr{and}$

$\left(2\right) : t = 20 + 15 \text{(because of further 15 days)=} 35 , L = 1200$.

$\left(1\right) \mathmr{and} \left(\square\right) \Rightarrow 3000 = 20 k + c \ldots \ldots \ldots \ldots \left({\square}^{1}\right)$.

$\left(2\right) \mathmr{and} \left(\square\right) \Rightarrow 1200 = 35 k + c \ldots \ldots \ldots \ldots \left({\square}^{2}\right)$.

Solving $\left({\square}^{1}\right) \mathmr{and} \left({\square}^{2}\right) f \mathmr{and} k \mathmr{and} c$, we have,

$k = - 120 , \mathmr{and} c = 5400$.

Then, $\left(\square\right) \text{ gives } L = - 120 t + 5400 = 120 \left(45 - t\right)$, as the

desired relation!

Feb 9, 2018

Capacity of the tank L = 5400 litres , rate of decrease of water - x = 120 litres per day

#### Explanation:

Given : L - total volume of the tank

$L - \left(x \cdot t\right) = l$ where L - volume of tank in litres, t no. of days, x rate of decrease of water per day and l - volume of water left in thetak after t days.

After 20 days $L - 20 x = 3000$ Eqn (1)

After (20 + 15) days, $L - 35 x = 1200$ Equation (2)

Subtracting Eqn (2) from (10,

cancelL - 20x cancel(- L) - ((-35x) = 3000 - 1200

$35 x - 20 x = 1800$

$15 x = 1800 , x = 120 l i t r e s$

Substituting value of x in Equation (1),

$L - 20 \cdot 120 = 3000$

L = 3000 + 2400 = 5400 litres#