Question

How to do this application question 3?

Answer 1

`L=120(45-t)`.

Explanation:

Let `L and t` be as defined in the Problem.

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

This means that, `(dL)/dt=k, k" const."`

`:. dL=kdt,` which is a separable variable type Diff. Eqn.

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

`intdL=intkdt+c, k" const."`

`:. L=kt+c........................(square)`.

To determine the consts. `k and c`, we use the given conditions.

`(1): t=20, L=3000, and`

`(2): t=20+15 "(because of further 15 days)="35, L=1200`.

`(1) and (square) rArr 3000=20k+c............(square^1)`.

`(2) and (square) rArr 1200=35k+c............(square^2)`.

Solving `(square^1) and (square^2) for k and c`, we have,

`k=-120, and c=5400`.

Then, `(square)" gives "L=-120t+5400=120(45-t)`, as the

desired relation!

Answer 2

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 - (x*t) = 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`

`35x - 20x = 1800`

`15x = 1800, x = 120 litres`

Substituting value of x in Equation (1),

`L - 20 * 120 = 3000`

L = 3000 + 2400 = 5400 litres#