How to do this application question 3?

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2 Answers
Feb 9, 2018

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!

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 - (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#