How to do this application question 3?

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2 Answers
Ratnaker Mehta
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!

sankarankalyanam
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#

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