00:01
Hello students, given a water tank as a rectangular base that let b is equal to 2 meters wide and given it was 5 meters long, let l be 5 meter.
00:16
And also given the height in meters of a water in the tank after tr says h of and with respect to time, we need to find the volume of the cubic meters of water in the tank after trs in.
00:35
The function v of t given the water level leaves the tank at a constant rate that is v of 2 is equal to 40 and v dash of t is equal to minus 3 given the conditions initial conditions therefore we know that volume of a rectangular box is given by length into breadth into height this implies we have volume is equal to 2 into 5 into h which gives the value value 10h.
01:08
And also we know that dv by d t is a constant value, let be c.
01:16
Now differentiating this with respect to t, we have dv by d t is equal to 10 into d h by d t.
01:26
Substituting dv by d t is equal to constant.
01:29
We have c is equal to 10 into d h by d t.
01:33
On variable separating, we have d h by d t is equal to c by 10 now integrating on both the side we get the value h is equal to c by 10 into t plus let be an integration constant d substituting in the volume we have therefore v is equal to 10 into c by 10 into t plus d this implies we have v is equal to c into t plus d this implies we have v is equal to c into t plus ten into d.
02:12
Now differentiating with respect to t, we have v dash of t is equal to c plus 10 d is a constant.
02:21
Therefore, it is zero.
02:23
And given that v dash of t is equal to minus three, therefore we have the value c is equal to minus three.
02:34
Substituting the value of c, therefore we have v is equal to minus 3 t plus 10 d.
02:41
And given v of 2 is equal to 40.
02:44
Substituting t is equal to 2 we have v of 2 is equal to minus 3 into 2 plus 10 into d...