# If the tub has 38.4, 21.6, and 9.6 liters remaining at 1, 2, and 3 minutes since you pulled the plug, what is the equation expressing liters in terms of time? The number of liters of water remaining in the bathtub varies quadratically with the number  of minutes which have elapsed since you pulled the plug.

Since the remaining water changes quadractically with time,

litres remaining can be written as ` L = at^2 + bt + c` where `t` is time since you pulled the plug.

At `t=1`  `at^2 + bt + c = a + b + c= 38.4`

At `t=2`  `at^2 + bt...

Since the remaining water changes quadractically with time,

litres remaining can be written as ` L = at^2 + bt + c` where `t` is time since you pulled the plug.

At `t=1`  `at^2 + bt + c = a + b + c= 38.4`

At `t=2`  `at^2 + bt + c = 4a + 2b + c = 21.6`

At `t=3`  `at^2 + bt + c = 9a + 3b + c = 9.6`

Reduce to Echelon form by subtracting row 1) from rows 2) and 3)

1) 1  1  1   38.4                  1)  1   1   1   38.4

2) 4  2  1   21.6           2)-4*1)  0  -2  -3   -132

3) 9  3  1   9.6            3)-9*1)   0  -6  -8   -336

Then multiplying row 2) by -1/2

1) 1   1   1     38.4

2) 0   1   3/2  66

3) 0  -6  -8    -336

Then adding row 2) to row 3)

1) 1   1    1     38.4

2) 0   1    3/2   66

3)+6*2) 0   0   1      60

` `

From equation 3) `c = 60`

substituting into equation 2)  `b + 3/2(60) = 66 implies b = -24`

substituting into equation 1) `a -24 + 60 = 38.4 implies a = 2.4`

Therefore the equation for remaining litres after t minutes is given by

L = 2.4^2 -24t +60

Check the solution at 1,2 and 3 minutes to confirm

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