Related Rates Problem

[shortman12012]

1. The problem statement, all variables and given/known data
Water flows into a cubical tank at a rate of 19 L/s. If the top surface of the water in the tank is rising by 3.7 cm every second, what is the length of each side of the tank?


2. Relevant equations
v=L^3


3. The attempt at a solution
so what I started doing was

dV/ds = 3l^2 dl/ds
changed 19 L/sec into cm^3 which is 19000
19000 = 3l^2(3.7)
19000 = 11.1l^2
divided both sides by 11.1
1711.71 = l^2
then took the square root of both sides to get
41.37 cm
however when i put in the answer for my homework, it says the answer is wrong, i have 10 tries, so i was wondering what am i doing wrong or what is the right way to do this problem

 

[Deleted member 4993]

1. The problem statement, all variables and given/known data
Water flows into a cubical tank at a rate of 19 L/s. If the top surface of the water in the tank is rising by 3.7 cm every second, what is the length of each side of the tank?


2. Relevant equations
v=L^3


3. The attempt at a solution
so what I started doing was

dV/ds = 3l^2 dl/ds
changed 19 L/sec into cm^3 which is 19000
19000 = 3l^2(3.7)
19000 = 11.1l^2
divided both sides by 11.1
1711.71 = l^2
then took the square root of both sides to get
41.37 cm
however when i put in the answer for my homework, it says the answer is wrong, i have 10 tries, so i was wondering what am i doing wrong or what is the right way to do this problem

You have square-bottom tank(which is constant) - the height of the water column (which is changing) is "h"

V = L² * h

\(\displaystyle \dfrac{dV}{dt} \ = \ L^2*\dfrac{dh}{dt}\)

Now continue......

 

[shortman12012]

but isn't the formula for the volume L^3 not L^2?