A lake has 100L of pure water. Water is added at the rate of 48L/year and pollutant is added at rate of 2L/year and water is taken out from the lake at 40L/year.
I have found the half of the equation.
dx/dt = something - 4x/(10+t).
The 'something' is 2 according to the answer book. How can I get it? I used
dx/dt =concentration in*rate in - cocentration out*rate out.
Thanks.
I have found the half of the equation.
dx/dt = something - 4x/(10+t).
The 'something' is 2 according to the answer book. How can I get it? I used
dx/dt =concentration in*rate in - cocentration out*rate out.
Thanks.
-
"Deep Thought " should think deeply....what does [ x ] represent ? if you don't know that then the problem becomes difficult....answer : liters of pollutant in lake
thus [ in - out ] ={ 1 } 2 L/y - { x/ [ 48 + 2 - 40 + t ] } 4L/y = 2 - 4x / [ 10 + t] = dx / dt
{ concentration / L } rate/y
thus [ in - out ] ={ 1 } 2 L/y - { x/ [ 48 + 2 - 40 + t ] } 4L/y = 2 - 4x / [ 10 + t] = dx / dt
{ concentration / L } rate/y