1. Flow rate

a) At a time t=0 a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time t is y. Show that there's a constant b<1 such that y=b^t.

b) Suppose instead that the tank contains one unit of water at time t=0, but in that addition to the water flowing out as described, water is added at a steady rate a>0. Show that

dy/dx-ylnb=a and hence find y in terms of a, b, and t.

2. Originally Posted by free_to_fly
a) At a time t=0 a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time t is y. Show that there's a constant b<1 such that y=b^t.
What you are told is that for some k>0:

dy/dt=-k y

which has general solution y(t)=e^(-kt) + C, where k and C are constants.
As y(0)=1, we have C=0, so:

y=e^(-kt)=(e^-k)^t = b^t

where b=e^-k.

RonL

3. Originally Posted by free_to_fly
a) At a time t=0 a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time t is y. Show that there's a constant b<1 such that y=b^t.

b) Suppose instead that the tank contains one unit of water at time t=0, but in that addition to the water flowing out as described, water is added at a steady rate a>0. Show that

dy/dx-ylnb=a and hence find y in terms of a, b, and t.
In part b you are told that:

dy/dt = -ky + a

or:

dy/dt + ky = a

but from part a you have that b=e^-k, so k = -ln(b) hence:

dy/dt - ln(b)y = a

RonL