mabasamunashe wrote:A colony of ants brings food to its hive at a constant rate of 500grammes per day.It consumes food at a rate of (x/2)grammes per day,where x is the amount of food present at time t days.taking x and t to be continuos variables write down a diffrential equition Ssatisfied by x andt to show the rate at which food is increasing in the hive.
i have said dx/dt=(inflow-outflow)/total amount available
Would not "total amount available" be, by definition, equal to "inflow, less outflow"? Therefore, what is the distinction between your numerator and your denominator above?
mabasamunashe wrote:that is dx/dt=(500-x/2)/x
This simplifies as dx/dt = (500/x) - (1/2). How does this relate to the posted exercise?