calculuas

A sack containing a liquid chemical is placed in a tank. The chemical seeps out of the sack at a rate of 0.1x litres per hour, where x is the number of litres of the chemical remaining in the sack after t hours. The chemical in the tank evaporates at a rate 0.2y litres per hour, where y is the number of litres of the chemical in the tank after t hours. If the sack originally contained 50 litres of the chemicalk, find differential equations for x and for y, and solve them. Find the greatest amount of chemical in the tank, and when this occurs.

When what occurs?

Greatest amount is 50L at t=0 Christ that was easy

more like calculass

I'm going to assume you mean "in tank, but outside the sack" when you say "in the tank" (your first statement was that this sack is in the tank).

Equilibrium is reached when evaporation = amount entering tank from the sack.  Before that, more is entering than is evaporating = + gain; after that, more is evaporating than is entering = - gain.  Therefore, that is the max.

x is independent of y -> dx/dt = -x/10
x = x₀ e^-t/10

dy/dt = x/10 - y/5
and I'm too lazy to solve this...

>>3
seconded, the sack is in the tank.

thanks 4tran heres one that i'm completely lost on
A will-o'-the'wisp is osillating in  astraiht line so that its desplacement rom the originat time t s a + bsinct, where a b and c are positive constants. it is chased by a kitten which moves o that its velocity at any time is equal to cy. where y is the displacement f the will-o'-the-wasp from the kitten. If x denotes the displacement of the kitten from the origin at time t, find a differential equationconnecting x adn t Show that , after some time,  x is approximately equal to a + (bsin(ct - pi/4))/sqrt(2).

actually don't worry just figure it out sorry

figured

how do you integrate
 bc e^ct sinct

(%i1) integrate(b*c*%e^(c*t)*sin(c*t),t);
(%o1) (b*%e^(c*t)*(c*sin(c*t)-c*cos(c*t)))/(2*c)

how do you get that

>>11
Christ, the * sign is annoying.