## anonymous 4 years ago pure water is poured at the rate of 3 gal/min to a tank containing 300 lbs. of salt dissolved in 100 gals. of water and solution kept while stirring, forced out at 2 gals/min. Find the amount of salt at the end of 1 hr. and 30 minutes?

1. anonymous

Tank loses 2 gal/min while adding 3 gal/min for a net of 1 gal/min $Volume = 100 + t$ Salt content in the tank starts at 300 and loses 2C lbs every min where C(t) is the concentration of salt at time t C(t) = Salt content/Volume $C(t) = \frac{300-2t*C(t)}{100+t}$ solving for C(t) $\rightarrow C(t) = \frac{300}{100+3t}$ Let S(t) be salt content at time t S(t) = C(t)*Volume $\rightarrow \ s(t) = \frac{300(100+t)}{100+3t}$ when t=90 $s(90) = \frac{300(190)}{100+270} = 203.57$

2. anonymous

thanks for the effort dumb cow, but i already got the answer. . it's 83.10 lbs.

3. anonymous

Can you please explain to me how you got that answer? Thanks

4. anonymous

this problem is actually related to mixture problem related to differential equation application. . have you encountered this one?

5. anonymous

no

6. anonymous

here, let me show you how i got it. .

7. anonymous

|dw:1328684363825:dw| we got 100 +t as denominator for the because the rate of going in minus the rate in going out (3t - 2t) resulted to t. .

8. anonymous

to make the equation--$dQ/dt=3(0) - 2(Q/100+t)$ we're finding the change of concentration

9. anonymous

|dw:1328684946025:dw||dw:1328685032254:dw| cancel ln.|dw:1328685126662:dw| this is now our equation. .

10. anonymous

now let's find the value for C or constant, given the initial number of salt (Q) which is 300 at t=0 |dw:1328685267814:dw| now that we have the value for constant, we can now find the remaining salt in lbs at t=90 mins.

11. anonymous

|dw:1328685463206:dw|

12. anonymous

that's how i got the answer. . kindly check if there are irregularities.

13. anonymous

No it looks fine, thanks for the explanation. i didn't see where i could solve dQ/dt explicitly like that

14. anonymous

aah, yea me either got difficulties in solving the problem. but anyways, thanks for the medal.