rates: liquid drains from tank at 4/(t+5) units/minute

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kpx001

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liquid drains from a tank at the rate of 4 / (t + 5) units per minute until empty. if the tank contains 3 units at time 0, at what time will it finish draining.

basiclly what i did was intergrate 4 / (t + 5) to get 4ln(t+5) + c . then i set it equal to 3 and plugged in 0 for t. at that point i get
R(t) = 4ln(t+5)+3 -4ln(5)

what do i do next? i plugged in a variable to solve for T and got T=5e^(-3/4) -5 , but im unsure if thats right cuz i dont know what to set it equal to. i sorta dont understand word problems. finish draining would mean when the 3 units -> 0 units, right?
 
Re: word problem involving rate.

1) It's a little confusing to use similar variable throughout. I switched to 'r', just to keep from confusing myself. Generally, using the integration variable in the integral limit is considered poor taste.

2) You are almost there.

\(\displaystyle \int_{0}^{r}\frac{5}{5+t}\;dt\;=\;4ln(r+5)-4ln(5)\;=\;3\)

And solve for 'r'.
 
Re: word problem involving rate.

the thing is we havn't learned the fundamental calculus theorom and im really confused now. like i understand intergrating but why do u intergrate again? should it not be 4/ t + 5 intergrated once, get the c or something? sorry but im clueless. this is the entire work i did

intergrate r(t) = 4/ t + 5
R(t) = 4ln(t+5) + 3 -4ln5
R(r) = 4ln(r +5) = 4 -4ln5
0= 4ln(r+5/5) +3
-3/4 = ln(r+5/5)
e^(-3/4) = r +5/5
5e^(-3/4) - 5 = r
 
Re: word problem involving rate.

kpx001 said:
the thing is we havn't learned the fundamental calculus theorom and im really confused now. like i understand intergrating but why do u intergrate again? should it not be 4/ t + 5 intergrated once, get the c or something? sorry but im clueless. this is the entire work i did

intergrate r(t) = 4/ t + 5
R(t) = 4ln(t+5) + 3 -4ln5<<< How did you get that?.


R(r) = 4ln(r +5) = 4 -4ln5 <<< What happened there - how did 't' become 'r' inside ln()? Where did that extra "=" come from?
0= 4ln(r+5/5) +3 <<< How did this come from above?

If you really have no clue - this class is not for you. You are putting in random operations - without any logic


-3/4 = ln(r+5/5)
e^(-3/4) = r +5/5
5e^(-3/4) - 5 = r
Click to expand...
 
Re: word problem involving rate.

R(t) = 4ln(t+5) + 3 -4ln5<<< How did you get that?. by solving for c after i intergrated 4 / t+5


R(r) = 4ln(r +5) = 4 -4ln5 <<< What happened there - how did 't' become 'r' inside ln()? Where did that extra "=" come from? because i am solving for r, because T is confusing variable.
retype R(r) = 4ln(r +5) + 3 -4ln5 ( on laptop)
0= 4ln(r+5/5) +3 <<< How did this come from above? i simplified from above using the log rules then set it equal to 0 because i want to find out when there are 0 units in the container.
 
Re: word problem involving rate.

kpx001 said:
R(t) = 4ln(t+5) + 3 -4ln5<<< How did you get that?. by solving for c after i intergrated 4 / t+5 <<< That is incorrect - Look at tkhunny's solution above.

R(r) = 4ln(r +5) = 4 -4ln5 <<< What happened there - how did 't' become 'r' inside ln()? Where did that extra "=" come from? because i am solving for r, because T is confusing variable.
retype R(r) = 4ln(r +5) + 3 -4ln5 ( on laptop)
0= 4ln(r+5/5) +3 <<< How did this come from above? i simplified from above using the log rules then set it equal to 0 because i want to find out when there are 0 units in the container.
Click to expand...
 
Re: word problem involving rate.

kpx001 said:
liquid drains from a tank at the rate of 4 / (t + 5) units per minute until empty. if the tank contains 3 units at time 0, at what time will it finish draining.

Assume,

Units of liquid in the tank = u

then

\(\displaystyle \frac{du}{dt} = - \frac{4}{t+5}\)

\(\displaystyle u = - 4\cdot ln(t+5) \, + \, C\)

at t = 0 , u = 3

C = 3 + 4 ln(5)

\(\displaystyle u \, = \, 4 \cdot ln\frac{5}{t+5} \, + \, 3\)

to find t = t[sub:30n6uqc8]1[/sub:30n6uqc8], when u =0

\(\displaystyle 0 \, = \, 4 \cdot ln\frac{5}{t_1 \, + \, 5} \, + \, 3\)

\(\displaystyle ln\frac{5}{t_1 \, + \, 5} \, = \, - \frac{3}{4}\)

\(\displaystyle \frac{5}{t_1 \, + \, 5} \, = \,e^{ - \frac{3}{4}}\)

\(\displaystyle t_1 \, + \, 5\, = \, 5\cdot e^{ \frac{3}{4}}\)

Now solve for t[sub:30n6uqc8]1[/sub:30n6uqc8] ....

basiclly what i did was intergrate 4 / (t + 5) to get 4ln(t+5) + c . then i set it equal to 3 and plugged in 0 for t. at that point i get
R(t) = 4ln(t+5)+3 -4ln(5)

what do i do next? i plugged in a variable to solve for T and got T=5e^(-3/4) -5 , but im unsure if thats right cuz i dont know what to set it equal to. i sorta dont understand word problems. finish draining would mean when the 3 units -> 0 units, right?
Click to expand...
 
Re: word problem involving rate.

The rate of change of a liquid in a tank depends on the rate a liquid is going into the tank by the rate a liquid is leaving the said tank.
Hence rate of change of a liquid in respect to time = dl/dt. dL/dt = rate in - rate out = 0 -4/(t+5)

Ergo dL/dt = -4/(t+5), separable ODE, therefore L(t) = ln[(t+5)^(-4)] + C, L(0) = 3, hence

L(t) = ln[((t+5)/5)^(-4)] + 3. Setting L(t) = 0, we have t =5e^(.75)-5 = 5.58500008306 minutes.
 
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