I just flopped this part of the exam because I tried to include the existing salt in the tank with my rate out. Since I didn't do that correctly, I couldn't perform the next 2 follow up questions. Practice, practice, practice.
Mindless practising isn’t going to help. Mathematics doesn’t work like that. Practise only once you understand and know the problem. do you think newton just practised really hard until F=MA magically appeared in his brain?…lol. Mathematics is an intellectual process. It’s not like lifting weights. Anyway, you need to understand the purpose of a differential equation in general. Its in the name. DIFFERENTial equation. It is used to describe things where its too difficult to immediately build an equation for a quantity. But its easy to see how the quantity CHANGES. so you should never try to include the initial values (existing salt) in the differential equation because at that point its not even a differential equation and if you can do that why bother with the differential equation to begin with? Look only at how a variable CHANGES. When you solve the differential equation you will Then havr a general solution and it is the purpose of the general solution to accommodate _any_ type of initial value you had. In this case, you could sub in the starting salt ok i’m done with my schizo rant
It seems that this process could be generalized. For example: The tank hold L liters of fluid, with K kg of salt, etc. Doing so would provide a general solution. I wonder how the general solution would look.
You absolutely can generalise this! Have a go yourself, simply follow the same process as I have done in this video, but replacing the numbers with letters.
If you differentiate ln |60 - y|, you will multiply the derivative of the function by the function inside, which is -1. Hence, in reverse differentiation, -1 is used. I know it is 4 years old, but I can’t help it.
Quick question, how much nastier does the problem get if the volume of solution is changing with time as well? Say we started with 1500L of brine and then let the tank fill to a max volume of 3000L. Volume would be a function of time too so would be an interesting ODE. I can think of how I'd solve this numerically, but a look at an analytical solution would be interesting.
A 1000-litre holding tank catches runoff from a chemical process. The tank initially holds 800 litres of water with 2 grams of pollution dissolved in it. Polluted water, containing 5 grams per litre of pollution, flows into the tank at a rate of 3 litres per hour. At the same time, the well-mixed solution leaves the tank at 3 litres per hour. When the amount of pollution in the holding tank reaches 500 grams, the inflow of polluted water is cut off and fresh water enters the tank at a decreased rate of 2 litres per hour, while the outflow is increased to 4 litres per hour. Determine the amount of pollution in the tank at any time 𝑡. HINT: Your answer will be a piecewise-defined function. Please solve for me.
Because the equation states y = 60 - 45e^(-t/300). As 45e^(-t/300) can never give a value lower than 0 (it's getting subtracted, so if it was negative it would be added to the 60), y can never be above 60.
This video explains the topic much better than my professor! Thank you for uploading this
You're very welcome! Thank you.
Oh my Jesus, thank you teacher Finally I understood what I have to do in nixing problem. I appreciate it.
very helpful, thanks. Im gonna show my mates now
Please do!
@@MasterWuMathematics had my vce specialist maths exam last week and this was on it
Thanks for explaining a problem to me 3:00 in the morning really saved my butt!
Wow, you should get a good nights sleep before an exam. But thank you!
Great tutorial!!!❤
Thank you!! 😊
very good explanation, and i think this problem is indeed quite tricky
Yes, you are right
Appreciate your hard work sir ❤
Nice presentation. Bonus points for plotting the tank concentration v time at the end. That was a nice way to wrap it up.
Thanks 👍
This rlly helps me out!
Thank u! Very helpful for JEE❤
Awesome video! Thank you!
Thanks for watching!
Appreciate the help
thank you soo much this really helped me
You're welcome! Glad to help!
Nice explanation
jesus christ brother you;ve solved it
Very helpful. Thanks so much
Thank you!!! This helped me a lot.
You're welcome!
THANK YOU THANK YOU THANK YOU!!!!!!!!!!!!!!
What changes if the inflow and outflow rates are different?
Thank you so much
Well explained! Thank you!
thanks mate
I loved how you demonstrated this concept! Thanks for the visual at the beginning.
Glad you liked it!
I just flopped this part of the exam because I tried to include the existing salt in the tank with my rate out. Since I didn't do that correctly, I couldn't perform the next 2 follow up questions. Practice, practice, practice.
Mindless practising isn’t going to help. Mathematics doesn’t work like that.
Practise only once you understand and know the problem. do you think newton just practised really hard until F=MA magically appeared in his brain?…lol.
Mathematics is an intellectual process. It’s not like lifting weights.
Anyway, you need to understand the purpose of a differential equation in general. Its in the name. DIFFERENTial equation. It is used to describe things where its too difficult to immediately build an equation for a quantity. But its easy to see how the quantity CHANGES.
so you should never try to include the initial values (existing salt) in the differential equation because at that point its not even a differential equation and if you can do that why bother with the differential equation to begin with?
Look only at how a variable CHANGES. When you solve the differential equation you will
Then havr a general solution and it is the purpose of the general solution to accommodate _any_ type of initial value you had. In this case, you could sub in the starting salt
ok i’m done with my schizo rant
THANKS , GREAT HELP FOR MY REVISIONS!!~
honestly thought it was JackFrags talking, a big gaming youtuber
why leave out /min?
dy/dt = 0.2[kg/min] - y/300[kg/min]
i think dt is time and /min too.
I don't understand how you got a '-1' under the 'ln (60 -y) / -1' ....
عاشت ايدك
It seems that this process could be generalized. For example: The tank hold L liters of fluid, with K kg of salt, etc. Doing so would provide a general solution. I wonder how the general solution would look.
You absolutely can generalise this! Have a go yourself, simply follow the same process as I have done in this video, but replacing the numbers with letters.
is there matlab solution for this?
Hi sir, what if the rate in and rate out is different? Can we use your method?
Yes, you can. I'll upload a sequel video with an example soon.
@@MasterWuMathematics tq sir 🥺
@@afiqhaiqal3706 My sequel video is up now: ua-cam.com/video/_DOlVJFP7c0/v-deo.html
Like it, but got a message that I disliked the video. I dont know what happened...I Like it. Thank you
i would not understand the graph clearly
What is it that you do not understand?
What if the rate in > rate out
See this tutorial: ua-cam.com/video/_DOlVJFP7c0/v-deo.html
why is there is negative 1 (-1) on 5:35
If you differentiate ln |60 - y|, you will multiply the derivative of the function by the function inside, which is -1. Hence, in reverse differentiation, -1 is used.
I know it is 4 years old, but I can’t help it.
Quick question, how much nastier does the problem get if the volume of solution is changing with time as well? Say we started with 1500L of brine and then let the tank fill to a max volume of 3000L. Volume would be a function of time too so would be an interesting ODE. I can think of how I'd solve this numerically, but a look at an analytical solution would be interesting.
See this video: ua-cam.com/video/_DOlVJFP7c0/v-deo.html
A 1000-litre holding tank catches runoff from a chemical process. The tank initially
holds 800 litres of water with 2 grams of pollution dissolved in it. Polluted water,
containing 5 grams per litre of pollution, flows into the tank at a rate of 3 litres per
hour. At the same time, the well-mixed solution leaves the tank at 3 litres per hour.
When the amount of pollution in the holding tank reaches 500 grams, the inflow of
polluted water is cut off and fresh water enters the tank at a decreased rate of 2 litres
per hour, while the outflow is increased to 4 litres per hour. Determine the amount of
pollution in the tank at any time 𝑡.
HINT: Your answer will be a piecewise-defined function. Please solve for me.
60 is our asymptote, why is that so ?
Because the equation states y = 60 - 45e^(-t/300). As
45e^(-t/300)
can never give a value lower than 0 (it's getting subtracted, so if it was negative it would be added to the 60), y can never be above 60.
.
(Y/3000)x(10) = y/300 ? Wtf?
it can also be seen as y*(1/3000)*10, which comes out to be y*(1/300) AKA y/300.