You are the main reason why i passed calculus 1, 2 and 3. Whenever i have a classmate that is struggling, i'd always tell them to watch your videos. Thank you so much, patrick!
@@stephendavis1837 Read your book and understand the example problems, then work problems from the book until you feel comfortable. That's how I got an A by teaching myself Diff EQs. I know you're probably done with the class now though.
I love that the main ad you have is "That Tutor Guy". I haven't even followed the link! For YEARS I have had great success with your videos. Nothing against the other guy. No One on the internet explains with the same clarity and consistency as you. Thank you!
May I say that you are AMAZING, omg finally I found someone that talks in my language. Professors usually skip a lot of steps assuming we know them!!! but your videos makes a lot of sense, that a lot, SUBSCRIBED
I've always wanted to know why calculus textbooks always had this big hard-on for big tanks of brine. My textbook literally has a problem exactly like this... with brine. Anyways, you quite literally are the best math teacher on RUclips in my opinion. You're the reason I get by in engineering. If I could, I would donate. Keep being awesome.
I've been struggling with the modelling of equations exactly like the one you explained in your video for quite a while. All thanks to your amazing teaching ability, I can now do mixing problems flawlessly. Thanks Patrick!
this was extremely helpful! We learned this in Diff. Eq today and i left class thinking it was so tough. You made it so much simpler. I swear im spending the rest of the semester studying your videos! Thanks!!
You're a great man! I've watched these vids all the way through calculus and now I am in diff eq on my way for a degree in physics. You're a huge help, thanks!
thank you sir. your explanation is so helpful especially in online class. its funny how i didn't get this when my professor discuss but understand it well in your video. keep it up!
My prof only taught us to solve the resulting diff eq using an integrating factor, so I've never thought to use separation. That's gonna save soooo much time haha. Thanks man
You can also skip the separation of variables by using the characteristic polynomial. A’ + 1/100 * A = 0 R + 1/100 = Root = -1/100 Y = ce^(-1/100) c = A naught = 15 15e^(-1/100) If it weren’t homogenous you could also use undetermined coefficients to find y particular and add it to the homogenous solution.
at 6:43 how multiplying by dt and dividing by a(t) = 1/A(t) da ?? because multiplying by dt gives just "da" left on the left hand side and then surely dividing by A(t) would bring the left hand side to = "da/At" so why is the answer 1/a da??? please help
Wonderfully worked out example. I have terrible handwriting so my notes are often useless and the book skipped several steps including the part where I messed up, multiplying by e.
I know you get a lot of these praise comments but I just thought I'd shoot you another. I actually have a really good calc 2 professor (unlike most of these people) but hearing another person work these problems helps a lot. It especially helps when I'm studying for my final and my prof hasn't covered this stuff in a few months. So thanks :)
If you were to differentiate ln(-1/100) + C (to check to see if your integration was correct), you would be get 0 since you are taking the derivative of a constant. The integral of (-1/100)dt is (-1/100)t + C and you can check by taking the derivative: -1/100 * t + C u=-1/100 v=t u prime= 0 v prime=1 u(v prime) + u prime(v) + 0 -1/100 * 1 +0*1 + 0 = -1/100
what happened to your rate in (10L/min), it seems that you left that out, shouldn't it be dA/dt +A(t)/100= 10? Making this a first order linear diff eq instead of an easier separable equation?
i think it's pretty clear cuz it's pure water flow in, water has no concentration, so when you multiply concentration it cancels out. watch video carefully, if it's salty water flow in, then that's the case you are talking about.
Thanks Patrick I was struggling substitute the initial condition but now I am ok many thanks please keep up can you also cover the complicated lakes problems please ? I could not understand them thanks
Into tankA: 12 L per minute pure water and 4L perminute water from tank B into tank B: 16 L per minute from tank A Out of tank A: 16 L per minute into tank B out of tank B: 4L per minute into tank A and 12L per minute drains out of the side
Why your C(out) is Mass/Volume of the tank? In my class, my professor told that the volume of the tank would change as a linear function. so your volume should not be constant.
Make sure next video ( If it is a similar problem with same RATEin / RATEout ) that using the given volume is not always the case. If the rates are different then there will be a function for the volume
I do some exercises of mixing problem like this, but the solutions write the denominator of rate out with the variable t. When the rate out part should have variable t? sorry for my poor english.
I have some question involving some mixture problems but I don't get it much: How many pounds of 35% salt solution and how many pounds of 14% salt solution should be combined so that 50lb of a 20% salt solution are obtained? I tried to solve this using different equations but I'd come up with some negative answers...
excuse me. Can u help me solve this problem? TQ - A tank with capacity limit of 1000 gallons initially contains 200 liter of pure water. Then at time t = 0, brine containing 5 lb of salt per gallon is allowed to enter the tank at a rate of 20 gal/min and the mixed solution is drained off at a rate of 10 gal/min. How much salt is in the tank at the point of overflowing? (it says that the tank INITIALLY contains pure water, and at the same time, at t = 0,they pump the salt) so what should I do? )
What if your A(0)=0? What if your initial condition states the amount of salt in the tank is 0? (pure water). How would I get the value of the C constant after integration? Thanks! Thanks for making this videos too! So very helpful. I even exponentiated each side to get rid of the ln[A], then I ended up with my new constant being 0, which cancels out the extremely important variable t. What to do?
When I plug in your final answer in the calculator, the more time passes by the closer to 15 it gets, shouldn't it be the other way? The bigger the "t" the smaller the quantity of salt it should be? (15e^((-1/100)t)=A
You are the main reason why i passed calculus 1, 2 and 3. Whenever i have a classmate that is struggling, i'd always tell them to watch your videos. Thank you so much, patrick!
I like that
dude doing the same, from high school till college
Same, and now hes the reason i'm gonna pass DifEQ too
@@stephendavis1837 Read your book and understand the example problems, then work problems from the book until you feel comfortable. That's how I got an A by teaching myself Diff EQs. I know you're probably done with the class now though.
I love that the main ad you have is "That Tutor Guy". I haven't even followed the link! For YEARS I have had great success with your videos. Nothing against the other guy. No One on the internet explains with the same clarity and consistency as you.
Thank you!
10 years down the line and, your material is still relevant !
Great work, Patrick.
Calculus is 400 years old, and still relevant. His videos will likely be relevant for the next 500 years, assuming youtube remains for that long.
@@MrNukenin16 hahaha makes a lot of sense.
May I say that you are AMAZING, omg finally I found someone that talks in my language. Professors usually skip a lot of steps assuming we know them!!! but your videos makes a lot of sense, that a lot, SUBSCRIBED
I've always wanted to know why calculus textbooks always had this big hard-on for big tanks of brine. My textbook literally has a problem exactly like this... with brine. Anyways, you quite literally are the best math teacher on RUclips in my opinion. You're the reason I get by in engineering. If I could, I would donate. Keep being awesome.
I've been struggling with the modelling of equations exactly like the one you explained in your video for quite a while. All thanks to your amazing teaching ability, I can now do mixing problems flawlessly. Thanks Patrick!
I wish I had known about your tutorials sooner! So much easier to understand than my professor! Thanks a ton!
Thank you so much for filling in the gaps left by my Professor.
Nearly 10 years on, this video is still really helpful. Thanks bro!
you and the khan academy guy should team up and be the kings of online education
professor leonard too ;)
don't forget professor dave, blackpenredpen and the organic chemistry tutor!
this was extremely helpful! We learned this in Diff. Eq today and i left class thinking it was so tough. You made it so much simpler. I swear im spending the rest of the semester studying your videos! Thanks!!
Your videos from 2010 are getting me grades in 2018. Can't thank you enough.
You're a great man!
I've watched these vids all the way through calculus and now I am in diff eq on my way for a degree in physics. You're a huge help, thanks!
This made a hell of a lot more sense than my textbook, thanks for the crystal clear explanations.
I just came across your channel and this section of differential equations has never been easy for me to understand until I watched this. THANK YOU
my professor completely blew through this stuff and my textbook is basically useless. Thanks for explaining it so clearly! you're a lifesaver =)
@ermitz90 cause i am lazy and it is easier to write A instead of A(t).
it is the same reason people write: y = x - 3 instead of y(x) = x - 3
Your explanation is very good. I'm only a sophomore in high school, and I understood you pretty well. Great video.
Thanks for making these vids patrickJMT. Keep it up!
he makes it look so easy
I know it’s been 8 years but thank you so much for this video! Much clearer than my ODE class
you are most welcome!
you are an amazing human being
Can't imagine my math classes without u ! Thanks man
this video is awesome still 13 years later thank you!
I just came across these. So much easier to understand now.
If I pass this test tomorrow; I owe it all to you. Thank you for your time.
This helped more than any other vid on the topic Thanks
thank you sir. your explanation is so helpful especially in online class. its funny how i didn't get this when my professor discuss but understand it well in your video. keep it up!
Bless you mate. You're saving my studies
Thank you for actually being a teacher unlike some of these theory professors that assume we can extrapolate from the book definitions lol.
Everybody kick in a few bucks to his Patreon, he deserves it more than most of our math professors!
My prof only taught us to solve the resulting diff eq using an integrating factor, so I've never thought to use separation. That's gonna save soooo much time haha. Thanks man
thanks a lot dude..i have an exam today..this will help me alot:))
Thanks for explaining this in such a simple way . My DE exam is in 12 hrs and I'm confident now that I won't miss the tank problem 😆
dude thanks i have a quiz tomorrow and this stuff might be on it!
Yes! the videos from 2006 are still helpful
Thanks for everything man. You've honestly helped make my life easier.
you saved my day and I'm sure for many people as well. Thank you for the videos, they are very helpful!
you are the man!!!..best math teacher out there!!,,:D
This video really helped me a lot!!! Thank you so much for this! You're such a blessing ❤
Thanks, you have made mixing problems a lot clearer to me.
I LOVE YOU! you are the reason why i'm passing!
thank you very much! I have my final exam on monday and this will certainly help
i love you and the way you teach hallelujah
Thank You Patrick .. Your videos have been so helpful !
wow thanks!!! I have an exam today on this material and it really helped :D
Thanks for this Patrick!
great job MR. AMAZING
Your the best. Thanks for helping me understand.
Patrick, thanks for the fantastic explanation!
Great video with clear explanation. Thanks!
Sir, @05:25, it is clear that dA/dt=-A(t)/100; hence on integrating why doesn't the R.H.S. becomes (t^2)/200?
You're my hero.
"It's a beautiful tank" I probably laughed harder than I should XD Thank you soooo much for the videos!
i love your vids! they help soooo much! THANKS :)
Beautifully done
my friend you are amazing as always
thank you for always helping me.. hope to meet you in person..take care
You can also skip the separation of variables by using the characteristic polynomial.
A’ + 1/100 * A = 0
R + 1/100 =
Root = -1/100
Y = ce^(-1/100)
c = A naught = 15
15e^(-1/100)
If it weren’t homogenous you could also use undetermined coefficients to find y particular and add it to the homogenous solution.
at 6:43 how multiplying by dt and dividing by a(t) = 1/A(t) da ?? because multiplying by dt gives just "da" left on the left hand side and then surely dividing by A(t) would bring the left hand side to = "da/At" so why is the answer 1/a da??? please help
Wonderfully worked out example. I have terrible handwriting so my notes are often useless and the book skipped several steps including the part where I messed up, multiplying by e.
Great explanation
imagine if patrick was your professor... every class could be 10 minutes
never!
nice, im doing diferential equations and it helped a lot. More videos pls :)
I know you get a lot of these praise comments but I just thought I'd shoot you another. I actually have a really good calc 2 professor (unlike most of these people) but hearing another person work these problems helps a lot. It especially helps when I'm studying for my final and my prof hasn't covered this stuff in a few months. So thanks :)
Great explanation, thanks
Thanks, this helped a lot!
If you were to differentiate ln(-1/100) + C (to check to see if your integration was correct), you would be get 0 since you are taking the derivative of a constant. The integral of (-1/100)dt is (-1/100)t + C and you can check by taking the derivative:
-1/100 * t + C
u=-1/100 v=t
u prime= 0 v prime=1
u(v prime) + u prime(v) + 0
-1/100 * 1 +0*1 + 0 = -1/100
Thank you so much! You're so amazing!!!
what happened to your rate in (10L/min), it seems that you left that out, shouldn't it be dA/dt +A(t)/100= 10? Making this a first order linear diff eq instead of an easier separable equation?
i think it's pretty clear cuz it's pure water flow in, water has no concentration, so when you multiply concentration it cancels out. watch video carefully, if it's salty water flow in, then that's the case you are talking about.
Thanks Patrick I was struggling substitute the initial condition but now I am ok many thanks please keep up can you also cover the complicated lakes problems please ? I could not understand them thanks
Thanks it helps me a lot!
Damn that sure is one mighty fine tank there, bud.
bless you patrick
when you talk it sounds like something from a sci-fi movie xD thank you!
Thanks so much! Awesome videos :)
Could you do a problem where the volume of water in the tank changes?
this was really good thank you!
Into tankA: 12 L per minute pure water and 4L perminute water from tank B
into tank B: 16 L per minute from tank A
Out of tank A: 16 L per minute into tank B
out of tank B: 4L per minute into tank A and 12L per minute drains out of the side
Awesome sir have enjoyed it
Why your C(out) is Mass/Volume of the tank? In my class, my professor told that the volume of the tank would change as a linear function. so your volume should not be constant.
damn this was soooo easy. Thank you sooo much again Patrick :D
Thanks man, really appreciate it! xoxoxoxo
Make sure next video ( If it is a similar problem with same RATEin / RATEout ) that using the given volume is not always the case. If the rates are different then there will be a function for the volume
Thankyou so much.❤❤❤
Patrick could you also do circuits. (differential equation RL and RC circuits)
very helpful. Thank you
I do some exercises of mixing problem like this, but the solutions write the denominator of rate out with the variable t. When the rate out part should have variable t? sorry for my poor english.
Great video, thanks.
straight from my calc book. siiiick :D
Brilliant!
@patrickJMT ...... How do benefit by making these videos?? It sure does benefit us, but how you??
hi
views=subscirbers=money
It is easy, thank you! ❤
Amazing video!!!!!!!!!!!!!!!!!!!!!! :)
Fire!
I have some question involving some mixture problems but I don't get it much: How many pounds of 35% salt solution and how many pounds of 14% salt solution should be combined so that 50lb of a 20% salt solution are obtained?
I tried to solve this using different equations but I'd come up with some negative answers...
excuse me. Can u help me solve this problem? TQ
- A tank with capacity limit of 1000 gallons initially contains 200 liter of pure water. Then at time t = 0, brine containing 5 lb of salt per gallon is allowed to enter the tank at a rate of 20 gal/min and the mixed solution is drained off at a rate of 10 gal/min. How much salt is in the tank at the point of overflowing? (it says that the tank INITIALLY contains pure water, and at the same time, at t = 0,they pump the salt) so what should I do? )
greeeaaattt explanation !!
What if your A(0)=0? What if your initial condition states the amount of salt in the tank is 0? (pure water). How would I get the value of the C constant after integration? Thanks! Thanks for making this videos too! So very helpful.
I even exponentiated each side to get rid of the ln[A], then I ended up with my new constant being 0, which cancels out the extremely important variable t. What to do?
When I plug in your final answer in the calculator, the more time passes by the closer to 15 it gets, shouldn't it be the other way? The bigger the "t" the smaller the quantity of salt it should be? (15e^((-1/100)t)=A
the best part is we have the same handwriting