They learned exactly the way I am currently learning with my 65 year-old calc3 professor who assumes that his 6 remaining engineering students have a PhD in mathematics... Proofs upon proofs upon tests with complicated proofs upon failure and shame. Last test I got a 70%... I was the highest grade.
I think it is very good how, around minute 7:45, you took the time to rewrite the derivative expression to set it equal to zero, as opposed to writing an equal and zero after the derivative. It might seem like a small thing to point out but I appreciate the extra step as it makes the flow of the solution more correct.
OMG YOU JUST POSTED THESE RECENTLY. MY test is tomorrow. Thank goodness for this. Have my babies I love you omg....Could not understand this at all even with the webtutor site where they have videos that explain to us, they sound like robots!
Thank you so much for the time and effort you put into these videos. Because of you, I received an A in Calc 1, 2, and now I'm sure I can get an A in Calc 3.
It is much more obvious what the max value is for the first two lines. For the third it's not easy to visual the line and therefore you find the critical points of that line and evaluate.
Thank you thank you thank you! I was struggling on how to do these kind of problems and you really made it clear and easy! I have tried looking at solutions manual and they always skip steps which made it very confusing! Thanks for your thorough explanation! Now I wish this will be on my test tomorrow! haha.
Amazing lesson,I missed a class in college and this helped me a lot.You're a great teacher,after watching this video,I'm sure I won't be attending any more math classes.I'll just use your videos instead! g
i want to add that you have to check the corners of the triangle as well. for this particular problem, the corners also give you 2, -2, and -2, so the corners didn't really matter. but other problems, like f(x,y) = 2 + 2x +2y -x^2 - y^2 in the 1st quadrant bounded by x = 0, y = 0, y = 4 -x will give you a max from one of the corners that you won't get if you just check the boundaries and the cp you get from setting fx and fy to 0 and solving for x and y.
For the first equation x = 1. After you plugged in 1 and solved. you got 2 - y. and said y can only be [0,4]. What about if you plug in 2 to the equation you will get 0 and that is a local min no?
At that point, Patrick was trying to find out what the maximum and minimum values along that boundary is. If you plug in 2 into the equation you are right, you get 0 but that is neither the maximum or minimum value you can get from f= 2-y as plugging in 2=0 will give a higher value and plugging in y=4 will give a lower value.
PATRICKKKKK!!!!! Why did you take the derivative of the third function? A lot of people have asked this question and I cannot find a solid answer, can you please explain. THANKSSS
+Karen Roman Try visualizing the functions. The maximum and minimums of the first two functions are the endpoints, because they are linear relationships. The function's don't curve, and thats why he can just use the endpoints. However, the last function is a quadratic, and its maximum and minimums aren't so easily visualized, and he takes the derivative in order to locate them.
Yes, You could take the derivative but would find it is a constant, and therefore Never zero. A linear function has a constant slope so its extrema are always at its end points. (Unless you have a horizontal line which has slope IS zero and has extrema EVERYWHERE)
Hey Patrick, what if we had to evaluate the max/min along a boundary thats not linear, like maybe the fn. x^2-x+1, would we just take the partials the function and basically carry out the same procedure for that boundary segment?
hey patrick, instead of putting the boundary points everytime and compare with what we currently have, can we just add them after we found other critical points?
In the single-variable case, it makes sense to set the first derivative to zero, but in the multi-variable case, why find partial derivatives, set them equal to zero and solve? How does it make sense? Secondly, what if the critical point you initially found by solving the partial derivatives lies outside the bounded region?
Wait, so why did you find the local max and min for the three functions if you only need to find the absolute values by plugging in the critical points into the function?
instead of derive and checking endpoints end points in the invervall -x^2+6x-7 , multiple with -1 x^2-6x+7 , use quadrule to write as (x-3)^2-9+7=(x-3)^2-2, now we see that minus is -2 and max is (5-3)^2-2=4-2=2
nice vid and you explain it really well, and i get this stuff, but damn these are the easy parts, you should have put world problems like applied max and min values, like find the dimensions of a cone and maximum volume that can be fit into a sphere of diameter 6cm
can we have an example of an f(x,y,z) function and how to determine the nature of the critical point. I dont know if you use the Hessian or what can you show an example. And if we do know how to put the hessian matrix together then how do we use it to find out if it is a saddle max or min?
You use the second derivative test to find *local* extrema (maximum if hill, minimum if valley, neither if saddle). To find absolute extrema, this information isn't strictly needed because you can just evaluate at all the candidate points (critical points and boundary) and just pick the largest and smallest value.
Extrema occur at the critical points or on the boundary. You must check them all since any one of them could be a maximum or a minimum. If you ignore a critical point, you might miss a maximum or minimum.
Hey Patrick! I noticed when you took the case x=1, and got -2 as a y, I noticed that that point was outside the boundary… a.k.a the triangle. With that, are we still allowed to use that point as a max or min even if it’s outside the boundary?
I dont follow why the -x+5 section had to have its derivative taken. were there no critical points in the other two? (I know the third equation looked harder but how do u know if/when to derive it and when there will be an additional critical point in there?)
7:59 Sir Plz guide me, u have sorted out the critical piont x=3,shouldnt be its y component there, have to write my symster paper on 24th may please help Athar (Pakistan) and thanks for the videos,luv:-)
nope, u r only interested in what the max/min values are and don't need to state where. as u can see at end of vid he shows z = 2,-2. But if u want u can add it's y component not gona change anything
i'll try.. the other two equations were obvious. Once we plug in the lowest and highest, it gives an easy result. However, when we substituted y=-x+5, it was not so simple. Differentiating will give quicker answer, whereas plugging each values to find the number will take longer (you can do that though, if you want).
Once again you have cleared up my confusion. My tuition should not go to my professors, it should go to you haha
My question is, how did people learn calculus before the internet?
The way profs/textbooks make things so complicated really makes me wonder.....
They learned exactly the way I am currently learning with my 65 year-old calc3 professor who assumes that his 6 remaining engineering students have a PhD in mathematics... Proofs upon proofs upon tests with complicated proofs upon failure and shame. Last test I got a 70%... I was the highest grade.
our professors are actual students of the our classes. We know how to apply meanwhile they learned math. Proofs upon proofs is math.
Check out this website:
betterexplained.com/cheatsheet/
"Intuitive" Math explanations. Just discovered it, pretty cool.
Justin LeBling thoughtful question
I ask myself the same question every time I'm doing my calculus homework.
x can't be anything it wants to be. That moment when you realise calculus is depressing :(
I feel so much safer when Patrick jmt is talking while i do my hw....somehow i feel like everything is going to be okay with Patrick.
If only my professor can be as fluid and organized as you are….you practically boosted me up from a B to an A. Thank you.
I think it is very good how, around minute 7:45, you took the time to rewrite the derivative expression to set it equal to zero, as opposed to writing an equal and zero after the derivative. It might seem like a small thing to point out but I appreciate the extra step as it makes the flow of the solution more correct.
You are amazing. When I graduate and find a nice paying job in engineering I will definitely show you my gratitude through that donation box.
OMG YOU JUST POSTED THESE RECENTLY. MY test is tomorrow. Thank goodness for this. Have my babies I love you omg....Could not understand this at all even with the webtutor site where they have videos that explain to us, they sound like robots!
Thank you so much for the time and effort you put into these videos. Because of you, I received an A in Calc 1, 2, and now I'm sure I can get an A in Calc 3.
wow
something take takes up to 30 min in lecture: is 10 mins over here!
U the bomb! Mr. Patrick!
International Chinese prof. crew checkin' in! You're a lifesaver man
dude i love you so much!!! i never write comments but when i do is because the video deserves it. you are an amazing guy.
Hi , why did you take the derivative of the third line and not for the first 2 lines ?
+abood nayfeh you could but its kinda pointless since you are going to use the endpoints anyways
It is much more obvious what the max value is for the first two lines. For the third it's not easy to visual the line and therefore you find the critical points of that line and evaluate.
visualize***
you just saved me hours from figuring this stuff out, thank you!
it's unbelievable how much YOU made me understand in like 10 minutes compared to my professors 2h lecture.. thanks ALOT!
Thank you thank you thank you! I was struggling on how to do these kind of problems and you really made it clear and easy! I have tried looking at solutions manual and they always skip steps which made it very confusing! Thanks for your thorough explanation! Now I wish this will be on my test tomorrow! haha.
I have a calculus final exam tomorrow and this will save me!! thanks you're the best :D
Amazing lesson,I missed a class in college and this helped me a lot.You're a great teacher,after watching this video,I'm sure I won't be attending any more math classes.I'll just use your videos instead! g
why did you take the derivative for the line y= -x+5 and not with the other boundries?
The man who saved me an hour before my exam thanks alot🙏
thanks patrick, you are the best!
@l21thl i will accept that compliment any time. bob ross rocks!!
thanks, this was very helpful, i was sick today couldn't go to class, and this saved my ass.
@bustachaina good luck on the test!
i want to add that you have to check the corners of the triangle as well. for this particular problem, the corners also give you 2, -2, and -2, so the corners didn't really matter. but other problems, like f(x,y) = 2 + 2x +2y -x^2 - y^2 in the 1st quadrant bounded by x = 0, y = 0, y = 4 -x will give you a max from one of the corners that you won't get if you just check the boundaries and the cp you get from setting fx and fy to 0 and solving for x and y.
he did that near the end of the video
youre a life saver. keep up the good work.
Thank you for the quick explanations.
you are my savior. on to lagrange multipliers...
Very neat and organized. Thank you. Clear to follow.
man!!!!!!!!! you are my hero 😭😭😭
Everytime he says back to semester 1 while I'm in semester one makes me worried of what's ahead😂
Im gonna be clear: ''Thanks sir.''
Thank you so much for this tutorial!! You explain it so Well!!
Thankyou so much, I will sleep in peace after this
clear and simple superb one!Thnx a lot!!
great video! So we'd use this method for straight edged boundaries? And unconstrained optimization for curved boundaries?
For the first equation x = 1. After you plugged in 1 and solved. you got 2 - y. and said y can only be [0,4]. What about if you plug in 2 to the equation you will get 0 and that is a local min no?
At that point, Patrick was trying to find out what the maximum and minimum values along that boundary is. If you plug in 2 into the equation you are right, you get 0 but that is neither the maximum or minimum value you can get from f= 2-y as plugging in 2=0 will give a higher value and plugging in y=4 will give a lower value.
Thank you so much. this helps me a lot.
PATRICKKKKK!!!!!
Why did you take the derivative of the third function?
A lot of people have asked this question and I cannot find a solid answer, can you please explain.
THANKSSS
+Karen Roman Try visualizing the functions. The maximum and minimums of the first two functions are the endpoints, because they are linear relationships. The function's don't curve, and thats why he can just use the endpoints. However, the last function is a quadratic, and its maximum and minimums aren't so easily visualized, and he takes the derivative in order to locate them.
Thanks, genius! :)
Yes,
You could take the derivative but would find it is a constant, and therefore Never zero. A linear function has a constant slope so its extrema are always at its end points. (Unless you have a horizontal line which has slope IS zero and has extrema EVERYWHERE)
go back to calc I
Why did you differentiate the function obtained after substituting (x, -x+5) and you didn't do that for (1,y) and (x,0)
Patrick,I love you!
patrick so what happens with the 1 we got from evaluating our function at (2,1)? what does that mean for our problem? is it not part of the answer?
Ugh, I absolutely hate these absolute extrema problems. Truly a PITA. But this video makes them a lot easier.
you the man patrick ily
Hey Patrick, what if we had to evaluate the max/min along a boundary thats not linear, like maybe the fn. x^2-x+1, would we just take the partials the function and basically carry out the same procedure for that boundary segment?
you are the best. thank you
hey patrick, instead of putting the boundary points everytime and compare with what we currently have, can we just add them after we found other critical points?
Thank you :)
In the single-variable case, it makes sense to set the first derivative to zero, but in the multi-variable case, why find partial derivatives, set them equal to zero and solve? How does it make sense?
Secondly, what if the critical point you initially found by solving the partial derivatives lies outside the bounded region?
Why is it you take the derivative of the y=-x+5 line but not the lines along the axes?
thanks a lot, Bro. I get it. It is very easy to understand what you points.
Wait, so why did you find the local max and min for the three functions if you only need to find the absolute values by plugging in the critical points into the function?
Thank you :) That helped a lot
Please do a video on conditional probability ,,,,
Teacher of the year!
instead of derive and checking endpoints end points in the invervall
-x^2+6x-7 , multiple with -1
x^2-6x+7 , use quadrule to write as (x-3)^2-9+7=(x-3)^2-2, now we see that minus is -2 and max is (5-3)^2-2=4-2=2
nice vid and you explain it really well, and i get this stuff, but damn these are the easy parts, you should have put world problems like applied max and min values, like find the dimensions of a cone and maximum volume that can be fit into a sphere of diameter 6cm
can we have an example of an f(x,y,z) function and how to determine the nature of the critical point. I dont know if you use the Hessian or what can you show an example. And if we do know how to put the hessian matrix together then how do we use it to find out if it is a saddle max or min?
When should you use the second derivative test? - why was it not used here to determine if they were max or mins?
You use the second derivative test to find *local* extrema (maximum if hill, minimum if valley, neither if saddle). To find absolute extrema, this information isn't strictly needed because you can just evaluate at all the candidate points (critical points and boundary) and just pick the largest and smallest value.
can you please explain whats the point of pluging in the critical point into the original function?
Extrema occur at the critical points or on the boundary. You must check them all since any one of them could be a maximum or a minimum. If you ignore a critical point, you might miss a maximum or minimum.
What should you do if your critical points are not within the boundaries?
Hey Patrick! I noticed when you took the case x=1, and got -2 as a y, I noticed that that point was outside the boundary… a.k.a the triangle. With that, are we still allowed to use that point as a max or min even if it’s outside the boundary?
Y is not -2, function value is -2. Y was 4 there
I dont follow why the -x+5 section had to have its derivative taken.
were there no critical points in the other two? (I know the third equation looked harder but how do u know if/when to derive it and when there will be an additional critical point in there?)
good job. well made
Thank you sir :)
Now I got it ty m8
7:59 Sir Plz guide me, u have sorted out the critical piont x=3,shouldnt be its y component there,
have to write my symster paper on 24th may please help
Athar
(Pakistan)
and thanks for the videos,luv:-)
nope, u r only interested in what the max/min values are and don't need to state where. as u can see at end of vid he shows z = 2,-2. But if u want u can add it's y component not gona change anything
Could you do an example where the Domain is something like 0
Thanks so much
thank you thank you thank you :)
you are an angel
If not already you should apply for a calculus professor at Harvard!
What's is the point of finding the critical point in this case?
To see if there's a greater or smaller value. If f(critical poin) were -3 then the absolute min value would be -3.
Thanks a lot
Thank you!!!
Thanks
Thankw so much its really helpfull....
You sir, are the fuckin' man.
sir can you please tell that what are the extreme values in this question???????. plz reply
Am I stupid or shouldn't it be y=-x+4 because you shift it up by 4 NOT 5???
@graciemaltry gla i could help you : )
Thanks!
ur a life saver :')
So what points f(x,y) correspond with the global min and Max?
i am in the exam room 😂😂😂
how did we find x from 2x=6 ??? i couldn't get it
just devide both sides by 2 and you get 3, why are you watching a global extrema video while that confuses you
thank you
omg thank you
You are like the GOD OF CALCULUS, you should definitely think about becoming a teacher .(if you're not already one)
If I sent you $5 every time you saved my GPA would you buy a Lambo or a Ferrari?
Thxxx alot
Can someone plz explain, why we have to apply Ist derivative test after substituting y= -x +5 in original function?
i'll try.. the other two equations were obvious. Once we plug in the lowest and highest, it gives an easy result. However, when we substituted y=-x+5, it was not so simple. Differentiating will give quicker answer, whereas plugging each values to find the number will take longer (you can do that though, if you want).
Jin Choi Thanks! It makes sense
fucking love you dudeeee thanks a lot
could we solve this using an easier method?
Why do we need to learn this?
@ExtremeEngineering91 He needs to show more wrist!
goat
I
this is tedious paprick
God i hate Calc 3...