Thanks Prof! This brings back so many great memories of my days in 1st year Maths. It's also really nice to have a Prof who calls you 'my friends' and doesn't treat you like an idiot like my lecturer did. If all your videos are like this I look forward to many wonderful and informative hours - cheers!
Thank you professor Leonard for actually explaining math in ways that makes it way easier to understand. Only wish I could have you as my actually professor.
The reactions of the class are so wholesome. My professor just post long lectures of only him in front of a white board, not bad by any means. But something about the reactions of the class really helps me, I think people need to learn socially.
Thank you for your lectures, I finished watching your Calculus 1 Lectures, now I'm on my way to finish Calculus 2 with you. Do you teach Linear Algebra and Multivariable calculus? I'm waiting for them.
@@charmendro does it really matter? The math doesn't really age. This playlist and Cal 1 is still worth watching in 2021 and has helped me in my studies.
@@davida99 ikr I too am watching it in 2021 and for the time being none of the ciriculum has changed even if it did prof leonard has included everything taught in calculus as of now
A serious conversation in Calculus II class today: Why are we paying thousands of dollars in tuition if we're all learning the material from Professor Leonard on RUclips? Something is broken in the education system. Luckily a solution can be found on RUclips with Professor Leonard.
For whatever reason we're taking this in our calculus 1 course, I haven't even been in college for a month and we're taking this.. and I'm still trying to figure calculus out and try not to be depressed from feeling really dumb. I'm really grateful for people like you because I almost don't understand anything in my lectures. Still lost I know a lot of basics are lost for me but for now they don't even give us the opportunity to try and learn that, always quizzes and stuff (from the second week) so I kinda have to try and understand the topics we're taking and it's so difficult :( It's all very overwhelming and I'd have probably changed what I'm studying if I didn't know it would help me reach something I really want
53:25 x−√x^2−1 won't be negative actually The reason why we can dispose of the minus sign is that, firstly, (x+√x^2−1)^(-1) = (x−√x^2−1) Therefore ln(x±√x^2−1) = ln(x+√x^2−1)^(±1) = ± ln(x+√x^2−1) But arcosh(x) is defined as non-negative, so the minus sign is not required
Professor Leonard truly is an amazing teacher - he is organised, he is relaxed, he is humble and helpful, he interacts a lot with the students, he erases EVERYTHING on the board without leaving any ugly dots, and he asks "someone" to answer his questions instead of forcing a specific student to do it.
Sir Leonard, at 40:24 we can actually use only 1 substitution if we choose the inside part of the function cosh(3x).. the output would be the same = 1/9*(cosh(3x)^3). Thank you so much. You are the best teacher I've ever seen.
thank you very much professor!! ive been stressing about hyperbolic funtions for days now and you have definitely helped me prepare for my exam in 3 days:] thank you very much and i will subscribe to your channel. keep up the great work!!
Honestly I love both, I watch Professor Leonard’s lectures to understand the math behind the theorems and really gain a deep understanding of the topic, and then I watch the OCT to practice solving problems so get the hang of the steps I need to use.
As a point of correction/clarification, when deriving the logarithm formula for inverse cosh, Professor Leonard states that the reason we need to take the positive root is that otherwise we'd get negatives. This isn't quite right, in fact the formula will still always be positive. The reason we need to take the positive root is that cosh isn't a one-to-one function, so we need to restrict the domain to find an inverse - the restriction we choose is to take cosh(x) where x≥0. The result of this is that we require the range of our inverse to be y≥0, so e^y≥1. Going back to the quadratic formula, choosing the negative root gives e^y between 0 and 1, so we must choose the positive root, which gives e^y≥1.
Thank you for this clarification, I was wondering about his explanation because he mentions 5 - sqrt(24) being negative but that gives a positive number.
Professor Leonard thank you for a solid Discussion of Hyperbolic Functions in Calculus Two. Hyperbolic Functions and Inverse Hyperbolic Functions are not difficult to understand however they can be problematic. An application of Hyperbolic Function is a hanging cable between two towers. The hanging cable makes a shape called a Catenary. An example of a Catenary is the Gateway Arch in St. Louis, Missouri. Professor Leonard, you are missing equal signs in your final example in this video. Please correct this small error in the video.
At 52:32 you mentioned that if you plug in 5 for x, you would get 5 minus the square root of 24. You said that you will get a negative value for e^y but I used my calculator in fact I got 0.101 which is positive. Please clarify thanks
yes, but for x-(x^2-1)^1/2 is always less than 1. so e raised to y is less than one. it is only possible if y is less than 0. Now cosh inverse x=y and x:[1,infinity) so cosh y =x where x:[1,infinity) now put value of cosh y that is e^x + e^-x whole over 2 and equate it to x so as to get minimum value of range of y. The minimum value you get is zero. So you cannot have y less than zero. but in x-(x^2-1)^1/2 , we get y less than 0. thats why this isn't right.
Sakaleshwar Singh thanks for the explanation I totally forgot that it has to be greater than 0. professor Leonard didn’t mention that but I guess we should know that by now!
I find it hilarious that on UND's online courses they just send you links to Professor Leonard. He got me through CLEP and looks like calc II is just the same thing
I love your content! At (41:41), you said that all hyperbolic trig functions are one-to-one. I'm certain you know that cosh(x) is the exception to this. Not sure if you correct it later on because I haven't watched that far yet. Would you mind adding a video note that cosh(x) is the exception at (41:41)? Thanks again for your amazing videos!
My best guess is that it is like the inverse trig functions where we have to limit the domain of the original trig function to get the inverse. In the case of cosh, the limited domain will be [0,∞) therefore the range of cosh^-1(x) will be [0,∞) I guess he doesn't say it and expects us to get that.
my doctor should of played your videos in class instead of his stupid explaining.Thank you very much, Professor Leonard for all of your great work . If you ever came to Saudi Arabia I'll be honored to have you as my guest sir.
At 49:43 where the quadratic is introduced there is suddenly a ey on the left hand side of the equation. Why? Gunter, Bonn, Germany, ardent follower of your excellent course!
I want to say thank you for the videos Professor Leonard. I haven't taken calculus ever since 2003-04 but Ive been using your videos to refresh my memory and have been helping me a lot. On the hyperbolic functions, around 6:50, I think you made a little mistake there. sec^2h (x), should it be sech^2 (x)? Great teacher though.
Actually, the derivative of tanh^(-1)(x) requires |x|1. Although their derivatives' formulae literally looks the same, they have different restrictions.
At about the 31:40 mark, Prof. Leonard overlooks an opportunity to use the 2sinh(x)cosh(x) = sinh(2x) identity to really clean up his answer. (The arguments to these functions were actually more complicated than x and 2x, but I didn't want to type them out in full here.)
At about 13:00, I'm not following the algebra. Wouldn't multiplying by 2/2 still result in a coefficient of 1/2 after reduction? You're multiplying 2/2 by 1/2, giving 2/4. Am I overlooking something?
@Christopher Williams we had the equation in form of : (e^x-e^-x/2)*(e^x+e^-x), and we mutiplied it by 2/2, and it gave us the equation in the form of: (e^x-e^-x/2)*(e^x+e^-x/2)*2. We know sinhx and coshx by definition are (e^x-e^-x/2) and (e^x+e^-x/2), respectively. Now, we have sinhx*coshx*2, and by bringing the 2 in front we have the identity "sinh2x = 2sinhx*coshx.
Hey Professor. I am a high school student from South Africa - in fact I just finished school about a week ago - and I've been really eager to learn more about maths as it gives me great joy, but I've come across a hurdle that I hope you can help me with (or anyone in the comments): In school, we learn about a function called a hyperbola in the form y = a/(x+P) + q, but upon researching sonic sections with the purpose of learning about them, I stumbled across another hyperbola function, as portrayed in your videos, that look vastly different, are essentially are, from the one I was taught in schoo. Could you perhaps inform me of the difference between the two? I am so confused and would really like to expand my knowledge. I'm not sure how they correlate with one another, despite possessing the same name.
You're talking about a hyperbolic function moved by a vector [p,q]. Now to answer you're question the function y=a/(x-p) + q or otherwise written is a special case of a hyperbolic function called a homography (The graph of it is called a orthogonal hyperbola). It can be written using the hyperbolic trygonometric functions, but does not include the entire set of all hyperbolic functions.
hey i think you are a great math professor my name is arthur adams and I had a rough start to my education especially with math because of my poor algebra skills. I genuinely love calculus and I have a fast paced calc 2 class at valdosta state university im taking in june for the summer your videos are extremely helpful although i need to be prepared for applications of integrals do you happen to have lesson on that ?
Amazing way of teaching I wish all teachers taught like this! Hey what book are you using? I see this chapter 6.6, but 6.6 in what book exactly? I would love it if I could buy it and follow along.
hi prof, I think you meant it's like the inverse trigonometric functions not like the HYPERBOLIC trigonometric functions when you're on the inverse hyperbolic function part..
cosh^2(x)-sinh^2(x)=1 That is just the regular Pythagorean theorem. The reason it's a minus is because there is a top secret 'i' in the sinh function. Same with the relativity stuff. We doing this wrong.
amazing teacher he does make it a lot easier but just quick question can you solve hyperbolic functions with just those formulas in the beginning or did I miss a class or somptin like that#home schoole problems
(Another way to prove the identity at 14:02 is by using the double angle sum> sin(α + β) = sin(α) cos(β) + cos(α) sin(β) (and substituting (x) instead of B
Professor Leonard: "Tell me the first thing I'm gonna do..." Me: "CHAIN RULE!" P.L.: "What do we do with this?" Me: "CHAIN RULE!" P.L.: "What is your first step?" Me: "CHAIN RULE!" All the other people in the coffee shop where I'm studying: 0_0 o_o O_O
Yeah the quotient rule still works its just faster the way professor Leonard explained it example you wouldn't do the quotient rule on x^2/1 you would just say 2x Even though doing the quotient rule will get you the same answer its just more convenient to use the power rule instead Hoped this helped
At 38:46 I picked "w" as a different letter before he said it OMG I can read minds. that or I can think like a mathematician from how much math videos I see LOL.
I get stuck when I have my du^3 and I don’t know how to get rid of the exponent. I never have the problem when I’m working with my tutor/in class (steered in the right direction) but on exams I keep getting stuck. When you manipulate the du to get the function rewritten are there any tips for then being able to still get rid of the du upon integration?
+Claudio Ferrer When he says that most numbers will result in the expression being negative, I think he was describing the derivation of the Sinh inverse function by mistake i.e. 5 minus the square root of (5 squared + 1), which would be negative. Cosh has a concave up shape and is not a one to one function. Therefore in order to derive a valid inverse function, the domain of Cosh needs to be restricted to positive values only. This corresponds to the range of its inverse having only positive values and, I think the reason why the positive square root term is used.
Exactly as RichOx100 said. ln[x+-sqrt(x^2-1)] is not even a function. What Professor means (I guess) is that we need y>0 -> Exp[y]>1->only x+sqrt(x^2-1) can be greater than 1. x-sqrt(x^2-1) is allways positive and always less than 1 in [1,inf)
Loving the lecture, one part caught my eye though and it's bothering me now: When proving that cosh^-1(x) is in fact ln[x+sqrt(x^2-1)], at 52:06 you mention that the domain of what we're working with is the domain of cosh^-1(x), but since we changed cosh^-1(x) into its inverse cosh(x), shouldn't we be using cosh(x)'s domain? Because cosh inverse's domain is actually cosh(x)'s range... If someone could clarify i'd be thankful!
No, cos^-1(x) is not reciprocal of cos(x), it is its inverse. So range becomes domain and domain becomes range. The range of cos(x) = [-1,1] so [-1,1] becomes domain of cos^-1(x) and vice versa. Similar argument can be put for cosh.
We aren't using the domain of cosh(x) we are using the domain of cosh(y) which is the same as cosh^-1(x) which says in the rule that cosh^-1(x)=y exists if and only if cosh(y)=x exists.
The explanation at 52:40 (about discarding the negative part) is not clear. For example x-√(x²-1) < 0 is not true for all x∈[1,+∞) Try x = 1.5 1.5 - √((1.5)² -1) = 1.5 - √(2.25 -1) = 1.5 - √1.25 = 1.5 - 1.118... >0
Professor Leonard Hi.... Sir can i ask you? best book for hyperbolic functions. i want to practice with examples. i need to preparation but i don't know from waht book do i prepare plz suggest me the book name anyone else tell me the name of book? sorry for my bad engish.....
Very common design for arches, because it economizes the use of material by keeping the load in the structure in either pure compression for arches and domes, or pure tension for hanging cables. Before people knew the analytic solution to this curve, they would build models upside-down with hanging cables and hanging fabric, and then studying the upside-down tensile shape to design the right-side-up domes and arches.
This sinch thing doesn't really work, does it? Much better is shin(e), cosh, than, coth, coshec, shec, et cetera. Just put the 'h' with the 's', unless there isn't one, then put it with the 't' instead. That way you do not need to add any random letters and change the sound each time, and then find you cannot do it in some cases anyway. Enjoy.
Professor Leonard, how does it feel to know that you're the best educator on this planet right now?
Facts
@@samuelcook7588 love it or hate it he is spiting straight facts.
@Charles Misael Stop the spam.
@Dee Macbeth who fucking cares about your girlfriend issues, we are here to learn the language of gods, which is math!
So far my top fans are Leonard and nancypi
I find myself raising my hand in front of my computer on occasion when he asks "Show of hands if you're with me"
me too.lol
Well i guess you‘re not the only one
Well, you're not alone
Professor Leonard, I keep raising my hand! You never pick me lol
Thanks Prof! This brings back so many great memories of my days in 1st year Maths. It's also really nice to have a Prof who calls you 'my friends' and doesn't treat you like an idiot like my lecturer did. If all your videos are like this I look forward to many wonderful and informative hours - cheers!
I am glad that the class says yes to Leonard when he asks: do you want to see a proof.
What is Math without proofs??!!
Thank you professor Leonard for actually explaining math in ways that makes it way easier to understand. Only wish I could have you as my actually professor.
The reactions of the class are so wholesome. My professor just post long lectures of only him in front of a white board, not bad by any means. But something about the reactions of the class really helps me, I think people need to learn socially.
Thank you for your lectures, I finished watching your Calculus 1 Lectures, now I'm on my way to finish Calculus 2 with you. Do you teach Linear Algebra and Multivariable calculus? I'm waiting for them.
i think he may have mutlivariable, but this is 4 years old sorryy
@@charmendro oof
@@charmendro does it really matter? The math doesn't really age. This playlist and Cal 1 is still worth watching in 2021 and has helped me in my studies.
@@davida99 ikr I too am watching it in 2021 and for the time being none of the ciriculum has changed even if it did prof leonard has included everything taught in calculus as of now
@@davida99 he's saying the comment is 4 years old, not the video
You’re a hero :) u made my engineering life so much easier right now
Certainly, you are the best math teacher of all time.
A serious conversation in Calculus II class today: Why are we paying thousands of dollars in tuition if we're all learning the material from Professor Leonard on RUclips? Something is broken in the education system. Luckily a solution can be found on RUclips with Professor Leonard.
For whatever reason we're taking this in our calculus 1 course, I haven't even been in college for a month and we're taking this.. and I'm still trying to figure calculus out and try not to be depressed from feeling really dumb.
I'm really grateful for people like you because I almost don't understand anything in my lectures.
Still lost
I know a lot of basics are lost for me but for now they don't even give us the opportunity to try and learn that, always quizzes and stuff (from the second week) so I kinda have to try and understand the topics we're taking and it's so difficult :(
It's all very overwhelming and I'd have probably changed what I'm studying if I didn't know it would help me reach something I really want
He's like the chadest and smartest of math professors out there
Happy new year professor, during Covid, your lectures saved a lot of university students.
Meanwhile we are learning Hyperbolic fnctn just to clear an entrance exam in class 11th:)
Best explanation ever!!!! if you agree 👍
53:25
x−√x^2−1 won't be negative actually
The reason why we can dispose of the minus sign is that, firstly,
(x+√x^2−1)^(-1) = (x−√x^2−1)
Therefore
ln(x±√x^2−1) = ln(x+√x^2−1)^(±1) = ± ln(x+√x^2−1)
But arcosh(x) is defined as non-negative, so the minus sign is not required
Thank you so much. The title of this video sounds scary, but you made it really easy and fun. I love your videos and your smile :)
Professor Leonard truly is an amazing teacher - he is organised, he is relaxed, he is humble and helpful, he interacts a lot with the students, he erases EVERYTHING on the board without leaving any ugly dots, and he asks "someone" to answer his questions instead of forcing a specific student to do it.
Thank you so much prof Leonard, today I have passed my calculus -1 with best grade because of you
Sir Leonard, at 40:24 we can actually use only 1 substitution if we choose the inside part of the function cosh(3x).. the output would be the same = 1/9*(cosh(3x)^3). Thank you so much. You are the best teacher I've ever seen.
1/9*(cosh(3x)^3) + C :)
thank you very much professor!!
ive been stressing about hyperbolic funtions for days now and you have definitely helped me prepare for my exam in 3 days:] thank you very much and i will subscribe to your channel.
keep up the great work!!
I’m here because @theorganichemistrytutor didn’t make a video on hyperbolic functiona
Jan Angel same 🤣
LMAOOO SAMEE
Honestly I love both, I watch Professor Leonard’s lectures to understand the math behind the theorems and really gain a deep understanding of the topic, and then I watch the OCT to practice solving problems so get the hang of the steps I need to use.
@@louloucs55 Professor Leonard is the goat for conceptualizing the math
@@louloucs55I see someone's discovered the meta
shout out to the greatest calculus teacher of all time after Newton
so clear explanation...thank you very much Sir for your video
Superheroes do exist in this planet. Thank you professor
As a point of correction/clarification, when deriving the logarithm formula for inverse cosh, Professor Leonard states that the reason we need to take the positive root is that otherwise we'd get negatives. This isn't quite right, in fact the formula will still always be positive.
The reason we need to take the positive root is that cosh isn't a one-to-one function, so we need to restrict the domain to find an inverse - the restriction we choose is to take cosh(x) where x≥0. The result of this is that we require the range of our inverse to be y≥0, so e^y≥1.
Going back to the quadratic formula, choosing the negative root gives e^y between 0 and 1, so we must choose the positive root, which gives e^y≥1.
Thank you for this clarification, I was wondering about his explanation because he mentions 5 - sqrt(24) being negative but that gives a positive number.
Professor Leonard thank you for a solid Discussion of Hyperbolic Functions in Calculus Two. Hyperbolic Functions and Inverse Hyperbolic Functions are not difficult to understand however they can be problematic. An application of Hyperbolic Function is a hanging cable between two towers. The hanging cable makes a shape called a Catenary. An example of a Catenary is the Gateway Arch in St. Louis, Missouri. Professor Leonard, you are missing equal signs in your final example in this video. Please correct this small error in the video.
My final is in five days and I think I'm gonna come out of it alive cause of this hahah Thank you, so much love honestly
At 52:32 you mentioned that if you plug in 5 for x, you would get 5 minus the square root of 24. You said that you will get a negative value for e^y but I used my calculator in fact I got 0.101 which is positive. Please clarify thanks
yes, but for x-(x^2-1)^1/2 is always less than 1. so e raised to y is less than one. it is only possible if y is less than 0.
Now cosh inverse x=y and x:[1,infinity)
so cosh y =x where x:[1,infinity)
now put value of cosh y that is e^x + e^-x whole over 2 and equate it to x so as to get minimum value of range of y.
The minimum value you get is zero.
So you cannot have y less than zero.
but in x-(x^2-1)^1/2 , we get y less than 0.
thats why this isn't right.
Sakaleshwar Singh thanks for the explanation I totally forgot that it has to be greater than 0. professor Leonard didn’t mention that but I guess we should know that by now!
@@sakaleshwarsingh7868 Thanks! that makes sense
@@sakaleshwarsingh7868 amazing proof..
Thank you so much Professor Leonard! You are awesome.
you're the best!
This is great work
Interesting lecture professor.Your course has enriched my knowledge .I'm a 12th grade student, by the way
Hi I think that this videos have been very helpful to me and is very cool that you share these videos with students that that are not in you class
3:19 and I'm already lovin' this video.
I find it hilarious that on UND's online courses they just send you links to Professor Leonard. He got me through CLEP and looks like calc II is just the same thing
I love your content! At (41:41), you said that all hyperbolic trig functions are one-to-one. I'm certain you know that cosh(x) is the exception to this. Not sure if you correct it later on because I haven't watched that far yet. Would you mind adding a video note that cosh(x) is the exception at (41:41)? Thanks again for your amazing videos!
could you pls tell me how cosh is not 1 to 1
@@mssafy2592 Cosh(x) is an even function and by that fails the horizontal line test. For all real x, cosh(-x) = cosh(x).
My best guess is that it is like the inverse trig functions where we have to limit the domain of the original trig function to get the inverse. In the case of cosh, the limited domain will be [0,∞) therefore the range of cosh^-1(x) will be [0,∞) I guess he doesn't say it and expects us to get that.
Feels like I'm in a proper math class :D
my doctor should of played your videos in class instead of his stupid explaining.Thank you very much, Professor Leonard for all of your great work . If you ever came to Saudi Arabia I'll be honored to have you as my guest sir.
this is a part of our jee advanced syllabus , just the difference is ; we aren't taught the names.
YOUR lectures really helped me a lot
At 49:43 where the quadratic is introduced there is suddenly a ey on the left hand side of the equation. Why? Gunter, Bonn, Germany, ardent follower of your excellent course!
Love how he says "nasty piece of junk" about the equations, lol
I often hear him say "this crap", lol.
I want to say thank you for the videos Professor Leonard. I haven't taken calculus ever since 2003-04 but Ive been using your videos to refresh my memory and have been helping me a lot. On the hyperbolic functions, around 6:50, I think you made a little mistake there. sec^2h (x), should it be sech^2 (x)? Great teacher though.
Thanks for the wonderful video
Actually, the derivative of tanh^(-1)(x) requires |x|1. Although their derivatives' formulae literally looks the same, they have different restrictions.
At about the 31:40 mark, Prof. Leonard overlooks an opportunity to use the 2sinh(x)cosh(x) = sinh(2x) identity to really clean up his answer. (The arguments to these functions were actually more complicated than x and 2x, but I didn't want to type them out in full here.)
عظيم
At about 13:00, I'm not following the algebra. Wouldn't multiplying by 2/2 still result in a coefficient of 1/2 after reduction? You're multiplying 2/2 by 1/2, giving 2/4. Am I overlooking something?
@Christopher Williams we had the equation in form of : (e^x-e^-x/2)*(e^x+e^-x), and we mutiplied it by 2/2, and it gave us the equation in the form of: (e^x-e^-x/2)*(e^x+e^-x/2)*2. We know sinhx and coshx by definition are (e^x-e^-x/2) and (e^x+e^-x/2), respectively. Now, we have sinhx*coshx*2, and by bringing the 2 in front we have the identity "sinh2x = 2sinhx*coshx.
we can break up the 4 by 2*2
this is the best class ever wish I was there
U r saving my life
Well done Brandon…Mike
Hey Professor. I am a high school student from South Africa - in fact I just finished school about a week ago - and I've been really eager to learn more about maths as it gives me great joy, but I've come across a hurdle that I hope you can help me with (or anyone in the comments): In school, we learn about a function called a hyperbola in the form y = a/(x+P) + q, but upon researching sonic sections with the purpose of learning about them, I stumbled across another hyperbola function, as portrayed in your videos, that look vastly different, are essentially are, from the one I was taught in schoo. Could you perhaps inform me of the difference between the two? I am so confused and would really like to expand my knowledge. I'm not sure how they correlate with one another, despite possessing the same name.
You're talking about a hyperbolic function moved by a vector [p,q]. Now to answer you're question the function y=a/(x-p) + q or otherwise written is a special case of a hyperbolic function called a homography (The graph of it is called a orthogonal hyperbola). It can be written using the hyperbolic trygonometric functions, but does not include the entire set of all hyperbolic functions.
hey i think you are a great math professor my name is arthur adams and I had a rough start to my education especially with math because of my poor algebra skills. I genuinely love calculus and I have a fast paced calc 2 class at valdosta state university im taking in june for the summer your videos are extremely helpful although i need to be prepared for applications of integrals do you happen to have lesson on that ?
Another approach to proving most of those hyperbolic identities is through the use of complex numbers.
53:11 e^y > 0 but if you plug 1 to x then e^y = 0, so minus sign does not work. Hope this helps!
thanks mr. leonard
Amazing way of teaching I wish all teachers taught like this! Hey what book are you using? I see this chapter 6.6, but 6.6 in what book exactly? I would love it if I could buy it and follow along.
hi prof, I think you meant it's like the inverse trigonometric functions not like the HYPERBOLIC trigonometric functions when you're on the inverse hyperbolic function part..
you are the best of the best
cosh^2(x)-sinh^2(x)=1
That is just the regular Pythagorean theorem. The reason it's a minus is because there is a top secret 'i' in the sinh function. Same with the relativity stuff. We doing this wrong.
After the Last few Videos i realized i have to watch some videos about how to memorize things quickly
But you could just substitute the inside of (cosh(2x))^2, right? Only one substitution. These videos are priceless, thank you. 38:00
choosing cosh(2x) for u that is
amazing teacher he does make it a lot easier but just quick question can you solve hyperbolic functions with just those formulas in the beginning or did I miss a class or somptin like that#home schoole problems
Thanks again!
(Another way to prove the identity at 14:02 is by using the double angle sum> sin(α + β) = sin(α) cos(β) + cos(α) sin(β)
(and substituting (x) instead of B
lmao, the subtitles at 9:30 say "I'm tired of riding to Mars".
1:04:19 I love this. Great Work!
Thank you sir thanks🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹❤❤❤❤🌹🌹🌹🌹🌹🌹❤❤❤❤❤❤
Professor Leonard: "Tell me the first thing I'm gonna do..."
Me: "CHAIN RULE!"
P.L.: "What do we do with this?"
Me: "CHAIN RULE!"
P.L.: "What is your first step?"
Me: "CHAIN RULE!"
All the other people in the coffee shop where I'm studying:
0_0 o_o O_O
Leonard, at 39:00 can't you just use a single substitution for cosh(3x)?
17:00 I tried the quotient rule and it gave the right answer
Yeah the quotient rule still works its just faster the way professor Leonard explained it
example you wouldn't do the quotient rule on x^2/1
you would just say 2x Even though doing the quotient rule will get you the same answer its just more convenient to use the power rule instead
Hoped this helped
When doing the problem listed at about 40:00, he could have used u = cosh(3x) instead and still going the correct answer.
At 38:46 I picked "w" as a different letter before he said it OMG I can read minds.
that or I can think like a mathematician from how much math videos I see LOL.
I get stuck when I have my du^3 and I don’t know how to get rid of the exponent. I never have the problem when I’m working with my tutor/in class (steered in the right direction) but on exams I keep getting stuck. When you manipulate the du to get the function rewritten are there any tips for then being able to still get rid of the du upon integration?
5 minus the Square root of 24 = 0.10102 positive at 52:43. Why does he say it's negative?
+Claudio Ferrer
When he says that most numbers will result in the expression being negative, I think he was describing the derivation of the Sinh inverse function by mistake i.e. 5 minus the square root of (5 squared + 1), which would be negative.
Cosh has a concave up shape and is not a one to one function. Therefore in order to derive a valid inverse function, the domain of Cosh needs to be restricted to positive values only. This corresponds to the range of its inverse having only positive values and, I think the reason why the positive square root term is used.
Exactly as RichOx100 said. ln[x+-sqrt(x^2-1)] is not even a function. What Professor means (I guess) is that we need y>0 -> Exp[y]>1->only x+sqrt(x^2-1) can be greater than 1. x-sqrt(x^2-1) is allways positive and always less than 1 in [1,inf)
since the domain is {1, infinity / x-sqrt x^2-1>1 ......> - sqrt x^2 - 1 >1-x ....>
sqrt (x^2-1) square both side of the inequality....> (x^2-1) x+1 1
Loving the lecture, one part caught my eye though and it's bothering me now: When proving that cosh^-1(x) is in fact ln[x+sqrt(x^2-1)], at 52:06 you mention that the domain of what we're working with is the domain of cosh^-1(x), but since we changed cosh^-1(x) into its inverse cosh(x), shouldn't we be using cosh(x)'s domain? Because cosh inverse's domain is actually cosh(x)'s range... If someone could clarify i'd be thankful!
No, cos^-1(x) is not reciprocal of cos(x), it is its inverse. So range becomes domain and domain becomes range. The range of cos(x) = [-1,1] so [-1,1] becomes domain of cos^-1(x) and vice versa. Similar argument can be put for cosh.
We aren't using the domain of cosh(x) we are using the domain of cosh(y) which is the same as cosh^-1(x) which says in the rule that cosh^-1(x)=y exists if and only if cosh(y)=x exists.
you awesome man 😉
At 35:27; can you use an identity for sinh(2x)? Because my answer was; -4csc^2(cosh(2x))•sinhx•coshx.
thanks from my deep hearts
Funny, was checking out someone in the math center and then he said "look at the board"
31:52
couldn't you factor out 6T into 6T(2sinh(3T^2+1)*cosh(3T^2+1) and then it become 6T*sinh(6T^2+2) or i'm just missing something
Am I the only one who cracks up whenever he says "just for funsies" ?
With so many many formulas, how do i use mnemonic to memorise all? My stomach feel very pain....
a video on Laplace transformations please.
I have realized that anything that involves Cosh is the opposite of cos for example cos2x=cos^2x-sin^2xbut the cothx^2 has (+) instead
The explanation at 52:40 (about discarding the negative part) is not clear. For example
x-√(x²-1) < 0 is not true for all x∈[1,+∞)
Try x = 1.5
1.5 - √((1.5)² -1) = 1.5 - √(2.25 -1) = 1.5 - √1.25 = 1.5 - 1.118... >0
Great work...out?
If Superman taught math.
In which video do you cover the area of a region between two curves prof?
See his Calculus 1 playlist.
Professor Leonard Hi....
Sir can i ask you? best book for hyperbolic functions. i want to practice with examples. i need to preparation but i don't know from waht book do i prepare plz suggest me the book name
anyone else tell me the name of book?
sorry for my bad engish.....
CHAIN RULE!!!
I used this to sleep. works well.
Sir plz solve this problem,
Cosh1/2x=√1/2(1+coshx)
Are you working with Mr hegarty by any chance
❤❤❤
W video
is this what Americans learn in high school or would this be considered college level for you guys? just curious
I am Turkish but I know the answer to the question you asked, the things taught in this video are taught at university in America.
the saint Louis arch is a cosh function
Very common design for arches, because it economizes the use of material by keeping the load in the structure in either pure compression for arches and domes, or pure tension for hanging cables.
Before people knew the analytic solution to this curve, they would build models upside-down with hanging cables and hanging fabric, and then studying the upside-down tensile shape to design the right-side-up domes and arches.
41:24 Yeah, obviously.
This sinch thing doesn't really work, does it? Much better is shin(e), cosh, than, coth, coshec, shec, et cetera. Just put the 'h' with the 's', unless there isn't one, then put it with the 't' instead. That way you do not need to add any random letters and change the sound each time, and then find you cannot do it in some cases anyway. Enjoy.
he's a very handsome man