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If you take the negative root of the equation you just get cosh(-x+c) which is the same! It's great because this is exactly the use of the properties of the hyperbolic functions :D
There is a general solution which accounts for this as well as for y = 1. More specifically, y(x) = cosh([cosh^-1][y(0)] + [sgn(dy/dx)]*x). sgn(dy/dx) = -1 when choosing the negative square root, but using the sgn function is a more rigorous method to choose the root. You can also choose the positive rot for sgn(dy/dx) = 1. Finally, this even covers the situation of y = c for when dy/dx = 0. This is more general than y = +/- 1. You can solve for the allowed sgn(dy/dx) or y(0) by substituting into the original y = sqrt[1 + (dy/dx)^2]. You find that cosh([cosh^-1][y(0)] + [sgn(dy/dx)]*x = sqrt(1 + [sgn(dy/dx)]^2*sinh([cosh]^-1[y(0)] + [sgn(dy/dx)]*x)^2). If sgn(dy/dx) = 0, then simply y(0) = 1 by force, and otherwise, you simply obtain the hyperbolic cosine-sine identity of squares. This also requires the definite integral method rather than using the indefinite integral.
y=1 works, and it is a function of x, if you define it to be. (I mean, everything is a question of a definition, so defining y=f(x), where f(x) =1 fir all x gets you there). In fact BPRP is still taking this solution into account until he divides by sqrt(y^2-1) not discussing what happens when this equals 0. Well, it happens that you lose this so to speak trivial solution y=1.
The constant function y = 1 is a trivial solution; You can see this from graphical inspection, or by realising that the equations shown in the video ( that is, y = √( 1+ (dy/dx)^2 ) and the equivalent equation dy/dx = √(y^2 - 1) ) have a valid solution when dy/dx = 0; namely, y = 1.
Cosh and Sinh are just sums of exponential functions, cosh(x)=(e^x+e^-x)/2 and sinh(x)=(e^x-e^-x)/2. He could easily write the answer of the last integral as (e^5-e^-5-e^2+e^-2)/2 instead of 70.576. Also, he divided away the trivial solution where y is a constant (y=1).
Next Question : Find a function g(x) such that the perimeter of the region bounded by the x-axis, x = a , x = b and g(x) equals the area bounded by the perimeter.
Why is the Arc length nearly Zero? When you approximate dirac Delta by other functions (dirac Delta is just an distribution) yes, it is true, that the function is Zero almost everywhwre, but the Peak around Zero is still "infinitely" high... I am Not certain if you are correct about the Arc length, but IT doesnt seam that trivial too me... Greatings, Jonathan
Arc length of practical dirac delta isn't zero...it's 1/a where a is the duration of the signal..... What you are saying is actually an ideality over concept where the width a is taken as limit a->(tending to)0......and hence the magnitude jumps towards infinity which is shown by an arrow.... Practically,we have a width for delta function....and we use it extensively in communication where we sample the analog signals to digital form.... Ideal sampling uses concept of ideal delta function but the practical situation is different.
How can you tell the arc length is nearly zero? You must calculate the derivative of the function to know the arclength, but I’m fairly certain the resulting equation has no closed form solution.
You can do it without solving a DE "the normal way". You want a function f(x) such that, when some other function g(x) is applied to it's derivative, returns the original function, i.e. g(f'(x))=f(x). If there is a solution, it's going to be an exponential function. As others have mentioned 1=exp(0) is the most obvious solution. Then if you examine what g(x) actually does, it looks awfully like a distance function, or something out of a Pythagorean theorem. I actually paused the video and tried doing it in my head, but I misremembered the formula for arc length and guessed y=sin(x) or y=cos(x). That didn't make sense because the arc length is strictly increasing and trig function integrals go to zero after each period... The point is, if it was 1-f'(x)^2, the solution would be obvious to anyone right away. Since you have a plus, you want a function that changes it to a minus, so you need to add an imaginary unit somewhere in that function. You want a real function, so something like i*sin(x) is out of the question, but you can put an imaginary unit in the function and the derivative will bring it out, but that's the same as using hyperbolic trig functions. From there, it's easy to find y=cos(i*x)=cosh(x), because sinh(x) is negative for x
y=1 seems works, take a range from x=a to x=b, area under y=1 = line segment length = b-a But, be careful, line segment is different from arc. But, should line segment be a subset of arc? This is a good question to discuss.
To clear up a bit, the d from the c and d at the end should not be confused with the d in dx and dy. The d from the dx and dy is actually the lower case delta.
Of course you can. Take a reasonable arc length(if it's too steep it might not work.), e.g. the shape in the video. Now slide it up and down. It won't change the arc length, but it will change the area. So just slide it down or up until the values are equal.
What about letting y = 1? therefore dy/dx = 0 so area equals integral from a to b of 1, which is (b - a), arclength is integral from a to b of sqrt(1 + dy/dx) which evaluates to integral from a to b of 1, again being (b-a). So another solution is letting y = 1
What about non smooth solutions, where y' alternates between being equal to sqrt(y^2 - 1) and -sqrt(y^2 - 1)? we can have a wibbly wobbly function that still fulfills the criteria. Well, okay, FINE!, I guess differential arc length won't be defined properly at the interface points. But still... :)
From a to b there is a corresponding height. However these heights have created a length which is longer than a-b , this is why volume is always smaller than surface area. Each height's length is invisible to volume unlike area 👍
Please can you tell me why am i not getting the correct answer in the following ques. •Find the derivative of Tan^-1[√(x+1)/(x-1)] for |x|>1 The solution is -1/2|x|√x^2-1 Method to use- I know one method where you put x=Sec2theta.. and solve it But when I solved it directly using derivative of Tan inverse then chain rule ,I didn't get the |x|, can you tell me why... please.. If the question is non. Understandable,please tell me I will paste a drive Link containing a photo of this question.
Funny that practically no one noticed that feet and square feet are not the same measurement, although the idea was to concentrate on only the numbers. However, he should have disclosed this at the end since failure to match units has failed many a groundbreaking discovery!
I think I can make it work on any function (I am not sure, please point out if I am wrong). Just take the interval to be [a,a+dx] and both, the arc length and area, would be literally 0. Here dx has its usual meaning.
Why do the integrands have to be the same? two integrals with the same bounds being equal doesn't imply the integrands being equal right? for example: integral{0,1}x dx=integral{0,1}1/2 dx=1/2, even though x=/=1/2
@@blackpenredpen You missed a solution (y=1). The reason for that is when you had dy/dx = sqrt(y^2-1), you divided by zero (sqrt(y^2-1) is 0 for y=1), and got dy/sqrt(y^2-1) = dx, at 3:33.
I am in doubt about this part 2:03 . If two definite integrals are the same, does that necessarily imply both integrands are the same too? I mean, it could be a solution but why can't there be other solutions to that equation? Someone mentioned below another solution could be 1. How do we know there are not more function that satisfy the equality? My knowledge in differential equations is little.
I love how you have up on saying cosine hyperbolic fungion and such after one go and said co-sh, and si-sh. Hahaha great video. Loveg the halloween theme, scary stuff jumping around screen.
you have taken equal and solved it....we need to prove it equal and not take it from beginning.....it is wrong ..... please correct it in the next video
Can you say something to Integrals that can not be solved analytically? ...though i guess it might not be that much fun :)... I stumbled across: Integral( exp(x) x^4 /(exp(x)-1)^2 dx) from 0 to some x_m and it says it cant be solved, but seeing all ur nice integration tricks and stuff, I was wondering how u can say, that here there will not be any such tricks, but instead its just not analytically solvable. (This Integral comes up when calculating the specific heat of a solid in the Debye-Model)
I don't think that [integral of f]=[integral of g] (both in the same interval) implies f=g. It looks to me that you just found one solution, but you have not solved the original problem in general. Am I right?
f doesn't need to be the same as g. For example. Integral of 0 from 0 to 2pi has the same value as integral of sin(x) from 0 to 2 pi. But sin(x) isn't just 0 for all x on that interval.
but it's 2 different units, the question itself is a trick question. the same as: can a rectangle have the same area as its circumference? 2 different units, no real answer
0:24 I like that mathematical proof.
logic 101
I love the dramatic sound whenever hyperbolic functions are mentioned. Was never able to memorise what they're all about.
They are the even and odd parts of e^x.
Memorise? where's the problem? Just a simple combination of e-functions.
If you take the negative root of the equation you just get cosh(-x+c) which is the same!
It's great because this is exactly the use of the properties of the hyperbolic functions :D
MrQwefty yes : )
There is a general solution which accounts for this as well as for y = 1. More specifically, y(x) = cosh([cosh^-1][y(0)] + [sgn(dy/dx)]*x). sgn(dy/dx) = -1 when choosing the negative square root, but using the sgn function is a more rigorous method to choose the root. You can also choose the positive rot for sgn(dy/dx) = 1. Finally, this even covers the situation of y = c for when dy/dx = 0. This is more general than y = +/- 1. You can solve for the allowed sgn(dy/dx) or y(0) by substituting into the original y = sqrt[1 + (dy/dx)^2]. You find that cosh([cosh^-1][y(0)] + [sgn(dy/dx)]*x = sqrt(1 + [sgn(dy/dx)]^2*sinh([cosh]^-1[y(0)] + [sgn(dy/dx)]*x)^2). If sgn(dy/dx) = 0, then simply y(0) = 1 by force, and otherwise, you simply obtain the hyperbolic cosine-sine identity of squares. This also requires the definite integral method rather than using the indefinite integral.
Of course y = 1 works also.
That was my first thought as well
That's not a function of x. In fact, it's a function of zero variables.
Ione Manuel yes it is, y= 1+0x, it's integral can still be taken with respect to x
y=1 works, and it is a function of x, if you define it to be. (I mean, everything is a question of a definition, so defining y=f(x), where f(x) =1 fir all x gets you there). In fact BPRP is still taking this solution into account until he divides by sqrt(y^2-1) not discussing what happens when this equals 0. Well, it happens that you lose this so to speak trivial solution y=1.
@@reIONEre Constant function is still a function.
The constant function y = 1 is a trivial solution; You can see this from graphical inspection, or by realising that the equations shown in the video ( that is, y = √( 1+ (dy/dx)^2 ) and the equivalent equation dy/dx = √(y^2 - 1) ) have a valid solution when dy/dx = 0; namely, y = 1.
of course. you can always moce the function up or down to achieve that
AndDiracisHisProphet
that's the secret
oh no, i spoiled it!
*evilish my friday evening cocktail sip*
Yes, for a specific a and b. But with some function it works for any a and b. Like y=cosh(x+C) or y=1.
no, it always works for any a and b. of course the shift changes if you change a and b
Doing it negative:
y = cosh(-x+C)
and cosh(-x) = cosh(x)
We can simply write y=cosh(x+C)
You can get
Y=1/sqrt{1-e^{2x+c}}
if you use trigonometrical substitution instead of the hypervololic cosine
.
Please do not ignore me :)
Got u!
I will remember you and will always respond to you!!!
@@blackpenredpen
ohh boy, thanks :,3
Cosh and Sinh are just sums of exponential functions, cosh(x)=(e^x+e^-x)/2 and sinh(x)=(e^x-e^-x)/2.
He could easily write the answer of the last integral as (e^5-e^-5-e^2+e^-2)/2 instead of 70.576.
Also, he divided away the trivial solution where y is a constant (y=1).
Next Question : Find a function g(x) such that the perimeter of the region bounded by the x-axis, x = a , x = b and g(x) equals the area bounded by the perimeter.
I'll just put down y, and you'll see y (why) lol 1:20
4:29 and let me just put down + c, and you'll c this c
... and know you c
Dirac Delta function has Area= 1. But Arc length nearly Equal to 0.
This is an interesting video. Loved it.
Why is the Arc length nearly Zero? When you approximate dirac Delta by other functions (dirac Delta is just an distribution) yes, it is true, that the function is Zero almost everywhwre, but the Peak around Zero is still "infinitely" high... I am Not certain if you are correct about the Arc length, but IT doesnt seam that trivial too me...
Greatings, Jonathan
„function“
Arc length of practical dirac delta isn't zero...it's 1/a where a is the duration of the signal.....
What you are saying is actually an ideality over concept where the width a is taken as limit a->(tending to)0......and hence the magnitude jumps towards infinity which is shown by an arrow....
Practically,we have a width for delta function....and we use it extensively in communication where we sample the analog signals to digital form....
Ideal sampling uses concept of ideal delta function but the practical situation is different.
How can you tell the arc length is nearly zero? You must calculate the derivative of the function to know the arclength, but I’m fairly certain the resulting equation has no closed form solution.
Judging from the music, this is a great video to watching on Halloween! 😂
Very nice video. Only problem is lighting glare on whiteboard in some spots.
You can do it without solving a DE "the normal way". You want a function f(x) such that, when some other function g(x) is applied to it's derivative, returns the original function, i.e. g(f'(x))=f(x). If there is a solution, it's going to be an exponential function. As others have mentioned 1=exp(0) is the most obvious solution. Then if you examine what g(x) actually does, it looks awfully like a distance function, or something out of a Pythagorean theorem. I actually paused the video and tried doing it in my head, but I misremembered the formula for arc length and guessed y=sin(x) or y=cos(x). That didn't make sense because the arc length is strictly increasing and trig function integrals go to zero after each period... The point is, if it was 1-f'(x)^2, the solution would be obvious to anyone right away. Since you have a plus, you want a function that changes it to a minus, so you need to add an imaginary unit somewhere in that function. You want a real function, so something like i*sin(x) is out of the question, but you can put an imaginary unit in the function and the derivative will bring it out, but that's the same as using hyperbolic trig functions. From there, it's easy to find y=cos(i*x)=cosh(x), because sinh(x) is negative for x
I love these dramatic sound effects
Love 2:30 when you move the camera up
I see that you've discovered sound effects :))
That background music....just wow....You are a mathegangster sir....awesome 😉👨🏫
y=1 seems works, take a range from x=a to x=b, area under y=1 = line segment length = b-a
But, be careful, line segment is different from arc.
But, should line segment be a subset of arc? This is a good question to discuss.
At what point in calculus will you need/find it convenient to know and remember hyperbolic trig functions?
Calculus 2 and 3
To clear up a bit, the d from the c and d at the end should not be confused with the d in dx and dy. The d from the dx and dy is actually the lower case delta.
Thank you so much sir...
Please upload these type of concept clearing topics...❤❤❤from india
Super simple, f(x) = 1, but the more difficult solution was super fun to solve for! Thanks for another puzzle
Andrew Thomas thanks for watching as well!!
does two integrals being equal mean that he integrands are equal?
Of course you can. Take a reasonable arc length(if it's too steep it might not work.), e.g. the shape in the video. Now slide it up and down. It won't change the arc length, but it will change the area. So just slide it down or up until the values are equal.
Physically, won't it be incorrect to equate two quantities with different dimensions? Can't wrap my fat head around it 😖
They're just measures of length and area, not length and area ! ;)
So it's ok to compare them
numerically equal
They do not have different dimensions. Dimensionless quantities can be made from every physical quantity.
This is math not physics
I can weigh 100 lbs. but can also be 100cm tall
What about letting y = 1? therefore dy/dx = 0 so area equals integral from a to b of 1, which is (b - a), arclength is integral from a to b of sqrt(1 + dy/dx) which evaluates to integral from a to b of 1, again being (b-a). So another solution is letting y = 1
Keith Allatt that works too!
In fact that the missing solution from my general solution. You can find it by letting sqrt(y^2-1) be 0. Since that is in the denominator.
Y =sin(x) + c from 0 to 2 pi is another. Choose c so c*2 pi equals the arc length.
considering the level curve f(x,y)=c where y=cosh(x+c) you find at c=-x you have that y=1 considering y=cosh(x+c)=(e^(x-x)+e^-(x-x))/2=2/2=1
Please work on zeta function....ramanujan's work, reimann hypothesis
What about non smooth solutions, where y' alternates between being equal to sqrt(y^2 - 1) and -sqrt(y^2 - 1)? we can have a wibbly wobbly function that still fulfills the criteria. Well, okay, FINE!, I guess differential arc length won't be defined properly at the interface points. But still... :)
Obviously f(x) = 1 also is a solution to the problem, but the solution doesnt seem to allow other answers than cosh(x). What am i missing here?
Negative version y= cosh(-x+c). This is the same function
What’s with all the creppy music at the beggening . BTW love your channel
I’m wondering if this is related to how cosh models a hanging chain
From a to b there is a corresponding height. However these heights have created a length which is longer than a-b , this is why volume is always smaller than surface area. Each height's length is invisible to volume unlike area 👍
You didn't deduct the most obvious case: y=1+0x.
Area[2,5]=3
ArcL[2,5]=3
Please can you tell me why am i not getting the correct answer in the following ques.
•Find the derivative of
Tan^-1[√(x+1)/(x-1)] for |x|>1
The solution is -1/2|x|√x^2-1
Method to use- I know one method where you put x=Sec2theta.. and solve it
But when I solved it directly using derivative of Tan inverse then chain rule ,I didn't get the |x|, can you tell me why... please..
If the question is non. Understandable,please tell me I will paste a drive Link containing a photo of this question.
It could have the same numeric value, but it can't have the same dimensions...length is in length units, area is in (length units) squared.
Funny that practically no one noticed that feet and square feet are not the same measurement, although the idea was to concentrate on only the numbers. However, he should have disclosed this at the end since failure to match units has failed many a groundbreaking discovery!
I think I can make it work on any function (I am not sure, please point out if I am wrong). Just take the interval to be [a,a+dx] and both, the arc length and area, would be literally 0. Here dx has its usual meaning.
Where's the video about deriving arc length formula?
cool audio effects
Why do the integrands have to be the same? two integrals with the same bounds being equal doesn't imply the integrands being equal right? for example:
integral{0,1}x dx=integral{0,1}1/2 dx=1/2, even though x=/=1/2
I found the solution using another way.. my answer was: (y + sqrt(y²-1)) = e^x
is it the same thing or another fuction that satisfies the condition?
Keep up the great work!
Oscar Troncoso thanks
@@blackpenredpen
You missed a solution (y=1).
The reason for that is when you had dy/dx = sqrt(y^2-1), you divided by zero (sqrt(y^2-1) is 0 for y=1), and got dy/sqrt(y^2-1) = dx, at 3:33.
1:14 "Let me put a "y" right here *and you will see **_y_* "
Y=1?
Y= -1
Damn the that dramatic music
y=1 feels discriminated here, so is y=0 when people talk about functions whose derivative and antiderivative is itself
I am in doubt about this part 2:03 . If two definite integrals are the same, does that necessarily imply both integrands are the same too? I mean, it could be a solution but why can't there be other solutions to that equation? Someone mentioned below another solution could be 1. How do we know there are not more function that satisfy the equality? My knowledge in differential equations is little.
He just wants to show us a possible solution.
where's the plus C after integrating the 1/sqrt(x^2-1) though?
Wow this is so cool!👍👍😺
what is the unit of arc length, and what is the unit of area. It's strange to put square meter equal meter ( m^2=?? m). Is it possible?
can you plz make a video proving the arc length equation?
Can you do a video on partial differential equations?
Specifically practice problems
I got scared at the end.
At 4:00 - supposedly Gauss used that same integration technique.
How about restricting the area to be the absolute value?
is this the only non-trivial function that satisfies this property? If not, then how may one find other functions?
And if it is, then what would be the best way to prove it?
yeah this is the only way, the trivial solution of course would be y=0.
WHERE'S THE BLACK PEN??? IS IT OKAY?????
That was a cool exercise. Thx for sharing.
first thing that came to mind is f(x)=1 (trivial but it works) my question is, why doesn't it appear on the solution to the diff eq.?
y=1 is also a solution
One question, what if the area is negative, will the arc lenght be negative as well, so the value keeps being the same?
Nuno Mateus hyperbolic cosine is only defined for y > 1 for all real values of x, so the area couldn’t be negative
Nice, thanks
even if the area is 'negative' the idea is to only integrate with ∫|f| so that the function is measurable
I love how you have up on saying cosine hyperbolic fungion and such after one go and said co-sh, and si-sh. Hahaha great video. Loveg the halloween theme, scary stuff jumping around screen.
The dum dum dumm part got me scared 😦😦😦
Is 2:10 a sufficient condition? Can there be other solutions where the functions are not equal?
Cool. Btw: Nah, new sound effect package has been dropped in to make the show more dramatic :D
Forgot constant solution of y=+-1 since this makes dydx =0
I think it’s wrong because integral from a to b of (y - sqrt(1 + (dy/dx)^2) can work with any function of y?
Nice Supreme shirt
Brother Banks thanks
Integral equations!
nice, how about an area under the curve that = the arc length squared? so like a straight at y = 2 line from 0 to 2
Damn, that was really cool.
Best thing i've seen today.
My intuition told be answer would be a parabola, so this is funny.
Awesome!
you have taken equal and solved it....we need to prove it equal and not take it from beginning.....it is wrong ..... please correct it in the next video
Nice more on this concept
f(x) = 1 :-D
arclength = x
area = x
Wouldn't y=1 work?
I worked with the negative and got the negative version. So much negativity.
Math On The Go : )
What’s up with the sound effects
dramaticfxpen yay!
It is very easy, just take any integral from n to n. 0=0.
Edit:
Oh, we want it for all values.
Can you say something to Integrals that can not be solved analytically? ...though i guess it might not be that much fun :)... I stumbled across: Integral( exp(x) x^4 /(exp(x)-1)^2 dx) from 0 to some x_m and it says it cant be solved, but seeing all ur nice integration tricks and stuff, I was wondering how u can say, that here there will not be any such tricks, but instead its just not analytically solvable. (This Integral comes up when calculating the specific heat of a solid in the Debye-Model)
Can't two definite integrals on the same interval be equal with the Integrands unequal?
Would y=1 also work?
Yes
y(x)=1 works just as well! :)
Before watching the video: Obviously yes
Lol it's 0:28 and he already answered
y=cosh(x+C)
Really really cool!!!
Wait, does the arc length formula come from the Euler's method video?
WHY DID SKIP THAT?
Is there any other functions ,where arc length is equal to area under curve ?? Or only coshx
First order differential equation have only one solution
@@hamsterdam1942 sinh?
@@simohayha6031 d/dx arcsinh (x) = 1/sqrt (x^2+1)
Nope
was expecting Ood at the end, haha
I don't think that [integral of f]=[integral of g] (both in the same interval) implies f=g. It looks to me that you just found one solution, but you have not solved the original problem in general. Am I right?
f doesn't need to be the same as g. For example. Integral of 0 from 0 to 2pi has the same value as integral of sin(x) from 0 to 2 pi. But sin(x) isn't just 0 for all x on that interval.
@Matthieu S that was convincing.
@@blackpenredpen thank you!
Is this solution related to the fact that e^x is its own derivative?
I understood everything as it was really well represented, but why use that music 😂😂
but it's 2 different units, the question itself is a trick question. the same as: can a rectangle have the same area as its circumference? 2 different units, no real answer
youre getting physics mixed up with maths here, they're both scalar functions, and this result is fine
I had a feeling that an exponential function was going to show up. I was pleasantly surprised it was a hyperbolic function.
cosh is essentially e^x with a fancy name
cosh is equal to (e^x+e^(-x))/2 so it's actually just an exponential function in disguise.