@@ycombinator765 jacobian is something that we use while changing the cartesion coordinates to some polar coordinate system or transforming to any other system.. it gives the amount of change that occurs in area after transforming. Eg: if cartesion coordinates are converted to spherical polar coordinates we have to substitute rho^2 sin(phi)d(rho)d(phi)d(theta) for dxdydz.
Dave Sir 👍🙏 Sir you are not just a Professor, in fact you are God sent Educator for all the students & Ex Students like me who studied Calculus 30 years ago ( 1991-1992 ) during my intermediate College days (11th Class ). From past two months i have been watching & already watched more than 40 Calculus lessons on your UA-cam Channel. Sir i Thank you & Salute You.
Calculus has been independently created/discovered at least 4 times (Pascal has a programming language named after him for his version) and thus we have 4 completely independent, yet internally consistent, sets of notating these concepts. Why did it get reinvented? Lack of local higher mathematics textbooks and education. Thank you for helping to fill this void in the current age, and making a 5th time unnecessary.
He taught in such a visual way... ! I couldn't "understand" this concept in my 6 months course of Partial Differential Equations. So I just crammed some formulas and just passed the exams.
Thank you for all your great videos! I'm studying for finals right now and your videos are infinitely better than my professor who lectures online from his bed. You make it so easy to understand in your great verbal and visual explanations, and your videos bring a lot of fun back into learning! I'm into engineering and I love learning through your videos on biology and astronomy and anything really, because learning is fun and awesome. You're a great teacher, thanks for doing such an amazing job :)
Profesor You are amazingly clear…like all my colleagues here say your 10:57 seconds video summarized hours of calculus classes.. my admiration to you. Thanks
4:10 if we are treating y^2 as a constant then why are we writing y^2? For example if we have this x^2(5) 5 is a constant so the derivative would be 2x since 5 is constant , and same goes for this question 4:10 , maybe it should have been 1 +3x^2 since y is constant? Instead of y^2+3x^2 I hope you understood my question
Yeah, if it's supposed to be constant, I thought won't that become zero?! It did bug me for a while but then I understood that the key point is that even though y is treated as a constant when differentiating with respect to x, the y^2 term does not become 0 in the final PDE equation. This is because we are equating the two partial derivatives, not just looking at the derivative with respect to x alone.
when taking the derivative in parts you treat y as a constant because your deriving with respects only towards x. say you derive 3x then answer is 3 same with y * x dx (dx = with respects to x) = y. So when taking the derivative of y^2 * x dx you get y^2. if it was y^2 * x^2 dx you would use the power rule only for x becuase your treating y as if its numerical so you would get 2 * y^2 * x. I hope this helped! (sorry if it was confusing)
also you are right, when deriving a constant without x attachted its 0 so in the equation x^2 + 2xy - y^2 deriving with respects to x you would get 2x + 2y - 0 becuase the y^2 is like a number and the derivative of a number without a pronumeral (that you are treating as a pronumeral not a number so not y in this case) is 0.
Actually, Product Rule is applied for this one. If we have x² + y² and differentiate with respect to x, variable y is a dependent term itself, therefore it becomes 0. However let's say we have x²y², it's understandable that we have two different variables in one single term. Applying the Product Rule, we'll have: u = x² v = y² *Note that we're still differentiating with respect to x vu' + uv' (y²)(2x) + (x²)(0) *v' which is the derivative of y² with respect to x, becomes 0 according to the constant rule itself We're left with (y²)(2x) + (x²)(0) (y²)(2x) + 0 = 2xy² And we can confirm that x²y² is 2xy² when differentiated with respect to x since we are only differentiating x² and treating y² as constant.
On the example given in the video, the xy² term consists of two variables; x and y. x is being multiplied to y² in other words. u = x v = y² *Differentiating with respect to x vu' + uv' (y²)(1) + (x)(0) (y²) + 0 = y² We can also confirm it by saying that differentiating xy² with respect to x only affects x which becomes 1 and treating y² as a constant by leaving it alone, thus giving 1*y² or simply y².
Even a 14 years old student would understand the gradient of a function with this video. I am not kidding I am 14 and I finally (after 5 days of search in internet) understood what gradient is.
Directional derivatives tell you what the slope will be, along a given direction among the input variables. Taking a sweep across all possible directions, you'll see that the maximum possible directional derivative occurs when the direction among the input variables is parallel to the gradient. To find a directional derivative, you form a unit directional vector, and take its dot product with the gradient vector. As an example, consider the function z = x^2/8 + y^2/4, at the point (1, 1). Suppose we're interested in a direction that is along the diagonal of a 3-4-5 triangle, that is roughly 37 degrees from the +x direction. Our unit directional vector (u) would therefore be given by u = . The gradient at this point is . So the dot product gives us 0.4 + 0.24 = 0.624. This is the directional derivative of this particular function. The maximum possible directional derivative at this point, will have the same direction as the gradient. Its unit vector will be . Taking the dot product with the gradient, and we get 0.75/sqrt(5) = 0.335. This is the maximum possible rate of ascent.
I have to admit that I've never understood why we have partial derivatives but not partial integrals. With the integral, the dx makes it clear which variable we're integrating and we don't need a special integral sign in addition.
Yeah, I get the gradient, but I am not sure I do total differential. You can also mention gradient of an error function of a neural network, as an example.
6:00 you confused me there f(x,y) is a 3 dimension function taking 2 inputs, f(x,y,z) is a 4 dimensions function with 3 inputs. How does the grad vector get expressed in the previous 3d graph if we can't calculate the partial derv in z (df/dz) with the k unit vect.
Hello, why does the grad f(x,y) have the component of z-direction? I mean if the gradient of f(x,y) points in the direction of maximum change, that would be a z-direction.
Gradients of a function of multiple variables, are limited to the space of the input variables. The gradient of f(x, y) only exists in the x-y plane. It represents stuff that is happening in the z-direction, when f(x, y) is represented as the z-position in a 3-D spatial coordinate system, but the gradient itself doesn't exist in the z-direction.
Curious - what does using the "curly d" really add here? Could we not have done exactly the same thing using the standard d/dx, d/dy notation? What (I think) I am trying to say is that since we *know* we're dealing with a multivariable function, is it even possible that the standard d/dx (etc) notation could be misunderstood as referring to the "derivative of the whole function" even if that made any sense? The gradient sort of does that, so if we're looking at derivatives w.r.t. x and y, what do we gain in the intermediate steps by changing to "curly d"s? Or am I overthinking this, and we use curly ds purely as a label to remind ourselves that we're in a multivariable problem? Seems odd, should you need reminding if you're at this level of calculus!? 🤔
@@GoBlue402 The vectors i, j, and k, are unit vectors that identify directions in the three cardinal axes of x, y, and z. Some books choose to hat the letters x, y, and z, to avoid a separate letter. It is an artifact of history that we call the unit vectors i-hat, j-hat, and k-hat. By definition, a unit vector has a magnitude of 1. This could be the axis unit vectors, or it could also be unit vectors in general. A unit vector's purpose is to identify direction, so it can give direction thru scalar multiplication to what otherwise would be a scalar quantity. Another application of unit vectors is in Newton's law of gravitation, and analogously, Coulomb's law, where r-hat is used as the radial unit vector, because the force acts radially along the line joining the two bodies.
Can someone help clear my confusion? When taking deriv wrt x of f(x,y), sometimes we say y is a constant so replace y with zero. Other times we say hold variable y as constant (and instead of replacing y with 0, we write down the y. This is so confusing!!! Here is clear example of my question. @t
Suppose you have f(x) = x^2 + 2. When you find the derivative you will get f'(x) = 2x. The plus 2 is a constant and doesn't affect the gradient of the curve. Furthermore if you have a function for example f(x) = 4x^2. The constant at the front (in this case 4) will affect the gradient so doesn't cancel like adding a constant. So the derivative will be f'(x) = 4*2x = 8x. For partial derivatives of x all you're doing is treating y as a constant. just like the '+ 2' and the '4 * ' in the two examples. So let's suppose you have f(x,y) = xy + y. From the first example you take the ' + y' as a constant which it's derivative is zero since this won't affect the gradient. While the constant for 'xy' will stay. Making ∂f/∂x = y + 0 = y. Hope this helps.
@@ProfessorDaveExplains Thank you for the reply. I think my confusion is whether x is a constant in the original function or being held constant say if you are working out the partial derivative with respect to y.
@@kevconn441 If you are taking the derivative of a function of multiple variables, relative to only one of the variables at a time, you treat all other variables as constants. So when using the d/dy operator, x becomes a constant in that particular differential operation. It is called a partial derivative when you do this, although the same principle still applies to differentiation in general.
The gradient points in the path of steepest ascent. This could mean that it points to a peak of a function, and not necessarily the global peak of a function. It could simply point to a local maximum. It could also mean that it points to a saddle point of a function, where the function has two opposite curvatures meeting. It could also point to a local maximum on the function that is a continuous line, rather than a point-maximum.
ooh that's explained earlier in the series, check out the ones on vectors in my mathematics playlist, a bit before the calculus content starts, or possibly after calculus and before linear algebra
Unit vectors. It's a vector with a magnitude of one. These particular unit vectors point in the X, y and z directions and give you another way to notate other vectors. For example, you could write the vector as 2i+5j+4k
@Diogenes TheDog I understand almost 90% of calculus(you know what I mean, even some difficult multiple integrals and relatively difficult problems on them) and even the last advanced maths video proffesor dave uploaded but i find vectors really difficult to understand so i have many queries like this one
Best 10 minutes and 56 seconds of my life. Such a clear explanation!
Fuck Intercourse, watching Dave is the real deal ❤
"The gradient is kind of like a slope for a higher-dimensional function" is just what I needed to hear to conceptualize this. Thanks!
This man carried me through general chemistry, calculus II and now calculus III. I can't thank you enough for teaching me math.
So good this explanation makes MIT lectures look like an overpriced DLC pack
🤣
ho can proffeser dave explains heart this comment?!
Man said "DLC Pack" lmao
same with Harvard lectures lmao
totally agree.. he explains algebra so nicely..i tend to sleep when i listen Glibert's lecture :-)
Gradient: A vector made up of all the partial derivatives of the function. Thank you!
Wtf is Jacobian then??
Pendulum theSimpleOne lmk when you get an answer
@@ycombinator765 jacobian is something that we use while changing the cartesion coordinates to some polar coordinate system or transforming to any other system.. it gives the amount of change that occurs in area after transforming. Eg: if cartesion coordinates are converted to spherical polar coordinates we have to substitute rho^2 sin(phi)d(rho)d(phi)d(theta) for dxdydz.
@@Darkev77 he got the answerrrrr seeee
@@ankitaaarya beast
This is awesome. Feel like I finally understand the gradient now.
Dave Sir 👍🙏
Sir you are not just a Professor, in fact you are God sent Educator for all the students & Ex Students like me who studied Calculus 30 years ago ( 1991-1992 ) during my intermediate College days (11th Class ).
From past two months i have been watching & already watched more than 40 Calculus lessons on your UA-cam Channel.
Sir i Thank you & Salute You.
Calculus has been independently created/discovered at least 4 times (Pascal has a programming language named after him for his version) and thus we have 4 completely independent, yet internally consistent, sets of notating these concepts. Why did it get reinvented? Lack of local higher mathematics textbooks and education. Thank you for helping to fill this void in the current age, and making a 5th time unnecessary.
That's cool to know. Thanks.
Thanks to you too Professor Dave.
He taught in such a visual way... !
I couldn't "understand" this concept in my 6 months course of Partial Differential Equations. So I just crammed some formulas and just passed the exams.
I just went down and liked every comment that was positive on this video. Its the only way to extend my appreciation! Thanks man!
this is what you call a video. One of the best teaching videos I've ever watched
I love how clear and concise your explanations are!
Thank you for all your great videos! I'm studying for finals right now and your videos are infinitely better than my professor who lectures online from his bed. You make it so easy to understand in your great verbal and visual explanations, and your videos bring a lot of fun back into learning! I'm into engineering and I love learning through your videos on biology and astronomy and anything really, because learning is fun and awesome. You're a great teacher, thanks for doing such an amazing job :)
You rock man! gracias amigo. Im so happy to finally understand it
Profesor You are amazingly clear…like all my colleagues here say your 10:57 seconds video summarized hours of calculus classes.. my admiration to you. Thanks
the coolest and prettiest explaining teacher I've ever seen!
Oh thank you so much! After 19 years I can finally picture it!!!
dude, you are capable of explaining multivariable calculus to a 15 year old so that he can actually solve questions, youre a God (I'm the 15 year old)
No he is human
No he is chemistry Jesus... who just happens to be teaching maths too.
Literally the best explanation. Trying to do a project where partial derivatives come up, and I needed a quick refresher.
Valeu!
Thank you sir for your dedication and for making this free! 🙏
This is highly underrated stufffff
Ngl I’m lucky i found this 🔥 I subscribed❤️
An eye opener video. Neat and tidy.
Thank you so much. Finally understood the real meaning of partial derivative and gradient. 😇
God!!!!!
You saved me !!!!!
I have test tomorrow on this topic !!!!
Hallelujah, just saved my calculus...
As always professor dave comes in clutch just before the exam 😁
happy teacher's day sir.....
Was the video time 11 mins😲I didn't have a feeling that 11 mins have passed by. it was deeply interesting.thanks sir🤗
4:10 if we are treating y^2 as a constant then why are we writing y^2?
For example if we have this x^2(5)
5 is a constant so the derivative would be 2x since 5 is constant , and same goes for this question 4:10 , maybe it should have been 1 +3x^2 since y is constant? Instead of y^2+3x^2
I hope you understood my question
Yeah, if it's supposed to be constant, I thought won't that become zero?! It did bug me for a while but then I understood that the key point is that even though y is treated as a constant when differentiating with respect to x, the y^2 term does not become 0 in the final PDE equation. This is because we are equating the two partial derivatives, not just looking at the derivative with respect to x alone.
when taking the derivative in parts you treat y as a constant because your deriving with respects only towards x. say you derive 3x then answer is 3 same with y * x dx (dx = with respects to x) = y. So when taking the derivative of y^2 * x dx you get y^2. if it was y^2 * x^2 dx you would use the power rule only for x becuase your treating y as if its numerical so you would get 2 * y^2 * x. I hope this helped! (sorry if it was confusing)
also you are right, when deriving a constant without x attachted its 0 so in the equation x^2 + 2xy - y^2 deriving with respects to x you would get 2x + 2y - 0 becuase the y^2 is like a number and the derivative of a number without a pronumeral (that you are treating as a pronumeral not a number so not y in this case) is 0.
Actually, Product Rule is applied for this one.
If we have x² + y² and differentiate with respect to x, variable y is a dependent term itself, therefore it becomes 0.
However let's say we have x²y², it's understandable that we have two different variables in one single term.
Applying the Product Rule, we'll have:
u = x²
v = y²
*Note that we're still differentiating with respect to x
vu' + uv'
(y²)(2x) + (x²)(0)
*v' which is the derivative of y² with respect to x, becomes 0 according to the constant rule itself
We're left with
(y²)(2x) + (x²)(0)
(y²)(2x) + 0
= 2xy²
And we can confirm that x²y² is 2xy² when differentiated with respect to x since we are only differentiating x² and treating y² as constant.
On the example given in the video, the xy² term consists of two variables; x and y. x is being multiplied to y² in other words.
u = x
v = y²
*Differentiating with respect to x
vu' + uv'
(y²)(1) + (x)(0)
(y²) + 0
= y²
We can also confirm it by saying that differentiating xy² with respect to x only affects x which becomes 1 and treating y² as a constant by leaving it alone, thus giving 1*y² or simply y².
What a Life saver, thanks so much, professor
Omg I finally got it, and I got the comprehension correct on the first try!! Can't believe it!!
Thank you....
happy teacher's day
Wow great explanation. Sucks to see us students pay for an education where profs have a hard time explaining clearly. Thank You!!
Even a 14 years old student would understand the gradient of a function with this video. I am not kidding I am 14 and I finally (after 5 days of search in internet) understood what gradient is.
Thank you very much! It is the best explanation of partial derivatives that I ever heard!
Del means gradient. Thank you for clearing that up for me.
Do you teach all subjects?👍good work btw....
Clear as water, helps me understanding Deep Learning!
Great explanation! Just one question: Why does the gradient point in the direction of maximum slope?
due to the addition of the partial derivative vectors using laws of vector addition
@@fineartpottamus9020 this was the final puzzle piece for me, now it all clicked together. thank you a lot.
Directional derivatives tell you what the slope will be, along a given direction among the input variables. Taking a sweep across all possible directions, you'll see that the maximum possible directional derivative occurs when the direction among the input variables is parallel to the gradient.
To find a directional derivative, you form a unit directional vector, and take its dot product with the gradient vector. As an example, consider the function z = x^2/8 + y^2/4, at the point (1, 1). Suppose we're interested in a direction that is along the diagonal of a 3-4-5 triangle, that is roughly 37 degrees from the +x direction. Our unit directional vector (u) would therefore be given by u = . The gradient at this point is . So the dot product gives us 0.4 + 0.24 = 0.624. This is the directional derivative of this particular function.
The maximum possible directional derivative at this point, will have the same direction as the gradient. Its unit vector will be . Taking the dot product with the gradient, and we get 0.75/sqrt(5) = 0.335. This is the maximum possible rate of ascent.
The lecture is awesome. Clear and precise. But the answer to the gradient of the function at (4, 1) should be (1/2, 0) cause ln(1)=0.
Great explanation, thanks! However, I didn't quite understand what i, j, and k represent in the equations around the 6th min
Wow You have explained very good Finally I understand this concept Keep it up.
Okay the content of this video is super but the Intro always have me like 😂😎
Excellent explanation!
Nice review of gradients! Thanks!
This is very helpful.
Why don't you do a video on higher order partial derivatives and total differention
Very clear explanation thanks
Thank you Professor Dave
Just got to see ur videos sr..... U are an super osm educator... Lots of love ❤😘
I watch this channel so much that I once had a dream and your theme song made a cameo 😂
2 e to the to z (too easy)thanks for this very well explanation!!!!!
But I don't understand why x axis + y axis vector will point to the direction of maximum change?
Fantastic explanation
I have to admit that I've never understood why we have partial derivatives but not partial integrals. With the integral, the dx makes it clear which variable we're integrating and we don't need a special integral sign in addition.
wow, amazing video. please keep making more ML videos!
These videos make me rethink my life choices. Uni is actually ass.
what do you mean ?
🤣🤣🤣
Thank you sir!
I wish you to make vedio on Total differentiation.
Please !
Please!
Please!
Please!!!!!!!!!
Yeah, I get the gradient, but I am not sure I do total differential. You can also mention gradient of an error function of a neural network, as an example.
Good explanation sir
6:00 you confused me there f(x,y) is a 3 dimension function taking 2 inputs, f(x,y,z) is a 4 dimensions function with 3 inputs.
How does the grad vector get expressed in the previous 3d graph if we can't calculate the partial derv in z (df/dz) with the k unit vect.
Mr Booshit He probably meant a function that varies with three different variables (as in dimensions or axes).
Yes it’s like when you have a component in the z direction that you care about, they are all still 3D I think
f(x,y) is a surface in 3D, not the 3D itself
Just awesome !!!
Thanks sir, clarified. Please do on line integrals and Greens theorem..
those are coming!
Thanks for making that so simple for me @ 59. Cheers!
Thankyouuu soo much professor….absolutely incredible explanation!!!!!
Omg now that I know what it s definition and what it means I can work out what these beatifull weird equations mean thank you!!!!!!!!
Hello, why does the grad f(x,y) have the component of z-direction? I mean if the gradient of f(x,y) points in the direction of maximum change, that would be a z-direction.
Gradients of a function of multiple variables, are limited to the space of the input variables. The gradient of f(x, y) only exists in the x-y plane. It represents stuff that is happening in the z-direction, when f(x, y) is represented as the z-position in a 3-D spatial coordinate system, but the gradient itself doesn't exist in the z-direction.
Really helpful video, thanks so much :)
Great teacher
Hope u make video on divergence and curl of a function, its goona make my visualisation much clear😁...by the way tq sir for the gradient video...😇
that's the next one!
Thank you.
very good explanation i am from india/
How would you find the gradient of f(x(s),y) is it still d/dx, d/dy or will the chain rule need to be applied?
Great video
Thank you! 😊
product: u' v + v' u
division: (u' v - v' u)/v²
I find it easier to memorize like this
u prime v plus v prime u has a ring to it
Curious - what does using the "curly d" really add here? Could we not have done exactly the same thing using the standard d/dx, d/dy notation?
What (I think) I am trying to say is that since we *know* we're dealing with a multivariable function, is it even possible that the standard d/dx (etc) notation could be misunderstood as referring to the "derivative of the whole function" even if that made any sense?
The gradient sort of does that, so if we're looking at derivatives w.r.t. x and y, what do we gain in the intermediate steps by changing to "curly d"s?
Or am I overthinking this, and we use curly ds purely as a label to remind ourselves that we're in a multivariable problem? Seems odd, should you need reminding if you're at this level of calculus!? 🤔
Please make a lec on real life application of matrix; projection of 3d image in eigen space and all that.
check my linear algebra playlist
Why this is so easyy???? Thanks..
thank you so much you are the best
3:10 so if I have a curvy D does that make me a partial man?
I'm only in calc 2, and we barely started with differential equations... Interesting
what do the i, j , and k represent in the grad f formula? (5:56)
the unit vectors along the three axes
@@ProfessorDaveExplains So does the length of the vectors stay the same along each gradient. Sorry if I am not understanding something correctly
@@GoBlue402 The vectors i, j, and k, are unit vectors that identify directions in the three cardinal axes of x, y, and z. Some books choose to hat the letters x, y, and z, to avoid a separate letter. It is an artifact of history that we call the unit vectors i-hat, j-hat, and k-hat.
By definition, a unit vector has a magnitude of 1. This could be the axis unit vectors, or it could also be unit vectors in general. A unit vector's purpose is to identify direction, so it can give direction thru scalar multiplication to what otherwise would be a scalar quantity. Another application of unit vectors is in Newton's law of gravitation, and analogously, Coulomb's law, where r-hat is used as the radial unit vector, because the force acts radially along the line joining the two bodies.
I think answer to second question is ?
Can someone help clear my confusion?
When taking deriv wrt x of f(x,y), sometimes we say y is a constant so replace y with zero.
Other times we say hold variable y as constant (and instead of replacing y with 0, we write down the y.
This is so confusing!!! Here is clear example of my question. @t
I also noticed it
Please let me know if you find the solution
Suppose you have f(x) = x^2 + 2. When you find the derivative you will get f'(x) = 2x. The plus 2 is a constant and doesn't affect the gradient of the curve. Furthermore if you have a function for example f(x) = 4x^2. The constant at the front (in this case 4) will affect the gradient so doesn't cancel like adding a constant. So the derivative will be f'(x) = 4*2x = 8x.
For partial derivatives of x all you're doing is treating y as a constant. just like the '+ 2' and the '4 * ' in the two examples.
So let's suppose you have f(x,y) = xy + y. From the first example you take the ' + y' as a constant which it's derivative is zero since this won't affect the gradient. While the constant for 'xy' will stay. Making ∂f/∂x = y + 0 = y. Hope this helps.
Thanks so here is how I now see it.
F(x,y)=x^2 + y.
F'( x)= d/dx (x^2) +d/dx (y)
=2x+0
F(x,y)=pi*x^2*y
F'(x)=pi*y* d/dx (x^2)
=pi*y*2x
Hi! just a trivial question, what does "In" in "2xIn(y)" stand for?
natural log, it’s logarithm of y with base e, (or which power would you raise e to , to achieve y)
Why are those videos private, in the playlist??????
i release them one per week
Hello sir!
When are you going to release the video??????
Is advanced math platlist multivariable calculus ??
Del? Aren't those called Nablas?
i believe that is synonymous but outdated
thank you sir.!!!!!!
Why do you say sometimes the derivative of the constant is 0, and the derivative of, say x, is x?
the derivative of any constant is zero, and the derivative of x is 1
@@ProfessorDaveExplains Thank you for the reply.
I think my confusion is whether x is a constant in the original function or being held constant say if you are working out the partial derivative with respect to y.
@@kevconn441 If you are taking the derivative of a function of multiple variables, relative to only one of the variables at a time, you treat all other variables as constants. So when using the d/dy operator, x becomes a constant in that particular differential operation. It is called a partial derivative when you do this, although the same principle still applies to differentiation in general.
I was geniuely so proud of myself that I could do the comprehension check lol
thank-you for this video!
Explained it.
thanks professer dava i d,not more engish but iunderstand
is it true that gradient always points to summit of function?
The gradient points in the path of steepest ascent. This could mean that it points to a peak of a function, and not necessarily the global peak of a function. It could simply point to a local maximum. It could also mean that it points to a saddle point of a function, where the function has two opposite curvatures meeting. It could also point to a local maximum on the function that is a continuous line, rather than a point-maximum.
This is great, I have exam on 13th, can you make some more videos public :D
Finally, prof got a haircut 😂💇💇♂️
That's a huge amount of knoweledge
4.5
respect
What are i, j and k ???
ooh that's explained earlier in the series, check out the ones on vectors in my mathematics playlist, a bit before the calculus content starts, or possibly after calculus and before linear algebra
Unit vectors. It's a vector with a magnitude of one. These particular unit vectors point in the X, y and z directions and give you another way to notate other vectors. For example, you could write the vector as 2i+5j+4k
@Diogenes TheDog I understand almost 90% of calculus(you know what I mean, even some difficult multiple integrals and relatively difficult problems on them) and even the last advanced maths video proffesor dave uploaded but i find vectors really difficult to understand so i have many queries like this one