Great video. It helped me prepare my lecture for my calc III class. I am going to let my students watch your videos. They are awesome. Please post more videos.
Hello Rhonda and Mukta! I'm a student studying at the Bsc level. Is there any youtube channel that you guys would suggest to learn partial differential equations that explains the concepts very well like this? I learned about ODE from Khan Academy but it doesn't have PDEs. Please help!
When I heard "next video" I instantly became excited for part 2. This has never happened before in all the tutorials I've watched. Please make videos on as many topics as possible!!! You are the best!
I believe that the Gradient vector is a two dimensional vector in a " plane tangent" to the surface of the function f(x, y) and not a vector in the x-y plane (i.e. it is in the tangent space of the surface). The "projection of the gradient vector ON THE x-y plane is actually perpendicular to the level curves.
the partial derivative slopes are in the so called tangent plane but their vectors are in x and y dir... so the resultant vector(sum of partial derivative vectors=gradient vector) will lie in the x-y plane and the gradient(slope) will lie in the so called tangent plane....please correct me if i am wrong....
I don't fully understand. If the gradient is (2,1). I need to go 2 in the x direction and 1 in the y direction. Therefore do I need to imagine there's an other coordinate system on the tangent surface?
I have a different opinion. I think the Gradient vector is actually in the x-y plane. First, It makes no sense to find a given Gradient vector, like grad f= xi+yj, in the plane tangent to the surface, doesn't it? There are twofold. First, the definition of Gradient is in x-y plane. Second, we haven not define a 2-D coordinate in the tangent plane. Apparently we cannot locate any vector before defining the coordinates, can we? Secondly, note that while Katya is skiing in the 3D space, it actually only needs two 2 variables , x and y, to determine her 3D projectile. This is simply because the 3rd value, z is determined by z=f(x,y). Therefore, when she decides to enjoy the max rate of descent, she only needs a 2-variable vector in her level plane, the -negative Gradient vector. The slope of Katya's actual projectile: is the Directional Derivative wrt a unit vector of the Gradient vector in xy plane. Mathematically, it equals to the |grad f|. So, even you find the vector tangent to the actual projectile, its projection onto the xy plane would not be the Gradient vector. It will be shorter than the Gradient vector. I hope this makes a little bit more sense.
wow ma'am, this was so helpful. I am pursuing graduation and do like mathematics, but not when it's only focused on solving problems rather than understanding. it feels great to be able to find a video that finally explains things. THANK YOU - a student from India
Madam why are we defining directional derivative as only in magnitude. Can't we multiply the magnitude with the unit vector and say the total vector is directional derivative...??
If you think of a sphere that is placed with its center at the origin. Parallel radii for the level curves of circles going up will be decreasing going up if think of going in towards the center. Parallel radii for level curves for circles going down will become smaller if you think of coming out from the center. Bingo! That's the idea. And for a sphere the gradient will be the same at all points.
I always thought of a series of vectors that had the shortest 2 dimensional length going up or down. That makes sense in terms of steepness. Do you know what I mean?
the (x0,y0) should be located with the space on X-Y plane inside the curve intercepting that plane, it is outside (minor issue though)! so its projection would never intercept the surface dawned.
let x^2 + y^2 + z^2 = 1, so that z = f(x,y) z = ( 1 - x ^2 + y ^2 ) ^ (1/2) z < 0 = - ( 1 - x ^2 + y ^2 ) ^ (1/2) z > 0 for z=0, use anything to make it continuous 1. prove the max value of gradient at (x,y,z) = (0,0,1) when the initial point is from (x,y,z) = (0,0,-1) 2. find the gradient at (x,y,z) = (1, 0, 0) from the problem1 , when the gradient becomes infinity if you will, change my questions to make it happen gradiently
You're an amazing teacher. When I was taking the course, at a well known university, the teacher of this class often seemed to be confused. Ugh! It was a weird class, but I did great. I understood the subject better than the professor at the time, and instead of complimenting me, he went in the opposite direction and acted like he hated me. You actually do understand this concept. I understood it years ago, but I had to deal with a lot of haters. What is wrong with some people?
@ 7:54: At the point K, where Katja is on the hill, there is a tangent-plane. Say, the point is (1,1,2) and the hill is represented by the function f(x,y) = -x^2 - y^2 + 4. How then can I calculate the tangent-vectors of that plane, when unit-vector is ?
Very interesting video again, but there's still something I'm puzzled with. If we take the unit vector to be negative, it can still have norm 1 but the direction will be that of steepest descent, because angle is pi now. I can visualize the arrow flippening the other way. But when you draw the slope, i.e the gradient, there is no direction at all in the sens that the gradient is not represented by an arrow itself. I mean what is said about unit vector u and the gradient is true if the gradient is an arrow pointed in the same direction as the vector u. If it's pointing the other way, u is not anymore the direction of steepest ascent. You'd tell me that it's because now the angle is pi so unit vector is in the opposite direction of the gradient. But how did you know in the first place the direction towards which the gradient is pointing ? the gradient is a vector and so a direction by itself, right ? But when we calculate it, I hardly see how we know the direction towards which it points and so why this direction is the steepest ascent. In fact when we are at the point where we calculate the gradient, and if the gradient is a slope, why should we draw the gradient as an arrow pointing towards the ascending side of the slope and not the descending one ? Thanks a lot !
Thanks for your thoughtful comments. If you take an example, say f(x,y)=x^2+y^2 and calculate the gradient at a point, say (1,1), you'll see that the gradient vector is a 2D-vector. If you move along the gradient in the xy-plane, the curve traced on the surface will take you in the direction of greatest ascent. The gradient will naturally point in the right direction. If you changed to the function f(x,y)=-x^2-y^2 and again calculate the gradient at (1,1),and do the same thing, you will move along a curve in the direction of greatest ascent. I think you may be confusing 2D and 3D gradients. The gradient in this case is not pointing up or down, it lies in the xy-plane. Hope that helps.
@@rhondahughes9648 Yes thank you very much, I studied your example and better see it is actually doing that. I'm just missing the geometric intuition of why, when we calculate this rate of change, it "naturally" points towards the ascent. To be clearer and very simple, if I calculate the gradient of x², I get 2x and I totally agree that when plugging a value in it, I will go where the function increases the most. I just don't see why it makes sens, why it's not the other way (decreasing) because when I calculate this rate of change, I can approach my point from the left or the right, and a rate of change indicates how much a thing changes but not necessarily in which direction. Anyway thank you very much, I still learned and visualized maany things thaanks to your videos, great work ! Best regards
You should consider making more videos. There are many topics in mathematics that such charisma could clarify for many people. Just a math teacher commenting.
Awesome video! I am a Civil Engineering Student at Cal Poly SLO currently enrolled in Calculus 4. Trying to pass my quiz tomorrow and learn a couple lessons in a day. Great examples! Thank you very much for sharing.
Why is the gradient vector represented by a partial derivative parallel to the x, y plane and not as an actual gradient[ tangent ] to the surface at a specific point or time as is the case in 2D ? I mean we don't consider the gradient in R2 as a vector parallel to the x axis. I just can't get my head around this and any insight or clarification would be gratefully received.
Gradient is a vector in space and we are here working with 2D space (x and y directions), while z is corresponding to values of function f(x,y), not third dimension of space. So, gradient gives you direction in which function changes the most, but to get there you must move in 2D space, i.e. in x and y directions and function will assume a value depending on the point in 2D space you are at. Therefore, "skiing down the function slope" story was unfortunate. In 3D space, it would look like w = f(x, y, z), where f can for example be a function that maps a temperature to all points in 3D space. In that case gradient would be a 3D vector, and function graph would be 4D. And if you instead use gravitational potential energy as a function od x, y, and z, you can start talking about skiing down the slope. Also, note that if you apply the gradient definition given here to 1D situation, i.e. y=f(x), the gradient _is_ a 1D vector parallel to x axis, and its length and direction define slope of the tangent at a given point, not the tangent itself.
I generally agree with this response, but don't think I say "skiing down the function slope" anywhere in the video. It's very clear that the gradient is the direction in 2D, and the rate of change along the curve is greatest in that 2D direction. That was the main point of the explanation:)
Please bear with me, mathematics is just a hobby of mine, I have no real qualifications in the subject. I do however find what I consider ambiguity in certain mathematical definitions leading to confusion.....at least on my part. I fully understand that the " gradient " is the magnitude of the sum of the two partial derivatives and it's direction is in the x, y plane. But as I understand it, the magnitude of the gradient is a real slope in R3 , however its direction is represented by a vector in R2. I mean if I am on a hill and it slopes one way East and one way West then the gradient gives me the maximum or minimum value of the slope of that hill in the correct direction in R2. depending on whether I wish to ascend the or descend. Why can't the actual slope in R3 be represented by a tangent vector whose tail is at the point of tangency and has a slope equal to the magnitude of the tangent vector. I mean I have to climb up the hill , not into it !Another definition I find confusing concerns the surface integral where it is used as a density function to calculate the mass of an infinitely thin surface........how can a surface that is infinitely thin have a mass? Is this a lack of comprehension on my part, or just poorly defined concepts? Once again any clarification would be gratefully received .
Очень интересное видео)) Вот что интересно градиент превращает скалярное поле в векторное, а дивергенция векторное в скалярное, т.е. F = grad(div(F)) , где F - векторное поле?
Excelent explain …you are a great professor…looking forward to par II and more Vector Cálculos videos from you.
Great video. It helped me prepare my lecture for my calc III class. I am going to let my students watch your videos. They are awesome. Please post more videos.
Thank you very much, Mukta!
Yes. Same here. I'm a teacher too and you described this really well. Thanks!!
Thank you!
I'm happy to hear that! You are very welcome.
Hello Rhonda and Mukta! I'm a student studying at the Bsc level. Is there any youtube channel that you guys would suggest to learn partial differential equations that explains the concepts very well like this? I learned about ODE from Khan Academy but it doesn't have PDEs. Please help!
When I heard "next video" I instantly became excited for part 2. This has never happened before in all the tutorials I've watched. Please make videos on as many topics as possible!!! You are the best!
Thanks so much, Dan!
Wow you're gifted. And most certainly talented! It finally clicked for me. Thanks
That is very kind, Yaman. Thank you!
Great video, 👌👌👌please make more videos
Thank you!
Thank you!
hey please maKe more videos
wow, katia looks nice. thanks
+Ander Stuhl Katya is nice!
This was absolutely breathtaking. Thank you
+PHaitiano Thank you so much!
I believe that the Gradient vector is a two dimensional vector in a " plane tangent" to the surface of the function f(x, y) and not a vector in the x-y plane (i.e. it is in the tangent space of the surface). The "projection of the gradient vector ON THE x-y plane is actually perpendicular to the level curves.
That's correct. Just trying to keep it simple, but thank you!
the partial derivative slopes are in the so called tangent plane but their vectors are in x and y dir... so the resultant vector(sum of partial derivative vectors=gradient vector) will lie in the x-y plane and the gradient(slope) will lie in the so called tangent plane....please correct me if i am wrong....
I don't fully understand. If the gradient is (2,1). I need to go 2 in the x direction and 1 in the y direction. Therefore do I need to imagine there's an other coordinate system on the tangent surface?
I have a different opinion.
I think the Gradient vector is actually in the x-y plane.
First, It makes no sense to find a given Gradient vector, like grad f= xi+yj, in the plane tangent to the surface, doesn't it? There are twofold. First, the definition of Gradient is in x-y plane. Second, we haven not define a 2-D coordinate in the tangent plane. Apparently we cannot locate any vector before defining the coordinates, can we?
Secondly, note that while Katya is skiing in the 3D space, it actually only needs two 2 variables , x and y, to determine her 3D projectile. This is simply because the 3rd value, z is determined by z=f(x,y). Therefore, when she decides to enjoy the max rate of descent, she only needs a 2-variable vector in her level plane, the -negative Gradient vector.
The slope of Katya's actual projectile: is the Directional Derivative wrt a unit vector of the Gradient vector in xy plane. Mathematically, it equals to the |grad f|. So, even you find the vector tangent to the actual projectile, its projection onto the xy plane would not be the Gradient vector. It will be shorter than the Gradient vector.
I hope this makes a little bit more sense.
👍
Thank you!
what a clear and impressive explanation!!
thank you so much anyway!!
Thank you very much!
Great work mam math is amazing
Thank you so much!
hi Rhonda.. wishing you well...however if you could upload newer videos ..it would really make a difference
Your videos are AWESOME. Please post more. Thanks so much!
Thank you!
wow ma'am, this was so helpful. I am pursuing graduation and do like mathematics, but not when it's only focused on solving problems rather than understanding. it feels great to be able to find a video that finally explains things. THANK YOU
- a student from India
Thank you so much! I'm glad this was helpful.
5:42 "At any rate..." Oh, sweet sweet music to my ears!
this is so much clearer than my Calc 3 professor's lectures... thank you so much!
+petergianf Thank YOU!
Madam why are we defining directional derivative as only in magnitude. Can't we multiply the magnitude with the unit vector and say the total vector is directional derivative...??
Please post more videos!!!!!!! Thank you!!!!!
Thank you!
You are god
If you think of a sphere that is placed with its center at the origin. Parallel radii for the level curves of circles going up will be decreasing going up if think of going in towards the center. Parallel radii for level curves for circles going down will become smaller if you think of coming out from the center. Bingo! That's the idea. And for a sphere the gradient will be the same at all points.
I always thought of a series of vectors that had the shortest 2 dimensional length going up or down. That makes sense in terms of steepness. Do you know what I mean?
Thank u mam. For our excellent and mind-blowing and easily understood lecture
You're welcome!
WHOA!
You saved me!
Thanks 😀
Good to hear! Thank you.
YOU ARE COMMITING A CRIME .. BY NOT MAKING VIDEOS these are 3 yrs old videos .. such a great ..explanation ..cant find anything comes closer to this .
the (x0,y0) should be located with the space on X-Y plane inside the curve intercepting that plane, it is outside (minor issue though)! so its projection would never intercept the surface dawned.
Agreed!
What a lovely couple of videos you've made! I wish there were more :(
Thank you so much!
This was very helpful now I can understand gradient descent
Thank you!
let x^2 + y^2 + z^2 = 1, so that z = f(x,y)
z = ( 1 - x ^2 + y ^2 ) ^ (1/2) z < 0
= - ( 1 - x ^2 + y ^2 ) ^ (1/2) z > 0
for z=0, use anything to make it continuous
1. prove the max value of gradient at (x,y,z) = (0,0,1) when the initial point is from (x,y,z) = (0,0,-1)
2. find the gradient at (x,y,z) = (1, 0, 0) from the problem1 , when the gradient becomes infinity
if you will, change my questions to make it happen gradiently
You're an amazing teacher. When I was taking the course, at a well known university, the teacher of this class often seemed to be confused. Ugh! It was a weird class, but I did great. I understood the subject better than the professor at the time, and instead of complimenting me, he went in the opposite direction and acted like he hated me. You actually do understand this concept. I understood it years ago, but I had to deal with a lot of haters. What is wrong with some people?
Thank you very much! Your last question is one we all try to answer. It's a great mystery.
the first part was great, the second part was too deep for me. Thanks all the same.
Thanks for trying!
Nice vedio. Can you please explain how gradient as a 3D vector works
@ 7:54: At the point K, where Katja is on the hill, there is a tangent-plane. Say, the point is (1,1,2) and the hill is represented by the function f(x,y) = -x^2 - y^2 + 4. How then can I calculate the tangent-vectors of that plane, when unit-vector is ?
This may help, Rhonda ;-) th-cam.com/video/um896UsFK-g/w-d-xo.html
Rhonda. Thank you for your eloquent explanation of this abstract topic. I love the visual demonstration. Please be safe.
Thank you! You, as well.
Very interesting video again, but there's still something I'm puzzled with. If we take the unit vector to be negative, it can still have norm 1 but the direction will be that of steepest descent, because angle is pi now. I can visualize the arrow flippening the other way. But when you draw the slope, i.e the gradient, there is no direction at all in the sens that the gradient is not represented by an arrow itself. I mean what is said about unit vector u and the gradient is true if the gradient is an arrow pointed in the same direction as the vector u. If it's pointing the other way, u is not anymore the direction of steepest ascent. You'd tell me that it's because now the angle is pi so unit vector is in the opposite direction of the gradient.
But how did you know in the first place the direction towards which the gradient is pointing ? the gradient is a vector and so a direction by itself, right ? But when we calculate it, I hardly see how we know the direction towards which it points and so why this direction is the steepest ascent.
In fact when we are at the point where we calculate the gradient, and if the gradient is a slope, why should we draw the gradient as an arrow pointing towards the ascending side of the slope and not the descending one ?
Thanks a lot !
Thanks for your thoughtful comments. If you take an example, say f(x,y)=x^2+y^2 and calculate the gradient at a point, say (1,1), you'll see that the gradient vector is a 2D-vector. If you move along the gradient in the xy-plane, the curve traced on the surface will take you in the direction of greatest ascent. The gradient will naturally point in the right direction. If you changed to the function f(x,y)=-x^2-y^2 and again calculate the gradient at (1,1),and do the same thing, you will move along a curve in the direction of greatest ascent. I think you may be confusing 2D and 3D gradients. The gradient in this case is not pointing up or down, it lies in the xy-plane. Hope that helps.
@@rhondahughes9648 Yes thank you very much, I studied your example and better see it is actually doing that. I'm just missing the geometric intuition of why, when we calculate this rate of change, it "naturally" points towards the ascent. To be clearer and very simple, if I calculate the gradient of x², I get 2x and I totally agree that when plugging a value in it, I will go where the function increases the most. I just don't see why it makes sens, why it's not the other way (decreasing) because when I calculate this rate of change, I can approach my point from the left or the right, and a rate of change indicates how much a thing changes but not necessarily in which direction.
Anyway thank you very much, I still learned and visualized maany things thaanks to your videos, great work ! Best regards
Thank you so much!
previous Video ?
Thanks!
Thank you soo much
Ur soo much better than my lect
+Jawaid Rezai I appreciate that!
I have a problem can u help me with the answers
You should consider making more videos. There are many topics in mathematics that such charisma could clarify for many people.
Just a math teacher commenting.
Thank you!
Awesome video! I am a Civil Engineering Student at Cal Poly SLO currently enrolled in Calculus 4. Trying to pass my quiz tomorrow and learn a couple lessons in a day. Great examples! Thank you very much for sharing.
Thank you, Jacob. Good luck with your studies!
WHYYYYYY YOU HAVE ONLY TWO VIDEO'S ??????
I love ❤ u
Thank you!
Awesome explanation
Thank you!!
Excellent tutorial . Pleas upload more.
Great video! Thx
Thank you!
good teaching.
You have disscuss it very conceptually. Great...
+Muhammad Zeeshan Khan You're welcome!
where Can Get quality stuff lIke that ????????
Thank you so much! You motivate me to do more:)
So gradient can give us max descent and ascend rite ?
Not only max ascent
Correct!
MY REGARDS TO KATYA
Why is the gradient vector represented by a partial derivative parallel to the x, y plane and not as an actual gradient[ tangent ] to the surface at a specific point or time as is the case in 2D ? I mean we don't consider the gradient in R2 as a vector parallel to the x axis. I just can't get my head around this and any insight or clarification would be gratefully received.
Gradient is a vector in space and we are here working with 2D space (x and y directions), while z is corresponding to values of function f(x,y), not third dimension of space. So, gradient gives you direction in which function changes the most, but to get there you must move in 2D space, i.e. in x and y directions and function will assume a value depending on the point in 2D space you are at. Therefore, "skiing down the function slope" story was unfortunate.
In 3D space, it would look like w = f(x, y, z), where f can for example be a function that maps a temperature to all points in 3D space. In that case gradient would be a 3D vector, and function graph would be 4D. And if you instead use gravitational potential energy as a function od x, y, and z, you can start talking about skiing down the slope.
Also, note that if you apply the gradient definition given here to 1D situation, i.e. y=f(x), the gradient _is_ a 1D vector parallel to x axis, and its length and direction define slope of the tangent at a given point, not the tangent itself.
I generally agree with this response, but don't think I say "skiing down the function slope" anywhere in the video. It's very clear that the gradient is the direction in 2D, and the rate of change along the curve is greatest in that 2D direction. That was the main point of the explanation:)
Please bear with me, mathematics is just a hobby of mine, I have no real qualifications in the subject. I do however find what I consider ambiguity in certain mathematical definitions leading to confusion.....at least on my part. I fully understand that the " gradient " is the magnitude of the sum of the two partial derivatives and it's direction is in the x, y plane. But as I understand it, the magnitude of the gradient is a real slope in R3 , however its direction is represented by a vector in R2. I mean if I am on a hill and it slopes one way East and one way West then the gradient gives me the maximum or minimum value of the slope of that hill in the correct direction in R2. depending on whether I wish to ascend the or descend. Why can't the actual slope in R3 be represented by a tangent vector whose tail is at the point of tangency and has a slope equal to the magnitude of the tangent vector. I mean I have to climb up the hill , not into it !Another definition I find confusing concerns the surface integral where it is used as a density function to calculate the mass of an infinitely thin surface........how can a surface that is infinitely thin have a mass? Is this a lack of comprehension on my part, or just poorly defined concepts? Once again any clarification would be gratefully received .
I'm so sorry for the delay in responding. I will think about your question, and respond soon.
Thanks a lot. . . Please continue making these videos.
+Engineerated I'll do my best! Thank you.
Why u stop making videos?You are far better then those khan acedamy tutorials on black screen...
Thank you! Eventually there will be more:)
Great Videos, keep on, you can explain clearly
Thank you, Lorenzo!
You lost me on dot product. Dont have those dialed
thank you
You are welcome.
You are the best! Very clear explanation, thank you ma’am!
Thank you so much!
Очень интересное видео)) Вот что интересно градиент превращает скалярное поле в векторное, а дивергенция векторное в скалярное, т.е. F = grad(div(F)) , где F - векторное поле?
Hmm i think your x0 en y0 must by in the cirkel of x en y axes. Now it's fault.
You are absolutely correct! Thank you!