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.
@rhondahughes96487 жыл бұрын
Thank you very much, Mukta!
@awolgeordie99267 жыл бұрын
Yes. Same here. I'm a teacher too and you described this really well. Thanks!!
@rhondahughes96487 жыл бұрын
Thank you!
@rhondahughes96487 жыл бұрын
I'm happy to hear that! You are very welcome.
@ankurc6 жыл бұрын
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!
@alejandroalmarza84472 жыл бұрын
Excelent explain …you are a great professor…looking forward to par II and more Vector Cálculos videos from you.
@ozzyfromspace5 жыл бұрын
5:42 "At any rate..." Oh, sweet sweet music to my ears!
@sukhmanikaurtoor95817 жыл бұрын
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
@rhondahughes96487 жыл бұрын
Thank you so much! I'm glad this was helpful.
@dangargiullo96345 жыл бұрын
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!
@rhondahughes96485 жыл бұрын
Thanks so much, Dan!
@Cheeriot5 жыл бұрын
Wow you're gifted. And most certainly talented! It finally clicked for me. Thanks
@rhondahughes96485 жыл бұрын
That is very kind, Yaman. Thank you!
@petergianf8 жыл бұрын
this is so much clearer than my Calc 3 professor's lectures... thank you so much!
@rhondahughes96488 жыл бұрын
+petergianf Thank YOU!
@karthiksirusiruvalam46855 жыл бұрын
Thank u mam. For our excellent and mind-blowing and easily understood lecture
@rhondahughes96485 жыл бұрын
You're welcome!
@wsmao11547 жыл бұрын
what a clear and impressive explanation!! thank you so much anyway!!
@rhondahughes96487 жыл бұрын
Thank you very much!
@keshavamurthy25853 ай бұрын
Awesome explanation
@rhondahughes96483 ай бұрын
Thank you!!
@guitarttimman4 жыл бұрын
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?
@rhondahughes96484 жыл бұрын
Thank you very much! Your last question is one we all try to answer. It's a great mystery.
@jacobacunavirgen14748 жыл бұрын
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.
@rhondahughes96487 жыл бұрын
Thank you, Jacob. Good luck with your studies!
@miltonweinhold68598 жыл бұрын
wow, katia looks nice. thanks
@rhondahughes96488 жыл бұрын
+Ander Stuhl Katya is nice!
@PhilV2058 жыл бұрын
This was absolutely breathtaking. Thank you
@rhondahughes96488 жыл бұрын
+PHaitiano Thank you so much!
@abcdxx10596 жыл бұрын
This was very helpful now I can understand gradient descent
@rhondahughes96486 жыл бұрын
Thank you!
@raafeh96018 жыл бұрын
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.
@rhondahughes96488 жыл бұрын
That's correct. Just trying to keep it simple, but thank you!
@sidravi38858 жыл бұрын
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....
@borisdragnev78756 жыл бұрын
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?
@dihan61306 жыл бұрын
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.
@muhammadzeeshankhan72518 жыл бұрын
good teaching. You have disscuss it very conceptually. Great...
@rhondahughes96488 жыл бұрын
+Muhammad Zeeshan Khan You're welcome!
@liesalllies5 жыл бұрын
What a lovely couple of videos you've made! I wish there were more :(
@rhondahughes96485 жыл бұрын
Thank you so much!
@guitarttimman4 жыл бұрын
Rhonda. Thank you for your eloquent explanation of this abstract topic. I love the visual demonstration. Please be safe.
@rhondahughes96484 жыл бұрын
Thank you! You, as well.
@Raynover3 жыл бұрын
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.
@rhondahughes96483 жыл бұрын
Thank you!
@andrewtey25872 жыл бұрын
👍
@rhondahughes96482 жыл бұрын
Thank you!
@Nitish-Rathod5 жыл бұрын
Great video, 👌👌👌please make more videos
@rhondahughes96485 жыл бұрын
Thank you!
@rhondahughes96485 жыл бұрын
Thank you!
@harshal1uplavikar5 жыл бұрын
hey please maKe more videos
@user-xt8xs3jp1n2 жыл бұрын
You are god
@wenchenshi97743 жыл бұрын
Please post more videos!!!!!!! Thank you!!!!!
@rhondahughes96483 жыл бұрын
Thank you!
@Festus20224 ай бұрын
Great video! Thx
@rhondahughes96483 ай бұрын
Thank you!
@yonasghirmay64733 жыл бұрын
Great work mam math is amazing
@rhondahughes96483 жыл бұрын
Thank you so much!
@user-xm9xo7jg4u3 жыл бұрын
Your videos are AWESOME. Please post more. Thanks so much!
@rhondahughes96483 жыл бұрын
Thank you!
@lorenzoservadei8688 жыл бұрын
Great Videos, keep on, you can explain clearly
@rhondahughes96488 жыл бұрын
Thank you, Lorenzo!
@paedrufernando23516 жыл бұрын
hi Rhonda.. wishing you well...however if you could upload newer videos ..it would really make a difference
@guitarttimman4 жыл бұрын
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.
@alejandrosanchez97095 жыл бұрын
Excellent tutorial . Pleas upload more.
@kina4288 Жыл бұрын
the first part was great, the second part was too deep for me. Thanks all the same.
@rhondahughes9648 Жыл бұрын
Thanks for trying!
@hqs95853 жыл бұрын
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.
@rhondahughes96483 жыл бұрын
Agreed!
@engineerated56278 жыл бұрын
Thanks a lot. . . Please continue making these videos.
@rhondahughes96488 жыл бұрын
+Engineerated I'll do my best! Thank you.
@monayin64795 жыл бұрын
You are the best! Very clear explanation, thank you ma’am!
@rhondahughes96485 жыл бұрын
Thank you so much!
@MS757-p3l4 жыл бұрын
Nice vedio. Can you please explain how gradient as a 3D vector works
@JJ-je4wl8 жыл бұрын
Thank you soo much Ur soo much better than my lect
@rhondahughes96488 жыл бұрын
+Jawaid Rezai I appreciate that!
@rish_hyun4 жыл бұрын
You saved me! Thanks 😀
@rhondahughes96484 жыл бұрын
Good to hear! Thank you.
@ankurc6 жыл бұрын
Thank you so much!
@jacobvandijk65254 жыл бұрын
@ 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 ?
@jacobvandijk65254 жыл бұрын
This may help, Rhonda ;-) kzbin.info/www/bejne/q56bammLqKt-Y8k
@guitarttimman4 жыл бұрын
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?
@ZinzinsIA2 жыл бұрын
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 !
@rhondahughes96482 жыл бұрын
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.
@ZinzinsIA2 жыл бұрын
@@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
@user-gc6we4co4f2 жыл бұрын
Thanks!
@TripedalTroductions8 жыл бұрын
WHOA!
@uhbarp81373 жыл бұрын
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...??
@abcdef20697 жыл бұрын
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
@userhdza22483 жыл бұрын
previous Video ?
@straycatgirl9 жыл бұрын
MY REGARDS TO KATYA
@harshal1uplavikar5 жыл бұрын
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 .
@naveennaidu90642 жыл бұрын
I love ❤ u
@rhondahughes96482 жыл бұрын
Thank you!
@barryhughes97647 жыл бұрын
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.
@oqardZ7 жыл бұрын
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.
@rhondahughes96487 жыл бұрын
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:)
@barryhughes97647 жыл бұрын
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 .
@rhondahughes96487 жыл бұрын
I'm so sorry for the delay in responding. I will think about your question, and respond soon.
@harshal1uplavikar5 жыл бұрын
where Can Get quality stuff lIke that ????????
@rhondahughes96485 жыл бұрын
Thank you so much! You motivate me to do more:)
@theunknown20905 жыл бұрын
So gradient can give us max descent and ascend rite ? Not only max ascent
@rhondahughes96485 жыл бұрын
Correct!
@harshal1uplavikar5 жыл бұрын
WHYYYYYY YOU HAVE ONLY TWO VIDEO'S ??????
@Saikumarreddypasam5 жыл бұрын
I have a problem can u help me with the answers
@bvenkatsai8 жыл бұрын
thank you
@rhondahughes96488 жыл бұрын
You are welcome.
@zahar_AI8 жыл бұрын
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?
@rohitagrawal13766 жыл бұрын
Why u stop making videos?You are far better then those khan acedamy tutorials on black screen...
@rhondahughes96486 жыл бұрын
Thank you! Eventually there will be more:)
@dagoninfinite6 жыл бұрын
You lost me on dot product. Dont have those dialed
@yveshelsen91688 жыл бұрын
Hmm i think your x0 en y0 must by in the cirkel of x en y axes. Now it's fault.