Khan Academy Video 1 (Gradient vs. Directional Derivative) #khanacademytalentsearch

under Hugh's coming to you from Winwood Pennsylvania with two of my favorite ideas for multivariable calculus I like to call this the tail that's tal e of the gradient versus the directional derivative now if you have never heard me speak before I don't usually sound exactly like this but I am losing my voice that seems to be an occupational hazard of trying to make videos for Khan Academy but we've got plenty of tea and we're going to soldier on and try and get the job done well I think most people would declare the gradient the clear winner here it is a mathematical celebrity it's known widely beyond mathematical circles by physicists chemists engineers even some economists now by contrast the directional derivative well kind of lives in the shadows that's a rain falling on the directional derivative it's much less widely known but my goal in this video is to thoroughly examine these two ideas and try and convince you that this is actually a true rivalry that much of the power that the gradient has is derived from properties of the directional derivative so in order to do that we're going to very briefly make a stop in two dimensions where we look at a function y equals f of X and just remind ourselves that if we graph that function and if it's graph is smooth which means no corners or cusps then at every point in the domain of the function let's say X naught if we plot that point there will be a tangent line that should only touch the curve at one point there'll be a tangent line to the curve at that point and the slope of that tangent line is none other than the derivative of the function at that point well that's all we need to remember from single variable calculus but our story takes place here in three dimensional Euclidean space we're now a function has two independent variables so let's say for example we take a point X naught Y naught in the X Y plane what could Z be well it might be the elevation at that point above or below sea level it could be the temperature at that point at a particular time or if Z is given by some mathematical expression it could be the number we obtain by evaluating that expression at the point X naught Y naught at any rate when we plot many points in the XY plane in general what we'll get is a surface a two-dimensional object unlike the one-dimensional curve that we get in single variable calculus now this is like a sheet blowing in the wind it might be flat but nevertheless it has two dimensions not one well I can immediately define our first major player in this saga I can tell you what the gradient is so this is the symbol for the gradient at the point X naught Y naught that's a Hebrew symbol it's a vector and the components of that vector it's a two-dimensional vector the components are the partial derivatives at the point partial of with respect to X and the partial of F with respect to Y well that's a two dimensional vector not a three dimensional vector so if I wanted to draw it where where should it be well it doesn't stick up like that it has no Z component so it is in the plane you're in a plane that's parallel to the XY plane so it might look like this I could draw it up here if I wanted to I don't know yet what direction it points in but nevertheless the important thing is that it's a two dimensional and not a three dimensional vector so now that we've mentioned the partial derivatives let's come back into three dimensions and remember what those partial derivatives mean at least geometrically we're not going to calculate partial derivatives we don't need to do that right now so let's start by drawing a line through the point X not Y not parallel to the x-axis and if you think about it for a moment this line is actually y equals y not because we're freezing Y but letting X vary now we can also extend this upwards and get a plane that's parallel to the XZ plane but I don't want to clutter up my picture too much but when we do that I like to think of this as kind of coming up and slicing through the curve or through the surface and what we get is a curve so now we have a picture very much like the picture that we had in two variables we have an axis and we have a curve above it now because I'm assuming that this surface is smooth and I am assuming that there's going to be a tangent line to that curve at that point and what is the slope of that tangent line well it's the partial derivative with respect to X at the point that's not why not okay now what about the partial derivative with respect to Y well we do the same thing but we draw a line now parallel to the y axis and if you think about that for a moment this is x equals x naught so this is the line x equals x naught we can bring it up into a plane and it too is going to make a curve and that curve has a tangent line and the slope of that tangent line is the partial derivative of F with respect to Y at the point X naught y naught now a perfectly reasonable question or thought at this point is why do we only have to draw lines parallel to the two axes why can't I draw any line through that point and do the same thing make a plane cut through the surface what would I get I would get another curve and I want to be able to talk about the slope of the tangent line to that curve so in order to do that and we're getting close to our second player here the directional derivative is lurking in the shadows at this point so what I want to do is think about this blue line and think about the blue curve and I want to figure out how to find the derivative of this curve at this point so how can we do that well first of all let's call this line L it's not quite like the X or Y axis we don't really know which direction is going to be positive but let's take a vector you and let's assume that U is a unit vector which means that it has length one that just makes certain things easier and let's let u have components a B now how do we write the equation of that line well it's a line in the XY plane so it should be able to write it in point-slope form or in slope-intercept form but the easiest way for our purposes is to write this line out using parametric equations and let me show you what those look like so T is a parameter now where does T range well it can be any real number but let me try and convince you that that describes this line well when T is zero we get the point X not Y not fair enough but what happens as T varies well when T is positive we're going to go in one direction negative we're going to go in the other you still not not be buying this but let me remind you that this is what we're calling X this is y and when you have parametric equations like this we could solve for T right we could solve for T in fact T is X minus X naught over a and then we could take that value of T plug it in here and just glance at what we get we get Y is equal to Y naught plus B times this well there's an X in there and if you rearrange things a little bit you will see indeed that we get an equation of the form y equals MX plus well I can't use a B because I've already done that but plus C so this indeed is a line through the point X not Y not in the direction of U well now we want to figure out how to find this derivative well let's think about that each value of T gives us a point here and here's our function f computed only at the points along that line so let's see what that looks like okay we have the variable T once we pick T we get two points X naught plus 80 y naught plus BT while we get two coordinates this is X and this is y and then we plug those in and we get f of X naught plus 80 y naught plus BT okay well what do I want to find I want to find DF DT the derivative of f with respect to T when T equals zero and that my friends is going to be the directional derivative in the direction of U of the function f at the point X naught Y naught well that might seem like a very hard thing to calculate but let's take away our picture which I hope has been helpful and remember that we have another very useful concept coming to us via single variable calculus and then multivariable calculus how do we find a derivative going from the top level to the bottom level at least that's the way I'd like to think about the chain rule well you have to pass through the middle layer okay the middle earth I don't know but we take appropriate derivatives along the way well what does that mean this is the partial of F with respect to X because F depends on two variables and then we get DX DT so we differentiate here and then here plus the partial of F with respect to Y dy DT but let's take a look at that we know what dx/dt is Hey and we know what dy/dt is it's B if you look at that for a moment you might think that looks a little bit familiar well that is a dot product the dot product of what two vectors the partial of F with respect to X partial of with respect to y dotted with the vector a B but that is none other than the gradient of F and it's at the point X naught Y naught because when we put t equal to zero we get that point here and a B of course is U so this is the definition the working definition of the directional derivative that's quite easy to calculate in most cases and there's our first player the gradient so closing down this first video what I've shown you is that well I haven't made my case yet have I because I have shown you that the directional derivative very much relies on the gradient R it's very existence but in the next video what I hope to convince you of is that there's action in this direction as well and that as we look at properties of the directional derivative we're going to see how it makes the gradient as useful as it actually is so I'll see you in the second video bye for now thank you

Channel: Rhonda Hughes

Views: 127,033

Rating: 4.8870811 out of 5

Keywords: Gradient, Directional Derivative, Multivariable Calculus, Vector Calculus, Level Curves, Partial Derivatives, Vectors

Id: NomUbVmmyro

Length: 14min 11sec (851 seconds)

Published: Mon Jun 22 2015