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Welcome to another week of vector calculus What I’m going to talk about today are gradiant vectors in tangent planes We have already talked about how to find a partial derivative with respect to x or with respect to y or with respect to any other variable It turns out because of that, there isn’t a derivative by saying the partial derivative What we’d much rather do is find the derivative like we did in Calc 1 when we just found the derivative It turns out that the gradient vector is going to be exactly what we need to talk about the derivative, but it won’t give us a number. It will give us a vector and that’s why we call it the “gradient vector” We will start out to introduce you to this idea of the gradient vector, we will talk about the directional derivative and that’s the idea of saying instead of going in the direction of x, say on a map, that would be going due east or instead of going in the direction of y, that would be going due north, maybe you want to go northwest or northeast or some other direction That is the directional derivative It tells us how to find the derivative of a multi-variable function in the direction of a vector After that, we will get into the big thing of the day and that is the gradient vector. We will look at using the gradient vector to find the directional derivative We will define the gradient vector in two different ways. One, we will define graphically and also we will define it graphically or with proper calculus Then we’ll look at the gradient vector’s properties and the things you can do with the gradient vector We’ll have a nice little application Then after that, We will get into tangent planes We have seen in first quarter calculus or calc 1, that if you have a curve then the big thing in calculus is what is that the derivative at a point. That derivative helps you find the slope of the tangent line It turnes out if you want tangent plane then we can kind of think of the derivative talking about not the slope of the plane because that’s not what we talk about, but the normal vector to the plane That’s the day as we go, so hopefully you will enjoy this lecture Let’s start out with the directional derivative I am lookinag right now at this hiker You can think about the hiker to have put his foot down and the hiker has lots of choices That hiker can go up or down or left or right in any direction he wants to go and if he goes, say, towards the left, he will be walking upwards He will walk up the hill. If he goes to the right, he will probably be walking downwards. We want to look and say if you pick a direction not just will he be going up, or will he be going down, but by how much? In other words, if he takes one little step, in one unit in the direction of a unit vector, how much will he be rising or dropping in elevation? And that’s what the directional derivative will talk about In particular, the directional derivative of a function: z = f(x,y) in the direction v, where v is some vector, that directional derivative measures how steep the surface is; not just how steep upwards, but how steep downwards also Is it going up? Is it going down? And by how much? In that particular direction. So how does a z change when moving along v? and as we’ve seen before when you have a function of two variables It doesn’t necessarily mean up or down in position. Sometimes the function might represent temperature. We want to say how much hotter is it getting if you go in this direction or that direction, or maybe it is electric field strength That is, how much is electric field strength changing one direction or another?

It could be all kinds of different things It talks about how much the dependent variable is changing if we move in a particular direction This picture shows several level curves for a function f(x,y,z) of two variables and a point right here that’s marked The question is, What are the directional derivatives in several directions? Let’s start out with the direction of the vector u = <1,0> That direction is due right or due east depending on how you want to look at it The question is, What is the directional derivative at that point in the direction of <1,0>? What we can think about is if you take one step in that direction how much is a function changing? The way you figure this out is we draw up a little arrow in that direction of unit 1 and we say that we move from the dot to the end of the vector, how much has the function changed? We can see we are staring at the level curve 10 and we are going towards the level curve 15 but not all the way It looks like it is about 3 in change in the f, or the function We can say that the directional derivative of this function in the direction of u = <1,0> is about equal to 3 Let’s look at another direction. Let’s suppose that u = <0,-1>. Now we will take one step due south and we want to find out how much is the function changing if you take one step due south Here, we go from 10 towards the number 5 And it looks like we have gone most of the way, but not all the way. So it looks like we have gone about 3 downwards in the function So we can say that the directional derivative of this function in the direction of u = <0,-1> is about equal to -3 Let’s look at one more example but suppose now we have the direction of the vector <1,1> I will draw a vector of length 1, and I want to see how much as the function changed if you move in that direction Now, instead of going due south or due east, now it looks like we are going northeast If you go northeast, it looks like you are going from this level curve 10 up to about 15. Now the directional derivative in the direction of <1,1> for the function f is about 5 Hopefully, this gives you an idea graphically what it means to talk about the directional derivative Again, these are just a approximations which is what you do with graphs, but for a lot of the time, that’s good enough Now let’s find out what is the gradient vector, because that will help os a lot to find directional derivative and it will do a whole lot more for us It’s the following. We have already had partial derivatives of a function with respect to x and y, What the gradient vector does is it creates a vector that lists those partial derivatives In particular, let’s suppose that f(x,y) is a differentiable function, then the gradient of f, we often write an upside down triangle to represent the gradient of f, is equal to the vector you can also write it in physics notation which is del f / del x i + del f / del y j In math, somethimes we use pointy brackets It’s is a little less writing We can also say that the directional derivative

in the direction of a unit vector u of a function f is the gradient of f dotted with that unit vector I won’t prove it but you could prove it using limits Use the definition of partial derivatives with limits the you show that this actually works You can look this up in a textbook or online and you can see a proof of this It should be somethat convincing that the directional derivative of f in the direction of a vector is always equal to grad f dot that vector Let’s look at some properties of the gradient What I have here is a topomap and this is one of the most beautiful places in the world. I am a little biased because I live real close to it. That is Emerald Bay There is some great hiking around there The question is, if we look at the function of two variables, the two variables now will just be the latitude and longitude, and the function will give you the elevation, then what is the gradient? If you have a level curve, the gradient is quite nice The gradient is always in the direction of maximum steepness That tells you that if you take one step in the direction of the gradient you will always be walking in this the steepest possible slope you can walk and that is the steepest upwards by the way Also, the gradient of f is orthogonal to the level curve of the function at that point. Here is level curve. And if you walk in a perpendicular direction to the level curve, then you’ll be walking along the gradient direction That’s really nice and really important and useful in all kinds of ways One thing it’ll tell you, for example, what direction water is likely to flow. It tends to flow in the steepest direction. It tells you what the temperature might look like. All kinds of different things. In terms of knowing the gradient of f, it is in the direction of maximal steepness That is because the directional derivative of a function, if you remember, is equal to the gradient of that function dotted with the unit vector The dot product of two vectors is the product of their magnitudes times the cosine of the angle between them But the cosine of an angle is a maximal when that angle is 0, and angles are 0 when the two directions are parallel to each other That tells us that the gradient is in the direction of maximal steepness because it is in the direction in which the directional derivative is the largest Now the see about finding the gradient Here’s an example Let’s let f(x,y) = e^(x + 2y) cos(y) Let’s find the gradient of f(x,y) To do that we find the two partial derivatives f_x, the partial derivative of f with respect to x, is equal to, now we take a derivative, we use the chain rule: the derivative of x + 2y with respect to x is 1, the derivative of the exponential is itself, we just get 1 times e^(x + 2y) and then cos(y) is a constant, so you just multiply by that constant cos(y) for the derivative with respect to y, now we have to use the chain rule. The derivative of x + 2y is 2 if it is with repsect to y There’s my 2, times the derivative of the exponential, that’s the exponential: e^(x+2y) times the second which is cos(y), I’m using the product rule to find this derivative because we have a product of an exponential and a cos and they both have y’s in them. I take the derivative of the first times the second and then plus the derivative of the second The derivative of cos(y) is negative that’s why it is a “-” times sin(y),

then times the first I can now put this together to get the gradient The gradient of f is equal to the vector e^(x + 2y) cos(y), that the partial derivative with repsect to x, 2e^(x + 2y) cos(y) – e^(x + 2y) sin(y) A lot of times we are interested in the gradient at a point. Let’s find the gradient of f at the point (1,0) That just means plug in one for x and plug in 0 for y Notice that e^(1 + 2*0) = e^1 = e cos(0) = 1 We get e*1 = e. So the first component is e For the second component, we get 2*e, remember we are plugging in the same value, times cos(0) and that’s 1. We just get 2e and then minus, well the sin(0) = 0, so it goes away. We just get 2e. We can say that the gradient of f at the point (1,0) is equal to the vector That’s how you find the gradient. You just find the partial derivatives and then if you need to, you just plug in the numbers after. This an example where we only have two variables but if we have three or more, it is the same idea for the gradient instead of just the partial derivative with respect to x and the partial derivative with repsect to y, you keep going and you take more partial derivatives in order Here is an example, a nice little application. It is predicting the wind direction Let’s suppose the barometric pressure today is P(x,y) = sin(x) + ln(x + 2y) The question is,Ffind the direction of the wind at the point (0,3) Let me give you a little bit of meteorology and what that is is that wind is caused by a change in pressure Wind always goes from high pressure to low pressure in the direction of the maximal change in pressure that turns out it will be the negative of the gradient, because the gradient goes from low to high and wind goes from high to low In fact that’s why if you’ve ever noticed, if we are in one weather pattern today and tomorrow we will have a different weather pattern, so maybe we’re going from a high pressure system to low pressure system so it is a change in barametric pressure that we will end up with maybe rain tomorrow because that’s what happens You’ll notice that when you have that change in the pressure system you have a windy day That’s because you have a high gradient it turns out Let’s find the gradient To find the gradient we take the partial derivatives The derivative of sin(x) is cos(x) For the derivative of ln(x + 2y), we use the chain rule: the derivative of x + 2y = 1 The derivative of ln is 1/(x + 2y), for the partial derivative with respect to x, we just get cos(x) + 1/(x + 2y) For the partial derivative with respect to y, sin(x) goes away. It has a 0 derivative Then lwe use the chain role. The derivative of x + 2y with respect to y is 2, then the ln tells us to bring it down stairs, so we get the derivative with respect to y is 2/(x + 2y) There’s our gradient. We want to plug in the point (0,3) We do that. cos(0) = 1. We have 1/(0 + 2*3) = 1/6 We get 1 + 1/6 = 7/6 Similarly, we get 2/(0 + 6) = 2/6 = 1/3 What we really want is the direction of the wind. So, let’s make that a unit vector To do that, I multiply everything by 6 and we get <7,2> Then find the magnitude of <7,2>. 7 squared is 49, 2 squared is 4, 49 + 4 = 53. We get < 7/root(53) , 2/root(53) >

That’s the gradient so now that tells us that the wind direction which is the opposite direction of the gradient will be < -7/root(53) , 2/root(53) > There’s our wind direction. Now let’s talk about tangent planes Tangent planes, as I mentioned, are really important and in fact in the old days they thought the earth was flat. Basically, they were looking at a tangent plane They didn’t have a satellite to go out and see this big spherical earth. They just saw the plane. We want to find the equation of the tangent plane to a surface at a point Let me remind you that in general to find the equation of a plane with normal vector v and point x_0, we take v dotted with as a vector and we set that equal to 0 Remember that v is a normal vector to the plane and x as a vector is , and x_0 as a vector corresponds to a point on a plane written in vector form Here’s the big definition. And what it says is that if f(x,y,z) = 0 is a differentiable surface at a point x_0, then the tangent plane has normal vector n equal to the gradient of f at that point That makes all the difference for us, because in order to find the equation of a plane you need a point and you need a normal vector The point is almost always given, and the normal vector is the gradient Here’s a picture of what is going on This is a surface. I like to think of a surface as a quilt blowing in the wind If we look at one little parallelogram of that quilt, It is a parallelogram because it is blowing in the wind, then if we point directly outward, that will be in the direction of the gradient of f and that is the normal vector to that surface Let’s look at an example of finding the equation of a tangent plane The question states: Find the equation of the tangent plane to the surface f(x,y,z) = 2x + yz^2 – 8 = 0 at the point (3,2,1) We need to find a point and we need to find a normal vector The point is just (3,2,1) For the normal vector, use the big definition which says the normal vector is the gradient. The gradient of this function is the vector of partial derivatives The partial derivative with respect to x is just 2, because 2x appears here and nowhere else is there an x We get 2. The partial derivative with respect to y is z^2 And the partial derivative with respect to z is 2yx. If we plug in (3,2,1), Then f_x = 2. There is no variable there so it is just 2. f_y is z^2. 1^2 is just 1 And f_z = 2yz. 2 * 2 * 1 which is 4 And the gradient is normal to the tangent plane So the normal vector is this vector: <2,1,4> Now that we have the normal vector <2,1,4> and the point <3,2,1>, we can find the equation of the tangent plane We can say that the vector <2,1,4>, the normal vector, dotted with is equal to 0 Now let’s take the dot product. Notice that we get 2x + y + 4z Then 2*(-3) = -6, 1*(-2) = -2, and 4(-1) = -4

-6 + -2 + -4 = -12 I add 12 to both sides and I get 2x + y + 4z = 12 and that is the equation of the tangent plane to this surface at the point (3,2,1) That’s all there is to it Let’s do one more example that will look a ittle different and that is the following: find the equation of the tangent plane to the surface z = x^2 e^(3y) at the point (2,0) Now this looks a little different because now we have a function of two variables and our point just has x and y The trick to this is just change this to a function of three variables and make it a point (x,y,z) In particular, we can subtract z from both sides of the equation and we can get x^2 e^(3y) – z = 0 Now we have the form that we want If we take a gradient by finding the three partial derivatives, we have f_x = 2x e^(3y) then it’s – 0, so 2x e^(3y), f_y is going to be, the chain rule says the derivative of the 3y is 3, we get 3x^2, the derivative of e is just the exponential, times e^(3y) and then – 0 and f_z is just 0 – 1 or -1 now let’s plug in the point (2,0) If I do that I get 2*2 = 4, times e^0, and that just gives us 4 We get 3 * 2^2 = 3*4 = 12 times e^0, and that just gives us 12 Then for the third component, it was just -1 There’s our normal vector For the poiint, we have a little more to do, because we are given (2,0), but it’s no big deal because z = x^2 e^(3y) so we get 2^2 e^0 which is just 4 So we can say that the vector x_0 is (2,0,4) Now let’s just put this all together We get the normal vector (4,12,-1) dotted with the x vector minus the point vector, so < x - 2 , y - 0 , z - 4 > is equal to 0 Now the rest is pretty basic. We just take the dot product We get 4x + 12y – 1z. That’s 4x + 12y – z. Now let’s do the arithmetic of the dot product. We get 4 * (-2) = -8 We have 12 * 0 = 0 and (-1)(-4) = 4 -8 + 0 + 4 = -4 We add 4 to both sides and get: 4x + 12y – z = 4 That’s all there is to it That’s all I have today to talk about the gradient vector and tangent planes If you have any questions, please let me know. I’m happy to help. Or if you have another instructor, please ask your other instructor and I am sure your other instructor will be very happy to help also Until next week, have a great week. Make sure you study and practice and practice and practice so you can get good at this Next week, we will have a new topic, so have a wonderful week

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