Hello. Our previous session tangent plane We have seen how to obtain. As we know that the equation surfaces covered outdoor or representations with parametric functions have achieved. So one third of the surface function before any can be shown were examined. Key issues to write the equation of a plane There is a great need for. And at that point of the surface, perpendicular to the plane vector needed. Since x is zero, because a point We choose Since the surface that other equations calculating vector. Here you will see both examples better concept of the tangent plane practice We will find the opportunity, as well as accounts operational To win, hands to dig habit we will have the opportunity. Now we start with a simple example. As open as a function of a given Consider the paraboloid. Is equal to two, on which the x, y Let's take a look. Already these two points, two arguments We easily find z given. x and y are two and Replace z is equal to one minus two really we find. We perspective view of a paraboloid If we draw, when such x and y is zero minus four and wherein z is Due to the downward pointing marks We know now that the paraboloid. In this way, a downward facing paraboloid We are working on the surface. These different methods that these fundamental issues find n given x is zero, because here given. N obtained from the equation of this surface that We will. Open to us at first as a function has been so negative z and y function. Where x and y at the left side subject to passing marks glance will be positive. x squared plus y squared plus z four of two Once we have left to sign this will change 're getting off the function. Off function when working with x, y, z equal value-weighted perpendicular vector giving it to him this off function we have a great show with f, x, y, z function of x, y and z respectively, we find the partial derivatives is required. This perpendicular vector at any point. Get short, derivative with respect to x two x, y Take the derivative with respect to four of y, z-derivative is taken in by a We see easily now. Choose this perpendicular vector at the point numerical when he calculated as the work of two to x, y a, z does not already appear here because it is a linear function came for. The perpendicular vector four, four, one We see that now. Four, four, with a perpendicular vector x minus x zero components When we separate the two, because x is x minus value at this point two. y because y is a negative value a. z because z value minus two plus two. As you can see by this inner product received four four x plus y plus z is happening. Then the numbers going. They structure the de've achieved the equation of the plane that is going on. So here's a curved surface that we choose the surface tangent at the point of We found the equation of the plane. Of course, a linear equation of the plane because it is a function of x, y, and z always only in terms of first-degree where the terms it is also just for a plane we see. Open when he calculated with the function of a gene we found in the previous session and we use As before, we use partial In determining the meaning of derivative saw. of x and y is the vector function which explicit function of x and y according to the partial derivatives minus and a third component selected from We know that. As you can see when their accounts to x derivatives by a minus two x's going to be here four years a happening. As you can see completely closed with the function We find that we obtain orthogonal vectors. Thereof according to x and y values of the subject We find when he calculated the same orthogonal vectors. The same perpendicular vector, x is the same in zero naturally obtained for the same tangent plane We will. This involved. Parametric representation of functions in the many There are many options. So this is a clear function given parametric functions to convert. We start with the easiest. After that, I give you an assignment. The second shows it with a parametric, the What it is I will show that we can do more than that There may be another indication. We're starting with the simplest. We call a parameter x, y a u parameters, there is what we call parameters. According to this u and v z is now self- involved. Squared minus four minus two times the square have so as xi two vectors u and v parameters We have shown as a function going on. This parametric representation of a surface. On this surface perpendicular vector, u and v in the direction of the tangent vector We find as a product. So has the surface. Have a vector direction on the surface. there are curves in the direction vector. They are the product of the vector perpendicular us giving. These courses have seen in the past, but such as formula. This is based on the derivative of the vector x When we received a coming here, there and v is independently from each other because u There is no change compared to. Because the independent variables are vectors here The second component of the zero vectors involved. when u take the derivative with respect to x in minus two turns out. As you can see here, I've got a vector. According to this same vector V x V is the partial derivative. According to V. Gene partial derivatives of u is zero. Because u and v independently of one another. v is a derivative. The term Indeed, from the derivative with respect to v turns out there is minus four. I repeat again on their surface V. When we hold constant u we obtain line. When we keep in constant we obtain along lines tangent vectors v is going on. We selected a point x in them, was two, has a year. This is because we also have x and y u v is giving the parametric values. Numerically vector xi, x v We find numerically vector. We know that the multiplication of the two vectors determinant of the i, j, k 're writing. We are writing the first vector, the second vector 're writing. When I hit this determinant component As you can see zero plus four, a sign from minus four four with the change. We are starting with j axis. j threw the column and row minus four involved here. This minus four're writing here. If k components, where k rows and turns throwing a column. You're writing. As you can see the page again in a prior we found four, four, union open function or equivalent closed in functions again as we found four, four, one we find. say that the same orthogonal vectors involved. According to the same point that we have chosen tangent plane will be the same. I told the function, paraboloid also another parametric representation I might. If we do here with circular coordinates easily by placing it you can find. This goes to the parametric representation. Wherein x is zero, M, x is zero the two where y is zero, A We find the value of z is zero at some point. When we give to x the two get here We are. Obtained using these r is zero, theta zero, Knowing sine cosine of theta We find zero. Here I give you guidance as such as There are two parameters r and theta here. When we get their derivatives r and gene by tangent vectors theta we find. This is the cross product of the tangent vector perpendicular vector We find and obtain the same perpendicular vector You will also see that we have. As a second example on a spherical surface We want to find the vector perpendicular and from here We want to find the tangent plane. Whether we choose three of the sphere radius. And also on this sphere we choose x zero and stem and root two to three years to get zero points. Of course we chose this point to the beginning of distance shall be three. Indeed, the length of this vector it account, if we Three of the square root of three, the square root of two frame was two to five. However when you add two to four in the frame Nine out of here. Of course, the square of the distance from It was. Get it by the square of nine provided that the radius of the square root of three We're going. Gene function in this indoor, outdoor function with representation and parametric functions We will find the tangent plane. We actually know from geometry. The vector perpendicular to the selected point, any a direction that connects the central point is that. Therefore in this example, the geometry we know that we can make sample accounts we have provided a foreboding accounts we choose we could be. Also very commonly used sphere is the model. Our ability to calculate on the sphere I chose this example because that causes. As a function of the sphere is now closed We know the function. x squared plus y plus z squared is equal to the square the square of the radius. We are here this off function We are producing. Vector perpendicular to the x to y of f z obtained by the partial derivatives We know that is. Just two of the derivatives x here, two years, two We find that z. Select this when he calculated at the point in the We find numerically the perpendicular vector. As you can see, this is a vector perpendicular two-fold multiplied by the number of condition. But the length of this vector is important to us it is not. After five times the size of one tenth albeit Important direction, direction. For this reason, these two simply throwing more We see the opportunity to work with the vector. This vector also really geometry We chose this point as we know it The same joining centers vector see that. So this geometry provided foreboding We're going. Of course, to find the equation of the tangent plane Want to be the two of you like this numerically to do without that other vector are writing. the components of x minus x is zero 're writing. Wherein x is zero the components of the root three, two and two radicals is that using Edit your current vektö by taking the inner product of the following we obtain the equation of the plane. Plane equation, because the x, y and z in terms of There are only first forces. At this point we use the perpendicular vector for tangent plane geometry equation is In terms provide. Still easy to open accounts with the function but a little more difficult to calculate. Why is that? Because of this sphere from the equation z we solve When we get something karekökl. This is not a difficult thing, but a little more account difficult. Other negative for the vector x, ie negative for x to y and partly to y We know that derivatives and consists of multiple, We learned in the previous session. So, according to the partial derivatives of x and y he gets, we'll get out of here. Partial derivatives of this function with respect to x pick up two forward slashes in front to a fall. This forces one to one-half of We're taking. Thus, a negative divided by two is going. And to take the derivative of. Derivative to the minus two x with respect to x. According to y minus two years. Here we arrange them, this We see that two of the shorter one another that simplify. Only minus x minus y and remains. If this occurs in the denominator of z itself. Because wherein the f, x, y, z itself. So that negative forces denominator for z is happening. Gene wherein X, Y, Z using we obtain orthogonal vectors. f x where x is the root of three, y's root of two. Wherein cons due wherein hence the plus signs going missing. We have a denominator. intermediate compound of the two. We put him. Here, too, but this one had two in the denominator. Therefore, this is a common denominator bring two going on. Yet it becomes simplified because you can get one-half length but significant, albeit the direction 15 perpendicular vector. We'd get the same perpendicular vector we see. This is already the zero point x, three components of the root, root of the zero point two and two in the x, circle, the center of the sphere in the same direction vector that connects are making. A general conclusion from the known geometry. Parametric representation of functions as well We need to learn a little bit, but about the sphere makes eye off operations because we need to become both in physics, chemistry courses both in practical in the sphere of engineering courses too a model used. For this reason, an account of the random instead of doing exercises that sphere Because we chose. Sphere equation in spherical coordinates such that in terms of two parameters, fi and theta terms. This fun and theta angle here. Ie from the northern axis Fi in geography We measure the angle from the north pole. The x-axis from a selected theta We measure the angle of latitude. So, geo, from geography to geometry we know on longitude and latitude angles. Here, given the more abstract case. These coordinates were previously interests. Would suffice to remind you here. The projection of this point on the x y plane We find that when we bought a small r. R is small with the same z-axis look. This is the distance of the point ka, coordinate relative to the center multiplied by the sine of the distance from f. After you find this r x and y components same circular coordinates as in, instead of radius r we have received for We said that the sphere of radius a sinus We have found multiply by f r. We hit a more teta'yl cosine of x found. Again we hit the radius r have found sinus. We hit it teta'yl sinis take, y found. z-component of this point to the center of the stratum connecting the projection on the z-axis vector time. When the cosine projection on the z axis future. r, to find the angle opposite the has length r. Then came the sinus. Here comes the cosine. It was a reminder. Our old course in it 've seen. So based on that given radius of x, o particular, fi and denominated variables theta vector function. Now here we have to do an additional process because you have given us zero point x but opposing it for zero, theta We do not know zero. This geometric variables u and v instead of I prefer to use. In general, because parametric functions a u and a pair of vector functions and parameters were defined as. We fi v is the geometry of the we see that theta. Its easy to get more monitoring he terms the geometric variables We will do our calculations. Please assume that the point of this first The geography Istanbul gave the coordinates, x and y of z. Which this latitude, longitude opposite which come from? You're trying to find, what we do equivalent. The shortest path is as follows:z has been provided for two basis. in the equation is just the cosine of z f. As you can see in the equation of x and y If both theta and have fun. Thus easier to start with z because is there an unknown, where two are unknown. Thus, step by step to find the plug ago We will find the back of the theta. This will make the process. In order to do this here again I wrote. x in terms of the theta and for a while in our two three while Once in a while our equation for the cosine Let us take instead of z and a. We obtain an equation as follows:This is no longer variable is not fun, fi at the point that we have chosen for him, here we put a zero indicator. An equation in one variable here for we find zero cosine. And of course the cosine of zero is found for sinus fi has zero immediately. We know the sine squared plus cosine squared is equal to nothing more than an equation it is not. So here we have found the plug. With this plug these values x Want Want y find theta using known, because Now we know the plug. This is obviously one of the two unknown. Let's select the root of x the value of x three We can say. Currently, we have achieved a certain sine here, This is the only known means cosine theta. I put the root x for three, a three, For sine fi, given here, the root-five divided by two here When we solve the cosine so this is now coming to the sine of theta structure fi Go to the bottom because three above five below the root here We find the cosine of theta zero. Cosine theta zero when a certain course sine theta has zero immediately. One minus the cosine square means for zero Find them all would have. I can say these things because, lets y We never use. Providing it is as follows:y the formula We know. So y is zero. a time for sinus, sinus theta. Here, here, take the value of sine for We know. We know the sine of theta here. See, as you can see we place Three of the more simple. The feeder roots stem back two more simple and remains. Indeed both radical as y We know that given from the beginning. We found a second that provide x, y, z If the values are correct thought, z particular, two squared, y here We found the root two means that two frames. x was the root of three. We collect them three the square root of three When you see five plus four of the nine spheres as means We provide the equation. This work was a preliminary, because the parameters We do not know the values of the selected point. Here we go, here back We wrote again to reduce the look. x clear. The equation for x and theta certain terms. According to these parameters will make again take partial derivatives. Get them in by a previous sample As fie Get derivative f showed longitude when changing f. Get this longitude Fi derivatives on the tangent vector, Take the derivative with respect to theta da lat on the tangent vector. This means that the longitude on the globe on vectors and latitude on We find these vectors by multiplying the We know that the vector perpendicular to these two. Derivative with respect to x, of course, the plug is fine easy. The following derivative of the sine cosine fi, cosine theta invariant staying here again the derivative of the sine plug cosine fi. The following derivative minus sine cosine fi. So here we find the vector. This is the first, and for the values of theta when positioned slightly t seem to mix, refine them by As we are getting a structure. Here even a little bit more We can simplify. Because you do not have roots in our holding five. Get it unless we obtain the following vector. When this derivative obtained by theta on the vector gives the longitude. The calculation is quite simple. Filar derivative is constant for the theta minus sine theta. The derivative cosine of theta-theta. Here the derivative is not zero theta. We find the corresponding vector work an account that is very difficult to but for certain sine, sine theta particular, certain cosine theta. All cosine theta particular, all when positioned here get it to zero, the third component We are. See this account of a little're strong Five of the more simple roots. Here there are three. Three is also more simple in the initial trilogy. It has a radius of three. You're getting now. He can do a checksum. Together with lengths of latitude on a sphere curves are steep. Therefore, the tangent must be perpendicular. Other means to be the inner product of two vectors be zero mean. This product really here when you get you can see six comes with a minus sign, such as the root, root of two times six, where the plus sign twice with six coming root. From here, because it is zero, zero We know that and has calculated that it is of the two other We're going, are making. This is the vector product of two vectors more We found the same orthogonal vectors obtained before We are. Therefore, the same orthogonal vectors and points in that the tangent plane to be We found an as needed before equation is happening. After this resolved the problem for you to do a little more account of what appears to be mixed, but no I'm giving a paper at no different. Here is a previous example of a sphere hes on tangent vectors, tangent We have calculated the plane. In the previous example, a paraboloid over here on the ellipsoid. You might say ellipsoid equation How do we know. For him, guiding it to you as promise. As well as a pretext ellipsoid equation can also be seen. A closed ellipsoid as a surface very nature technology problems and in problems It is a model used surface. a and b is O, A, B and As you can see, although c equal the equation of the sphere would occur. Paraboloid, ellipsoid equation As you can see here. Buradan da x vektörü fi ve tetalar cinsinden. Aynı küreninkine benziyor. Yalnız kürede a ortaktı. Burada a, b, c olarak bu elipsoidin a boyundaki, x ekseni boyuncaki yarı eksen uzunluğu or y doğrultusundaki yarı eksen uzunluğu b. z yönündeki yarı eksen uzunluğu da c olarak bu. Buradan bu teğet düzlemi herhangi bir noktasında değil de hadi somut olsun diye de a, b ve c'leri üç, iki, bir seçerek paraboloid, elipsoidin yüzeyindeki noktanın parametrelerini de ilave bir hesaplamaya gereksinim olmasın diye doğrudan veriyoruz. Bunlan bu paraboloi, elipsoid üzerindeki teğet düzlemi bulmanızı expect. Bu kapalı fonksiyon gösterilimi ile de elde edilebilir. Açık fonksiyon gösterilimi ile de elde can be. Ama paraboloidlerle burada elipsoid over it ilave bir çalışma yapmanızı bekliyoruz. Bugünkü beraberliğimizi burada bir ara vermek iyi olacak çünkü teğet düzlemi hesaplamayı öğrendik. Şimdi teğet düzlemde neler yapabiliriz konularına geçeceğiz. Buna geçmeden önce biraz konuları çalışmanız, üzerine düşünmeniz için bir ara veriyorum. Bundan sonra teğet düzlemi kullanarak daha ileri konular göreceğiz ve bunların We will see the application. Goodbye.