¡Descarga Understanding Contour Lines and Their Role in Representing Functions of Three Variables y más Apuntes en PDF de Cálculo diferencial y integral solo en Docsity! Contour lines What are contour lines? Contour lines are the intersection of the graph z=f(x,y)z=f(x,y) with a plane z=kz=k (where kk is a constant). In practice, we equate the function to a constant: f(x,y)=kf(x,y)=k I mean, contour lines is when you equate a function to a constant, but what do we want to see this for? where does the "curve" part come in? Well, that's what we want to find out! When we equal f(x,y)f(x,y) to a constant, a curve will form, and that curve will always be in the x,y)x,y) plane. Know that each value of kk represents a plane in x,yx,y (you will understand this in the following example). On the other hand, the representation of many contour lines is called: Contour Map. Example: f(x,y)=x2+y2 ------√f(x,y)=x2+y2 Note: this surface is the top of a circular cone. Contour lines f(x,y)=k x2+y2−−−−−−√=k x2+y2=k2f(x,y)=k x2+y2=k x2+y2=k2⇒ ⇒ ⇒ ⇒ Then, at a certain height z=kz=k, the intersection between the cone and the plane z=kz=k is a circle of radius kk. It is easy to understand what happens with the graph, so let's see a bit: Subscribe to DeepL Pro to edit this document. Visit www.DeepL.com/profor more information. If we take the red curve and project it onto the xyxy plane we will have a contour! To get all the contour lines, we need to vary the value of kk. You've probably noticed, but as we vary the value of kk, the plane will go up or down, making the circle formed at the intersection larger or smaller. Note: note that in the example, kk must be greater than or equal to zero because f(x,y)f(x,y) is always greater than or equal to zero (since it is the top of a cone). If k=0k=0, the contour curve reduces to the point (0,0)(0,0). Remark 1: You may also encounter these questions: "Find the level of the contour line of the function f(x,y)=x2-y+3f(x,y)=x2- y+3 passing through the point whose coordinates (x,y)(x,y)(x,y) are (1,2) (1,2)". In this case, simply replace the values in the function. f(1,2)=1−2+3=2f(1,2)=1−2+3=2 That is, this is the level 22 curve of the function. Let's imagine we have the following contour map: The only restriction on the function is the logarithm: the term within it must be strictly positive. That is to say: 1−x2−y2−z2>0 x2+y2+z2<11−x2−y2−z2>0 x2+y2+z2<1⇒ ⇒ Therefore, the domain of ff is the interior of the sphere of radius 11 centred at the origin. The point is that the spherical envelope itself is not part of the domain, because in the inequality we have "< < < " and not "≤≤≤". This is an important thing to clarify. The drawing looks like this: A small tip: generally for these exercises the shapes you will have to draw are the same (spheres, cones, cylinders, etc). What changes is how the exercise is set up. We are going to study what happens when we mix contour lines with the parameterisation of curves. For this reason, it is essential that they handle well the parameterisation of the most classical curves (straight lines, ellipses...). It is also good to know how to draw a curve when faced with its parameterisation, but that will come with practice. Parameterised curve contained in a contour line It is common to encounter the following problem: you are given a parameterised curve of the type γ(t)γ(t) and asked to prove that its image is contained in a contour of a given function ff. Basically, what we have to remember is that in a contour line the function is constant. In other words: f(γ(t))= constant f(γ(t))= constant And that constant is precisely the level of the contour line. In practice, what we have to do is put the parameterisation expressions into the function and see if we get a number from them. A bit confusing, isn't it? Don't worry, with this example you will understand it. For example, consider a curve γ(t)γ(t) and a function f(x,y)f(x,y): γ(t)=(sent-1,cost+2)γ(t)=(sent-1,cost+2) f(x,y)=(x+1)2+(y−2)2f(x,y)=(x+1)2+(y−2)2 It is enough to do: f(γ(t))=f(sent-1,cost+2)f(γ(t))=f(sent-1,cost+2) What it gives: (sent−1+1)2+(cost+2−2)2=sen2t+cos2t=1(sen t−1+1)2+ (cos t+2−2)2=sen2 t+cos2 t=1 Basically, what we did was to take γ(t)γ(t) as if they were points in (x,y)(x,y), and we substituted them into (x,y)(x,y) of the function. For the rest, we solved as we usually do. Therefore, the curve γγ is contained in the level curve 11 of the function ff. We have already seen that contour lines are excellent tools for graphing functions of 22 variables because by putting a few of them together we can "see" the final graph. Contour lines are the equivalent of contour lines for functions of 33 variables. These give an insight into the behaviour of the function, and to calculate them one only has to equal the function to a constant. Therefore, the level surface kk of the function f(x,y,z)f(x,y,z) is given by: f(x,y,z)=kf(x,y,z)=k So if we want to describe the level surfaces of the function. f(x,y,z)=x2+y2+z2f(x,y,z)=x2+y2+z2 Equal ff to a constant kk: f(x,y,z)=k→x2+y2+z2=k=(k√)2f(x,y,z)=k→x2+y2+z2=k=(k)2 It is nothing more than the equation of a sphere of radius k√k centred at the origin, whenk>0k>0. But if k=0k=0, the level surface is the point (0,0,0) (0,0,0) and if k<0k<0 there is no level surface. The latter is quite intuitive: there is no such thing as an object that has "negative dimensions". As you solve a couple of level surfaces you will notice that the sphere gets larger and larger, so the graph will be:
