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Scicntia are studylng the temperature ditant planet Thexfro that the surface temperature at or t locajion rrmorratugr Occreaes_ Celslus each Ucmeler You Ircxn the *...

Question

Scicntia are studylng the temperature ditant planet Thexfro that the surface temperature at or t locajion rrmorratugr Occreaes_ Celslus each Ucmeler You Ircxn the *Kiece450 Cels u5 . Iney also find that therraresere the emperalle Evadon usinc the axe5 elowCelsius} erd lettrc ncight atavc thc sunforekiameters'_equation relatng Tto H, anJ men [email protected] your

Scicntia are studylng the temperature ditant planet Thexfro that the surface temperature at or t locajion rrmorratugr Occreaes_ Celslus each Ucmeler You Ircxn the *Kiece 450 Cels u5 . Iney also find that the rraresere the emperalle Evadon usinc the axe5 elow Celsius} erd let trc ncight atavc thc sunfore kiameters'_ equation relatng Tto H, anJ men [email protected] your



Answers

Let $T(x, y, z)$ denote temperature at each point in space. Draw level surfaces (also called isotherms) corresponding to the fixed temperatures given. $$ T(x, y, z)=x-y+2 z, T=0,1,2 $$

We are given the temperature in three space as a function T of X, Y. And Z. And were given different values of temperature T. And were asked to draw Aisa therms corresponding to these temperatures. So the temperature injury spaces T of X, Y Z equals X squared over four Plus y squared over nine plus Z squared Okay. Okay. Mhm. Yeah. Now to draw these ISA Therms will simply plug in different values of the temperature and then grab the corresponding equation in three space. So just for example, if T equals zero, Well, this gives us zero equals x squared over four plus Y squared over nine plus Z squared. We see that in fact all of these ice with terms they're going to be ellipses Or now ellipses but ellipse sides since there in three space except for T equals zero. In this case we can solve for Z. And we get that Z is equal to plus or minus the square root of negative x squared over four plus y squared overnight. Thank you this company. And this is the equation of a cone. Mhm. Yeah. So we have our three axes X, Y&Z. Yeah. Mhm. Mhm. Now, as a matter of fact, this is a very special kind of cone because we see there's only really one solution to this equation which is the point X, Y and Z are all zero. And so this is the cone which is really just the origin. And therefore our graph is a single point shall draw in red the point of the origin. So there's only 1.3 space which has a temperature of zero. That's what the ice of Durham tells us. Now when T equals one, I'll draw the resulting graph in green. Well then the equation becomes one equals x squared over four Plus why squared over nine plus z squared. And they see that this is the equation of an ellipse oid centered at the origin. So we draw X, Y and z axes. Yeah. Now along the X axis we have a radius of Square to four or 2. Yeah. And so we have points here and here along the y axis we have The radius of three. So we have points, we have points here and here as well. And then finally along the Z axis we simply have a radius of one. So we have points at 001 and 00 -1 connecting these points. We obtained the ellipsis. Mhm. Mhm. Yeah. And so the ellipse oid looks a bit like this. It's kind of a rough drawing but this makes a little bit clearer and you sort of can see it now. Yeah. Okay. Uh huh. And finally we'll draw the ice of term corresponding to T equals four. And I'll draw this in blue. So this gives us the equation four equals x squared over four plus Y squared over nine plus Z squared Dividing both sides by four. We get the standard form, one equals x squared over 16 Plus y squared over 36 plus Z squared over four. And in this form it's easy to see it. This is lips oid centered at the origin. Mhm. Now, from our equation we see that we have a radius of four along the X axis. So we have points at 400 and negative 400 is located about here and here and then along the Y axis. Sorry this isn't blue, not green. We have a radius belt, six along the y axis. We have points at 060 and zero negative 60 Mhm. So these are located about here and here. Finally, we have a radius to along the Z axis. So we have points at 002 and 00 negative two. But you're located about here and here, connecting these six points we can obtain the graft part looks oid okay, yeah. And the graph of the ice a therm looks something like this. This is an approximation of the flip side and the size of term represents all the points in three space where the temperature has The measure of four.

For this problem, we were asked to draw the level surfaces corresponding to the fixed temperatures. Given for the function T. Of X, Y Z equals x minus y plus two Z. Where we have to equal +01 and two. So I have here plotted out in full three D. Our ISO Therms, we can note that because we have a pretty simple linear function that we get out of that. Just simple planes and each one of the planes or is a thermic planes we could say are parallel to each other. Um Where A is corresponding to the sort of burgundy plane down there. Um So that's when it equals zero or t equals zero. Uh The T equals one is the orange, N. T equals two. Is the blue. Now for drawing the eye. So Therms, the best uh best advice that I could give you is to try to approximate this sort of um perspective here. Uh and sort of try to make it clear, like I would say that's what you want to do is try to get these sort of corner lines, the edge lines there. Um but aside from that, it is going to be a challenge to draw this.

For this problem you're asked to draw the level surfaces or I. So. Therms for the function X squared plus Y squared minus Z, T, T. And T equals +01 and two. So I have plotted out here in three D. What that should look like where we can note that if we look down onto the xy plane we have a series of concentric circles with radius. Um Yeah yeah we have a serious series of concentric circles with starting off with Essentially Radius zero. So we have a point there then radius one then radius route to or if we look at the full plane, which is what you should be trying to do. We can see that we have a series of these sort of rounded cone shapes. Um It is going to be a little bit tricky to draw this but I'd say that essentially just trying to focus on getting that sort of parabola shape for each one of these consecutively and making sure that it's clear that you have the points being negative to negative 10 should be enough to reproduce this decently. Well, try to get this sort of angle or perspective on it.

For this problem we are asked to draw the level surfaces or iso. Therms for the function TF x Y Z equals x squared minus y squared plus z squared. Where we have to do I actually need to make a correction here. We want to do that. 40 equals +012 As well as negative one and negative two. So I'll add those in here. Uh Let me just make sure that this is coming out correctly here. So let's change that to -1 and -2. All right. So we can see that we have our green and our purple or blue or not purple um sort of burgundy ish pinkish plots coincide exactly. So The difference between the two is nothing. Um But we can see that we have corresponding to t equals zero. Would be the one with the sort of sharp um Sharp approach to zero where it equals zero at the origin and then the sort of concentric um hourglass shapes around that corresponds to the increasing or decreasing values of T. Now for drawing this I'd say that trying to get an angle along these lines would be the best idea and the thing to focus on would essentially be trying to get the yeah the profile in this. The profile as you're looking down on it sort of correct. So trying to get the diagonal lines coming out and then you'd have to do a little bit of shading to indicate that it is curving. It's going to be a bit tricky to draw this but making sure that you're getting that sort of parabolic shape or straight line into parabolic shape um as you go along X would probably be the thing to focus on for re re creating this on paper. Yeah.


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