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ChapterHigher Dimensions and Other J16 Coordinatet Because differentiation with respect to X,y or gives the with respect t0 2, and because = samne outside the integ...

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ChapterHigher Dimensions and Other J16 Coordinatet Because differentiation with respect to X,y or gives the with respect t0 2, and because = samne outside the integral the boundary 5 resekt inwt , (12), we find that UTNRY $ dz = 02v ax?and satisfies the two-dimensional heat equation azv 8-v Ju 0 < x < a, Dy2 k at 0 < Y < b, 0 < [ (See the exercises for details and for boundary and initial conditions } If z-variation cannot be ignored, we could try to get rid of by introducing an

Chapter Higher Dimensions and Other J16 Coordinatet Because differentiation with respect to X,y or gives the with respect t0 2, and because = samne outside the integral the boundary 5 resekt inwt , (12), we find that UTNRY $ dz = 02v ax? and satisfies the two-dimensional heat equation azv 8-v Ju 0 < x < a, Dy2 k at 0 < Y < b, 0 < [ (See the exercises for details and for boundary and initial conditions } If z-variation cannot be ignored, we could try to get rid of by introducing an average in that direction, the Yavariaua w(x, 20=b J ulx, Y,2,t)dy. From the boundary condition (13) we find that du Ju 8yzdy (x,b,2,t) _ (x, 0,2,t) [(Tz u(x, b,2,t)) + (Tz ulx, '0,2,t)11 . If b is small-_-the parallelpiped is more like plate_we approximation u(x_ may b.2,t) + u(x, 0, 2,t) = Zw(x, 2,t) , which accte" t from the expression above, would Iak: Ody = 2h b bk (Tz w(x,2,t)) _ After applying the averaging process to Eqs: (10), (11), (12) and (I4) weobzz the following two- ~dimensional problem for W. 02w 92w 2h dw axl bk (T-w) = k au 0 < x < a, 0 < 2 < €, 0 <t / w(0,2,t) = To, w(a,2,t) = T;, 8w 0 < 2 < €, 0 < t (x,0,t) = 0, dw d2 dz (x,c,t) = 0, 0 < x < a, 0 < t w(x,2,0) = 6 k" f6xsy,z)dy, 0 < x < a, 0 < 2 < €



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If $S$ is the surface defined by a function $z=f(x, y)$ that has continuous first partial derivatives throughout a region $R_{x y}$ in the $x y$ -plane (Figure $16.49 ),$ then $S$ is also the level surface $F(x, y, z)=0$ of the function $F(x, y, z)=f(x, y)-z$ . Taking the unit normal to $R_{x y}$ to be $\mathbf{p}=\mathbf{k}$ then gives
$$
\begin{aligned}|\nabla F|=\left|f_{x} \mathbf{i}+f_{y} \mathbf{j}-\mathbf{k}\right| &=\sqrt{f_{x}^{2}+f_{y}^{2}+1} \\|\nabla F \cdot \mathbf{p}| &=\left|\left(f_{x} \mathbf{i}+f_{y} \mathbf{j}-\mathbf{k}\right) \cdot \mathbf{k}\right|=|-1|=1 \end{aligned}
$$
and
$$
\iint_{R_{\mathrm{xy}}} \frac{|\nabla F|}{|\nabla F \cdot \mathbf{p}|} d A=\iint_{R_{\mathrm{xy}}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y
$$
Similarly, the area of a smooth surface $x=f(y, z)$ over a region $R_{y z}$ in the $y z$ -plane is
$$
A=\iint_{R_{y x}} \sqrt{f_{y}^{2}+f_{z}^{2}+1} d y d z
$$
and the area of a smooth $y=f(x, z)$ over a region $R_{x z}$ in the $x z$ -plane is
$$
A=\iint_{R_{\mathrm{xz}}} \sqrt{f_{x}^{2}+f_{z}^{2}+1} d x d z
$$
Use Equations $(11)-(13)$ to find the area of the surfaces in Exercises $39-44 .$
The surface cut from the "nose" of the paraboloid $x=1-y^{2}-z^{2}$ by the $y z$ -plane

Solving party of this problem here I can write develop physical to a T. Is equal to R V by am also I can write the value of DB by D. T. Is equal to minus RV by am negative sign because velocity is decreasing integration of me not to be D V V I P is equal to minus are by m integration of zero to T. T. T. No taking on solving it, I get long he by being not is equal to minus R. T. By am going forward and simply buying it further so I get physical to be not into the power minus Rt by Um this is a question number one when Physical 20 m/s then he tends to infinity now going to solve part B. So here also a physical to R P is equal to our people, I am B D P by dx is equal to minus our people, I am negative sign indicates velocity is decreasing integration of me not to be T V is equal to minus are by am integration um period to S the X. Going forward and solving it. Whether I can write a B minus B not is equal to minus are by AM X on possess simplification, I can write physical to peanut minus are by AM X at final position At final position he is equal to zero. So putting this in the application, I get the value of X is equal to M. By our be not now solving further so just look at it carefully in the next part I can write the value of B is equal to be not by and so the note by an is equal to be not to the power to the power artie by AM so I'll bring it for the long and is equal to RTB I am on for the simplification I can write develop two is equal to and by our loan and putting physical to be known by an so be by N is equal to be not minus are by em X. Acts is equal to peanut minus be known by end multiplication and by our average velocity average velocity is equal to displacement by time which is equal to X. Y. T. Here on for the simplification I can write deep into an average velocity. We are very difficult to peanut minus B, not by end multiplication and by our bye and by our long and which is equal to be not and -1 by and loan and as the final answer

We're told that the temperature t on some of lips and the X Y plane is given by G of X y and the lips is defined by these Parametric equations can see that this will make in a whips that looks like this in the X Y plane. Where the maximum Why value a spirited too? And the map Some X value is two times the square root of two, and our parameter runs from 0 to 2 pi So were first asked to locate the maximum and minimum temperatures on the Ellipse So will first find possible maximum and minimum by checking the first derivative in time, which is d by b the capital to you by the little T and I'm going to say time here. But t he's just the parameter. Um, this will give us the possible maximum minimums and then we'll analyze the second derivative with respect to t. In order to find the the weather to Max. Were men given to further pieces of information which here that T X, which just stands for partial T partial X we're going to use that notation equals why and t why is equal to X just so we're clear T sub x is just able to dp be pecs. So to get this, we're going to need to to use the chain rule so you can see that our X and our wire only functions of tea. So we're going to end up with total derivatives like here. But we're going to have partial derivatives in X and y So we work this out, we get well, we get be T d X, but D T d X is just equal to G d. X, so I don't dread This is G g X and then a total derivative here the x d t plus the g d Why de y e t and so you can see where we end up with total derivatives. We have total derivatives all here. If our X's and y's dependent on more than one parameter say t and s, we would have partial derivatives. Okay, so let us actually work out what this is So this is equal to be g d X and then the X DT is well going to get a minus sign Going to get a minus to root que comes the sign of tea and then we're going to get for why we will get route to G. De. Why times the co sign of tea. So we want to solve this equation right here. We want to find the teas that satisfied that equation and that will in turn give us the the X's and y's on the ellipse, where the temperature is at a max or men. So if we look over here, we see that T X, which is just partial G partial acts. It's just a why, and b g y or t y is just an X. So we're going to have a why, right here and an X right here. But this is just equal to this number right here. The thing X is just equal to that and the why it's just equal to that. So you can solve this equation. Now if we plug in, those things will have only tease first. We can't tell every twos. And then we say that that X so the equation gonna end up with his ex e g d y minus to sorry x times the co sign of tea, minus two. Why sorry in of tea equals zero. So I switch the order of the terms. This one went here and the other one went to the other side. But think you can see what what happened and then we can plug in our exit are wise. So this tells us that that que square root too co sign square of tea minus to and then plug in. Or why square root Teoh sine squared of tea equals zero which then tells us that her sine squared of tea has three equal to sign squared of tea because we can't lease and moved to other side. Well, this will happen whenever we have some. We look on the unit circle at any 45 degree or pi over four radiant mark. The signs and co signs along these lines will be the same. So the signs wrong in red. Then the co signs drone in green because they're squared. So since their squared, we get that the equivalence between these ones that would normally be negative. So this tells us that t is equal to pi over four plus in pi over two. Because these air shifted by 90 degrees and in is just equal to 012 and three. So at those points t our function is maximized. But we can figure out which X and y's thes are actually. And so when When in equals zero we get that X is equal to well, X will be equal to To To get that I just plugged in pi over four right there. And similarly, why will be equal toe one when in equals one. Then we're going to be looking at this one right there. This piper for Mark multiple of pi over form. Sorry. And at that point, all that changes is that X equals minus two. Whereas why still equals one and then moving to this one right here we get that but X equals two and why X equals minus two. And why equals minus two than in equals three. We have a positive X and sorry that should not be minus two, right that she and minus one and why equals minus from so these coordinate payers right here defined places where they're either maximum or minimum temperature values on this ellipse. But now we need to analyze the second derivative and the second derivative this in order to find the maximums and minimums. So we know that these on the left right here either maximum men of four minimums, but don't know which one. If we get a positive value for one of these points for the second derivative, then we know that the point is a minimum. And if it's a negative value than the point is a maximum. So we have that T t DT was once again equal to G the X, the x d t and then DDG do you? Why do you want why That should be a total derivative. Do you? Why dp? But then we need to take the derivative off this full thing again. So this is going to involve using some some chain rules. But we can actually make things a little bit easier on ourselves by first substituting in some things. So if we take no, maybe that can wait. Okay, let's just go straight forward with the calculation. So But now we need to figure out if these are Maxima or minima. So to do that, we look at the second derivative and if this is greater than zero, then it will be a minimum. And if it's less than zero, it will be a maximum. And we evaluate that at each of the points in the bottom, left hand corner of the screen. So we have this right here. We want the second derivative, but this is just equal to this. But we've already said, as we said before that, that is just equal to T X in the x e t see why? Why anything? And as we saw on the previous page right here, well, this will just be equal to D by d t. Of T X was just equal toe Why? Which is square root of two sign of tea and then dx DT It's just equal to times to root two minus sign of tea and then same for the other one. T why is just equal to square to to to route to you her sign of tea then d y d t just route to her sign of p. So now we can simplify this and we get D by d t. Uh, well, these three numbers together along with the negative sign, just become minus four sine squared of tea and then on the right, we get the same thing for co sine squared of tea and this ends up just giving us Steve, I d t of four times co sine squared of p minus sine squared. Sorry. Probably can't really see that to well, over in that corner. But Okay, so now we just need to take the derivative of that So we can note that this right here by a trick identity is just equal to the co sign of two t. So we have that d squared key by the little T squared is equal to D by D key times for where comes the co sign of two teeth. Well, we take this derivative, we get that it is equal to minus eight times the sign of two t so we can now move to new page where we get that d squared T ever DT squared is equal to minus eight times the sign Two teeth but river that are possible maximums and minimums were t equals pyre for plus in by over two where in equals 012 or three. So we can evaluate each of these possibilities. So when in equal zero, we get t equals pi over four. Consequently, the argument of the sine function is pie or two. So the sign is just one. So we get that the second derivative is just equal to minus eight. Then when in equals one, we get that t is equal to three pie over four. And here we get that the second derivative just gonna draw these two dashes. But because I don't want to keep writing the second derivative out for I guess I'll just go ahead and do it. But all right, it is t prime Prime just means the second derivative teeth is equal to positive eight. And then when in equals two, we get that t equals five pi over four. And the second derivative is equal to once again minus eight. And then when in equals three we get key equals seven pi over four. Which gives us that the second derivative is equal to eight. So we get maximums when the second river is negative. So we have maximums here and here. So, Max, and then we get minimums here and here. So, men So for our Max mums, we get when in equal zero x y is equal to x y is equal to 21 And when in equals two X y equals minus two minus one. And then for our minimums when he in equals one X Y is equal to minus two. And then when equals three X Y is equal to minus two Sorry, Q minus one. So these correspond to our maximums and minimums on the ellipse for the temperatures. So if we just take a quick look at the ellipse again just as a reminder Sorry for that squiggly ellipse. These were the points that correspond to get the color right Are maximums are here and here and then our minimums are here and here For the second part of the problem, we're explicitly given the temperature function on the lips were told that T is equal to x. Sorry x times why minus two Now we could And then we're told to find the maximums and minimums of tea again. But we could do it like last time. We could plug in x and y here and get the function in terms of tea and then take to derivatives. There's a much easier way to do it. Remember that we were given the general conditions last time that were that T of X has vehicle tow. Why and t of why equals X And there's the only things that we were told about the function. So everything we derived previously will apply to any function that satisfies these differential equations. But as we can see, this function here does satisfy that. If we take the partial of tea with respect to X, well, then we just get why. And if we take the partial of tea with respect for why we just get at consequently, we will have the exact same maximums and minimums as before, because everything that we worked out for the previous problem applied to any function that satisfied these general conditions. So Max and men are same as before, but just to justify that really quick just to show that this is indeed true. Well, if we plug in, I'm going to do this somewhat quickly. But we would get t equals four times the co sign of tea time, the sign of tea minus two. And then if we work out the derivative of this, we would get that P Prime is equal to four times the co sign of two t get that to the product rule, and then we'll get that T double prime is equal to minus eight times the sign of two T, and these were the exact same 1st and 2nd derivatives we got in when we worked out the problem previously, So it's just a little justification for the more direct method that was done up here.

You have? Why two Vehicle to 4? My Nets said. So this implies for every respect to as a partial derivative. Oh, that's It's equal to zero. No pasha directly with respect to were zeke respect to set, It's equal to -1. So then this implies that the square root of F X squared plus F Z x squared plus one is going to be square roots of suit. So then this implies that our area our area will be called to the Dublin seeker over our X. Z Square root of two. The mm. And this is equal to Jessica From 0 to 2. Yeah, from 0-4. My next ex squared Square roots of two, the X D. Z. And this is equal to the square root of two as a constant integer from 0 to 2. You have four minutes, Z x squared the Z. And this is equal to 16 square which of two divided by three itself. I know

Russia they receive of f respect why? What I said, It's equal to -2. Why? And with respect to z Will be equal to -2 Zeke. So then this implies that the square root of F Y squared, F Y squared plus upset squared plus one is going to be square roots of four way squared plus four Z squared plus one. So this implies that our area will be equal to the w C gra over our you have square roots of four y squared plus four Zach squared plus one D. X. The Why? So in polar coordinates This is going to be the anti girl from 0 to 2 bikes You have from 0 to one. So do I. It is A. So this is the white, is it the way the zig? Okay, so you have 0-1. Have square roots of four hour squared plus one. Are the are the sita. And this is equal to by divided by seats. You have five square weeks of 5 -1 as a final answer.


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