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Consid e Hu Pxoblem min [6, ~o 5)*+ (11+0.5)'] Subjcc bo 4 X , +2x 2 4 0, 1, + %2 2 1 ) X 2 2 0 Ducribe Ku tangent Conc t #u fesibk Se ct Ku petut (-1,2). Vs...

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Consid e Hu Pxoblem min [6, ~o 5)*+ (11+0.5)'] Subjcc bo 4 X , +2x 2 4 0, 1, + %2 2 1 ) X 2 2 0 Ducribe Ku tangent Conc t #u fesibk Se ct Ku petut (-1,2). Vs< Hu nece S ax] (ondhen bp hmali ty Axe Mue Cond; hons ulso Sukniuent } form nlat e Ku duc pRoblem. dlt is 'k ophmal Value 06 Hu duau ty relbou hold Phovi de Saddlc Poiut 4 Mu Lrjtanaie an

Consid e Hu Pxoblem min [6, ~o 5)*+ (11+0.5)'] Subjcc bo 4 X , +2x 2 4 0, 1, + %2 2 1 ) X 2 2 0 Ducribe Ku tangent Conc t #u fesibk Se ct Ku petut (-1,2). Vs< Hu nece S ax] (ondhen bp hmali ty Axe Mue Cond; hons ulso Sukniuent } form nlat e Ku duc pRoblem. dlt is 'k ophmal Value 06 Hu duau ty relbou hold Phovi de Saddlc Poiut 4 Mu Lrjtanaie an



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Find the value of $k$ so that the given differential equation is exact. $$\left(6 x y^{3}+\cos y\right) d x+\left(2 k x^{2} y^{2}-x \sin y\right) d y=0$$

Either. In this problem, we're have to find a units Tandon Fekter For this director very valued function are so to do that, we will first find attention vector By taking the derivative of our And remember, we can just take the derivative of each component separately. So the derivative of 66 is just a constant. So the derivative 60 that's her. I component I should write. This is a plus. Okay. The driven of co sign three t is minus sign three t and then by the change will be multiplied by the derivative of three T, which is three. So that's RJ component. Finally, the derivative of three signed 40 with three is a constant, so that stays multiplied in their derivative of Sinus co sign. We leave the inter function for tea, Then we multiply by the derivative of fourty which is for so that's our K component. Ah, In other words, on debt switch to angle brackets. Now our first component is zero our second component. We CIA's minus three sign of three t and our third component is 12 times co sign of 40. So this is a tangent factor for any value T. It's not a Units Tangent vector, though, So to get the unit Tangent Vector, we're going to have to figure out the length of the derivative we just found. That's the square root of the sum of the squares of our components. Zero squared plus minus three time. My three times signed three t squared, plus 12 co sign 40 all squared, and there's not a whole lot we can do here except for right of what we have. So we end up with that C nine sine squared of three T That's just this term squared. Plus 12 scored is 144 Coastline Square to fourty. That's our second Karen there squared. So that's the length we really cannot do anything to simplify that. And so our final answer. Our units tangent Vector uh, for any point t it will simply be won over this length. So one over this big square root I was going to re copy it, and from the last step 14. Good. So that's the length, and that's what we multiplied by our attention. Vector from above. Zero minus three. Sign three t 12 co signed 14. So that is our final answer. We're done, and hopefully that helped

The probability that the state with energy e is occupied A temperature T is given by P O V E eyes he called to one or exponential energy minus for me. Energy divided by Katie. So this is all over here. Plus one where case the bonesmen constant for about a day. Trivedi off a p off e will be then p d off right by e is equal to minus one or well, he exponential minus e f where me energy divided by a Katie plus one last one. This whole power to then did he write you off the exponential respect to e Just be energy minus e f Oh, our Katie. This gives us ah dp or d to be, um, minus one over exponential E minus e f derided by Katie plus one. So this will be this will come here. Hold square times one over. Katie. So if we take the data radio for this part here we get one over Katy times that financial itself, which is e minus e f divided by Katie for easy equal to e f far me energy. Um, we can. Then you let the above integral. So So the daily rate you d off p divide the d off E that he's a cool to see if simply we substitute the value off e to B e Every the question that we obtain earlier for the daily rate You then the answer we get here is minus one over for Okay, T for a part to be off the problem. The equation off line is we know why is it going to slope X minus X Nord ran Emmys slope here, which is a minus one over four. Katie, Um, ex lord is the X intercept, Um, is X in Durst's inter, except which which is required to obtain for this. But, um, it is clear that the P off E off F is equal to one over two. So our equation off a line even ready that X is a cool to e f becomes so this will become one or two. So one over two is equal to minus one over four Katie times E f so e f minus X nod, which leads to extend onto physical, too, which is X intercept e f plus to K T.

Right. They want us to find the third Taylor polynomial for F of X is equal to coast X squared at equal 05 or four equals 05 I guess not at. And then they want us to find the maximum possible size for the error bound with X equals 0.6. So let's start with the tailor polynomial first, because thes derivatives congee a little bit messy, so hum our very first derivative. For this, we would have to take the derivative of coast first. So the derivative of Coast's negative sign. And then we have to take the derivative of the insides of the X squared, which is two X. So our first derivative is negative to x sign X squared. So it doesn't seem too bad just yet. Now we get into our second derivative. So now we've got the product. Will we have to do so? We have to deal with the derivative of the two X first, which is a negative too, that gets multiplied by sine X squared. And then we have to add that onto the derivative of sine X squared, which is Cose X squared, then take the derivative of the X Squared, which is two X and then time's up by the original function that was there, which we wrote as negative to X. So what we have here, cleaned up, is negative to sign X squared, you know, that we would have minus for X hosts X squared. That is our second derivative. How we have to find her third driven of in order to finish off the tailor polynomial So going through that Ah, for the 1st 1 Luckily the derivative of negative to a zero. So that solves it's that one. So we just have to do the derivative of sine X Square, which is coast X squared times to x and then times by the original number there. That was negative too. That one's done. Now we have to do the next one. So the derivative of four X is negative for exes. Negative four times by what was there, which was our coast X squared. Then we need to add on that derivative. So the derivative coast X squared is negative. Sign X squared derivative of X squared is to x and then multiply by what was already there, which was a negative for X So cleaning that one up, I end up with negative for ex coasts X squared minus four coast X squared. And then on our last one Here we end up with negative Who? Positive? In the end, you get even a negative. Positive positive eight X squared side X squared. So that is plus our care Double check with this because it does get a little bit messy. So mmm, about zero sign X squared is Cossacks Square. Take the druid of the inside, which is X squared times. Negative, too. Yeah, that's good. Do our next one's the Druid of that is negative. Four times Coast X squared. Yep, that's good that do the next one. So cose X squared sine X squared times two x times What was there originally? So that should be good to go. Here is a error in the work here. I totally forgot to put a squared here to x times to access for X squared, which is going to change just lightly here. This derivative when I go to do it, it's not four anymore. It is a eat X. When I move it down so it becomes a next coast x squared, and then this would be a four x squared that I have there. So put in these together, I can actually put these together a little bit more and simplify it out because, huh? This guy here is a negative eight x so that actually simple eyes to these guys could go together. So negative 12 ex coast X squared plus eight x squared sine x squared. Oh, sorry. That's some eight X cube signed X squared. Because for what? That's the squares that makes that cute. That makes that one cute. All right, fixed everything up. So do you watch that? It's very easy to make a mistake very quickly. And I have messed up what you've got. So that gives us that we now have to find our values at 0.5 so thes we're just gonna right thumb, uh, using coast. They're not going to give us a rational number, so we'll leave them as is. So our 1st 1 there, Cyril 10.5 squared is your 0.25 So would be whatever coast of quarter is. And then for our first derivative negative. Two times 1/2 is negative one, and so we'd have negative ones are negative. Sign of 0.25 and then for our second derivative, which is somewhere in here. Oh, gosh. Well, this right here 0.25 We're using Negative too. Uh, much negative too. Sign 0.25 I went around around. I'm just gonna move this to the bottom. Thio, move these guys to the bottom. So that on that, our second derivative works out to native to sign of 0.25 Ah, zero point daughter. 0.5 squared is 0.25 times negative for Just works up to negative one. So minus coast of 0.25 and our third wanda 0.5. Ah, half of 12 is negative. Six coast of 0.25 and Oh, gosh, That one cubed eight times. The cube of 0.5 works out to one, so we end up with sign of 0.25 All right, so we need to do the third Taylor polynomial. So this is going to be a mess. It's very long, but at least it gives us an exact answer. So our constant term is that 0.25 coasts and then our X. We're using 0.5 for this. So our X minus 0.5 term is negative to sign 0.1 That's our 2nd 1 negative sign of 0.25 times X minus 0.5. Have that are X minus 0.5, squared over two factorial term. If you're still keeping this street works out to native to sign 0.25 minus coasts 0.25 um, times that by X minus 3.5 squared over, too, and I still have to do 1/3 1 Ah X minus 0.5 squared over the cube. Sorry works out to native six coasts, 0.25 plus side, 0.25 times X minus 0.5 cubed all over six. So now we have to try to write that all in one line. It's a mess, but we could do it. So our third Taylor polynomial, which is going to probably take us multiple lines too right is Coast 0.25 plus or minus Sign 0.25 times X minus 0.5 My ous to sign 0.25 because I took the minus out front. I'm gonna write. This one here is a plus. Uh, Times X minus 0.5 squared all over two. And I almost had a room. I just have to write my 3rd 1 Um well, just read it as plus it'll write the sign. First sign 0.25 minus six coasts, 0.25 times X minus 0.5 cubed all over six. So there is the third Taylor Polynomial. We're not done yet because they also want us to find the error. The maximum possible error from 0.5 to 0.6 as our interval. So we're solving this'd equation. Now for the interval. Um, 0.5 20.6. So we have to find her cape Diviner. Kay. That means we have to go back through that mess. And we have to play the fourth derivative because we know that the absolute value of the fourth derivative at its maximum form is Okay, So we're doing the derivative of a mess, if you remember, right. Our third derivative ended at 12 ex coast X squared and eight X Cube signed X squared. So, working through that derivative, we have 16 x to the power of four. Um, wait. Now that's simplified out. We'll go back and we'll write it all out to make, like, a little bit easier to follow. So going through that we head up with negative 12 coasts X squared, plus negative side X squared times the driven of the inside, which is two x times by what was there originally, which is native 12 x plus 24 x squared sine x squared. Then we have to do that one. So we end up with plus coast X squared. Take the druid of inside So two x times what was originally there, which was eight x cute. Okay, so let's clean that up a little bit. Negative. 12 coast X squared, Um, two X times negative 12 exits The negative 24 x squared Tubbs, A native side gives us a positive 24 x squared sine x squared. Uh, although we end up with 24 x squared sine x square, too. Can't do much to that one. And then our last one here to x times feet. X cube gives us a 16 x to the power for Coast X squared. So lastly, the only thing I could really do is I could put these middle guys together to make it slightly smaller. So I had it was 16 extra power for coast X squared plus 48 x squared sine X squared minus 12 coast X squared. So the absolute value of that, the maximum value of the absolute value of that function will give us what we want. So I went ahead and I already entered that into the graphene program and through the bounds on it, just to save a little bit of time and effort and looking at where we've got intersections here, uh, intersection. It looks like my intersection for there is at 0.5. So that's our max value, and it gives us a K a value of about 7.69 So the back of the textbook definitely says that they said 10 is acceptable. 10 is really high when our maximum values about 7.6. So I would actually say, um or acceptable answers boat eight or even 7.5. But that's what they would give you that for their, and then you would fill out your values accordingly. So whatever you choose is cake because it does say, make arrest a rough estimate. You can use whatever value you want there that makes sense. As I said, the back of the textbook uses 10 which is way off from what we can actually see so filling out the rest of that formula. Then, whatever you've chosen for Kay, you would have 0.1 to the power of four over four factorial. So again, with this one, it's about 7.7. That would give you your air bound. But since the back of textbook says you can use up to 10 feel free to use any value from about 7.7 again, I would roughly go with eight because it's closer to your value. So that's really as much as you can do with this formula.

We're given the vector valued function R f t equals to t to the fourth times vector I plus 60 to the three house times Victor J plus 10 over tea time sector and were asked to find the tangent vector of are at the point t equals one. So to find this, we need to take the derivative of our FT. And to do is we take the derivative of each of the components. So the derivative of T to the fourth are two to the fourth is eight t to the third. The derivative of six t to the three has becomes 92. The 1/2 time in sector J and then 10 over tea is really 10 t to the negative one. And so now we can use the power will here to get negative 10 t to the negative too. Okay. And since we're finding this AT T equals one, we're going to substitute in one for each of our components, and we get 8 10 Flynn to the third plus nine times one to the 1/2 plus negative 10 friendless sees, uh, negative 10 times one to the negative, too. And each of these simplifies to eight I plus nine j minus 10-K And that is our attention factor at the


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