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#X (8 pts) Aivpot 6 is 300 Mi the fpm aixpot A + fiqute) A P: Gt ~ishin] to beavind N3o € (see Aivpox t B tla /~ A to k east af 209 Mi mbtekelthe flies dae J9...

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#X (8 pts) Aivpot 6 is 300 Mi the fpm aixpot A + fiqute) A P: Gt ~ishin] to beavind N3o € (see Aivpox t B tla /~ A to k east af 209 Mi mbtekelthe flies dae J9o6 ty 30 hi ninutel, nhen ewok: Motite) (a) How for 1) the P-Gt tm 60 Zotinatin c the time he notites the eWoy 7 Awpot A (L) Uhat beovid , shol -( ke GeacQ t, he plane aw ive a bYclev ct awpst B ? ~) 24/.96+i, N23"E , (5) 278.676;, Ns6 E (c) (98mi , N 22.1 € (c) 2 GksY w;_ N (2. 89 " E (e) non € thexe

#X (8 pts) Aivpot 6 is 300 Mi the fpm aixpot A + fiqute) A P: Gt ~ishin] to beavind N3o € (see Aivpox t B tla /~ A to k east af 209 Mi mbtekelthe flies dae J9o6 ty 30 hi ninutel, nhen ewok: Motite) (a) How for 1) the P-Gt tm 60 Zotinatin c the time he notites the eWoy 7 Awpot A (L) Uhat beovid , shol -( ke GeacQ t, he plane aw ive a bYclev ct awpst B ? ~) 24/.96+i, N23"E , (5) 278.676;, Ns6 E (c) (98mi , N 22.1 € (c) 2 GksY w;_ N (2. 89 " E (e) non € thexe



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An airplane flying north at $200 \mathrm{mi} / \mathrm{hr}$ passed over a point on the ground at $2: 00 \mathrm{PM}$. Another airplane at the same altitude passed over the point at 2: 30 P.M., flying east at 400 mi/hr (see the figure). (a) If $t$ denotes the time in hours after 2: 30 P.M., express the distance $d$ between the airplanes in terms of $t$ (b) At what time after 2: 30 P.M. were the airplanes 500 miles apart?

Alright for this exercise, We're considering the motion of two different planes. So we have a Cessna. I'm going to label see heading south at 120 MPH. And we have a Boeing 7 47 heading west Boeing. I'm labeling with B that's heading wet. Whoops. Wrong direction that should be here heading west at 600 MPH. Yeah, eso They're flying towards the same point at the same altitude. We're also told that our Cessna is a distance of 100 miles from the point, while our mowing is a distance of 550 miles. So obviously our little sketch there is not to scale. So what we're going to do for this problem here is first of all, we want to set up equations that will describe the positions of these two planes. So what I'll do is we'll have C equals zero. I had plus 100 minus 1 20 t j hat. If that notations unfamiliar, it's sort of just like a vector notation or a particular form of vector notation. So, alternatively, I could just write it as pair of coordinates. So we're saying that the Cessna is in line with the point, so it's distance horizontally will always be zero. It starts 100 above, but it's getting closer at a rate of 120 per hour than for B. We are starting 550 miles west. We're sorry we're starting 550 miles east of the intersection Point, so we start out at 5 50 and, because we are heading westwards, will be subtracting from our position. So we'll have 550 minus 600 t and then there's no vertical motion. So that's just going to be zero for Part B. We want to figure out on equation that will give us the distance between the two planes. So we know that we have our Cessna. You have our point. What's called that? Oh, it's called that C. It's called E. The distance between the two planes. It's going to be the high pot news of this right angle triangle that's formed here so the total distance will be equal to I'll put this as brackets with to the power of one half outside, rather than doing this square root over everything. Um, there will be the distance from B to the origin square, plus the distance from sea to the origin squared. MEPs squared all to the power of one half. So in this case, you would have 100 minus 1 20 t squared plus 5, 50 minus 600 t squared all to the power of one half. So one moment as the inside of the square root, expanding everything out and simplifying, we should get 374,400 t squared, minus 684,000 T plus 312,500 still all to the power of one half. Now, for part C, we want to plug this into a graphing utility and figure out what the minimum distance between the planes are. So I'm just going to jump over to a graphing utility. So loading up into days, most app, you can also run this just a Web page. Plug in. We'll have everything under square root. Then we have 374,000 374,400 rather X squared, minus 684,000 x plus, 312,500 which should give us our plot of the distance. Plugging this into the graphing utility is actually part C. And then Part D is answering. When did our how close do they get and when do they get that close? So we can see that the closest that they'll get is at about 0.9135 and they get about 9.806 miles away from each other as last part. We want to simulate this by plugging in the formulas that we derived for part A and plotting them simultaneously. So the way that we can do that using Dez most here is what I'll do. So plug in the coordinate form. So we had for C was zero plus 100 minus, we would have won 20. All right? Actually, I believe Okay, Yeah, so we would write 120 t. But for the sake of being ableto see how the planes are moving relative to each other Instead, I put in 120 a, uh, which makes Dez most put in a slider for a So that means that basically, I'll be able to slide a long time. We'd be able to see the emotion of the two planes. So label that Cessna and I'll also adjust the time range that that a varies over. So we'll be going from zero upto. One should be able to see the starting position for assessment. Yet there we go. And then similarly, for the Boeing you would have it was 5 50 minus 600 a and then zero. The vertical axis enable a label of the Boeing and need to adjust our view Here. There we go. So we have our Boeing in our assessment being plotted at the same time. And now we can watch. Of course, it looks like the Cessna is moving faster, but that's just due to a distortion. Because of the scaling, we see that the Boeing is in fact going much faster, But it will pass by that, um, intersection point of their flight paths a little bit after the settlement passes by. So no air tragedy crisis averted

Welcome back to another cross product problem. This time we're looking at what happens if we want to find the distance from a point. Let's call it P to a plane. We know we can define a plane by three different points. Let's call these Q, R and S. And so if we define a couple of vectors, let's say a between Q and r. Be between Q and S. And say see between Q and r. Point. Then what we can do is try and figure out the distance from the point to the plane. Using some things that we know about. Triple products. Specifically the volume of the triple product of the parallel pipe ed between a B and C is defined as the magnitude of across B times the magnitude of C. Cosign theta. And so rearranging this a little bit. If we write C dot a cross B divided by the magnitude of a cross B, then this is going to be the magnitude of C. It's the length of that vector C times Cosine Theta where theta is the angle between our vector and perpendicular. This is theater right here. Now, since cosign Theta is adjacent over hypotenuse, that means the length of this perpendicular lines. The distance from the point of the plane, going to be the magnitude to see times. Cosine Theta. We can use this if we want to calculate the distance from say point P 214 to the plane divided by Q, R and S. In order to do that first, we need to calculate projectors defining the plane. So A is Q. R. AR -Q Sierra -1 To -00 0 and Q. S. That's s minus cues. Again zero minus one, zero minus zero, three minus zero. We'll also need to see that's the vector QP p minus q is two minus one. One minus zero, four minus zero. So if we want to calculate the magnitude of c dot a cross B, we can do that in one step using our triple product. If we plug in C. A and B into our matrix here, I'll point out this isn't the only way we can do this. Um But we'll talk about another way in just a second. So let's plug in 114 -1-0 And -103. And using the formula from our textbook, remember we ignore the first column And look at two times 3 minus zero times zero, Going to be 6 0. And normally we would multiply by I We're actually gonna multiply by one minus and ignore the second column -1 times three zero times negative one. Eight of three minus zero times not J but times one again us. And then we ignore the third column negative one times zero minus two times negative one will be zero minus negative too, Which is 0-plus two times four. And since we don't have any eyes jay's or k's this is not a factor but just a number six times one minus negative three, That's plus three times 1 Plus two times 4. And so we're looking at six plus three plus eight is 17. The other thing we need from our formula remember is the magnitude of a cross B. And so we can get that by calculating the cross product of A and B. Once again we're looking at six minus zero, I minus negative three minus zero jay plus zero minus negative too. Okay, Giving us the vector six three two. If you want to calculate the magnitude of a Crosby, that'll just be the magnitude of 63 two. Which is the square root of six squared Plus three squared plus two squared or The square root of 36 plus nine plus four. And that's the square root of 49 or just seven. Since we determined that the distance is the ratio of the two numbers that we just found, we get the distance is let's go back to that cross product Or the triple product with 17 over seven. I said earlier, this wasn't the only way we could have found this. Since we already know the cross product a Crosby, you could have the magnitude of that and then found the magnitude of C. Got a cross B. Just using the vector C. And a dot product. That would avoid having to do the cross product twice. Thanks for watching.

Okay, so we want to use the map to write a the new equation that gives the distance Why in killer is remaining in the flight in terms of the number of minutes x since 8 a.m. So let's see, we're on a plane flying from Montreal to my Miami. So let's notice that our line is a critical line. So let's recall that if you have a vertical line, our equation is this X is equal to some value. So we're given that the plane leaves want your all and 8 a.m. And then that's 9:15 a.m. We're flying over Washington. Okay, so let's assume that if we just have eight linger equation, we have wise you put too, um, extra put speed. So at, um, 8 a.m. That is when our time ex is Earl. We have a Y value, which is our distance is also zero. Okay, So if we like this and we have zero is equal to, um, time zero plus b. So we see that B is equal to zero. And now we're given that, um when X is equal to, um, our exes in minutes sort of difference between this is one hour and 15 minutes. So that's 60 minutes, which is in our post 15. About 75. We are at Washington throughout. This means that Okay, so now we just have wise equal to M X. So if we plug in our Ekstrom, you have why is equal to, um, times 75? Okay, so, actually, let's note that we have a vertical line. So our equation this is just going to be X is equal to 75.

Okay, So in this program, we want to find the, uh, distance the smallest distance from a point under some constraint equation to the origin. That means we want toe finally a minimum point for the function if it was to x squared plus y squared. Plus these where and the constraint the equation is X plus tau one minus 12 equals zero and x plus y minus six equals zero. Since we have two constraints, we use LaGrange multiplier method Brilliant off every equals to London times Brilliant off G time A plus new times Gregan of age. So it gives us a system of equation two x equals Lambda plus view to why equals two meal on A to Z equals two to lambda So we have we can really express x y v in terms off new on the Lambda, for instance in importance, you know? Okay. And we plug in this into the constraint the equation. For example, if we plug in X y into G, we have one half lambda plus me equals +26 And if we plug in X y into, uh so this is the not one. So if we plug it X Y into age. We have, um, five Lambda plus meal equals to 24 and then we can solve this Lambda equals toe me, um equals to four. And the corresponding point is X equals to fall. Why close to two and Z equals toe for and for this point, if equals the 36. So the distance to the origin people's too six mhm.


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