any-curve-with-the-property-that-whenever-it-intersects-a-curve-of-a-given-family-it-does-so-at-an-a

Question

Any curve with the property that whenever it intersects a curve of a given family it does so at an angle $a 
eq \pi / 2$ is called an oblique trajectory of the given family. (See Figure 1.1.7.) Let $m_{1}$ (equal to tan $a_{1}$ ) denote the slope of the required family at the point $(x, y),$ and let $m_{2}$ (equal to tan $a_{2}$ ) denote the slope of the given family. Show that
$$
m_{1}=\frac{m_{2}-	an a}{1+m_{2} 	an a}
$$
[Hint: From Figure $\left.1.1 .7, 	an a_{1}=	an \left(a_{2}-aight)ight] .$ Thus,
the equation of the family of oblique trajectories is obtained by solving
$$
\frac{d y}{d x}=\frac{m_{2}-	an a}{1+m_{2} 	an a}
$$
(FIGURE CAN'T COPY)

Michael Jacobsen · Accepted Answer

instead of dealing with orthogonal trajectories in this problem, we&#x27;re gonna deal with it trajectories that intersect at angle A which is not equal to pi or two. We&#x27;re going to let the blue curve represent the given family of curves, and we&#x27;ll let the green curve represent the family of curves that we seek, which will always intersect in that angle. A. Furthermore, will let em one denote tangent of a one where a one is indicated here and m to denote tension of a to were aces up to Is this single indicated here to start this process of showing that M one is equal to end to my extension of a divide by one plus two times tension of A. We&#x27;ll begin by writing tangent of the quantity a one and working with this quantity, a one knows that tangent of a one can also be expressed as taking the larger angle a to so we&#x27;ll have tangent of a to subtracting off the smaller angle. A so tender day one is equal to this tangent of the difference of angles, but that allows us to use the falling formula from trigonometry that whenever we have a difference of Ingles. It can be broke broken down into this format. So now we can express that attention to of a one is equal to tangent of a To with a minus will use a minus in the numerator. So my extension of a divide by one plus the product of these same tangents. So tangent day to times tangent A for the next step of this process will make some substitution. Recall that tangent of a one was meant to be m one. We obtained m one on the left hand side and tension of a two was meant to be m two. So this tangent of M two or a two can be expressed as m two mice The tangent of a divide by one plus tangent of a two which is m to itself times, tangent A And this is expression that we were looking to find

Samuel Hannah · Answer

says this problem. We have a differential equation. Equation. 1.8 points, 16 on We need to solve it for the family of curves F one. Now it&#x27;s dependent on the soap of families Curves F two. So the first step is different. JP equation for F to in terms of x discuses two x plus two y d Want the X is equal zero of a slope is the y the X So we need to rearrange this equation, which gives us that d y over the X equal to minus X over. Why? And this is the slope to now we can substitute this into the equation for F one. But before we do that, we also need to find out what the value of town of al Forests town of Alfa is just the town of pyre before from the question so tired of five before is just one. We now have everything we need to substitute in the equation. 1.8 points 16. Let&#x27;s differential equation redes de y over. The X is equal to minus X over Y minus one all over one minus x over. Why rearranging this office is equal to X plus y So the X minus y. Now, at first glance, this is a homogeneous equation because all the components on the right hand side or if the same degree but is there of degree won&#x27;t. Therefore, we can make the variable transformation the Eagles y over X or, in this case, why equals b. Times X have a left hand side. Either product rule becomes V plus X d v D. X. On the right hand side becomes X plus V X over X minus V X. Now each component in V knew it and denominator has a common factor of X, so we can cancel this out and we&#x27;re simply left with one plus V over one minus three. Now, if we subtract fee from both sides, we can rewrite the right hand side in terms of a single fraction, and this gives us that X D. V. D. X is equal to one plus B squared over one minus feet. Now this is a separable first order differential equation two separate. We bring the V variables to the left hand side, and the ex variables to the right hand side discusses that one minus V over one plus B squared Devi equal to one over x the X We integrate both sides to get a solution. We noticed that the left hand side, we need to split up into two component forms to be able to integrate easily. So the left hand side becomes the interval of one plus one 1/1 plus feet squared minus V over one. Plus. He&#x27;s glad TV on, and this is solved by known integration solutions. First part of this integration simply Arc Tana V, and the second is 1/2 natural log one plus V squared. Now this is equal to the right hand side of the equation, which is the integral of one over X, and that is simply the natural log of X. So putting this all together the solution of the differential equation is the Ark town of the minus. 1/2 natural log of one plus B squared is equal to the natural log of X, close to the integration constant C. Now this is a solution for the but we want the solution for why so we need to make the substitution being equals y over X. So the Ark 10 term becomes our town of Y over X and then get minus a horse. I was not to log. Have one plus y squared over X squared, this sequel to the natural log of X plus C. Now this is a solution for why, but we can actually simplify it by rewriting the natural long term. On the left hand side, we noticed that one plus y squared over X squared is equal to X squared plus one sweat the X squad on. If we were to take the natural log of this turn on the right hand side for red equation, the natural log can be split up using the log laws into the natural log of X squared plus y squared minus the natural log X squared. Now we also have a factor of 1/2 so we can put the half in front of both terms. Well, when we have a coefficient in front of the lock natural love rhythm, we can simply make that into power. So we get that the natural log on the right hand side of 1/2 natural log of X squared is equal to the natural log of X. And so if you put all this into our solution for V. We now find that are 10. Why of Rex? Minus half the natural longer of them have X squared plus y squared. Plus the natural logarithms vex is equal to the natural log riven of X plus C. But we noticed that we have a natural logarithms of external on both sides of the equation. Civvies can be subtracted and this gives us our final solution for wide, which is that the Ark town of y over X minus 1/2 times natural. Other evidence of X squared plus y squared is equal to some constant safe.

Samuel Hannah · Answer

this problem. We&#x27;ve been given a family of curves F two and we want to find the family of curves intersect This angle a equals pi over four. We do this by using equation 1.8 points. 16 which gives us the differential equation of a family of curves F one. And this depends on the slope left to fun slope enough to need to differentiate with. Respect to X gives us two X plus two wine you want. The X is equal to to see have a sleep is just Do you want the X so we re arrange to get the Y DX equal to C minus x over white. But this isn&#x27;t a form that we can really use because we would like all the terms of D. Y D. X to be of why or X. We don&#x27;t like this seater, so what we can do is go from the original cars equation and rewrite this for C and C. But C is equal to X squared. That&#x27;s why squared over two x their fault weaken. Substitute this into the soap, which gives us that D Y. The X is equal to X squared plus y squared over two x minus X, and then this is all over. Why? And we re arranges to give us that y squared minus X wed of the two. X Y is equal to the soap of F two, so we can substitute this in to the differential equation. 1.81 16. So the differential equation for F one is given by D Y. The X is equal to y squared minus X glad over two x Y minus one over one plus was where minus x squared both two x y. But if we rearrange its right hand side to a single neat fraction, we get that it&#x27;s equal to y squared minus two x y minus X squared all over. Why squared plus two x y by the sex sweat. This is a differential equation. Family thrives. F. One Looking at the right hand side, we can see that this is a homogeneous deflation so we can make the variable transformation the equals y over x or y equals a few times X By product rule. This turns left hand side into V plus X DVD X, and by substitution, the right hand side. It was V squared X squared on this to V X squared, minus X squared all over eastward X word plus to the expert minus X quit. We have this common factor of X squared in all the terms in the numerator under denominator service cancels and with make further simplification by subtracting the from both sides, the right hand side we can rewrite. This is a single fraction, which is minus V cubes minus B squared minus V minus one all over B squared plus do V minus one. We can make this slightly simpler by noticing that the numerous on the right hand side fact arises, and we get a first order separable differential equation. Just X DVD X is equal to minus B plus one times three squared plus one over B squared plus two B minus one. Teoh Want to solve this by separating the variables and bring the left to the left the V variables. So we get that B plus one. Sorry, he&#x27;s glad. Plus to the minus one. Oh, the the plus one times V squared plus one TV is equal. Tu minus one over x The X now to solve this by integrating both sides. Now the integral off the right hand side is simply minus natural log of X. But for the left hand side, we would like to split up the fraction involved. We do this by method of partial fractions, and for this we find that the squared plus do the minus one. The plus one times T squared plus one is equal a TV over B squared, plus one minus one over the plus one, so integrating we can integrate these fractions separately. Now the integral off the second fractional term, it&#x27;s simply the natural log V plus one on city into the role of the first fraction turn. We know from substitution that the integral of the over B squared plus one is 1/2 the natural logarithms of V squared plus one. But this is much flanked by two, so that factor of 1/2 simply disappears. So the left hand side is equal to the natural log V squared, plus one minus for natural log of D plus one, which we can combine using the locals. There&#x27;s natural log that piece bad plus one over D plus one. So if we bring this all together, be integral for a differential equation. It&#x27;s sold by not to log B squared, plus one over V plus one that&#x27;s equal to minus the natural log of X. And then we&#x27;ve had an integration constant, which we call the natural log of some constant C. So if we bring the natural work of X to the left hand side and combine it, this equation becomes the natural log is X Times feast glad plus one over B plus one is equal to natural look of some constancy and then we can take the exponential of both sides, which removes the natural log room. This is a solution for the differential equation in terms of e What we want this in terms of why so we make the substitution that the equals y over axe Onda The left hand side becomes this we get the X is times why squared over X squared plus warm and this is all multiplied by why over X plus one. Now we can factor out this one of wrecks in the denominator. This gives us the denominator is now why plus X on. Bring an extra factor of X to the new Morita. Do we get X squared times y squared over X squared plus one. And so the numerator becomes. Why squared plus X squared on denominator is why plus X. So our overall solution, in terms of why is given why Sled plus X squared over Y plus X is equal to come some constant c. Now suppose we let this constancy equal to times another constant K We can vent. Multiply both sides by y plus X When we get that y squared. Plus X squared is equal to two k y plus two K X. If we bring the two K axe and two k y to the left hand side, this becomes X squared plus y squared minus two K Y minus two. K X is equal to zero, so this form doesn&#x27;t seem very useful. But we can actually fact arise this formula into X minus K squared. Plus why minus k squared, as long as we add a to K squared to the right hand side and not to get squared compensates for the extra terms generated when we take this factory ization of disk Wes. And this is the final solution. The why of our differential equation

Chris Trentman · Answer

we&#x27;re told two curves are orthogonal, their tangent lines are perpendicular at each point of intersection. Were asked to show that the given families of curves are orthogonal trajectories of each other. That is every curve in one family is orthogonal to every curve in the other family. Then we&#x27;re asked to sketch both families of curves in the same axes. The first family of curves is x squared plus y squared equals r squared. These are circles centered at the origin and the second family of curves is A. X plus B. Y equals zero. These are lines through the origin. Intuitively this seems like orthogonal trajectories. Check this. However, when do three things first will differentiate the equation of the family of curves. Then we&#x27;ll paint a differential equation and then we&#x27;ll solve this differential equation. So first well differentiate both sides of our first family with respect to X. So we get two X Plus two y times. Dy DX equals zero. Very common. And therefore solving for dy dx we get dy dx equals negative X over Y. Now, in our second step I&#x27;m gonna replace dy dx Oh yeah with negative dx Dy in this way will generate curves which are orthogonal to our original family, X squared plus y squared equals r squared. So we get negative dx dy equals negative X over Y. Therefore dx dy equals X. Y. This is a separable differential equation which is easy to solve. So we get D X over X equals dy over why. We split up the derivative into differentials, integrating both sides. We have the integral of D X over X equals the integral of Dy over why. And therefore the natural log of the absolute value of X equals the natural log of the absolute value of Y plus some constant C. Hi guys now combining terms and using log properties, we get the natural log of absolute value of X divided by Y equals C. Using inverse functions we have absolute value of X over Y equals E. To the constant C. Right. And therefore we obtained the X over Y equals plus or minus E. To the sea. Yeah. Now we&#x27;ll let plus or minus E to the C. Which is a constant. Be B over a. Sorry, negative B over A. Then it follows that X over Y equals negative B over A. And therefore A X plus B. Y equals zero. Or thanks. We have shown that these two families of curves right, X squared plus Y squared equals R squared and A X plus B. Y equals zero are orthogonal trajectories. I love beef jerky. That was, you know, okay. So I was vegan but I did eat beef jerky.

Frank Lin · Answer

So for this problem again with computed derivative by imprisoning depreciation and as you and we beautify, there are Flo. And we can use these to evacuation to chance things because on their interception on both, the question should be satisfied so we can use those equation to swap variables. And then here we go the Internet before means is to ah, lines are ah is two lies are prevented from each other, thes to tension. And so these are all over them curves.

Samuel Hannah · Answer

So for this problem, we have a set of concentric circles given by the family of curves F two on. We&#x27;ve been asked to find the family of curves intersect this angle a Now we&#x27;ll use it. Use equation 1.8 points. 16 to do this and so first need find slope of the curves F two. So if you differentiate with respect to X, this gives us that two X plus two y t y dx is equal to zero Now the slope is simply d y the x So if we rearrange just we find d Y d X is equal to minus X over. Why not? Is m two. So if we go to Equation 1.8 points succeed, substitute em two n we find that d y the X is equal Tu minus X over Y minus M over one plus, sir. One minus and X over. Why now? This m is just kind of a but as a is the m firsthand of them be time functions, kinds of each other out and we&#x27;re just left with that. So if we rearrange this right hand side, this becomes minus X minus M y Oh, there. Why? Minus m x. And if you just bring the minor scientific denominator, this becomes X plus m. Why? Oh, there Amex minus y. So this is a differential equation that is homogeneous. This means we could make the variable transformation the equals y over x Oh, equivalently Why equals v times X By the product rule, the left hand side becomes V plus X DVD X and the right hand side becomes X plus and the X over an ex minus v X cancel the common factor of excel in each term. In the numerator and denominator, this becomes one plus M V over and monos fee. We could do further simplification by bringing the V from the left of the right hand side by subtracting. And if we rewrite this over a single fraction, this becomes one plus B squared. Oh, the M minus V. This is a separable first order differential equation which we consoled by bringing the the variables of the left hand side. This gives us that M minus V over one plus V squared TV is equal to one over x The x You sold this by integration and the integration for the right hand side is quite simple and it&#x27;s simply for natural logarithms of X, not forget integration On the left hand side, this integration can be rewritten as the integration of em over one plus B squared minus in school of the over one plus B squared. Now the first of these terms is simply m times the ark time of the and to the second term we&#x27;ll see what happens. Went by substitution. So for the integral, the over one plus v squared we sub in that one plus V squared is equal to T. Then Devi transforms is DT equal toe to vdv And so the integral becomes 1/2 time central of one over t DT. The integral of one over tea is simply a natural log of tea. And once we make the reverse, substitution physicals are half times the natural log of one plus the squared. Therefore, this entire Antrel on the left hand side of their French equation becomes the art town of the right. This is 10 to the minus one 10 to minus one V times M minus 1/2 natural log, long plus B squared. So that&#x27;s the left hand side We bring this all together we get the differential equation is sold by 10 to the minus one to be times m To put that in the front minus 1/2 natural log long plus V squared. That&#x27;s equal to the natural log of X Awesome integration. Constant C So it will do some more, um, simplification. But first we want to go back into our Y variables. We do this by surfing. V equals the wave. Rex, you get the M times the inverse tan. Why ever X minus, huh? Natural log. His one plus y squared over X squared is equal to the natural log of X plus C. Now, if we bring the natural law terms all over to the right hand side, this gives is the M. I was 10 minus one. Why have Rex is equal? Using the low close is equal to the natural log is X squared times one plus why squared over X squared plus c and how we&#x27;ve done this is we&#x27;ve multiplied everything through by two to most if I ever multiply every front free by two. And on the right hand side this Ellen of X ghost a lead of X Squared. So just need to put me to hear. And then this simplifies because the most part they exploit freedom bracket and this becomes to m. I&#x27;m Steve Ember first hand. Why have Rex and that&#x27;s equal to be natural Log vex word plus y squared. Let&#x27;s some constant Now we can move into polar coordinates by using the fact that tan to most want of whatever ax is feet er and X squared plus y squared is our squad. So this goes to two feet. Er is equal to the natural log of r squared, plus some constant c on If we divide fruit by to once again we can. But the power of 1/2 to office clad and this becomes that M feta is equal to the natural log of r plus some new constant K. We just take the exponential of both sides. This gives is that oh is equal to some constant K £1 the exponential of em feta, where we&#x27;ve just brought the constant term over to the left hand side. And this is exactly in form, which we wanted to show our differential equation solution had. So what we can do now is draw Cem sketches of all solution which have done buffet That&#x27;s up. So here the black circles of Justin Concentric circles on Be for a Equals pi over six. We have Theo. Absolution is one over the square root of free. This means that the intersecting curves are given by r equals some constant K times the exponential of feet over square it free on this gives the spirals here.

Problem

An object is released from rest at a height of 10…

Problem 23 Medium Difficulty

Answer

$m_{1}=\frac{m_{2}-\tan (a)}{1+m_{2} \tan (a)}$

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$m_{1}=\frac{m_{2}-\tan (a)}{1+m_{2} \tan (a)}$

Differential Equations and Linear Algebra

Heather Z.

Caleb E.

Samuel H.

Michael J.

Integration Review - Intro

Definite vs Indefinite

Stephen W. Goode, Scott A. AnninDifferential Equations and Linear Algebra

Heather Z.

Caleb E.

Samuel H.

Michael J.

Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra