find-the-equation-of-the-orthogonal-trajectories-to-the-given-family-of-curves-in-each-case-sketch-s

Question

Find the equation of the orthogonal trajectories to the given family of curves. In each case, sketch some curves from each family.
$$x^{2}+9 y^{2}=c$$

Michael Jacobsen · Accepted Answer

in this problem. We&#x27;re trying to find the orthogonal trajectories to the families of curves given by the equation X squared plus nine y squared equals C. Our first job in finding those a thaw Grenell trajectories is to differentiate both sides of that equation with respect to X implicitly. So the set up will be that D by D. X of X squared plus nine Weiss Word equals D by the x of the right hand side. See, differentiating a place implicitly on the left gives us two X plus 18. Why? But because Y is a function of X, this will be multiplied by de y dx through the chain rule. The right inside is the driven of a constant and so we obtain just zero. Our next step is to solve for that derivative. D y DX will obtain that 18 y times d y. The X is negative. Two X dividing by 18 y gives us de y dx equals negative two X divide by 18 y or more simply, de y dx equals negative X over nine y, so the slope of the original curves is always equal to negative. X over nine y Barca was to find orthogonal trajectories, and we know from our training and algebra that we find slopes are negative. Reciprocal. Then these curves will be orthogonal. So it&#x27;s right out that expression to the right that we&#x27;re seeking a family of curves such that de Y. D X is equal to the reciprocal nine y and the opposite over X for this derivative that we found. Now we have a differential equation that can be organized by first dividing both sides by y we obtain one of her Why de y DX equals nine over X And now we have a helpful trick that tells us that D by D. X of the natural log of y is always equal to one over. Why times de y dx. So it&#x27;s like use that equation Here we have d by D. X of the natural log of y is now equal to nine over X. This is an equation that could be integrated on both sides, and it would be an integral the left hand side with respect to X as well as integral of the right hand side with respect to X. So it&#x27;s solved both anti derivatives we know first, the anti derivative of the derivative is just the inter grand itself natural log y. So it&#x27;s right on the left side. The natural log rhythm of why is now equal to the anti derivative of nine times the natural log of X plus a constant C who were. To solve this differential equation, we can convert to a Basie exponential equation to write that why is now equal to e times or to the power of natural log of X to the power of nine by using properties of logarithms, plus a constant using rules of exponents. This can also be written as why equals e to the power of C Times E to the power of natural log of X to the power of nine. With this arrangement, we also have a cancellation of the natural auger and the function with the Basie Exponential, which allows us to write this solution in the form Why equals? Let&#x27;s call this constant C Times X to the power of nine, where see is any rial number. So this is the family of curves that&#x27;s or if agonal to the original family that was given

Michael Jacobsen · Answer

in this problem. We&#x27;re looking for the family of orthogonal trajectories to the curve given to B y squared equals two x plus e. Due to the power of to in this equation, we must differentiate both sides implicitly. If we differentiate the left hand side with respect to X, we&#x27;ll obtain to why times de y dx equals on the right hand side. We obtain simply a two since the constant C will go to zero in differentiation. We can solve this equation now for D Y dx by divine pull sides by two y so that D Y t X will be to over two. Why or de y? The X is one over Why now? Defying the family of our doggerel trajectories, we formed the differential Equation D y DX, which is equal to the negative reciprocal of the expression just found. So now D Y d X is equal to negative y. Let&#x27;s divide both sides of this differential equation by why to obtain that one over. Why Times D y. The X is equal to negative one. Then we can use the formula that the derivative with respect to X of the natural logger them of why is equal to one over Why Times de y dx. In other words, what we have over here in our differential equation can be substituted in for D by D. X of the natural log rhythm of why, although that doesn&#x27;t look like much, progress, will be able to do some cancellations on the step just after Sultry writes that D by D. X of the natural log of why is now equal to negative one. The next step is we can integrate both sides of this equation with respect to X. Let&#x27;s do that immediately at this step, and we noticed the cancellation immediately. The left hand side is looking for an anti derivative to the derivative, so all that&#x27;s left is natural log y on the left hand side, giving us a large simplification. So now natural log of why is equal to Negative X plus a constant Let&#x27;s call it upper case C. We have a log rhythmic equation. If we convert this to a base e exponential equation, we&#x27;ll obtain that Y is equal to each of the power of negative X plus C, then using properties of exponents, why is now equal to U to the power of C times each of the power of negative X Another, since substitution will allow us to write our answer in the form y equals D times each of the power of negative X, where d equals e to the C. And that&#x27;s the format for this trajectory of orthogonal curves.

Michael Jacobsen · Answer

in this problem, we&#x27;re looking for their orthogonal trajectories to the equation. Y equals C times X squared. We start by differentiating both sides of this equation implicitly. With respect to X, the left hand side will be de y DX, and the right hand side will be too time. See Time&#x27;s X to the power of one fix expression for why Prime or D y DX involves the constant C. But see himself depends on both the function. Why, as well as an independent variable X buff we saw for C, We know that C is equal to why divide by X squared. Let&#x27;s substitute that back into our expression for the derivative D y. The X we&#x27;ll obtain that d y. The X is equal to two times of expression for C, which is why divide by X squared times X and upon simplifying this right hand side of the derivative D Y DX, well obtained to why divide by X, the next step of this process is to now solve the differential equation de y dx, which is equal to taking their original expression for the derivative and multiplying it by negative one. As well as taking the reciprocal. So we have negative X divide by two y is equal to D y DX. Now the salt this differential equation. I&#x27;m going to collect all expressions involved. Why the D. Y and the two y and all expressions involved X and bring them to opposite sides of this equation will obtain to why de y is equal to negative x DX. And if we integrate both sides of this expression, then we&#x27;ll obtain that Weiss word is equal to negative 1/2 of X squared, plus a constant of integration. Let&#x27;s call this one upper case C. The equation is essentially solved, but we can simplify it further by adding 1/2 X squared double sides. So half X squared plus y squared equals a constant C. If we&#x27;d like to simplify further, multiplying both sides of the equation by two gives us the family of curves. Ex word plus two y squared equals. Let&#x27;s use a new constant D where de is equal to two time see, and this is the format for the family of curves that&#x27;s orthogonal to the original family. Given

Banhishikha Sinha · Answer

Well, here we are solving the push Number 30 We have given X squared life. Boy square? Yes, they called stool. I don&#x27;t see a way that&#x27;s differently. They questions implicitly with respect to X, which will be two x less so. Why the life? My DX is because you to see. Do you like baby? It&#x27;s becomes the wife. Yes, X My guess why market as a question one? This contains constant C So it is not the difference immigration off the given family so we can express the value off X on the bill in question. But this this one so you can read it is see is a question exclude? No. Why? Square divided by the way. Now substituting this an equation one you get Oh, well, why my two years? Yes, he called Stool x. A group? Yes. Why Square do Why? I mean, it&#x27;s a tropical store X squared plus y squared to square Modify boy, which becomes door. That&#x27;s why stroll square minus my square. From this equation, we get the difference Still equation for the family of portable tragic story. This is the wife ideas The coastal my Nestle exclude minus y script divided by dual X y. Can the witness do you? Why my idea? There&#x27;s pickles toe minus one minus. Why buy Hex? We go. Do you know why? Thanks now Substrate puting. Why speak with stool? Thanks. Be thanks. You convey d y by D it? Yes. Equals to be less x b by the X we&#x27;ve close. Do you be by D X close to one, minus this square by Do we? No, thanks. Nick will be on the right side. Next. Davies idea. Nicholas, stool, maybe square my next one after it so operated form you will be by three b squared minus one. This sheik was still give me a yes. It goes to the X by X now in to get both the sites of the question left side by B and they would say by X like the some instigation will be three square in this one. They request to integration one by eggs in your smart get is in question Took so the integral off this side it is the left side. Let&#x27;s put previous governments one is equals to think which we come six me Do you me some cold stew about into girls becomes intimidation. My B Dave Days Stool The Natural Law Day. Let&#x27;s see. Stop shooting back. The value for these questions do becomes three. Natural law No. Three Squared minus one. Is it still natural? Next messy that comes through this gloominess. One. We&#x27;ll kill. We&#x27;ll still see one. Knicks No substituting back or B is a cold too. Live by s. We get dream. Why buy eggs? Hold square minus one. My cue Because do anyone necks? No, this is cool. 23 way square. My news It&#x27;s good. Hope you close to see line to the bar for So this is a question of the family of portable and tragic stories. Oh, next let&#x27;s drew it. This is 100 another 100 No y Axis 200. This 100 someone to use Bring one for the family of what I wanna trajectories for the oneness because to 90 which way? Who and these and I&#x27;m raising that one for six was 2 20 True. Yeah. Move this. No

Samuel Hannah · Answer

So for this problem, we&#x27;ve been given this trajectory of curves, and we&#x27;ve been asked to find the trajectory of people for Ganahl curves and then sketch a few of those. So first thing to do is we need to work out the trajectory off the off the purpose that we&#x27;ve been given. And we do that by differentiating this equation with respect to X. When we do this term by term, so different shading with respect to X first term becomes two x second term becomes too. Why then multiplied by d Y, the X and the right hand side becomes to see times d y the X. Now we can rearrange us to get a differential equation which governs the trajectory off the curves that we have been given on this differential equation. Rearranges to give us that he wanted the X It&#x27;s equal two x over C minus y. Now, in this current form, we can&#x27;t say that this explicitly governs extract Aries because we still have this constant time. See, we want this entire equation to be in terms of X and y only. So what we need to do is need to go back to our original equation here. Rewrite this for C. So if we do this here, we find that C is equal to x squared plus y squared over to why? Where we&#x27;ve just divided for by two. Why for all equation This thing gives us that the one of the X equals X over exc lead plus y squared over two. Why minus one And then if we most play through on this side by two Wine This is equal to two x y over x squared plus y squared minus 21 squad and then we have some cancellation on the bottom. So final equation that governs of shek trees for given curves is two x y over x squared minus y squared. Yeah, now for you or for colonel curves. So if we say that our family of curves that we have given by coordinates x one and wild one your for colonel curves must satisfy But they&#x27;re equation d y to the X two. When multiplied with dia de y want the X one is equal to minus one and this is satisfied in all case by the curves. If b y the x that eagles two minus X squared minus y squared over two x Y. And what we&#x27;ve done here is we&#x27;ve just flipped the solution we found earlier and then put in minus sign in front. So this is the differential equation, which we want to solve now. Looking at it, it looks like a homogeneous equation, and indeed it is because we can rewrite it as follows. So do you want the X equals minus X squared times one minus y squared over X squared well over X squared times two y over X, Be common factor of X squared on the top and bottom cancels, and we are left with a function that&#x27;s completely dependent on why over Rex, this means we can make the variable substitution off B equals Y over X, which is again equivalently given by y equals v Times X. This means that D Y DX is equal to V plus X DVD X by product rule. So if we transform this differential equation into the V variables, we have that the plus X DVD X is equal. Tu minus one minus v squad of the TV Andi, if we just bring the minus sign into a fraction this is B squared, minus one over to be. Now we can subtract the from both sides, but it&#x27;s minus fee here and then bring this all over. One fraction on this becomes free V squared minus one over to V. So the first head in the last 10 are the separable differential equation for the or for colonel trajectories. Now, to solve this, we separate the V variables to the left hand side on the X variables to the right hand side. So we have the two V over free V squared bonnets. One TV is equal to one over x the x To solve this movie, we need to integrate. Now the integral of two V over free V squared minus one, could be solved using with substitution. U equals free v sled minus one. Andi, this transforms Devi as the U equals six v Even putting this into the integral, we find that the interval of two V of a free V squared minus one TV is equal to eventual afraid over you. You on this simply has solution free times, natural log of you which back in all the variables is free house natural log the B squared minus. Sorry if natural log of free V squared minus one city over Integral theatre. One of Rex by simply the natural log of X. So our entire solution reads free. That&#x27;s natural. Log free V squared minus one is equal to the natural log of X times. Plus the integration constant, which we will write is the natural log is okay so we can combine the log laws here on the left hand side. So this is now equal to Ellen XK, and then we can just remove the natural log. So because we have it on both sides Andi, when we get on the left hand side, we need to take the free that&#x27;s in front of it. Power service gives us the natural log free V squared, minus one, well cubed. And then we remains the natural locks. So the solution for V is given by free V squared minus one. Cubed is equal to X K. Where K is the integration constant. The solution for why we simply need to substitute back in that why sled is equal. The square V is equal to y correct. Sorry on. And this gives is that free. Weiss word one sex squared cubes is equal to X multiplied, the extra four multiplied. Okay. And what we&#x27;ve done there is We&#x27;ve substitute multiplied fruit by X cubed different get in any form. So this is our solution, the vehicle funnel trajectories. We&#x27;ve also been asked to sketch and I&#x27;ve sketched some earlier up here. So what we have is the affordable trajectories, but given by he&#x27;s curves have four branches, one in each quadrant. Andi, I have sketched two calves. The red curve has an integration constant, which is lower than the integration constant of green curve, and we see that it therefore has a lower Grady int.

Michael Jacobsen · Answer

in this problem. We&#x27;re looking for the orthogonal trajectories to the family of curves described ABS y equals a constant C divide by X. To find those trajectories will differentiate the left side of the equation, obtaining D Y DX and the right side of the equation. D by D X of C over X. So the left hand side is just the derivative of why. So just call that the white the excess before the right hand side will turn into negative see divide by X squared. So we have an expression for the derivative. But I expression involves a constant C well, so we&#x27;ll have some work to do to determine what the drifter will look like after we saw for C, we solve for C by going to the original equation. Multiply pull sides by X and we get C is equal to X times y so we&#x27;ll obtain negative X times. Why divide by X squared, which simplifies into the expression de y DX is equal to negative. Why divide by X So the family of curves like we&#x27;ll see Divide by X has derivative. Wyche will excuse me. D y d X equals negative white divide by X to now find the orthogonal trajectories will take the negative reciprocal zone. The right hand side network with an expression de y DX equals X divide by Why to solve this differential equation, I&#x27;m going to involve bring all parts involving Why, to the left hand side, that would be the y and why and all parts involving X to the right hand side. Our differential equation will then be why times D y is equal to X times DX. Let&#x27;s next integrate both sides and find the anti dread of each side. The left hand side will be, ah, half of y squared, and likewise, the right hand side is also half of X squared, plus a constant of integration. Let&#x27;s call that upper Casey. The next step we can take is to subtract both sides by well half X squared to obtain negative 1/2 of X squared, plus 1/2 of Weiss. Word equals the constant C. If we multiply both sides of the equation by negative too, the equation now becomes X squared minus twice. Word equals a new constant D, where de is equal to negative two times see, which is still in arbitrary constant. So this equation represents the family of curves orthogonal to the original family

Problem

Find the equation of the orthogonal trajectories …

Problem 11 Medium Difficulty

Find the equation of the orthogonal trajectories to the given family of curves. In each case, sketch some curves from each family.
$$x^{2}+9 y^{2}=c$$

Answer

$\frac{-x}{9 y}$

Related Courses

Differential Equations and Linear Algebra

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Samuel H.

Joseph L.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra

Grace H.

Catherine R.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$\frac{-x}{9 y}$

Differential Equations and Linear Algebra

Grace H.

Catherine R.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

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

Grace H.

Catherine R.

Samuel H.

Joseph L.

Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra