find-the-orthogonal-trajectories-of-the-family-of-curves-sketch-several-members-of-each-family-ym-x

Question

Find the orthogonal trajectories of the family of curves. Sketch several members of each family.
$$y=m x$$

Matthew Allcock · Accepted Answer

in this video, we&#x27;re gonna go through the answer to question number five from chapter 16.3. So testifying to the orthogonal trajectories off the curve, why is it was m facts where m is a constant. So curves Why was m X are gonna have Grady in the Y The x of i m so that four. Well, I thought, you know, trajectories, I&#x27;m gonna have Grady in. You are The X is equal to negative Pacific, all minus one of them. Integrate this to find the family of curves that&#x27;s gonna give us why is equal to minus one of them ex plus a constancy.

Matthew Allcock · Answer

in this video, we&#x27;re gonna go through the answer to question number 10 jumped 16.3 to rest if I into the orthogonal trajectories of the family of curves. Why is equal to eat the K X work? A is a constant diverse. Once find the Grady in off the cars is equal to u k E X. We know that the author no trajectories are gonna have ingredient Pribadi X given by the next reciprocal this That&#x27;s equal. Teoh minus one of Okay, tons e tu minus kayaks. We can solve this just by direct integration. You&#x27;re gonna have Why is equal to minus one of the K that we&#x27;re gonna have to times that by minus one of the K E to the minus k X plus a constant. Let&#x27;s call it. Let&#x27;s go. See, that&#x27;s one of the case squared E to the minus K X. Let&#x27;s see

Matthew Allcock · Answer

in this video, we&#x27;re gonna go through the answer. It&#x27;s question of benign from chapter 16.3. So for us to find the orthogonal trajectories, the family of curves given by Y is equal to see a constant times e minus x. So let&#x27;s differentiate this. Expect X to find out what the grading of this kind of is. Dwight D. X. You have equal Teoh minus c times e sex. The orthogonal projector is I&#x27;m gonna have girls getting by the negative, reciprocal reckless to do y the x. It was equal to one of the sea times e to the X. So just integrate this to find why is equal to one of the sea He&#x27;s the ex Aziz the actress integrates too. So plus a constant let&#x27;s call it be

Matthew Allcock · Answer

in this video, we&#x27;re gonna go to the Ansett&#x27;s question about six, which have a 16.3 to ratify the orthogonal trajectories to the family of curves and my wise you want to see X squared? Let&#x27;s find the brilliant off this from your curves. So just differentiating there&#x27;s gonna be to see of X so that for uh oh, I thought, you know, trajectories are gonna have Grady. And given by the Y, the X is equal to negative is typical of this so we can integrate that. Ah, see, we can interpret that using separation variables Philippines great left inside like so I would take a a constant minus 1/2. I wanted to see as integral one of acts the X And let&#x27;s add to that a constant of integration was Call it G integration Left inside, we get why that&#x27;s equal to minus one over to see integrate one over X and you get the natural log of X. That&#x27;s like the great Innovex. Sorry, the National Ogilvy Absolute value of X plus day as that&#x27;s the, uh, all the books directories

Matthew Allcock · Answer

in this video, we&#x27;re gonna go through the answer to question number seven from jumped 16.3 to rest to find the orthogonal trajectories to the family off curves Given my k X squared plus y squared is equal to one. Did everyone see differentiate this equation with speculative divine d y dx different photo differentiation the first time we get to X? Okay, the second time we get to Why do you want the X is using the train will for that term Example 20 Okay, so these curves have radiant given by do why the X is equal to minus K x over why lots of that Foley on the orthogonal trajectories I have great and given by do why the X is equal to the negative is typical of this. That&#x27;s gonna be why of a K X. This is a separable differential equation. So we can so bit by integrating one over. Why do you Why equal Teoh one of a k and take our constant in school of one of X. The X was the constancy. It&#x27;s great one over why you got the natural log we actually value of why and again integrate one of expect the absolute value. It does not look of the absolute value of X plus C so we can take Eastern power both sides it&#x27;s the left hand side is easy. That&#x27;s just gonna be why the right hand side we can write this is e to the sea times e to the natural log Thanks. The power of one of the K so eats the scene. We can just call up that you know, the name is called Capsule A for A and you constant than e and natural log invest functions cause this is just gonna bay X to the power of one of the K So that describes our family of curves. We could write that in a different way, as X is equal to a let&#x27;s call a one quality from constant times Why, okay,

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 orthogonal trajectories of the family of…

Problem 5 Easy Difficulty

Find the orthogonal trajectories of the family of curves. Sketch several members of each family.
$$y=m x$$

Answer

Related Courses

University Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Catherine R.

Anna Marie V.

Michael J.

Joseph L.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 16

Video Transcript

Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Michael J.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

University Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Michael J.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Joel Hass, Christopher Heil, Przemyslaw BogackiUniversity Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Michael J.

Joseph L.

Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals