find-the-equation-of-the-orthogonal-trajectories-to-the-given-family-of-curves-in-each-case-sketch-5

Question

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

Michael Jacobsen · Accepted Answer

in this problem. We&#x27;re looking for the thaw gurnal trajectories to the curve given to be why equals C times the exponential function each of the X. The first step is to differentiate both sides of that equation. So we&#x27;ll have that D y t x will be equal to the derivative of a constant C times an exponential What that would give us C times either the ex itself. Since our exponential is to the base e To find the full expression for this derivative, we need to substitute out the constant C since see, it depends on both X and Y from the original family of curves. We know that C is equal to why divide by either the ex. So we&#x27;ll substitute that back into our derivative for our next step. D y d X is now equal to in place of see we have why divide by you to the X times, even the X. That gives us a quick cancellation in this problem, or the base of the X is cancelled from the numerator denominator. And now we have that D y d X is equal to why itself next to find orthogonal trajectories, We&#x27;ll take that D Y DX must be equal to the negative reciprocal of the expression we have just found. So our differential equation becomes D Y D X equals negative one over why and to solve this differential equation, let&#x27;s collect all portions that involved why toe left inside and all the portions involved x to the right hand side. Upon doing that, we have that why d y will be equal to negative one times DX. Then we can integrate both sides the left hand side with the respect to why the right hand side with respect to X to obtain that 1/2 a y squared is equal to negative X, plus a constant. Let&#x27;s call this one upper case C. If we add extra pill sides, the equation is now X plus. 1/2 of y squared is equal to upper case C, and then we can simplify this equation further by multiplying both sides by two. If we do that, the equation becomes two. X plus y squared equals a constant will call this one D where d equals two times Upper Casey, and this is the equation for the orthogonal trajectories off the original family

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

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

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

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 the thaw Grenell trajectories to the family of curves, described as why equals C Times X to the power five. To find those families. Let&#x27;s differentiate both sides of that equation. With respect to X, we&#x27;ll obtain that D Y. D X is equal to five times. See Time&#x27;s X to the power for the constant C needs to be expressed in terms of X and y, and that could be done from going to the original equation in dividing both sides by excessive power. Five. We obtained that C is equal to white divide by X to the fifth power than are derivative. De Y T X can now be re expressed as it&#x27;s equal to five times why divide by X to the power five times X to the power for upon simplifying we obtained. Now that D Y D X is equal to five times why divide by X. Now we can take this expression for the derivative and obtained the following differential equation. Orthogonal trajectories will satisfy the differential equation. That D Y D X is the negative reciprocal of the expression just found. So are negative. Our right hand side will be negative. X divide by five y And to solve this differential equation, I&#x27;m going to collect all terms involving why the D Y in the five y toe, left hand side and all other terms involving X to the right hand side. So our differential equation now takes the form that five y de y is equal to negative X dx. We can integrate both sides of this equation the left side with respecto Why and the right site with respect to X. To obtain that five halves of Weiss word is equal to 1/2. Excuse me, negative 1/2 of X squared, plus a constant of integration. Let&#x27;s call that upper case C. We bring ah half of X squared to both sides of this equation. We obtain 1/2 X squared plus five halves of y squared is equal to that constant C, and upon clearing denominators, the family of orthogonal trajectories is X squared. Plus five y squared equals a new constant D where de is equal to two times upper Casey. And that&#x27;s the expression for the orthogonal trajectories to the original given family

Problem

Find the equation of the orthogonal trajectories …

Problem 15 Easy Difficulty

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

Answer

$2 x+y^{2}=D$

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Stephen W. Goode, Scott A. Annin

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$2 x+y^{2}=D$

Differential Equations and Linear Algebra

Grace H.

Anna Marie V.

Kayleah T.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

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

Grace H.

Anna Marie V.

Kayleah T.

Samuel H.

Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra