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3. Find the equations of the system of curves on the cylinder Zy x? orthogonal to its intersections with the hyperboloids of the one-parameter system Xy = 2 + c...

Question

3. Find the equations of the system of curves on the cylinder Zy x? orthogonal to its intersections with the hyperboloids of the one-parameter system Xy = 2 + c.

3. Find the equations of the system of curves on the cylinder Zy x? orthogonal to its intersections with the hyperboloids of the one-parameter system Xy = 2 + c.



Answers

Let $C$ be the curve of intersection of the parabolic cylinder
$x^{2}-2 y$ and the surface $3 z-x y .$ Find the exact length of $C$
from the origin to the point $(6,18,36) .$

All right. In this problem, we want Teoh find the length of a certain curve. This curve is described as the intersection between X squared equals two. Why? Which represents a parabolic cylinder and a surface three z equals X Y S O R. Curve that were interested in is the intersection of these 23 dimensional shapes. And we want to find the length of C from the point 000 Teoh 6 18 36 So one of the ways that this is different than finding the length of a curve that's already represented in vector form is that we are not given the parameter ization of that curve. So we need to primer, parameter rise, See, in order to solve this the way that that we know how to find the length of curves. So two parameter, I see what I'm gonna do. One of the easiest ways is toe let our first component function for X just equal t. And that makes it really nice to figure out what intervals he is going to be on and to solve for the remaining component functions. So then I could look at our first equation that were given for the parabolic cylinder and I can plug in T for X and then I can solve it for why? And when I get is 1/2 a t squared as my second component function. And then lastly, I'm gonna use the equation for the surface and plug in our values for X and y now. So what I'm gonna have is three z equals X is represented by T and why is represented by 1/2 a T square. And if we solve this for Z, we're gonna end up with 1/6 times t cubed. So we need to figure out well, first. So now we have our curve represented by these three component functions, so I could write it as t 1/2 a t squared and 1/6 t cute. But the only thing that we're missing now is the interval that t is on. But since we set t equals X, what we can dio is we can just look at the first component of the points that were given that were interested in for the length of our curves. So here we would be looking at the interval from zero 26 so Now that we have everything represented in vector form, we can use the equation in blue on the top right of the screen to find the length of the curve on the interval from 0 to 6. So I'm gonna do is I'm going to first take the derivative of our vector. So we're gonna have a derivative of tea. Is one derivative of a 1/2 a T squared is going to be just tea and derivative of 1/6 t Cubed is going to be 1/2 a t squared and then we want to find the magnitude of that derivative, which is going to be one squared plus t squared, plus 1/2 a T squared quantity squared, which we can right out as one plus t squared, plus 1/4 times t to the fourth power. And we want to be able to integrate this function so it's in our best interest to get a simplified as possible. So let's keep going with this. We can actually rate that as one plus 1/2 a T squared quantity squared and you can you can check to make sure that that expands out to the same expression under the square root sign. But what this is going to give us is one plus 1/2 a T squared, and we know that it's a positive square root because tea is a positive number. So now we want to find the length of that curve that represents the intersection between the parabolic cylinder and the surface. We were given eso the length of that curve again. We're going from 0 to 6. It's going to be the integral of one plus 1/2 a t squared d t. So if we integrate that, we're gonna end up with T plus 1/6 T cubed from 0 to 6. That means using the fundamental theorem of calculus. We're gonna have six plus 1/6 times, six cube minus zero. So then we end up with six plus 36. We simplify about second term, and that gives us 42. So the length of the curve that represents that intersection between those 3 to 3 dimensional objects we were given is going to the length of that curve is going to be 42

So for this problem we are given to curbs. One is an ellipse with equation three X squared, just y squared is equal to five and the other one is some random curb with equation Y cube is equal to expert. It was even in the problem that they intercepted at the following points, one comma one in negative one comma one. And what we wanted to identify or to know in this problem is to see whether these two curves are talking about and for them to be said to be or talking about they should have, they should intercept for particularly at each of the point of intersection. So first thing you wanted to do is to identified the expression for the slope of tangents for each of the two curbs. So for the first one for the lips we'll have six XDX plus For Y. Dy is equal to zero or simply dy over dx is Neither six x over for white were negative three x over two y. For the curve will have three Y squared. Dy is it was two X dx simply be why over T X. S two X over freeway screen. Now we wanted to find how the slow. So let's call this as M one this one as M two. So we want to evaluate the slopes for each of these curves at the given point of intersection. So let's start with .101. So M one should be equal to negative three times one. All over two times one. We're simply equal to negative three hubs and you will then become two times 1/3 times one squared or simply 2/3. So they know that for two tangents line to be perpendicular, the relationship of their slopes should be should obey this equality. And looking at these two slopes here, we could say that they are in deep perpendicular. Now, considering the other point of intersection, hopes should be negative 0.0.1. Let's evaluate what M. One and M. Two is would have negative three times negative 1/2 times one It's equal to 3/2ves and two becomes 2 times two X. Two times negative one. All over three times one square. So we'll have negative twitter strike here and disobey this equality as well. So I can say that it is perpendicular also. Or these two herbs intersected perpendicular early at this point of intersection. And since it is perpendicular at both of the point of intersection, we could say that these two groups are said to be. We're talking then

Were given to surfaces were asked to find a vector function that represents the curve intersection of these two surfaces. Two surfaces are the cylinder X squared plus y squared equals four, and the surface Z equals X Y. Notice that since explicitness, Weiss Creek was forced. If we take X to be to co signed data and why to be to sign of data, then we have the expert. Plus y squared is equal to full. So these are valid. And then if we take ze to simply be x times y, which is when you before co signed data signed data, then we have that this is parameter ization for curb intersection, and so is the vector function can write it as our tea equals to co sign T to sign of tea and four co sign T Sorrenti.

We know that two curves are orthogonal at a point of intersection. If their tangents at that point cross at right angles. So we want to show that the given um curves are orthogonal at 1 1 and one negative one. So what this is gonna look like is When we take the derivative will get four x. Class six wine. That crime equals zero. And then we'll get um two Y. Y. Prime. Okay equals three X squared. So now we're going to plug in X equals one, Y equals one. So that will be coming for this would just be this right here And then this will be one. Why will equal one? So when we divide this we end up getting 3/2ves and in this case we'll subtract and we'll get negative four or six which will end up being a negative two thirds. So we see that the slopes are opposite reciprocal of one another. So they are north.


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