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# 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 $(6,18,36)$ .

## $$42$$

Vectors

Vector Functions

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##### Lily A.

Johns Hopkins University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

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

Campbell University

#### Topics

Vectors

Vector Functions

##### Lily A.

Johns Hopkins University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp