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# Find the vectors $\mathbf{T}, \mathbf{N},$ and $\mathbf{B}$ at the given point.$$\mathbf{r}(t)=\left\langle t^{2}, \frac{2}{3} t^{3}, t\right\rangle, \quad\left(1, \frac{2}{3}, 1\right)$$

## $$\textbf{T}=\bigg\langle\frac{2}{3},\frac{2}{3},\frac{1}{3}\bigg\rangle, \textbf{N}=\bigg\langle-\frac{1}{3},\frac{2}{3},-\frac{2}{3}\bigg\rangle, \textbf{B}=\bigg\langle-\frac{2}{3},\frac{1}{3},\frac{2}{3}\bigg\rangle$$

Vectors

Vector Functions

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

in this problem, we want to find the unit Tangent unit normal and by normal vectors of the curve represented by the vector function are with component functions T square to third its teeth T cube and T specifically at the point one 2/3 comma one. And if we look at our third component function for the Z coordinate, we can see that this corresponds to a value of T equals one. So in order to find these three vectors, I want to split it up into three parts. So let's look at the unit Tangent vector team. So the first thing that we need to do is take the first derivative of our which gives us to tea to t squared and one and then we need to find the magnitude of that vector. So magnitude of our prime, which is going to be the square root, uh, for T squared plus four t to the fourth power plus one, which we can write as the square root of to t squared plus one quantity squared, which we know is the absolute value of two t squared plus one. So that means that and because we're evaluating this for tea was one. Ah, we know that this quantity inside is going to be positive. So again, specifically because we're looking at the quantity t equals one a value T equals one. We can drop her absolutely value bars. So that means that our unit tangent Vector T can be written as one over to t squared plus one That's the one over bagged itude of our prime times. Our prime, which is to t to t squared one. And what we want to do is evaluate that at won t equals one. So this was going to be one over two times. One squared, just one plus one. Then we'll have two times one, which is to two times one squared, which is also two and one. So what this is going to give us when we simplify it is 2/3. 2/3 1/3. So this is our unit Tangent Vector T. So there is the first part of her problem. Now we want to go on Teoh, find the unit normal vector, and And before we do that, I'm going to write, um because we need our unit tangent vector to compute this. I'm going Teoh, go ahead and write that in a little bit more of a concise way. So that way it's easier to take its derivative. So what? We're gonna have us to you two times t times two t squared plus one to the negative First power as our first component function, our second is gonna be to t squared times two t squared plus one to the negative first power and then to t squared plus one to the negative first power. What that's gonna do is make it a little bit easier to take the first derivative of or vector t so t prime is going to be got to use your product rule here. So we end up with two times two t squared, plus one to the negative First power plus to T at times negative one times two t squared, plus one to the native to power times for T. Taking the derivative of the inside there our second component function. We get four times T times two t squared plus one to the night of first power course to T squared times native one times two t squared plus one to the negative second power times 40 When our third component function get negative to t squared, plus one to the negative to power times for tea. Let's go ahead and simplify that drastically. So what we'll get it's too over to t squared plus one minus e t squared to t squared plus one squared second component function. We get four t over to t squared plus one minus eight t cubed over to t squared plus one squared and negative 40 over two U T Squared Plus one squared as our third component function, and we're gonna keep simplifying away here. So at for my first component function, if I multiply that first term by two t squared plus one over to T squared plus one, what we're going to get is the common denominator of two t squared plus one squared. And then in our numerator, it's gonna simplify to be negative for T squared, plus two, her second component function again. If we multiply that first term by two t squared plus 1/2 or to t two t squared, plus one over to T squared plus one, we're going to get the common denominator to t squared plus one quantity squared and the numerator is going to simplify to four t Then our third component function stays as is. And then we want to take the magnitude of T prime. So magnitude of tea rhyme I was going to equal the square root of these three component functions squared and added up, Since they've all got the same denominator, I'm gonna go ahead and write that as a single fraction. So we have to t squared plus one, uh, to the fourth power because we're squaring it, all right. And then in our numerator, we're gonna have a negative four t squared plus two squared plus for t squared plus negative for T squared. This is going to simplify to be 16 Teoh fourth power 16 to the second power plus for 16 t squared plus 16 t squared all over to t squared, plus one to the fourth power. We've got some light terms that will cancel out in that new marine. Or so it was here What we're gonna end up with. It's four times two t squared, plus one squared over Jude. He squared plus one to the fourth power which will simplify to two over absolute value of two t squared plus one. Which again, Because we're looking at the value T equals one, we're gonna be able to drop those absolutely value signs. So then computing the unit normal vector, it's going to be t prime over the magnitude of T prime, which is going to give us to t squared plus 1/2 times the vector with three component functions that we found above. It's right here. And if we want, we can combine those. So get is negative to t squared plus one over to T squared plus one to t vertu ti squared plus one and negative to t over to t squared plus one. All right, so there's our unit Normal vector, um, function, but we want to evaluate it At T equals one. So what we're going to get is negative two plus one over two plus one to over two plus one and negative to over two plus one, which gives us the nice factor. Negative. 1/3 2/3 negative. 2/3. All right, so there is our unit normal vector. Last thing we need to do is find our by normal vector. So we know that the buying normal vector B is equal to the cross product of the unit Tangent Vector t crossed with the unit normal vector end because we want to evaluate this. I found it easier to go ahead and take the cross product of the two vectors rather than the vector functions. So we're gonna be looking at the cross product of T of one cross with and of one which taking the the vector that we found for T of one earlier. We've got 2/3 2 3rd 1/3 crossed with negative 1/3 2/3 native, 2/3. And you write this, we take that cross product and you might want to review that, Um, if you need to. But what we're gonna end up with is negative for nineths, minus two nights as the first term negative negative four nights plus one night's one night was the second term and four nights plus two nights as third, which simplifies to be the fraction negative 6/9 or negative 2/3 as our first component function. Positive 1/3 because we have negative 3/9 cir negative 1/3. But then we take a negative against you get a positive and then positive six nights or 2/3? Is that third affect your component? So this right here is the by normal vector evaluated at the point ROOTY equals one.

Campbell University

#### Topics

Vectors

Vector Functions

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp