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# (a) Find the unit tangent and unit normal vectors $\mathbf{T}(t)$ and $\mathbf{N}(t)$ .(b) Use Formula 9 to find the curvature.$$\mathbf{r}(t)=\left\langle t, \frac{1}{2} t^{2}, t^{2}\right\rangle$$

## $$\kappa=\frac{\left(25 t^{2}+1+4\right)^{\frac{1}{2}}}{\left(1+5 t^{2}\right)^{2}}$$

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

all right. In this problem, we want to find thes unit tangent and unit normal vectors for are empty as well as the curvature of this curve. Written by the victor are 50. So let's start with the first part of our problem and look at the unit Tangent Vector. So the first thing we want to do is take the derivative of our so relegate our prime of tea which, taking the derivative with respect to t of each of our component functions, we're gonna get one t and to Tiu and then finding the magnitude of our prime, we're going to have square root of one squared or just one plus T squared plus four t squared, which is going to give us a square root of one plus five t square. So when we find our unit tangent vector T, we're looking at our prime of tea over magnitude of our prime of tea which is going to be one over route one plus five t squared times won t to t which that could be written. Weaken, expand that out. Our sorry distribute that term. So we would end up with one over swear e one plus five t squared comma t over square root one plus five t squared, uh, and to t over that swear e so we can stop here and say that this is our unit Tangent Vector. Now what I am going to do for the remainder of the problem to find the unit normal vector is I'm gonna write this, uh, with all of our component functions written as products instead of quotients. And it's just gonna make things a little bit easier to deal with Iran? I think so. We're gonna have one plus five t squared to the negative 1/2 power tee times one plus five t squared to the negative 1/2 power and to t times one plus five t squared to the negative 1/2 power. Okay. And so that's our first part. We found our unit Tangent Vector. Now we want to focus on our unit Normal vector. So do that here. Okay, so let's start by taking the derivative of the unit Tangent Vector. So we're looking at T prime. All right? We have to use the chain rule for each of these. So, for our first component function, we're gonna have is negative 1/2 times one plus five t squared to the negative three halfs power. And we have to take the derivative of the inside serenely multiplying that by 10 t are second component function. We have to use product and chain rule. So our first component is going to be our first term is gonna be one plus five t square to the negative one half's power because we took the derivative of tea, which is one. And then we're gonna add first function times drove of of the second so t times negative 1/2 times one plus five t squared to the negative three house power. The time is derivative of the inside, which is still 10 t here. All right, And then let's take the derivative of that third component that we've got eso first function times derivative of the segment. So we're gonna have, uh I see. Oh, sorry. Let's dio let's keep the same consistency here. So we have derivative that the first which is a derivative of two team or tee times or or to excuse me Times one plus five t squared to the negative 1/2 power plus first function times driven of the seconds we have negative 1/2 times one plus five t squared to the native three halfs power times derivative. The inside, which is 10 t That's a little bit unwieldy to look at, but we can go ahead and simplify. It will get a little bit better. We're gonna have negative five tee times one plus five t squared to the negative three House power men will have one plus five t squared to the negative one halves power minus. So multiplying all these terms here we're gonna end up with minus five t square times one plus five T squares to the negative three House power. It's our second component function, but our third is gonna be too Times one plus five t squared to the negative one House power and we're gonna multiply these coefficients out here so we're gonna get negative. 10 t squared times one plus five T square to the negative three halves power. Okay, so we can probably right this a little bit differently. Let's see. How do we want to do this? I am going. Teoh, leave our first tournament as is so we're gonna stay with native five tee times one plus five t squared to the negative three halfs power. What I'm gonna do for my second component function is I'm gonna factor out a one plus five t squared to the negative one halfs power. And what we end up with for that is one minus five t squared times. It's gonna be one plus five t squared to the negative one Power. I'm gonna go ahead and put that in the dominator. Hurry! And that I'm gonna do the same thing here except running a factor a two times one plus five t squared to the negative one. House power, which is going to you also give me one minus five t squared over one plus five t squared there. Okay, so we're starting to see some things that can simplify a little bit more nicely. I'm gonna leave my first component function alone. What we do you see is that our second and third component functions have this common term here, which I'm going right off to the side. If we were to write that one with as a fraction with the same common denominator is gonna put this off in a box years. That way we don't get it mixed up with the rest of our problem. So if I wrote this first term as one plus five t squared over one plus five t squared, now they have the same denominator. My fire t squared and negative five t squared. And the numerator are going Teoh add up to be zero. So we're gonna be left with one plus five t squared. So I'm gonna go ahead. And right now, my my vector components in that condensed form. So we're gonna have one plus five x squared to the negative 1/2 power times 1/1 plus five t squared and two times one plus five t squared to the negative one house power times 1/1 plus five t squared. Okay, it's starting to look a little bit better. Still a little bit unwieldy, but that's okay. We will persist. Uh, I think I'm going to do one last thing and write everything in terms of products again just because I think it's gonna make it a little bit easier to square it later on. Um, having it is questions for that second and third term That was easier to work with us for a simplifying it. But I think now, just for consistency sake, we're gonna put everything back in the same format. So I'm gonna have one plus five t squared to the negative one power and same thing here who? That should be in the negative One power almost there. OK, so let's have negative five tee times one plus five t square to the negative three house power. And we see that these both have the same base. So we're gonna be left with one plus five t squared to the negative three halfs power and then two times one plus five t squared to the negative three halfs power, and that is going to be t prime. Now we have to find the magnitude of that vector, and having it in the most condensed form is going to be helpful to us. So let's take the square root. We're gonna square each of our component functions and add them all up. So square reversed. 1st 1 ends up being 25 t squared times one plus five t squared to the negative third power. Plus we're in a square. That second component functions we're gonna have one plus five t squared to the negative bird power and 3rd 1 squared is gonna be four times one plus five t squared to the negative third power and we are almost done. So we see that both of these are like terms that can be added together. So we have 25 t squared plus one. They're times one plus five t squared to the negative third power plus five times that seem term, which we can then factor out r one plus five t squared to the negative third power leaving 25 moves t squared plus five. And we are so close. If we also factor out of five from this here you left with five. I was one plus five t squared to the negative third power and that's actually going to be hoops. Let's see, factoring out of five. So we're left with five p squared plus one which hey, those air the same. Right? So we can right this as square five times one plus five t squared to the negative too. And that is going Teoh, give us square to five. If we take the square root of this term, we're gonna end up with one plus five key square to the negative one power or one plus five t squared in the denominator. All right, That was a lot of algebra work. But what it will give us is our unit normal vector and which is going to be t prime over absolute. That are magnitude of T prime. So we're going to have one plus five t squared over route five times. See, this is our t prime right here. So I'm gonna go ahead and multiply that and hopefully we're gonna see some terms combined pretty nicely. Okay, so then if I distribute this one plus five key square term into each of my component component functions, we're gonna end up with Well, and we're gonna distribute the root whatever re five as well. We're gonna end up with Native five over route five t. And then, um, we're gonna have one plus five t squared to the negative 1/2 power, and we're also gonna have went over square five times one plus five T square to the negative 1/2 power and then Teoh over square to five times one plus five t squared adding our exponents together we have negative 1/2 power, and if we want, we can write that a little bit more concisely. So we're gonna end up with thinking of five t over. I'm gonna put my negative 1/2 power term in the denominators. So that way we can have the positive square root of five times one plus five t squared, then when over square to five times one plus five t squared and to whoever square to five one plus five t squared. And if you want, you can distribute that five within the square root sign. But I'm gonna go ahead and leave it as is, and we end up with this as our unit normal vector. So, like I said, lots of algebra work, but we were able to find it. Now let's find one last component of our problem. Part B asks for the curvature Kappa of tea, which we know is the magnitude of t prime over the magnitude of our prime magnitude of t prime. We found right up here, so that's going to be square to 5/1 plus five t squared. Then let's screw up and remind ourselves what was the magnitude of our prime? And we see that right here, square root of one plus five t squared. So if we simplify that complex fraction, what we have is square to 5/1 plus five T square to three halves power, and that represents the curvature.

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