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Problem 13 Medium Difficulty

(a) Find the unit tangent and unit normal vectors $\mathbf{T}(t)$ and $\mathbf{N}(t)$ .
(b) Use Formula 9 to find the curvature.
$$\mathbf{r}(t)=\langle t, 3 \cos t, 3 \sin t\rangle$$

Answer

$$\text{(a) } \vec{T}(t)=\bigg\langle\frac{1}{\sqrt{10}},-\frac{3}{\sqrt{10}}\sin(t),\frac{3}{\sqrt{10}}\cos(t)\bigg\rangle, \vec{N}(t)=\bigg\langle0,-\cos(t),-\sin(t)\bigg\rangle$$
$$\text{(b) }\kappa(t)=\frac{3}{10}$$

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Video Transcript

okay in this problem were asked for a couple of things were asked for the unit Tangent Vector, the unit normal vector and the curvature of this curve are of t represented as a vector in three component functions. So what I've done is over off to the right. I've written the formulas for the three different things that were asked for, Especially when you're learning these problems. That's really helpful to have your formulas handy as you're getting more familiar with him. So let's go ahead and tackle the unit tangent Vector first. And this is part a of our problem. So to find the unit Tangent Vector, we are going to need to take the derivative of R of tea with respect to T. So what we get is one for our first component function. Negative three sign of tea for the second and three co sign of tea for the third. And then we need to take the magnitude of that function. So magnitude or the magnitude of the vector skinny is the magnitude of our prime of T is going to be one squared plus nine sine squared of tea plus nine curse signs squared of tea. I'm gonna go ahead and factor out a nine from the second and third terms. So we end up with co sign square or sine squared of T plus coast and squared of tea. And we know from the jaggery and trick identity that that is equal to one. So we end up with square root of one plus nine or squared of 10 as our magnitude. So then, if we want to find the vector T, which is our unit Tangent Vector, We look at our prime if t over magnitude of our prime of tea it so we're gonna get one over square of 10 times that vector one comma, negative sign three side of tee times, three co sign of tea. And then if I go ahead and distribute that one over squared of 10 what I end up with is one over squared of 10 negative three over route, 10 times sine of t and three over Route 10 Crow Side of T. So that is going to be our unit tangent vector. Since the first thing that we're asked for, we're also asked for the unit normal vector. Let's find that next scroll down a bit here. So what we want to have is the derivative of the vector tea and the magnitude of that vector. So first, we're gonna go ahead and take the derivative of the vector T. So what? We get zero for first component function Negative three over squared of 10 times co sign of tea for our second component function a negative three over squared of 10 times sine of t for our third. And then we want to go ahead and take the magnitude of that vector magnitude of t prime. I was going to be zero squared for our first, uh, terms. That's gonna just stay zero. We don't need to add it in Let's square our second component functions who we get. 9 10 9/10 co sine squared of tea plus 9/10 sine squared of tea by factor out at 9/10 on left worth co sine squared plus sine squared which fire without Korean identity. It's one. So we end up with the square root of 9/10 or we can simplify that to three over square 10. So then we can use that to find our unit normal vector and which is going to be t prime over magnitude of t prime. And what we're gonna get is route 10/3 times zero comma, negative three over Route 10 co Sign of tea, comma negative three root, 10 sine of t. And if we distribute that route 10/3 anti each of the component functions. We have zero negative co sign of tea since we're multiplying reciprocal Z and negative sign of tea. So this is our unit Normal vector. So you've got to the things that we were asked for now, last but not least, we just need to find the curvature using the third formula that I wrote up in the top right hand of the screen. This is part B and defined the curvature Kappa. We look at the magnitude of t prime over the magnitude of our prime. Unfortunately, we have already found those two things in are in our problem. So we look at the magnitude of T prime, which is here. We see that we have three over Route 10 her name a raider. Then in our denominator, we screw up a little bit. You can see we found our prime of tea here that's gonna be squared of 10. Which, if we reduce that fraction or simplify that fraction, we end up with 3/10. So this is our curvature.

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