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Numerade Educator



Problem 14 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.
$$\mathbf{r}(t)=\left\langle t^{2}, \sin t-t \cos t, \cos t+t \sin t\right\rangle, \quad t>0$$




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Video Transcript

Alright. In this problem, we want Teoh. Find the unit Tangent Unit and unit normal vectors for the vector that the curve represented by the vector Our which is T squared scientist e minus t co sign T and co sign of T plus tty sine of t from on the interval from T greater than a T rooted in zero so t from t zero t increasing And we also want to find the curvature. So let's start off with our unit tangent Vector for pert A So first thing we need to do is find the derivative of R. So we're gonna look a our prime of tear and taking the derivative of our first component function gives us to tea derivative of our second component function gives us co sign of t minus. We need to use your product rule so we have minus co sign of tea for the first component plus t sign of tea. Matters are second component function and then our third component function Ah, we have negative sign of tea first, uh, derivative of the first term and then again using our product rule, we end up getting sine of t plus T's I'm Times Co sign of tea for the product using the product rule on that third component function on. We can simplify that a little bit, right? We see that we have some like terms that will add up to be zero so co sign of T minus co sign of tea will be zero and so will negative sign of t plus sine of t. So we end up getting to t t sign of tea and t co sign of tea as our three component functions for the derivative of the vector are now, let's find the magnitude of that derivative, which is gonna be the square root of two t squared plus t times sine of t squared plus t terms co sign of tea squared. So if we ah go one step further, get rid of our exponents will have four times t squared plus t squared times sine squared of tea plus t squared times co sine squared of tea. We see that our second and third terms have a t squared in them that we could factor out and we're gonna be left with sine squared plus co sine squared which no by the Pythagorean identity is equal to one. So we're gonna be left with four t squared plus t squared under our square root sign or the square root of five T squared, which can be written as the square to five times t. Since we know that T is greater than zero, we take the positive square read. So now, finding the unit tangent vector teeth We have our prime of tea over the magnitude are prime of tea, which means that we're gonna be multiplying one over square to five times t times to t t sign of t t co sign of tea. And if we distribute that we end up with two over Route five as our first component function one over route five times sine of t for our second component function and one over Route five times Co sign of tea for a second component function or third component function. So here we have the unit tangent Vector t. Now let's go ahead and find our unit normal vector next. So we know that we're gonna need the derivative of the vector T and the magnitude of that. Let's roll on a little bit further came. So if we look at t prime, we're gonna take the derivative of each of those component functions with respect to T. So we're gonna have zero for the first component functions, since that's just derivative of a constant one over route five times co sign of tea for the second component function, and then negative one over route five. Sign of tea for that 3rd 1 And then let's take the magnitude of that derivative. So zero squared is gonna be our first term under the square roots on. We don't have to write it, um, squared. If we square the second component function, we end up with 1/5 co sine squared of tea, and then we square that second component function or third component function. We end up with 1/5 sine squared of tea again. 1/5 is gonna factor out, leaving us with co side squared plus sine squared. That's gonna be one soon. We're left with one over Route five as that magnitude. Which means that our unit normal vector n it's going to be t prime of tea over magnitude of t prime of tea. It's a second formula that I had written in the top right corner at the beginning of the video. This is going to be route five times two over square to five. One over route five times sine of t and one ever read five co sign of tea. Oh, sorry. I'm looking at the wrong function here. I was looking at Ah t rather than t prime. Should be looking at this one here. So 01 of a Route five co sign of tea and negative one of every five sign if t ever you a little bit better. Okay? And let's go ahead and distribute that square to five. So we end up with zero are coefficients end up multiplying to be one. So we're just left with co sign of tea and then negative sign of tea for that third component function. That is our unit Normal vector. Last thing that we need is the curvature. You're gonna right over to the side. We've got plenty of room. This is part B of our problem, sir. Curvature Kappa of tea is going to be the magnitude of t prime over the magnitude of our prime. Unfortunately, what? You found both of those two things already. We have all the ingredients we need for our formula. So if I look at magnitude of t prime, I see that right here as one over Route five. And then if I scroll up some Aiken, see where I took the magnitude of our prime, which is squared five t. So what we're left with for our curvature is 1/5 t.

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