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Problem 40 Hard Difficulty

Find the vectors $\mathbf{T}, \mathbf{N},$ and $\mathbf{B}$ at the given point.
$$\mathbf{r}(t)=\langle\cos t, \sin t, \ln \cos t\rangle, \quad(1,0,0)$$

Answer

$$\textbf{T}=\langle0,1,0\rangle, \textbf{N}=\bigg\langle-\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}}\bigg\rangle, \textbf{B}=\bigg\langle-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\bigg\rangle$$

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Video Transcript

in this problem, we want to find the unit Tangent unit normal and by normal vectors, T n and B of the vector function Represented by are of T equals co sign of tea, Sign of tea and Natural Log Co sign of tea. And specifically, we want to find those vectors at 2.100 which we can figure out what t value this corresponds to by looking at the component functions. So we know that CO sign of T needs to equal one, uh, and that occurs AT T equals zero. All right. And we can check evaluate t equals zero at other component functions and see that it checks out. So first, let's find her unit Tangent Vector. All right. First, we need to take the first derivative of our our prime is equal. Teoh negative side of t her sign of tea and using the channel rule, we end up with one over co sign of tee times, negative sign of tea which we can simplify that last component function and right, it is negative tangent of t. And then let's take the magnitude of our prime, which is going to give us negative sign T Quantity squared, which is going to be signed squared of tea plus co sine squared of tea plus tangent squared of tea. And we know that sine squared of People's Co. Sanskrit of T is equal to one by the factory and identity. So we end up with one plus tangent squared of tea. And that is the same thing as seeking squared of tea or absolute value of sequent of tea, which, because we're looking AT T equals zero where seeking of zero is equal to one. We can drop her absolute value signs because we know that this is going to be positive for the value of tea that we're gonna be plugging in so we can then find our unit tangent vector t by looking at one over second of tea times the vector No sign of t. No sign of t a negative tangent. Lefty, remember that one over second is equal. Teoh co sign. So go ahead and rewrite this. Distributing one of her secret seeking of tear distributing co sign of t end up with negative sign of t co sign of tea co sine squared of tea and negative sign of tea as our third component function. So it's because if one oversee Kindt of tea is co sign of tea, it's times tangent, which is negative. Sign of tea over who sign of tea. So for that third component function, we're just gonna get negative sign of teeth. All right, that we want to evaluate this Oppa point t equals zero. So t zero Well, to, uh, negative sign of zero times co sign of zero sign of zero is going to be zero. So we're gonna have zero as the first Ah, component purse. I'm squared. Zero is going to need one and negative sign of 00 So here we have our unit Tangent Vector. All right, let's go ahead and find our unit Normal vector, which is represented by n All right. So first we want to take the first derivative of your unit tangent Vector t so t prime Milic it this form event right here. What we're gonna get is a negative co sine squared of tea plus sine squared of tea. What's applying the product rules? So that first component function, second component function. When we take the first derivative, we're gonna get negative two times sine of t co sign of tea The man third component function Take the derivative of that will get negative co sign of teeth and I'm taking the absolute value or sorry, the magnitude of t prime What we're going to get. I'm gonna go ahead and square our first component function here. So we're going to get sign of T to the fourth power minus two times sine squared of TV Times co sine squared of tea plus ho sign t to the fourth power plus second competitive function squared is gonna be four sine squared t care sine squared of tea They're component functions gonna be cose on squared of tea. All right, that's are like terms. This is going to give us sign of T to the fourth power used to terms we're gonna combine. I'm gonna write my co sign to the fourth power of tea first. Ah, the two that I just underline what's gonna give us to sine squared of tee times co sine squared of tea plus Carson squared t Oh, let's see, How could we simplify that a bit? So these right here I recognize that we can write as sine squared of t plus co sine squared of tea When that some is going to be squared, adding her sine squared of tea This by the Pythagorean identity is one. So we end up with square root of co sine squared t plus one, right. So then, for you, normal vector, it's going to be t prime over magnitude of t prime, which means we're gonna have one over the square root of co sine squared of tea, plus one times the vector negative coastline squared of T plus sine squared of tea. Negative two times sign FT co sign of tea. Negative curse sine of t. Right? And then we don't necessarily have to simplify that further cause we're just gonna go ahead and evaluate this AT T equals zero. So what we get is one over the square root co sign of zero is one. We have one squared plus one, which is two, and our denominator, our first component function, is going to give us negative one. We evaluated AT T equals zero second component function will give us the felt You 0/3 component function will give us negative one. So when we distribute the one of her route to you. Get negative one over route to zero. A negative one over route. You, right? That is our unit. Normal vector. Last but not least, we need to find our by normal doctor. We know that by a normal vector is equal to the cross product of the unit Tangent Vector and the unit Normal vector. Since we wanted evaluated at T equals one or T equals zero. Excuse me. We're gonna go ahead and across the T vectors that we found. So we're looking at 010 crossed with negative one over route to zero negative one ever route to. And when we take that cross product, what we end up with is negative. One of her squared of to zero positive one ever swear to to that is thereby normal vector.

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