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(a) Show that $d \mathbf{B} / d s$ is perpendicular to $\mathbf{B}$ .

(b) Show that $d \mathbf{B} / d s$ is perpendicular to $\mathbf{T}$ .

(c) Deduce from parts (a) and (b) that $d \mathbf{B} / d s=-\tau(s) \mathbf{N}$

for some number $\tau(s)$ called the torsion of the curve.

(The torsion measures the degree of twisting of a

curve.)

(d) Show that for a planc curve the torsion is $\tau(s)=0$ .

$$

\begin{array}{c}{\text { For a plane curve the torsion }} \\ {\tau(s)=0}\end{array}

$$

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in this problem, we're gonna prove four different things about the by normal vector as it relates to the unit tangent, vector of the unit Normal vector on the derivatives, some of the derivatives. So let's start with part A, which is to show that the derivative of the by normal vector with respect to our length s is perpendicular Teoh the vine normal vector. So there's a couple of things that we want to know or that we wanna think about. So the first thing that the by normal vector is a unit vector, which means that if we take the dot product of the by normal vector with itself, it's equal toe one. So then what we could dio is we could take the derivative of both sides of this equation with respect s using the product rule. So on my suit, for the first term, we're gonna have d B over de s dotted with B. We're going to have be dot d b over D s and then the derivative of the right hand side. It's just derivative, a constant which is zero. We know that the dot product is community of, so I can right these terms and either order, which means that we end up with two times D B over DS dotted with B, which is equal zero. We divide both sides by two. We end up with D B over DS, dotted with B equals zero. Now, remember that the dot product of two vectors if that dot product is zero, that means that the two vectors are perpendicular. So this tells us that do you be over? DS is perpendicular to be because that dot product zero So there's part one. You're part a for part B. We want to show that derivative of the by normal vector with respect to us is perpendicular to t the unit tangent vector. So one of the things that we know is that the unit tangent vector on the by normal vectors be anti are perpendicular So we know that the by normal vector dotted with the unit Tanja Vector is equal to zero. So what I'm gonna do is take the derivative of both sides of this equation with respect to s. So I'm gonna put zero. Uh, I'm gonna sweep the flop the order of those two just for, um a consistent notation here. If I take the derivative of this using the product rule, what we end up with is DB every day. Yes, Again. This is derivative with respect to s Got it with tea plus my normal vector dotted with T t River DS. Now, in a previous problem, we showed that this quantity here is equal to the curvature Kappa times a unit normal vector n So we can rewrite this as DB over ds dotted with tea plus be dotted with Kappa times end. We can write caba of tea as out in the front because if we're talking about with respect to t, that could be treated as a scaler. And we know that the by normal vector and the normal vector unit normal vector If I dot those two together not cross product, adopt those two together, we're gonna get zero because those are also orthogonal to one another or perpendicular. So this term equals zero and we end up with Do you be over? De s dotted with t is equal zero, which tells us that db over ds and t are perpendicular. There's the proof of part B. All right, so for part C Then what? Two years A and B to show that D B over DS is equal to negative towel of s times the unit normal vector end And here televisions Just some number. And that it represents Tor Shin as something called Tor Shin. So here's how we can reason there This we know that be and and t are all Weerasak it'll to one another. We also know that d B over ds from parts A and B is perpendicular Teoh T and perpendicular to, uh, be so What that means is that d b over DS must be parallel with the unit Normal vector and and if it's parallel, that means that it has to be a scaler multiple of the unit normal vector. So if we define negative towel of s as a scaler because it's some number again representing portion we can represent D B over DS as the scaler times the unit normal vector. So that's how we can reason three part C. Last but not least, we want to show that for a plane curve, the Tor shin is equal to zero. I found it helpful when reasoning through this problem to draw a picture. So let's say that we have a plane and we have some curves that lies along that plane. We know that our unit Tangent Vector might look something like this, and it's going to be parallel to the plane. There's going to be contained in the plane. There's tea. We know that our unit normal vector is worth ogle to that. And then our by normal vector would come up off the plane something like this, and it's or five channel to these other two. So if we think about this, it's going to be perpendicular to the plane and its A unit vector. And that's for any point along our curve. So we can actually reason that the by normal vector is a constant unit vector. And because our curve is contained in our plane and our unit Normal and Unit 10 defectors are contained in the plane, the by normal vector is always going to be perpendicular to that plane, which is why we could say it's a constant. So if we can treat it as a constant, then that means that Davey over DS is equal to zero. Well, we said from part, see that Devi over DS is equal to negative towel s times the unit normal vector n. However, whenever her unit normal defector is defined, it cannot be zero so and is not equal to zero where it is defined. And thus we have to conclude that negative tower of s or just tell ofhis is equal to zero for a plane curve and that concludes airport.