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Prove that the curve rit) = {a btP c + dt?,e + #P/, whereand are real numbers andpositive integer; has zero curvature Give an explanation.What must be shown to prov...

Question

Prove that the curve rit) = {a btP c + dt?,e + #P/, whereand are real numbers andpositive integer; has zero curvature Give an explanation.What must be shown to prove that rlt) has zero curvature?It must be shown that the dot product, a *V is Zero, or that the acceleration,is constant:must be shown that the velocity; and the acceleration; are constant: must be shown that the magnitude of the cross product, laxvl; is zero or that the unit tangent vector; T is constant: It must be shown that the cr

Prove that the curve rit) = {a btP c + dt?,e + #P/, where and are real numbers and positive integer; has zero curvature Give an explanation. What must be shown to prove that rlt) has zero curvature? It must be shown that the dot product, a *V is Zero, or that the acceleration, is constant: must be shown that the velocity; and the acceleration; are constant: must be shown that the magnitude of the cross product, laxvl; is zero or that the unit tangent vector; T is constant: It must be shown that the cross product axv is constant. In this problem, show that the magnitude ofthe cross product, laxvl; is zero. To do so, first find the velocity; bt" ,€ + dt" e + #P) Find the acceleration, Compute the cross product, axv: axv This result for xv demonstrates that the given curve has zero curvature_



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Zero curvature Prove that the curve $$\mathbf{r}(t)=\left\langle a+b t^{p}, c+d t^{p}, e+f t^{p}\right\rangle$$ where $a, b, c, d, e,$ and $f$ are real numbers and $p$ is a positive integer, has zero curvature. Give an explanation.

You should. It is, I guess. Let's off then. Unless this That this question might be yes. If I ds Yes, Yes, this is okay.

Were asked to find an equation for the curvature of a plane curve in terms of the angle phi between sectors, tea and I. So the angle of inclination of the tangent line Well, you know that for a plane curve, that's your tea. This is same as the magnitude of tea times cosine of five I plus the magnitude of tea times the sign of Fi J where I and J span the curtain. The plane which is equal to so it follows that derivative tea with respect s by the chain rule. This is the derivative of tea with respect to fi times the derivative of five with respect to s from the above. If we say that the magnitude of tea is one so T is normalized, then we have a derivative of tea. With respect to fi, this is going to be the vector negative sign. Bye I plus co sign of fi j times Thebe, Areva tive defy ds and therefore the magnitude of de TBS is the magnitude of negative sign Phi I plus co sign Phi J times The magnitude of defy Yes, and the magnitude of the first factor is simply once This is the magnitude of defied. Yes. Therefore, for a plane curve, the curvature Kappa, which is the same as the absolute value or the magnitude I mean of D TVs is the same as that's the value of defy DS.

Yeah. Fight Brady s actually get one point or a question mark over a point of the post. What's deep? So are tasty. Jacob. Are you so because it was six so Beyonce.

Okay so we need to show that the curvature over playing the plane took off is given by copper equals absolute value of the fee. Ds. Um Where fee is the angle between T. And I. So I this is just Another way of saying 10. And in order to do this. And also I need to mention that the fee is the angle or inclination of the tangent line. Okay so let's let's define the tangent. Thanks to yeah we just need to entries X. I'm gonna call it exurbs and why? Oh Bess and I'll do a little trick in order to produce of course I in sign what I'm gonna do is to fucked her out. Exurbs. Square plus Wild guess square. And then entry should be divided by X. Over square plus Wild west square. And then while Bess or bar the same denominator. Okay so this is the link. Okay. Oh bess. Right Which is which is one. And then the X. Entries can be written as T. O. B. S equals course I of feel best. And then yeah. Why a bus or a bar the length of your best. This can be written as sign. Oh feel this Okay soy mm we can write ups as course I of for your best and sign of your best. Now feet. This is the angle between zero. Uh No this is this is the angle between T. O. B. S at the at the point as and the and the access all the X axis 10 The way you can confirm this is to use the property The one God be too. They see cause The one the rank of the one Bank of the two course I or V. Where V is the angle between the angle between the vectors V. One and V. Two. So when you take the doctor product between todos and 10 which is the x axis, you get call sign of feel this with our new expression. So this the the angle fee. This is the angle between Between the vector todos and 10. Okay, so now the original definition of the coverage of function this is given by length of DT ds. Right? So when we when we take the derivative or about a new expression of tms we get the bar Ds T O. S. Because D. D. S. Of course I oh feel this sign of fields and you can use the chamber to conclude that this is sign negative sign or the U. S. And course I. Or feel best and fee primal best. When we take the link of the derivative of tangent vector we get absolute value of the prime with us times square root sign square or feel best plus call sign of square. I feel best. And as we know, using the trick property, we can say that this equals one. So now the problem best. This is the fee or body. Yes, So we have the results.


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