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University of North Texas

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Problem 52 Easy Difficulty

Show that the circular helix $\mathbf{r}(t)=\langle a \cos t, a \sin t, b t\rangle$
where $a$ and $b$ are positive constants, has constant curva-
ture and constant torsion. [Use the result of Exercise
51$(\mathrm{d}) . ]$

Answer

$=$ constant

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{'transcript': "to show our curvature, and our tours are both constant. We can just go ahead and find over derivatives, take some cross product stock products, etcetera, uh, and then just plug everything into this equations down here. So this was the equation that they give us from 63. 63 D? This is 63 D, and then this one is just in the chapter or are curvature. Uh, so let's first, uh, so we need to find up to the third derivative. And then after that, we just plug everything in and go from there. So let's do that. So our first derivative is going to be so our prime, and I'll just start writing our prime as opposed to our private team, just lazy and remember to set the drills of each of these components. So this would be negative, a sign of t that would be a co sign of T. And then this is just going to be, uh, be second derivative. That would be negative. A co sign of T is going to be negative, a sign of tea, and then that's going to be zero. Because remember A and B or just arbitrary constants. And then our third derivative is going to be, um so it be a sign of t and then this is going to be negative. A co sign o t. And then we saw zero for that last position. Okay, so for our curvature, So Kappa here, we need to take the cross product of the first derivative and the second derivative. So let me go ahead and just do that first. So we want to look at our prime cross our double prime. So this is gonna be i j and K. So I put our prime first. That would be negative. A sign t This would be a co sign t. Then I have B and then it would be negative a co sign t negative a sign t and then zero, uh, and to do the cross product. But I like to do is rewrite the first two columns and then I do like down spice up so you could do like, the What's that old? The co factor expansion. But I always end up messing something up, doing it that way. So that's why I just prefer to do it this way instead, um, negative a sign. So what I'm gonna do is first multiply going down the diagonals like this, and I'm going to add all those up. So actually, let me do this one at a time, so I don't accidentally write something down, so I'll underline them instead. So I'm going to multiply all those together first, so it's just gonna be zero. And actually I mean, like that. Back up. Let me skip this down here. So I have more room to write, so that is going to be zero. And then we're gonna multiply all of these. So that would just be negative A b CO sign t j. And then we're going to do all those with the babies. So the negatives cancel. And it would be a squared sine squared tee times K. And I guess all they should have little hats on them also. Right now, I'm going to do minus the diagonals, going upward like this. So let me just go ahead and do this. So first we have those three multiply together, so that would be negative. A square co signs where t k that, uh And then we would have these threes. Who would be negative. A b sign t I had and then lastly, zero times all of this. So it's just gonna be zero. Yeah, um, we can go ahead and distribute the negative. So those just cancel out and then we can collect are like, terms. Uh, so let's see, um, our eyes, the only I have is this. So I'm just gonna write a B sign t and then the jays. The only J we have is this right here. So it would be negative a B co sign t. Uh, then we have these two case here, but it's a squared sine squared, plus a squared coastline squared. And that just becomes a square due to Pythagoras. We have a squared there. Okay, so this is a cross product. What else do we still need? So we need to take the magnitude of this and then divide this by the magnitude of our prime. Um and then we cube that. So let me just go ahead and write. This is our prime of X double prime, and I'll go ahead and just take the magnitude of this since we have it here. Uh, and that is going to give So a squared B squared side square t was a squared B squared co sign square teeth plus a to the fourth all square rooted. And then, just like before this a squared B squared, sine squared plus same thing. But coastline square will just be the coefficient. So this is going to be a squared B squared plus ace word or expert eight to the fourth. Yeah. Okay. Now we need to take the magnitude of our prime. So let me, actually just scooped all of this over a little. So we have some space to write that and actually, just go ahead and move this down. So the magnitude of this now is going to be so to be a squared sine squared T plus a squared CO signed square T plus b squared, and then we square root all of that. And then just like four a squared sine squared, plus a square coastline square is just gonna be a square. So this is going to be a squared plus B squared square root. Okay, now we have everything we need, um, to plug in. So let me scoop this down. So that is what we have in the numerator. So this is going to be equal to the square root of a squared B squared plus a to the fourth all over. And then our magnitude of r squared again was the square root of a squared plus B squared, so it would be the square root of a squared plus b squared. But then we cube this whole thing. Okay? I guess if you want, you could go ahead and pull this to be one square root. So just be the square root of a squared B squared plus a to the fourth all over a squared plus B squared. Cute. Um, that's one perfectly good way to write it, Capa. And so remember, A and B are all constants a B Constance. So then this whole thing, if I just add, subtract, multiply raised powers, take the square root. That's still going to be constant. So then this implies Kappa Constant. So we've got Kappa down. And now for our torgyan, Um, what do we need? So we need the cross product of actually, let me go ahead and collect everything really fast. So I'm gonna move some things around, and I'll be right back. Okay, now I've collected all those. I guess I was in an instance, for you, but that took me a minute to kind of move everything around. Um, so these are just the three things that we need. Uh, we have the cross product of the 1st and 2nd derivative, which we already found the magnitude of this, which we also have there. And then we also have the third derivative, which we found there. So we first need to take the dot product of these two. So let's go ahead and do that. So we're going to do our prime X across our double prime dotted with our triple prime. And so we just multiply these coordinate y so that would be a square to be sine square teeth. That would be a squared b co sine squared t. And then that would just be plus zero. I don't know why I put a bracket here and a squared B sine squared t plus a squared B co sign square teeth again. The science where plus co sign is just one multiply. So that would be a squared B. All right. And now the magnitude of this cross product. Well, we can just come over here and square this and then that's going to give us a squared B squared plus 8/4. And then we just divide those two. So our torgyan is going to be equal to so a squared over be divided by a squared B squared plus a to the fourth. And again, a. B. Constance implies that towel is a constant. So let me go ahead and come up here and grab Kappa. And that way we can just have them side by side so we can see what both of these are. As you can see, though, in each case we have where both of these are going to be a constant."}

University of North Texas
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