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Problem 19 Medium Difficulty

Use Theorem 10 to find the curvature.
$$\mathbf{r}(t)=3 t \mathbf{i}+4 \sin t \mathbf{j}+4 \cos t \mathbf{k}$$

Answer

$$
\kappa(t)=\frac{4}{25}
$$

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Video Transcript

in this problem 20 years here on 10 to find the curvature. Uh, the curve represented by the vector are of tea which could be written as three tee times Vector I plus four sine of t times vector J plus for co sign of tea times vector k. I'm gonna go ahead and write that in my angle bracket notation as the re t four sine of t and for co sign of tea. So let's start off by taking the first derivative of our prime, which will need for the numerator of Are the numerator and denominator actually of our curvature function by theorem 10. So what we'll have is three, four times co sign of tea and negative four times sine of t. And that's just taking the first derivative with respect to t of each of the component functions in our and then we're gonna take the derivative again to get the second derivative of our so are double prime. We'll have zero negative four times sine of t and negative four times co sign of tea. We want to take the cross product of R prime and our double prime. So our prime if t crossed with our A double prime of tea. We can write that as the determinant of a three by three matrix. So we're gonna have by 30 in our first column J four times Co sign of tea and negative four times sine of t in our second column and vector K negative four times sine of t and negative four times co sign of tea in that third column and then we'll go through. And what we'll have for our first term is we're gonna look in Collins, J and K to get the two by two determinant four times carry science of tea, negative four times sine of t negative four times sine of t and negative four times co sign of t is the appropriate elements and that two by two determinants gonna be multiplied by vector I Then we're gonna subtract a two by two determinant times vector J So we're looking in the eye and K columns who will have three zero negative four sine of t a negative for co sign of tea kind of specter j then plus the two by two determinant of the items and vector in columns I and J So we're gonna have 30 or co sign of tea and negative for sine of t. And that's gonna be times factor K. Let's go through and take those two by two determinants. So for our 1st 1 we're gonna have a negative 16 co sine squared of tea minus 16 sine squared of tea. That's time sector by. We're gonna subtract negative 12 times. Co sign of tea minus zero. We don't have to necessarily right the huh Put us negative 12 times sine of t minus here on again. We don't have to write that to save some space. So it's simplified that a little bit not going to go ahead and write it in. Uh, the angle bracket notation noticed that we can factor and negative 16 out of this term up here. And then we're gonna be left with negative 16 times co sine squared of t plus sine squared of tea which we know by the factory and identity is equal to one. So we're just gonna have negative 16 is that first component function. You will have positive 12 co sign of tea in the second component function and negative 12 sine of T as the third component function. There's our cross product. We don't want to take the norm of that cross product so norm of our prime cross with our double prime. This is going to give us the numerator for the coverage or function for the M 10. So we're gonna have the square root, um, 16 square negative 16 squared, which is positive. 2 56 Quest 12 co sign of t quantity squared, which is gonna give us 144 times co sine squared of tea plus 144 times sine squared of tea. And what I can do is I can factor out the 1 44 from the second and third terms under the square root sign. And then we're gonna have that same thing happen where we have the coastline squared of t plus sine squared of tea which is just going to equal one. I'll go ahead and write that out. Just that way we can see square root of 2 56 plus And factoring out 1 44 we're gonna have co sine squared of tea plus sine squared of tea. That is going to be equal to one. So all we have left is squared of to 56 points 1 44 or square of 400 which is gonna give us nice number 20. So then, for the denominator of our curvature function, we also need to take the norm of our prime, which we have our prime written right up here. So we're gonna have us three squared or nine plus four co sign of t squared or 16 curr sine squared teeth plus 16 times sine squared of tea. Again, we can factor out of 16 from the second and third terms were left with sine squared plus co sine squared, which is going to give us one. So this is going to become nine plus 16 or squared of 25 which is just five. So then applying theorem 10 and our curvature function, we have the norm of the cross product as our numerator and then the norm of our prime cube which is going to be five cute. That gives us 20/1 25 which reduces down to 4/25. So the curvature of the function represented by the vector are is going to be for over 25

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