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Use Theorem 10 to find the curvature.

$$\mathbf{r}(t)=t^{3} \mathbf{j}+t^{2} \mathbf{k}$$

$$

\kappa(t)=\frac{6 t^{2}}{\left(9 t^{4}+4 t^{2}\right)^{\frac{3}{2}}}

$$

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Campbell University

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

all right. In this problem, we want to use their and 10 which is a the top right corner of the screen to find the curvature of the curve represented by the vector are of tea uh, t cubed times vector J plus T squared times factor k. I'm gonna go ahead and write this in the angle bracket notation so we would have zero. Since we have no vector. I component that's written t Cubed and T squared is our third component function. So Teoh use theorem 10. We need to have the first and second derivatives of the vector are so I'm gonna take the first derivative so we'll have our prime of tea, which is going to be zero as our first component function. Second component function is going to be three t squared and third component function is going to be to t. Then let's go ahead and take the second derivative as well. So we're gonna have our double prime of tea. Just going to be zero 60 and two. Should I get the numerator? We need to take the cross product of those two vectors of the first and second driven. It's so to take our prime crossed with our double prime and we're gonna do is we're gonna write it in this notation. So that way, it's easier to take the determinant. So on our first rule, we've got 03 t squared to t our bottom row. We have 0 60 and two. So to take the determinant of that for the component function that goes with Vector I we're gonna look at the columns for J and K, So we're gonna have three t squared and two t in the top row 60 and two in the bottom row, and that's times vector I. Then we subtract the vector J component. So we're gonna look at our I and K column So will have zero zero and our first column and two t two in the second column. Then we're gonna add the K component functions. So zero three d squared. Excuse me. Component Vector 0 60 squared And feel for you to look back at how we take cross products and, um, right out this determinant structure when we're computing it. So for our i component vector, we're gonna have six t squared minus 12 t squared times. Vector. I they would have minus zero minus zero times Vector J plus zero minus. Here again, Vector K. And if we want to write this in the angle bracket notation we end up getting is negative 60 squared as the first component function and then zero and zero is there second and third component functions Then we want to find the norm of that cross product. So we're gonna be looking at norm of our prime crossed with our double prime, which is gonna be the square root of ah, native 60 squared quantity squared plus zero squared plus zero squared. And we don't have to write that necessarily. But for consistency sake will do that s o that were we refresh on how to take norms and then this is just gonna simplify to being 36 t to the fourth power or 60 squared. So there is our numerator. Then in the denominator, we need to also have the norm of our prime. So the norm of our prime of tea, which we have written right up here, it's gonna be the square root of our first component function squared zero squared plus three t squared, quantity squared, plus two t quantity squared, which is going to give us 90 to the fourth power plus four t to the second power all under the square root. So then, if we want to used their intent to find the curvature, we're gonna have cap of tea is equal. Teoh in our numerator, we're gonna have the norm of our prime crossed with our double prime all over the norm of our prime. Cute. So in our numerator, we're going to have 60 squared and our denominator We're going to have the square root of 92 the fourth plus four t squared, all raised to the third power or if we want to, we can write that as 60 squared over nine t to the fourth power plus four t to the second power raised to the power of three halves. And that is our curvature. Funk, shirker curvature function for the curve represented by the vector are of tea