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# Find the curvature of $\mathbf{r}(t)=\left\langle t^{2}, \ln t, t \ln t\right\rangle$ at thepoint $(1,0,0) .$

## $k(x)=\frac{\left|12 x^{2}-4\right|}{\left[1+\left(4 x^{3}-4 x\right)^{2}\right]^{3 / 2}}$

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in this problem, we want to find the curvature. Uh, a function represented by vector function are at a certain point. So here are of tea is made up of the three component functions T squared natural log of tea and t times natural log of tea, and we want to find the curvature at the 0.100 So to find the curvature, I'm gonna use the earned 10 which is on the top right corner. So we know that we'll need to take the first and second derivatives of are. So I'm gonna get our prime, which is going to be to t one over tea and using the product rule, you'll get natural log of t plus one is that third component function and then taking the second derivative are double prime of tea. We're gonna have to negative one over t squared and one over tea. So then we want to find the cross product between our prime and our double prime, which we can write is the determinant of a three by three matrix. So our first column will have to t to second column. Will have won over tea and negative one over T squared Third column will have natural log of t plus one on one over tea. Serena We can evaluate that by looking at the determinant of the Matrix one over tea, negative one over T squared natural log of T plus one from one over tea that determinant that two by two matrix time. Inspector I. We're going to subtract the determinant of the Matrix with to tea and two in the first column natural log of T plus one and one over tea. In the second column, Inspector J plus the determinant of the two by two nature X with to tea and two in the first column and one over tea. Negative on over T squared in the second column times Dr J. There back to Kate. Excuse me? Serve and let's go ahead and evaluate that. And I'm gonna go ahead right at my angle bracket notation just because I get ah a little bit more concise. So what we'll have is one over T squared minus negative one over T squared, which we just radios, plus D squared times, natural log of t plus one, and then here will have negative to tee times one over T, which is just going to minus two times natural log of t plus one. And when our third component function is gonna be to tee times negative one over T squared, which is going to give us negative to ever t minus two times one over tea or to t, and we can simplify about all out a little bit here. If we pull one out of T's one of her t squared out of each term, we're gonna end up with natural log of T plus two. And then the second component function will simplify, too. Two times natural log of tea when our third component function will simplify too negative for over tea. All right, so there we've got our press product and then we want to take the norm of that cross product so norm of our prime across with our double prime, which is going to give us the square root of first component functions. Squared, which is one over T to the fourth power times natural log of T plus two quality square, plus four times natural log of tea, quantity squared plus 16 whoever t squared and we don't necessarily have to simplify that very far. Soon. Then we also need to have the norm of the first derivative. So that way we can use it in our denominator. So and up with we're taking the norm of this guy right here. Square root. Ah to t quantity squared or fourty squared. Plus one over T squares class natural log of tea plus one squared. So then we want to, uh, used the rim, tend to find the curvature. We're gonna end up with a cap of tea being equal to one ever t to the fourth power times natural log of T plus two All squared, plus four natural log of tea squared plus 16 over t squared. All that's under the square root sign. In the our denominator, we're going to have four t squared, plus one over t squared, plus natural log A T plus one squared on. That entire quantity is gonna be raised to the three halves power rather than writing it as the square root. Cute. But we're not done yet. We want to be able to evaluate this at a certain point. So we were given the 0.1 00 We need to figure out what value of tea that corresponds to. So if I look at, I'm gonna look at my why coordinate here? So I know that the natural log of tea or why component is equal to zero. And if we think about that, the only value that we can plug in to natural I need to get zero is one. So we're looking at T equals one. And then what we can do is we can plug that in, and you are curvature function. I'm gonna do it up here at the top, so I do it at a different color suit, stands out a little bit more. So here. We're looking at Kappa of one for tea. Again, we're using this function down here in our numerator. What will get for that first term? We're gonna have 1/1, So that's not gonna be very interesting. Ah, so one times this value here, natural log of one, is going to be zero. So we really just gonna be left with two squared or four. Is that first term this term? If we plug in t equals one is going to be zero. This term here is gonna be 16 Lenin. Our denominator. This first term was going to be four. The second term is going to be one, and this last term is also going to be one, and that's raced to the three halves power. I won't go through the algebra city. This. It's not too bad to simplify, but what you're going to get end up with is square root of 20 20 over 216 which will simplify to the square root of 5/54. So that is the curvature of are at the 0.1 comma, zero comma zero or T equals one.

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