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Find the length of the curve.

$$\mathbf{r}(t)=\mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k}, \quad 0 \leq t \leqslant 1$$

$$

\frac{1}{27}\left[13^{3 / 2}-8\right]

$$

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Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

and this probably would evaluate the length of the curb, as described by the vector function are Tee, which is equal. Teoh Doctor, I plus he squared Time Inspector J plus T cubed times Vector K, which is evaluated on the interval from 0 to 1. First thing we want to do is take the derivative of our vector functions. So we've got our prime of tea, which is going to be equal. Teoh, they're the coefficient for the vector. I is one so that it reverted. That is going to be zero just for consistency. I'm gonna write that as zero times vector component I plus the derivative of T squared with respect to t is to tee time is vector J and the derivative of the third component function. T tube is three t squared times better K. We want to take the magnitude of that derivative. So look at the square root. Uh, zero squared plus two t's quantity squared plus three t squared, quantity squared and simplifying that. We end up with four t squared ls nine t to the fourth. We're gonna simplify this as much as we can. Um, I'm going. Teoh Ah factor out a t squared under my square root sign. And because we know that T is between zero and one at tea is non negative, so we can actually pull a positive t the positive square root t squared out of our square root function, which is going to make it easier for us to integrate dysfunction. All right, so now let's use our formula for the link. The curve. We're looking on the interval from 0 to 1. Ah, that magnitude of our prime of tea, which is tee times four plus nine t squared duty. And it's gonna be easiest for us to use u substitution here. So I'm gonna rewrite this real quick as tee times four plus nine t squared to the 1/2 power beauty, and I'm gonna do my work for you substitution over here on the right hand side. So I'm gonna let you be this quantity here. So you is going to be equal to four plus nine t square and then do you taking the derivative of both sides is going to be, uh, 18 times t d t. Ultimately, I want something that I can hear is in my original integral. I see that we've got a t d t here. So what I'm gonna do is divide both sides of this, um, equation by 18 song gonna have 1/18 to you is equal to t d t. And then I can plug that right into, uh, the function. So we're gonna have now, and I'm gonna leave my bounds of integration the same because ultimately, I'm gonna substitute, um, four plus 90 squared back into you once I've integrated, But what we end up having is you to the 1/2 power. Uh, excuse me. 1/18 times you to the 1/2 power d you. And if we go ahead and integrate that what we end up with is 1/18 times 2/3 you to the three halfs power. And if we go ahead and substitute, um, our function for you back into this expression on and simple fire coefficients what we end up having is 1/27 times four plus nine t squared to the three halfs power we're gonna evaluate from 0 to 1. So using fundamental theory of calculus got one over 27 on the outside. If I plug one into, uh, one in for tea. I get four plus nine times through her to the three halfs power, and then I'm gonna subtract four in the city plus 90 squared t zero. So 74 plus zero to the three halfs power. And as usual, I'm gonna simplify this a bit. This will be 1/27. Four plus nine is 13. What's gonna be to the three halfs power and, um, four to the three halves power? I can think about four cubed, which is 64 then taking the square root of that, which is eight. This is gonna be 13 to the three halfs power minus eight. And I believe it, like that is our exact answer. So this represents the length of the curve. Um, the length of the curve described by the vector function, or if t on the interval 0 to 1