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Numerade Educator



Problem 5 Easy Difficulty

Find the length of the curve correct to four decimal places, (Use your calculator to approximate the integral.)
$$\mathbf{r}(t)=\left\langle t^{2}, t^{3}, t^{4}\right\rangle, \quad 0 \leq t \leqslant 2$$




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Video Transcript

All right. So for this problem, we want to find the length of the curve represented by the vector are of tea, which could be written in terms of its component functions as t squared, t cubed and T to the fourth power. And we wanna look at the curve from zero to to. So what we're gonna do is we're gonna first take the derivative of that vector. So our prime of tea and then we're just gonna take the derivative with respect to t of each of the component functions it will have to teen three t squared and four t to the third power. And then we want to find the magnitude of that vector. So magnitude of our prime of teeth And that's gonna be the square root. Uh, our first component functions squared plus three t squared, quantity squared, plus 40 cute quantity squared. And we're going to simplify about just a little bit before we put it in our calculator to evaluate length so we'll end up with four t to the fourth power plus nine, Group 14 to the second power. Scuse me. Plus 40 90 to the fourth Power plus 16 t to the six power. And then we're gonna use the equation in the top right corner of the screen to find the length of the curve on the specified interval. So we're going to say that the length is evaluated from 0 to 2 of this function square a fourty squared plus 90 to the fourth plus 16 t to the sixth with respect to t. So to evaluate that we're going Teoh, plug it into our calculator. I've got a t I 84 here. Turn it on, clear out. Everything s so if I go to my y equals menu, what I'm gonna do is I'm gonna go ahead and put that function. Um, that is the argument for our integral into Why one So we're gonna have a square root of a four variable squared, and here it'll show on X. But it could be I can represent he as well plus nine x to the fourth power plus 16 x to the sixth power. Alright, So I've got my function in there and I do second quit to clear out of the screen and then sick be integral. I'm gonna go to my math menu and scroll down until I see fn i n t which is option nine in my calculator. I'm gonna put zero as my lower limit and two is my upper limit. And then, since I already have the function in why one I'm gonna go to my variables menu, Why variables click function and then select? Why one and we're integrating with respect to t and what we wrote out. But because our calculator stores that variable is X, select that here. Then I'm gonna click Enter. And what we get is, um, is this the value here for that? Integral. And we're gonna go ahead and rounded to four decimal places. So what we'll get is that the length of our curve on the interval from T d T equals zero to t equals two. We have a length of 18.6833 units

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