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Problem 6 Easy Difficulty

Find the length of the curve correct to four decimal places, (Use your calculator to approximate the integral.)
$$\mathbf{r}(t)=\left\langle t, e^{-t}, t e^{-t}\right\rangle, \quad 1 \leq t \leqslant 3$$

Answer

$$
L=2.0454
$$

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Video Transcript

alright. In this problem, we want to evaluate the length of the curve represented by the vector R t with the three component functions t e to the negative t power and T times e to the negative team. And we want to evaluate that on the interval one, just three. So first thing we're gonna do is take the derivative of that vector. So our prime of t taking the derivative with respect to t of each of our component functions, we're gonna have one negative e to the negative t power. And then we have to make sure that we use our product rule for our 3rd 3rd component function there. So what we'll get is e to the negative t plus, uh, actually minus t t e t times teach that tee times e to the negative teeth. Our And if we want, we can, uh, even simplify that last term a little bit. Um, by factoring an e to the negative are negative e to the negative TL, which will leave us with t minus one. Um, it doesn't really matter, but that just might be another way of writing. It s then we're gonna take the magnitude of that derivative, which is gonna be the square root of one squared plus negative ive the negative t squared. Plus, we're gonna square each of those components since their multiplied by one another so negative e to the negative t squared and t minus one squared. And we can, if we would, like, just simplify things out a little bit. So that way, when we enter it into our calculator to estimate it'll be or to evaluate it will be a little bit easier. But you've got one plus e to the negative to t plus e the negative to tee times T minus one squared and then went to evaluate the length of the curve represented by that vector that original vector. So we've got l is the integral from 1 to 3. Just our interval for teeth are, uh, the magnitude of our have the derivative of the vector. You hear that expression we just found above? We want to integrate that with respect to t. So I'm going to evaluate that using my graphing calculator. What I'm gonna dio is I'm going to go to my y equals menu and put in the argument for the interval, um, into our functions, So do the square root of one plus e to the negative to t, which is gonna show up as an ex. And our calculator plus beer to the negative to t times t minus one squared. It's now that we've got that in there, we're gonna do second mode to exit out of that menu and to integrate. We're gonna go to our math menu. Ah, scroll! Teoh Option nine we're gonna put in are lower and upper bounds. So one and three my function is in. Why one? So I'm gonna go to my variables menu, go to buy variables and select function. Why one we're integrating with respect to the variable X and our calculator, which is representing our variable t We're gonna click enter and we get a nice approximation of our interval and we're gonna go ahead and round that to four decimal places. So we have that the length of our curve is equal to 2.4 54

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