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# Reparametrize the curve with respect to arc length measured from the point where $t=0$ in the direction of increasing \$t .$$\mathbf{r}(t)=2 t \mathbf{i}+(1-3 t) \mathbf{j}+(5+4 t) \mathbf{k}$$

## $$\vec{r}\left(t(s)\right)= \frac{2s}{\sqrt{29}} \vec{i}+\left(1- \frac{3s}{\sqrt{29}}\right) \vec{j}+\left(5+ \frac{4s}{\sqrt{29}}\right) \vec{k}$$

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buried in this problem. We want Teoh re parameter rise a curve with respect to its arc length from from T equals zero with t increasing to the curve that we want. Teoh re Pramuk tries is currently represented as our f t to tease our first component functions to tee times Vector I plus one minus three tee times Vector J plus five plus 40 times factor K. And I'm gonna go ahead and write this in vector notation just cause sometimes it's a little bit easier to work with. Um, the alternative vector nok dictation that just lists are component functions. So to t one minus three T and five plus for teeth. So in order to re parameter rise this curve, we first need to find DS every GT, which is represented in the box in the top right corner, that first equation. So what we can do is we can, uh, find the magnitude of the derivative of this vector. Say, first, I'm gonna find the derivative with respect to t. We're gonna end up with two is our first component function negative three as our second and four as our third and then we want to find the magnitude of that. So we're gonna have magnitude of our prime of tea equals square root of two squared plus negative three squared plus four squared. Just gonna be the square of four. Class nine plus 16. Which gives us the square root of 29. Matt is DS over DT. Then we want Teoh find the Ark link. So we're going Teoh, use the second equation in the top right corner and here s or s a T. It's going to be evaluated from in this case, we wanna look from t equals zero with t increasing. So we're gonna look zero to t, as are the bounds of integration. Then we want Teoh function that were integrating is the magnitude of the derivative of our So I would integrate that with respect to you. So we're gonna do is we're gonna have zero to t We found this to be squared of 29. We're just using you, um, to distinguish it from teach that we don't get things confused. So if I integrate square root of 29 I end up with square root of 29 You we're evaluating that from zero to t using her fundamental theorem of calculus. We end up with square root of 29 times T minus squared of 29 times zero, which, of course, is zero for that second value. So we get squared of 29 teeth. Now we want to be able to reap rammer re Pramuk tries the curve. So what we're trying to do is replace um the teas in in our original vector function in our original component functions with with the new expression that involves s the arc length. So what I can do is I can take this resulting relationship between S and T, and I could just solve it for T. So what I end up with is that t equals s over squared of 29 and then I can use that Teoh rewrite my original Ah, Victor, my original curve. So but I'm now gonna have is our of t of s. I'm going to start by writing what I had originally so two times. T one minus three times t five plus four times t And now I'm gonna substitute s in for all of those. Substitute the expression that involves s now. So what we're looking at is to s over a squared of 29 one minus three s over squared of 29 and five plus four s over square root of 29. And if we want to write that using the notation that we were given to represent our, uh our team, we can do so we would have two s over squared of 29 times Vector I plus one minus three s over squared of 29. Time is vector J plus five plus four s over squared of 29 Who was Vector K? So now we've just rewritten our curves In terms of the arc length s

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