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Numerade Educator

# Reparametrize the curve with respect to arc length measured from the point where $t=0$ in the direction of increasing \$t .$$\mathbf{r}(t)=e^{2 t} \cos 2 t \mathbf{i}+2 \mathbf{j}+e^{2 t} \sin 2 t \mathbf{k}$$

## $$\mathbf{r}(t(s))=\left(\frac{s}{\sqrt{2}}+1\right) \cos \left[\ln \left(\frac{s}{\sqrt{2}}+1\right)\right] \mathbf{i}+2 \mathbf{j}+\left(\frac{s}{\sqrt{2}}+1\right) \sin \left[\ln \left(\frac{s}{\sqrt{2}}+1\right)\right] \mathbf{k}$$

Vectors

Vector Functions

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

in this problem who and re Pramuk tries occur as represented by the vector our tea with respect to its arc length. So go ahead and write out our function are if t here e to the power of to tee times co sign of duty. The second component function is just to and the third is e to the power of duty Sign, uh, two teeth. So we know that we need Teoh. We want to use the expression the equations on the right hand side and the top right corner to be able Teoh find s in the first place. So that way we can re premature eyes The curve in terms of it, remember s is the arc length. So first we want to be able to take the derivative of the vector are so we're gonna look at our prime of tea and we have to use our product rule to find the derivative of the first and third component functions so derivative of our first component function is going to be two times either the power of to t Times co sign of two t my s two times e to the power of T T time sign of two t derivative of the second component function is just going to be zero and derivative of the third component function is going to be two times e to the power of to t sign of two T plus two times e to the power of to t Times CO sign of two team. And if we want, we can make this just a little bit nicer to look at. I would like to factor out uh, the common term two times either the power of to t from my first and third component functions. So we're left with that term Times co sign of two T minus sign of two t second component function is still zero and factoring two times even to the power of to t we're left with sign of to t plus co sign off duty. I'm gonna go ahead and switch the order of those since its addition and the order doesn't matter just so that way. It's consistent with our with the order of sign and co sign in the first component function that will become a little bit easier later on toe simplify things. All right, so now that we have the derivative. We need to find the magnitude of that resulting vector, which is gonna be the square root of each of our component functions squared and added together. So for first component function squared, we have four times either the power of fourty times co sign of two T minus sign of two t not quantity squared. Plus C R squared is just zero and then plus third component functions squared so four times e to the power of 40 times co sign of two t plus sign of two t all squared, and we can simplify that further. So I noticed the common term four times each of the power of fourty. I'm gonna factor those out of both of them. And in fact, I'm gonna pull them right out of the square root sign. So we have two times e to the power of to t and then under the square root sign. We have co sign of two T minus sign of two t quantity square plus CO sign of two T plus sign of two t all squared. So, Teoh, simplify this down a little bit. What I'm gonna do is represent this as a simpler expression just for the sake of being able to simplify things faster. So I'm gonna let co sign of two TB a and sign of two t equal. Be so off to the side. I'm gonna write with that expression under the square root sign would be if I express it in terms of A and B. So what we would have is a minus B squared, plus a plus B squared. If I expand, boost both of those terms. I end up with a squared minus two, a B plus B squared for the first term and a squared plus two a B plus B squared for the second term. If I add my like terms together my negative to a B and my plot positive to a B and to be zero and then I'm left with two times a a squared plus two times B squared or two times a a squared plus B squared. Now we can go back to our original expression and right our value under the square root sign, as in terms of the simplified expression that we have in bread over on the right, so we would have is two times e to the power of to t times the square root of two and substituting co sign of two t and sign of two t back in for A and B we would have co sign of to t squared or co sine squared of two t plus I sine squared of two t And that becomes really nice because by that the factory and identity one of our trig identities. We know that this expression is equal to one. So what we're left with then is two times the square root of two times E to the power of two teeth, and that is the magnitude of the derivative of our. So what we can do then is we can use the expression s or s of t equal to the inner rule in the top right corner in that form. So we know that we're integrating based on the problem statement from zero in the increasing directions there from zero to t. And then I'm going to write my, uh, expression that I just found the magnitude in terms of you just a dummy variable to take the place. So that way we don't get it confused with the tea, Uh, in our bounds of integration, I'm gonna integrate that with respect to you. It's what I end up with is square it of to times e to the power of to you going from zero to t If I use the fundamental theorem of calculus to evaluate that I end up with squared of to times eat of the two teeth minus the square root of two times E to the 20 r or just zero We know that is equal toe one. So we end up with squared of to times E to the power of two key minus squared too. So we want to now solved for key because we want to be able to write the original parameter ization of the curve in terms of s now. So we have to substitute an expression that's equal to T. So what? Weaken Dio? We can look at the fact that right now we're at s is equal. Teoh swear it of two and on a factor that square to two out through 22 times e to the power of to t minus one. So I want to solve that four t So I'm gonna divide both sides by squared of to and that I'm gonna add one to both sides and then if we take the natural long of both sides of this equation we have on the right hand side is the natural log which is a log a rhythm based e of e to the power of to t. So we end up with not your log of e to the power of two teeth which we know simplifies to to t We're almost there now we just have to divide both sides by two and what we get is that T is equal to 1/2 times natural log of s over squared of two plus one. So now that we have an expression for tea, what we want to do is plug that back into our original privatization of of the Curve. So scroll down a bit. So the original way that we wrote our of tea was eat of the power of to t co sign of two t to e the power of to t sign duty. So now if we write this in terms of t of s right, we're thinking of the parameter t Now, as a function of the arc length s, we're gonna go ahead and we're gonna substitute that value in for tea now. So for first component function, we're going to simplify this quite a bit. We're gonna write it, um, as it would originally come. We have two times. T the 1/2 is gonna cancel out. So our numerator is just gonna be natch. Are exponents gonna be natural? Log s over squared of two plus one. That's gonna be times the co sign of two t again to and the 1/2 are going to cancel out. So we end up with co sign of natural log of a super squared of two plus one That is our first component function. Second component function is just to, and that one does not change. And then, lastly, we have e Teoh the power of to t again R two and R 1/2 are gonna cancel out. So we're left with natural Log s over squared of to plus one time sign of to T or two times 1/2 which cancels out natural log s squared over a squared of two plus one. You got a lot going on there, but fortunately, some things were gonna simplify out for us. We have an e raised to the power of natural law, and we know that these two can essentially cancel out one another. So what we get is s over squared of two plus one times the co sign of the natural log of s over squared of two plus one. Choose our second component function on our last component function Again E and natural log cancel each other out. So we have s over squared of two plus one all times. Sign of natural log of s over square of two plus one. And that is the re prayer characterization of our curves Now, in terms of the Ark link.

Campbell University

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Vectors

Vector Functions

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp