đź¤” Find out what you don't know with free Quizzes đź¤”Start Quiz Now!

# Evaluate $\int_{C} x d s,$ where $C$ isa. the straight-line segment $x=t, y=t / 2,$ from (0,0) to (4,2).b. the parabolic curve $x=t, y=t^{2},$ from (0,0) to (2,4).

## (a) $4 \sqrt{5}$(b) $\frac{1}{12}\left(17^{3 / 2}-1\right)$

Integrals

Vectors

Vector Functions

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

Okay, folks. So in this video, we're gonna take a look at this problem from number 19 which, which is a problem about lined into girls over plane curves. So we're given this, uh, integral right here that we need to evaluate along the curve c where C is. Um, Well, there's really two parts, so there's part and there's part B, um in each park corresponds to a different a different curve. See? So that's that's first solved for part eight first, um, and I have drawn out the curve for party because the curve for parties really simple is really just a straight line segment. Um, from 00 to 42 this is this is the X axis and desist the Y axis. I'm reevaluating the lining Negro along. You know this straight lying segment right here. Okay, So and and along the curve X and wire both parameter rised by another variable t um, and were given a function. But we're given to functions one for excellent, one for why? And they're both functions of tea. So X is just tea. And why, as a function of T is T Hafs. Okay, So so that we wouldn't do. This is just the usual way of doing lining girls. We're first going to rewrite this in fantasma line segment DS as yes equals ex prime squared plus y prime squared D t. Okay, And where? Ex Prime White Primer, of course. Just the derivatives of X and y with respect to t. Okay, so So that's the value of X prime. And why prime? Well, ex prime excess t So ex prime is this one. So one plus why Prime squared What wise 1/2 times. T y promise 1/2 1 half square. It is 1/4. Um, all right, so that's Ah, this thing right here is DS, and we see we have a constant here, which we can pull out of the integral. So we end up with square root of 5/4. Uh, x DT, but excess t so I mean, every right X s t multiplied by D t. Let's figure out what the what? The limits of integration is well, when x zero wise, you know this This is this right? Here is the point that we start with when that point corresponds to t being, of course, zero because extra liquidity and x zero. That means t zero. So we start off from zero. What about, um, the, uh the destination s so to speak? Well, the destination is for X equals four. So four equals X equals T. That means that to use for Okay, so So we have figured out the, uh, the limits of integration here. All we need to dio is evaluate Had the rest of the integral. So we have roots. 5/2, 1/2 A T squared, zero to fourth. Okay, so we have route 5/4 times 16 minus zero, which I'm going to ignore. Um, so we have four root five as Thean surfer part A. This right here is Thea answer for party. Um All right, so that's the answer for party. Let's figure out what part B is. Well, part B is a parabolic curve. Where X um So we're going from 00 2 to 4. Where x as a function of t is t. And why, as a function of t is t squared. Okay, well, simples, we're gonna do it the same way as we get for part a where we, uh we verily the integral as, um x d s. But where we also have to Ah, rewrite the DS as as a function of t. So we have X as a function of tea is just tea multiplied by X promise 111 square to still one. So I have one plus why, as a function of teased, he squared Well, why Prime is to t Okay, so we have four t squared more supplied by DT. Okay, now, now we're gonna figure out the limits of integration. Well, when X Let's look at where we start with that we start from this point 00 while ecstasy with a T. But x zero, that means t zero. So we start from zero. Um, let's look at the end 00.24 Well, um, the at the end point ecstasy with a two x s t. That means tea is too. So we have successfully figured out our limits of integration again. Um, but now let me write this integral like this. We have t multiplied by one plus four t squared a t. Now, this is gonna be a little bit trickier than part A Because we have a square root, which we, uh, we're gonna have to get rid of the way we want to get rid of it is by doing a substitution where I'm gonna define, I'm going to define a new variable. You I would define it as 40 square two plus one, which is exactly was under the's square root sign. And do you? As you can see, it's really just eight. I can't write eight t d t. Okay. Um, yes, I know I couldn't take this integral, and I can rewrite it in a different way. So I have, um, route you t d t is really just his religious. Do you over eight? Um, well, and then let's figure out the limits of integration. Well, when t zero, that means you is one. I'm just left sitting back in here when t is 22 squared is 44 times four is 16 16 plus 1 to 17. So now I have successfully converted the team to grow into a you integral, which is much easier to do. Um, so I'm gonna pull out the constant, and then I have you to the power of 1/2 d'you, which could be rewritten as you three. Half over three. Half from 1 to 17. No, um, I have one day, Times two or three. Um, this is ah, 17 to the power of three. Half minus one to the power of 3/2 which is one. Um, no, I could simplify this a little bit more. I have 1/12 um, 17 to the power of three. Half minus one. Okay, you can plug this into a calculator and and get a more, uh, concise are, I guess, expression for her for this, for whatever this is. Um, but yeah, this is the answer for part B. It's one of her. 12 times, 17 to the power of 3/2 minus one. All right, we have partying. We have Barbie, and I think that's it for this video. Thank you. Bye.

University of California, Berkeley

Integrals

Vectors

Vector Functions

Lectures

Join Bootcamp