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Problem 19 Medium Difficulty

Evaluate $\int _ { C } x d s ,$ where $C$ is
a. 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 $( 2,4 )$

Answer

(a) $\frac{5}{2}$
(b) $\frac{1}{12}(17 \sqrt{17}-1)$

Discussion

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Video Transcript

Okay. What we want to dio is walk through how to evaluate, um, the integral over a curve of X d s, um, and where see, is gonna be defined to separate ways. So the first time is gonna be a straight line segment. Um, where x equals t, why equals t over to and we're going from 00 24 two. Okay, so the first thing we need to do is to to find our a t. So r of t is gonna be t I plus t over to k. And now we need to, um, determine, um, the interval of tea. And so, if ty's defined to be X X goes from 0 to 4. So t has be going from 0 to 4. Um, and then what we needed you is now fine, be of tea, which is gonna be the derivative of our to this and be I plus 1/2 J. And so the magnitude of e of t is gonna be the square root of one squared plus 1/2 squared, Uh, and so this is going to give me the square root five over to, um, because we get one plus 1/4 which gives me 5/4. And so this is gonna be the scroll to 5/2. And so now what we need to d'oh! And of course, D s is equal to the square to five over to t t. And so this becomes the integral from 0 to 4 of tea because that's what X is times at D s, which is gonna be this grow to five over, too. T t. So this could be the square root of five over to, um, times, um, 1/2 t squared, evaluated at four. And then at zero. Um, and so we get, um, we get four, Route five when we do that. Okay. And so now, the second time we're gonna do this is, um, this time it see is a parabolic curve defined as X equal to t. And why equal to t squared? And this time, we're going from 00 22 comma four. Okay. And so now our tea is going to be, um t I plus T Square, J and T is going to go from 0 to 2 because it's still based on what X is okay. And so now are integral on. So now we need to take our derivative of our which gives us that V of tea. So that's gonna be I plus two, tea j. And then, um, the magnitude of B of tea is equal to the square root of one squared plus two t squared, which is going to give me the square root of one plus four t squared. And so D s is going to be that square root of one plus 40 squared times t t into the integral from 0 to 2 of, um x and X is t times the square root of one plus four t squared D t. Okay. And so you recognize this, um, going to have to take do a use up and so we're gonna let you, um, be one plus 40 squared. So d'you is equal to eight. T d t. Well, all I have is a t d. T. So 18 d'you is equal to t t t. And so this becomes equal to. And let's go ahead and change our upper and lower limits. A swell. So this could be the integral If, um, t zero use one if t iss two we get, um, we get, um, equal to 17 as you. And so this is gonna be a 18 out here and a you to the 1/2 to you. And so this becomes 1/8 times 2/3 you to the three half's evaluated at 17 and then at one. So we get on this equal to 1 12 times this 17 times the square root of 17 minus one, and there we have it.

University of Central Arkansas
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