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# Mass of a wire Find the mass of a wire that lies along the curve $\mathbf { r } ( t ) = \left( t ^ { 2 } - 1 \right) \mathbf { j } + 2 t \mathbf { k } , 0 \leq t \leq 1 ,$ if the density is $\delta = ( 3 / 2 ) t$

## $2 \sqrt{2}-1$

Integrals

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

Okay. What we wanna step two is we want to find the mass of a wire that lies along. Um, a curve represented by r T is equal to t squared minus one J plus two t k. Um, where t is going to go from zero toe, one inclusive. And if my density, um, is equal to three over to tea. Okay, um and so what we know is that our, um, or mass, um is given by the integral over the curve of our, um, density. Um, and we're integrating with respect to, um s and so s d s is actually gonna be dependent upon our of tea. And so the first thing we need to do is we've got to find what d s is. An D s is equal to the magnitude of this v of tea time C t. And so, um, V f t is the derivative of Artie. So that's the first thing is to take the derivative, um and so that's gonna give me to TJ plus two. Okay. And so the magnitude of e of t is gonna be the square root. Ah, to t squared. Plus two squared. So that is going to give me a MME two times the square root of T squared plus one. And so D s is equal to two times the square root of t squared plus one times t t. Okay. And so now what we have is, um our mass is equal to the integral from 0 to 1 of our density, which is three have tea times two times the square bit of t squared, plus one integrating with respect to t. And so we're in Take out the three. So the mass is going to be equal Thio three times the integral from 0 to 1 of t times the square root of T squared plus one TT. And so you noticed We're gonna have to do use up on this. So let's go ahead and switch over. So the mass is equal to three times the integral from served one of t times the square root of T squared plus one and we're integrating with respected T. So I'm gonna go ahead and let you wth e t squared plus one. Therefore, d'you is equal to two tea t t and all I have inside is a T d t so I'm gonna have to divide by two. And so I can replace T d t with the 1/2 to you now. And so now my integral becomes three halves, Um, times in a cool of you to the 1/2. Do you? And I'm gonna go ahead and change my upper and lower bound. So if tea is zero, you will be one. If t is one, you will be a two. And so now when I integrate, I am integrating. Um, three have times 2/3 you to the three halves, evaluating it to end at one. And so this gives me to raise to the three halves minus one, which we Cal to write his two times a screw to two minus one.

University of Central Arkansas

#### Topics

Integrals

Vectors

Vector Functions

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp