🤔 Find out what you don't know with free Quizzes 🤔Start Quiz Now! # Find the mass of a thin wire lying along the curve $\mathbf{r}(t)=\sqrt{2} t \mathbf{i}+\sqrt{2} t \mathbf{j}+\left(4-t^{2}\right) \mathbf{k}$$0 \leq t \leq 1,$ if the density is (a) $\delta=3 t$ and (b) $\delta=1$.

## (a) $4 \sqrt{2}-2$(b) $\sqrt{2}+\ln (1+\sqrt{2})$

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### Video Transcript

Okay, folks. So in this video, we're going to take a look at problem number 35 where were asked to find a mass of a thin wire lying along the curve. Um, and given by the function are of tea, okay. And so there's really two parts of this problem, so we're gonna look at party first. So for part A, we have a density function. We have Delta. My hand writing is living off Delta of tea is three t. So we have this delta of tea, and the way you find the maras of the wire is well, the way you find the mass of anything is just density multiplied by, ah, volume element for three D object were like allying element evident has malign segment for a wire. So for this problem, we're gonna do Delta DS the D s stands for ah, really small, you know, infinite hospital life segment. And so that's the mass. We're gonna be integrating this over. I mean, along the wire. All right, so So let's plug in in delta as a function of t. So we have three t for part A and D As while the S could be rewritten, Um, in terms of x, y and Z, But X, Y and Z, here are all functions of tea. So let's go ahead and rewrite ds as, um ex dot squared plus why dot squared plus z dot squared multiplied by D. T. And the integration limits is between zero and one for t. All right, so this now, I should let you know that that I'm using a notation here and that you might not be very familiar with the dot On top of the letters means time derivative or derivative with respect to T. Okay, so this is this is really just dx DT squared plus B y did he squared plus the z DT squared and were given along the curve while the curve, the the usefulness of the curve are of tea lies in the fact that that this is basically just three functions for for X and Y and Z respectively are of t which is a vector. Basically, gives gives us the re functions. Okay, so we have our of t. For example, the X component of our of t is gonna be a function, so we have route to t. That's the X component and the y component of his route to tea as well. And the Z component is this four minus t squared. Um, and when you take the time derivative of three of these functions, all three of them and you plug it in, you're going to get 01 three t two plus two plus minus two t squared DT. Okay, I skipped through the part where I'm supposed to take the derivative because I think that's too trivial and you can handle that yourself. All right, so we have 30 won t four multiplied by a squared plus one, the t. And this is trivial. Buy. It requires, you know, a little bit of a special technique Not so special, because I'm not because I'm sure you guys have all heard of it. We're gonna do a u substitution. Where I defining you, variable you and I'm going to define it as t squared plus one. So now I'm gonna have a d. You being equal to two t DT and we have a t d. T here. So this could be rewritten as, uh, Let's see. Six zero won t ah t squared. Plus one DT, which is just six integral u to the power of a have do you over to between one and two. And when you crank out the numbers, you get two times to the power of 3/2 minus one. Okay, so that's for part A. And let's go ahead and start doing part B. The same strategy applies, but But the only difference here is that we have a different density function, which is now a constant. And this one So we have Delta equals one for part B. So mass of the wire is just dealt a DS. But Delta is one. So it's out of the game. So we have integral Yes. And we're going to use some of the previous results that we have obtained. Um, DS can be written as two times t squared plus one DT. And now I am going to the way I'm gonna evaluate this Inter crow is I'm going to define T as a function of fada. They didn't the new variable. So now I have indeed t equal sequence word of data de theater. And now I'm gonna right, I'm gonna plug it in here. So now I have two times seeking squared of data multiplied by Seacon Square, the data d theta. And this is really just integral of seek and to the power three de Seita and the integration limits. When you plug in the numbers and see for yourself, you get zero and pipe, Okay? And this is where things get tricky. I suggest you, ah, either Google or look it up. If you don't know what intro of seeking Cube is because I don't know what it is, I had to look it up. Um, but basically, this is, um this is 1/2 seeking data Tangent data plus 1/2 natural log seeking data plus tangent data between zero in pi over four. This is really this is you know, all of all of the technicalities here, um, has been ah, treated. And all that's left is just painful ruling algebra. So I'm gonna skip through a few steps and I have co sign pi over four. This is just plug and chug plus natural August one over. Coz I'm Piper four plus one minus zero. And if you crank out theology where you get route to plus natural log Route two plus one. And that's it for Let's see. Yeah, that's it for Part B. And so we're done for both parts of the problem. We have problem. 35. That's it for this video. Thank you for watching. See the next video. All right. University of California, Berkeley

#### Topics

Integrals

Vectors

Vector Functions

##### Top Calculus 3 Educators ##### Catherine R.

Missouri State University ##### Heather Z.

Oregon State University  Lectures

Join Bootcamp