Find the line integral of $f(x, y)=\sqrt{y} / x$ along the curve $\mathbf{r}(t)=t^{3} \mathbf{i}+t^{4} \mathbf{j}, 1 / 2 \leq t \leq 1$.

You must be signed in to discuss.

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

Okay, folks. So in this video, we're gonna have a interval alignment girl to evaluate. So that's ready. Down here we have a line integral along the curve C Well, not curve. See you. But time by the girls are which is privatised by Ah, by the variable t Here. Um, we're integrating the function f which is a function of vaccine. Why along the curve. Ok, so now let's copy down. That's right down the function F but not as a function effects and why? But as a function of t because both x and wire functions of tea that's ready down, we have square root of why which is really the square root of tea. Fourth, divided by the divided by X but because too cute OK, do yes is really just ex prime skirt. Why? Prime squared multiplied by D d. Ex prime is three t squirt squared plus y partners guard. But why? Prime is four t to the power of three for T to the power three squared multiplied by DT. Okay, um and then the limits of integration is between 1/2 and one. Okay, so let's crank this out. We have, uh, one over t here for this term because we have t to the power to over a t to the power three. So that's one of routine multiplied by the but this thing right here, 92 the power fourth pause. Ah, 16 t to the power of six BT. Okay, um, now I'm gonna have t to the power of negative one, because that's what whenever it he is multiplied by t to the power of to nine plus 16 t to the power of to Okay, but now I'm gonna multiply these two terms together. So now I have to get our one multiplied by 16 t square plus night. The key. Um, so now is now is the time. Now is the time for us to ah, do a little bit of a U substitution here. I'm going to define the variable. You as, uh, this term here inside of the square Root 16 teas group last night. Okay, so the U is 30. 30. 32 t DT. Okay, um, I'm gonna substitute this back in here. I have square root of you, which is just U to the power of one have multiplied by it multiplied by T D t. Which I'm going to write as d you over 32 as you can. Clear clearly. See her here. Um, now I have 1/32 of U to the power of three. Half over 3/2 um, evaluated at the at the end. Points. I'm gonna figure out what the endpoints are. Well, when t is you, the one have t. It went easier. The 1/2 you is equal to 16 to the power 16 times. 1/4 plus night, which is 13. And when t is equal to ah one use equal to 25. Okay, you just have to substitute that back in here. Okay, so now I'm going to Ah, plug in EU limits of integration. Here we have from 13 2 25 Okay, so I have ah, two or three times 1/32. That's for the constant term. Multiplied by 25 to the power of 3/2, minus 13 to the power of 3/2. Well, I know I know about this term, but for this term, I know that the square root of 25 5 and five to the power of three is really just 1 25 Okay, I hope I did that math correctly. Um, for the constant term I have This is a two. This is 32 so I have 16. So one of her 48 1 25 minus 13 to the power of three have. Okay, this is the answer for for problem 24 here. And we're done for this video. Thank you for watching by.

University of California, Berkeley