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Numerade Educator



Problem 27 Easy Difficulty

(a) Evaluate the line integral $\int_{c} \mathbf{F} \cdot d \mathbf{r},$ where
$\mathbf{F}(x, y)=e^{x-1} \mathbf{i}+x y \mathbf{j}$ and $C$ is given by
$\mathbf{r}(t)=t^{2} \mathbf{i}+t^{3} \mathbf{j}, 0 \leqslant t \leqslant 1$
(b) Illustrate part (a) by using a graphing calculator or computer to graph $C$ and the vectors from the vector field corresponding to $t=0,1 / \sqrt{2},$ and 1 (as in Figure 13$)$ .


a) $\int_{C} \mathbf{F} \cdot d \mathbf{r}=\frac{11}{8}-\frac{1}{e}$
b) see graph for answer


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Video Transcript

Yeah. Okay. So let's have a look at problem. 29. So what is this problem talking about while in this problem we want to evaluate the integral F D R. Okay, where All right. So uh well, f dot er where we have f is this? We have see the curve is given by this. And so let's let's evaluate it. Okay, so we know this is equal to 10 f of x of T Y of t dot our prime t DT. So this is gonna be 10 of E to the t squared minus one, I plus t squared t cubed jay. And then we need to multiply it by the derivative of this vector function. So this is gonna be to T I Plus three T Squared J. And then we have our DT. So yeah, now we have to T E to the two t square E to the t squared minus one plus three T. to the seven DT. Yeah. Okay, so this is gonna be 1 to 0 to T E to the t squared minus one DT plus 10 of three t. to the seven d. T. So we can so we're going to evaluate each of these separately. The second integral is very easy. The first one is also fairly simple. We just need to use the substitution. So we're going to use the substitution U equals two T squared minus one. So then D U D T is gonna be to t. So then do you is equal to two T D T. So then we have this integral here is given by one squared minus one. So that's just gonna be zero on this terminal and then the other 10 squared minus one. So this is gonna be minus one. So this is gonna be E. To the U. And then we have two D. Two T. D. T. It's just do you and then we have the second integral Which is going to be three T. to the eight on 8. And then this is going to be 1-0. Okay so then this is gonna be E. To the U. Zero negative one plus. So this is just gonna be three on eight. So then this is going to be one minus E. To the negative one plus three on eight. So this is just going to be 11 on eight minus one on E. Mm. Okay and that is part A. So this question also asked us to uh have a look at the vector field. So let's have a look at that. So here's a picture of the curve C. And here's a picture of the vector field. Notice that the vectors here when you compare them with the tangent vectors to the curve, the dog product between the vectors in the vector field and the tangent vectors to the curve will always be positive. And this is reassuring considering that our final answer is positive. Okay so if it wasn't positive we we would know something's very wrong. So now they asked us to draw the vectors. So the vectors corresponding to the vector field at 01 on T zero T. Is one on square two and one. Okay, so A. T equals zero T equals half and T equals one. What are the corresponding X values? Well X values are given by T squared. So that is X zero X is half and X is one. So the vectors that they want are simply given by the vectors here. So that would be this vector and then at half it would be given by this vector roughly. And then at one it would be given by this vector. So once again I just want to stress that this vector field All the vectors have been scaled by .25. That means the actual vectors in the in the vector field are four times longer than they appear here. But of course the direction is still the same. The reason I did this is so that we can actually see the vector field better. Whereas if you draw them at full length the arrows are so long, they all cross each other and it's very jarring to look at. Okay, and that's it. Thank you