Vector Calculus

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So here we're going to do an application problem thinking more specifically about line integral XYZ and the amount of work being done. Uh, so where we're going to start is saying that, um, we can start by remembering that the gravitational force is gonna be conservative, So thief force by the gravity on the man in the can is equal to the negative force by the man against the gravity. So to find the work done, we want to look at the magnitude of gravitational force acting on the man on the can of paint. So we have 160 plus m pounds, then the mass of the man is constant, and that's the 160. But the mass of the can is constantly decreasing because it's letting out all that paint. So the mass of the can is ÂŁ9 less by the time that the top is reached. So based on that, we can say that em is equal to its initial ÂŁ25. Minus nine are because it's losing those ÂŁ9. Um, over S s is the height of the silo, and R is the distance covered, um, during some time, so based on that. We can look at the fact that f dot d r. Is now going to equal this portion right here. 1 85 minus nine are over s. And the 1 85 comes from the 1 60 plus the 25. And then that's gonna be D R. So the work done as we see is just gonna be this integral right here. So we take the line integral. We're gonna take the center, go here and it's from zero to s from the beginning of the time until we get to the top. So, since we know the height to be 90 ft s can become 90. So what we do when we evaluate this integral is 1 85 R minus nine R squared over two s evaluated at zero and s, or, in this case, zero and 90. So we evaluated at 90 and what we end up getting is that this is equal to 1 80 0.5 s, which is equal toe 1 80.5 times 90. And that's going to give us our final answer of 16,245 ft pounds. That's gonna be our final answer for the work done

California Baptist University

Vector Calculus

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