C∫(5x+3y+3z)ds, where C is the line between A (1,2,3) and B (-3,-2,2)
#Hey, I need some help with some calculus questions.
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C∫(5x+3y+3z)ds, where C is the line between A (1,2,3) and B (-3,-2,2)
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scenic sigil
C∫(5x+3y+3z)ds, where C is the line between A (1,2,3) and B (-3,-2,2)
I'll use dl instead of ds, as that's too easily confused with dS.
(5x + 3y + 3z)dl
Let's parametrize the line between A and B with 0 ≤ t ≤ 1.
x = 1 - 4t
y = 2 - 4t
z = 3 - t
So:
5x + 3y + 3z = 5(1 - 4t) + 3(2 - 4t) + 3(3 - t) = 20 - 35t
Then, let's deal with dl. We need the derivatives first.
dx/dt = -4
dy/dt = -4
dz/dt = -1
So:
dl = √(dx^2 + dy^2 + dz^2) = √((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)dt = √(4^2 + 4^2 + 1^2)dt = √(33)dt
Thus:
(5x + 3y + 3z)dl = √(33)(20 - 35t)dt
So, now you just need to integrate it from 0 to 1.