The Arc Length of a Vector Function
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MORE MULTIVARIABLE VIDEOSSS !!!!
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Please some help here. I have 30 exercices this hard:
Arc lenght of:
(3cos(2t), 3sin(2t), 3t) [0.1]
Can you please help me?
The best I can do is get to a monstrous radical which i have to integrate trough wolframalpha.
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(3cos (2t), 3sin (2t), 3t) [0.1]
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dfdfs
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Great explanation as always!
I'm glad you finally decided to switch to a whiteboard - It killed me to see you throw away 3 unused printer papers because of marker-bleed stains.
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^^That is a true statement, a true statement indeed.
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thanks for saving my ass.....again!
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Taking an integral of that involved integration by parts.
I came up with 12 as the solution. Unless I made a mistake somewhere :-(.
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For a non-parametrioc function on the form of y = f(x), the arc length is calculated by taking an integral from a to b over sqrt(1+(f' (x)^2)) (That's f'(x) = f prime sub x = drivative of f(x) in there).
In your case, where it's x = f(y), the arc length is calculated by taking an integral from a to b over sqrt(1+(f' (y)^2)).
I tried this problem and got f'(y) (f prime (y)) = (y-1) / (2*sqrt(y)).
sqrt(1+(f' (y)^2)) = (y+1) / (2*sqrt(y)).
(cont.)
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thanks to your videos i got 105/100 in my cal 2 and 98/100 in cal 3 :D
the best youtube channel
choconiel 1 year ago 2
@choconiel that is very good : ) keep up the great work!
patrickJMT 1 year ago 2