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compute the length of the curve from 0<= x <=3 of the function y = 1/3(x^2 +2)^3/2
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just grab a measuring tape :DD
arc length = \(\large \int_0^3 \sqrt{1 + (\frac{dy}{dx})^2} dx \)
find the derivative and plugin
\(\large y = \frac{1}{3}(x^2 +2)^{\frac{3}{2}}\) \(\large \frac{dy}{dx} = ?\)
ok... here goes! sorry, I skipped it and got caught up in another.
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yah
\[\frac{ dy }{ dx }= (\frac{ 1 }{ 3 })(\frac{ 3 }{ 2 })(x^2+2)^{1/2}(2x) = \frac{ 1 }{ 2}(2x)(x^2+2)^{1/2}\]
well also the 1/2 cancels with the 2x
Looks good bro. Plug it all in. Jam through it.
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