the length of the loop of the curve 6y2=x (x-2)2 is
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Answer:
\L9y2=x(x−3)2
\Ly=x(x−3)29−−−−−−√=x√(x−3)3
Take derivative and square it:
\Lx√(x−3)3dx=x−12x√
\L(x−12x√)2
Now use the arc length formula:
\L\\\int_{0}^{3}\sqrt{1+\left(\frac{x-1}{2\sqrt{x}}\right)^{2}dx
That's just the section of the loop above the x-axis. Because the loop is symmetrical about the x-axis, multiply by two.
\L\2∫30(x+1)2x√dx
HOPE U UNDERSTAND
Step-by-step explanation:
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