Question

the length of the loop of the curve 6y2=x (x-2)2 is

Answer 1

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: