have problem to find length of loop 9y^2=x(x-3)^2
- Thread starter Smily
- Start date
Are you studying the arc length formula?.
\(\displaystyle \L\\9y^{2}=x(x-3)^{2}\)
\(\displaystyle \L\\y=\sqrt{\frac{x(x-3)^{2}}{9}}=\frac{\sqrt{x}(x-3)}{3}\)
Take derivative and square it:
\(\displaystyle \L\\\frac{\sqrt{x}(x-3)}{3}dx=\frac{x-1}{2\sqrt{x}}\)
\(\displaystyle \L\\\left(\frac{x-1}{2\sqrt{x}}\right)^{2}\)
Now use the arc length formula:
\(\displaystyle \L\\\int_{0}^{3}\sqrt{1+\left(\frac{x-1}{2\sqrt{x}}\right)^{2}dx\)
That's just the section of the loop above the x-axis. Because the loop is symmetrical about the x-axis, multiply by two.
\(\displaystyle \L\\\2\int_{0}^{3}\frac{(x+1)}{2\sqrt{x}}dx\)
\(\displaystyle \L\\9y^{2}=x(x-3)^{2}\)
\(\displaystyle \L\\y=\sqrt{\frac{x(x-3)^{2}}{9}}=\frac{\sqrt{x}(x-3)}{3}\)
Take derivative and square it:
\(\displaystyle \L\\\frac{\sqrt{x}(x-3)}{3}dx=\frac{x-1}{2\sqrt{x}}\)
\(\displaystyle \L\\\left(\frac{x-1}{2\sqrt{x}}\right)^{2}\)
Now use the arc length formula:
\(\displaystyle \L\\\int_{0}^{3}\sqrt{1+\left(\frac{x-1}{2\sqrt{x}}\right)^{2}dx\)
That's just the section of the loop above the x-axis. Because the loop is symmetrical about the x-axis, multiply by two.
\(\displaystyle \L\\\2\int_{0}^{3}\frac{(x+1)}{2\sqrt{x}}dx\)
Hello, smily!
galactus did a marvelous job explaining the procedure.
We have: \(\displaystyle \L\int^{\;\;\;3}_0\frac{x + 1}{x^{\frac{1}{2}}}\,dx \;= \;\int^{\;\;\;3}_0\left(x^{\frac{1}{2}} \,+\,x^{-\frac{1}{2}}\right)\,dx\)
galactus did a marvelous job explaining the procedure.
Don't let that integral scare you . . .
We have: \(\displaystyle \L\int^{\;\;\;3}_0\frac{x + 1}{x^{\frac{1}{2}}}\,dx \;= \;\int^{\;\;\;3}_0\left(x^{\frac{1}{2}} \,+\,x^{-\frac{1}{2}}\right)\,dx\)