parametric eqn: find length of x= cos t, y= t+sint, 0<t&l

[shivers20]

Find the length of the of the curve for the paramteric equation.

x= cos t, y= t+sint, 0 < t < pi

(dx/dy)^2 =sqrt[(-sin t)^2 + (cos t)^2 dt]

sqrt [sin^2t + cos^2t dt]

I end up getting 1 for some reason. Did I miss a step somewhere? How do i go about finishing up this problem. I forgot how to plug in the pi. Thanks for any help.

 

[galactus]

\(\displaystyle \L\\\int_{0}^{\pi}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}dt\)

\(\displaystyle \L\\\int_{0}^{\pi}\sqrt{sin^{2}t+(1+cos(t))^{2}}dt\)

\(\displaystyle \L\\\sqrt{2}\int_{0}^{\pi}\sqrt{1+cos(t)}dt\)

 

[soroban]

Re: parametric eqn: find length of x= cos t, y= t+sint, 0&lt

Hello, shivers20!

Only a few errors . . .

Find the length of the of the curve for the paramteric equation.

. . \(\displaystyle \begin{array}{cc}x & = &\cos t \\ y & = & t\,+\,\sin t\end{array} \;\;0\, <\, t\, <\,\pi\)

Formula: \(\displaystyle \L\:L \;=\;\int^{\;\;\;b}_a\sqrt{\left(\frac{dx}{dt}\right)^2+\,\left(\frac{dy}{dt}\right)^2}\,dt\)

We have: \(\displaystyle \:\begin{array}{cc}\frac{dx}{dt} & = & -\sin t \\ \frac{dy}{dt} & = & 1\,+\,\cos t\end{array}\)

Under the radical, we have: \(\displaystyle \-\sin t)^2\,+\,(1\,+\,\cos t)^2\;=\;\sin^2t\,+\,1\,+\,2\cos t\,+\,\cos^2t\)

. . \(\displaystyle = \;(\sin^2t\,+\,\cos^2t)\,+\,1\,+\,2\cos t \;= \; 1\,+\,1\,+\,2\cos t \;= \;2\,+\,2\cos t \;=\;2(1\.+\,\cos t)\)


From the identity: \(\displaystyle \,\cos^2\left(\frac{\theta}{2}\right) \;=\;\frac{1\,+\,\cos\theta}{2}\;\;\Rightarrow\;\;1\,+\,\cos\theta\:=\:2\cos^2\left(\frac{\theta}{2}\right)\)

. . we have: \(\displaystyle \:2(1\,+\,\cos\ t)\:=\:4\cos^2\left(\frac{t}{2}\right)\)


Our integral is: \(\displaystyle \L\:L\;=\;\int^{\;\;\;\pi}_0\sqrt{4\cos^2\left(\frac{t}{2}\right)}\,dt \;=\;2\int^{\;\;\;\pi}_0\cos\left(\frac{t}{2}\right)\,dt\)

Go for it!