evaluating length of curve: x = t + 1/t, y = ln(t^2)

[Erin0702]

The problem states to find the length of the curve from x=t+1/t y=ln(t^2) and t is between 4 and 1. I know to use the arc length formula which is:

sqrt((dx/dt)^2 + (dy/dt)^2)dt and then evaluate it over the integral. I got that the derivatives would be:

dx/dt=1+lnt
dy/dt =1/(t^2)*2t=2/t

so by plugging it in i get:
(sqrt( (1+ln(t))^2 + (2/t)^2))dt

from there i got
(sqrt (1+2lnt+(lnt)^2+(4/(t^2)))) dt and i got that from squaring each part.

From here I do not know where to go to integrate it. I can't use u substitution and I can't think of any way to simplify it. Please help to integrate it, I can do it after I get that far.

Thanks!

 

[pka]

You missed the first derivative:
\(\displaystyle \L
x = t + \frac{1}{t}\quad \Rightarrow \quad \frac{{dx}}{{dt}} = 1 - \frac{1}{{t^2 }}.\)

 

[Erin0702]

thank you!!!

 

[galactus]

\(\displaystyle \L\\ln(t^{2})dt=\frac{2}{t}\)

\(\displaystyle \L\\(t+\frac{1}{t})dt=1-\frac{1}{t^{2}}\)

Square each and you get, respectively:

\(\displaystyle \frac{4}{t^{2}}\) and

\(\displaystyle \frac{(t^{2}-1)^{2}}{t^{4}}\)

So, we have:

\(\displaystyle \L\\\sqrt{\frac{(t^{2}-1)^{2}}{t^{4}}+\frac{4}{t^{2}}}\)

Which boils down to:

\(\displaystyle \L\\\frac{t^{2}+1}{t^{2}}\)

Now, you have:

\(\displaystyle \L\\\int_{1}^{4}\frac{t^{2}+1}{t^{2}}dt\)

You can take it from there, can't ya'?.

 

[Erin0702]

yes...thank you!