Couldn’t decide if I should put this in the calculus or general math forums but...(adsbygoogle = window.adsbygoogle || []).push({});

I’m studying for a final that’s coming up this Wednesday and I’ve been looking at some past quizzes with the steps to finding the solutions that my instructor has posted online. Given the problem:

1. Compute the length of the curve~r(t) = (t^2, 2t, ln(t)), from t = 1 to t = e.

I understand that I need to take the integral from 1 to e of the sqrt( (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2) but unfortunately I’m having trouble taking the sqrt.

I get down to… *EDITED*sqrt( 4t^2 + 4 + ( 1/t^2 ) ) and I get stuck even with the answer and the steps in front of me. In the next step the problem gets simplified so as to have everything over a t^2 and then the sqrt is done and I’m left with the integral from 1 to e of (2t^2 + 1)/t.

I’m hoping someone could explain to me how sqrt( 4t^2 + 4 + ( 1/t ) ) becomes (2t^2 + 1)/t.

I’m ashamed I don’t know this but I don’t want to get stuck trying to do algebra on a calculus final.

Thanks in advance.

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# Trouble taking a Square Root

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