<p>ok. So here's the thing. On my BC Text book there's questions on parametric differentiation with the length of an Arc. I solve some all the way till i get to the equation sqrt(t^2 + 1) in an integral. I don't know if this would be an AB technique because I totally forgot. But if there's a way to do this problem... PLEASE RESPOND BACK!!!</p>
<p>I've been trying to do this for the past 30 mins.</p>
<p>could you possibly use "u" substitution?</p>
<p>well if i sub-u this sucker, I get a 2t for the deriv and there's nothing else outside of the integral. so I don't think that's legal</p>
<p>Its not really working out all that good for me but have you tried trig substitution with that being the hypotenuse.</p>
<p>oh yeah, wait that looks a lot like trig subs using tangent...let x = tan(0)</p>
<p>Triangle method, possibly?</p>
<p>Hey I had a quick question for u. I also have a BC math problem that has been making me struggle for all of spring break. How do u integrate 5sqrt of
2(1-sintsin5t-costsos5t)???? from 0 to pie over 2???</p>
<p>Also how do u find the integral of convergence for the power series of y= (3x-1)/(x squared-1)</p>
<p>please reply i have to do it tonite! Thanks!</p>
<p>natasha: if i remember correctly, the integral seem like it could be simplified with the trig addition formula, then the double angle formula</p>
<p>wat rv bks u guys usin for ap calc bc</p>
<p>WPI is using Edwards and Pennys for Calc 3, which is basically 3.</p>
<p>To M1st3rmarbl3s: there must be 2 equations for parametric. You only gave us one, how can I do it w/ just this one???</p>
<p>To natasha: <a href="http://putfile.com/pic.php?pic=4/9220444552.jpg&s=x11%5B/url%5D">http://putfile.com/pic.php?pic=4/9220444552.jpg&s=x11</a> that's the solution</p>
<p>Would this be a calculator problem..? In any case the answer involves ln</p>
<p>nope, unless the problem states "a graphing calculator is required", you can't use the calculator at all.
To Natasha: sorry, I didn't know there's one other problem too >_<
<a href="http://putfile.com/pic.php?pic=4/9315481890.jpg&s=x10%5B/url%5D">http://putfile.com/pic.php?pic=4/9315481890.jpg&s=x10</a></p>
<p>um, natasha's second question is not an integral, it's asking for interval of convergence. and frankly, (3x-1)/(x^2-1) always diverges. if you cancel everything except the highest power, which yields 3/x, and then use integral test, you find that it diverges to infinity.</p>
<p>well, then I need some kind of a form in terms of n and x. There is no "n", only x, so that's y I thought it was an integral problem.</p>
<p>well for the problem, since it's in the form of 1 plus a variable within the square root, you can use the parametric equation for arc length. that would be the intergral from a to be of the square root of [(dy/dt)^2 + (dx/dt)^2]. from this you know that dy/dt is equal to 1 and dx/dt is t. and then you can use that.</p>
<p>I still have no idea</p>
