<p>I somewhat answered this on the other thread but my reasoning was a little bit shady. I said that you could pull out the ^2 of the absolute value part to get (abs(x-3))^2<1 since both the absolute value and the ^2 will make the output positive. Then you just have the abs(x-3)<1, -1<x-3<1, 2<x<4 - the interval of convergence is between 2 and 4 (open interval since the ^2n of the original series always makes it positive so it just becomes the harmonic series) and the radius is 1. But like I said, I don’t know if you can pull that ^2 out of the absolute value condition set <1.</p>
<p>^ Thanks. Well, I didn’t know what to do with the ^2 because it would make the whole thing positive all the time.</p>
<p>I have a question about the parametric free responses. There are some asking for the distance the particle has travelled and others for displacement whats the difference in the calculation you perform?</p>
<p>Who wants to explain logistics here??? PLEASE</p>
<p>distance is overall how far the particle travelled. Displacement is the difference in position form the start to the end. Say I were to hit a home run, and the distance between the bases is 50 feet. My distance travelled would be 200 feet, but my overall displacement would be zero, as I start at home plate and end at home. You see the distinction?</p>
<p>Somebody answer this!
'Cause I have noooo idea. lol. Actually we didn’t even learn this stuff in my class. for some reason.</p>
<p>And I disagree with TheMan66 about lagrange error. The error is just the next term in the series. If you’re asked to approximate the value of the series when x=whatever using the first 3 terms, the fourth term is the value that your error is less than or equal to. The interval for the correct answer made possible BY that error bound is your approximated answer +/- the error.</p>
<p>For difference between distance and displacement, the calculus is that you’d do integral of the velocity for displacement, and integral of the ABSOLUTE VALUE of velocity for distance, I think, I always get it backwards lololol, but that way makes sense to me.</p>
<p>Mesquite: For the logistic curve, you basically have some differential where the rate of growth depends on (P-K)/K, where P is population and K is carrying capacity. When the population reaches carrying capacity, it is 0/K=0, so zero growth. When the population starts off, it is basically exponential. So, the population graph for logistic growth starts as an exponential and then flattens out at the carrying capacity.</p>
<p>For the test, I highly recommend you remember the differential equation AND the solution for such problems as logistic growth and linear or squared drag and stuff. At least look them over today, so you have some idea if your solution for the differential is correct.</p>
<p>Oh ok, how do the formulas for calculation differ. SO displacement is the integral of velocity, what is the formula of total distance?</p>
<p>I know that in the cartesian plane system, the integral of the absolute value of a velocity function gets you total distance while integrating just the velocity function will get you displacement. But is there a way to calculate displacement of a particle with parametric equations? I know that the integral from a to b of ((dx/dt)^2 +(dy/dt)^2)^1/2 dt is the total distance, which is the same as arc length, but how do you do displacement? or is it the same since the squaring of the derivative will always yield a positive value?</p>
<p>I believe that you are correct INVIENANVIEM.</p>
<p>So, total distance is the integral of the absolute value of velocity, otherwise known as speed. Like in my baseball example above, if you run the bases at 10mph (which is speed, NOT velocity), your distance would be the integral of 10dt. But, your velocity is zero, because V=dx/dt (where x is displacement). So, the distance would be the integral of 0dt.</p>
<p>formula of total distance: integral from a to b of ( squ.rt ( (dy/dt)^2 + (dx/dt)^2) )</p>
<p>@ PKWsurf, I thought the arc length is integral from a to b of (squ.rt (1 + (dy/dx)^2)) ? Or maybe these two are the same equation?</p>
<p>His is the parametric form. It’s basically the same thing, you start with</p>
<p>ds = \sqrt{dx^2 + dy^2}</p>
<p>and instead of distributing a dx^2 out like you do for rectangular form, you multiply by dt/dt and distribute 1/dt into the radical</p>
<p>question from the 2003 MC: What is the coefficient of x^2 in the Taylor series for 1/((1+x)^2) about x=0. Choice are 1/6,1/3,1,3,6. I said it was like the 1/(1+x) squared but that didnt work. I cant obtain the answer of 3. Does anyone know?</p>
<p>I have that, for a parametric equation, arc length/distance = int from a to be of sqrt([dx/dt]^2+[dy/dt]^2).
For a regular cartesian function, arc length = int from a to b of sqrt(1+[f’(x)^2])</p>
<p>I dunno about switching between the two types of equations with that though, lol, I wouldn’t try it.</p>
<p>Do you guys think the series q’s will include the basic tests, or other ones like the root test or the negative one?</p>
<p>x^2 term is the second derivative of it</p>
<p>so f ‘’(x)/2 is the coefficient</p>
<p>6(1+0)^{-4}/2 = 6/2 = 3</p>
<p>@ sixthsense, I just wrote out the Taylor series for 1/ (1+x)^2. It worked…</p>
<p>
</p>
<p>I didn’t mutate the 1/(1+x) series, but if you just use the way to make a taylor series (meaning… fN(a)(x-a)^N/N!) it works… you get the series 1-2x+(6x^2)/2!
So the x^2’s coeff is 6/2!=3.</p>