<p>Good luck to everyone on tomorrow's test!</p>
<p>Did we come to a consensus on whether or not Newton's Law of Cooling is going to be on it?</p>
<p>I don't think it'll be on it.</p>
<p>can someone explain slope fields? also, what is lim x--> -∞ (x^2 / e^x) ?</p>
<p>We know that (e^x) increases much faster than (x^2), therefore, the limit = 0.</p>
<p>barron's says the limit is -∞ (BC Exam #1 Section I) tho</p>
<p>For slope fields, just plug in the x and y values at each "tick" into the differential eqn and graph the slope that comes out.</p>
<p>For logistic growth:</p>
<p>All models have two stable points where the limit approaches a horizontal asymptote. Since you know that the derivative at flat points is zero, just set the logistic equation to 0 and solve to obtain your answers.
Ex: The previous example:</p>
<p>dP/dt = (P/5)(1-P/12)</p>
<p>If p(0) = 3, what is the limit of P(t) as t goes to infinity?</p>
<p>Solve the differential equation for 0 and you obtain P = 12 and 0. The larger term is correct when going to infinity so the answer is 12.</p>
<p>maclaurin for e^x = 1 + x + x^2/2! + x^3/3! + ... + x^n/n!</p>
<p>cosx = 1 - x^2/2! + x^4/4! - x^6/6!....</p>
<p>sinx = x - x^3/3! + x^5/5! - x^7/7!...</p>
<p>1/(1-x) = 1 + x + x^2 + x^3 + ....</p>
<p>1/(1+x) = 1 - x + x^2 - x^3 + ....</p>
<p>tan^ -1(x) = x - x^3/3 + x^5/5 - x^7/7</p>
<p>ln(1+x) = x - x^2/2 + x^3/3 - x^4/4</p>
<p>I think that's all of them. Should be right as well.</p>
<p>ya, but how do you determine what equation the slope field is? just try to connect the points and see what kind of graph it is? i'm guessing then we have to memorize equations for hyperbola, parabola, ellipse, etc.?</p>
<p>No, you don't connect the dots. Just make little slivers that show the slope at the point. When you overlay the actual equation over the slope field, they should match.</p>
<p>and btw fullmoondragon, wrong answer.</p>
<p>x is approaching NEGATIVE infinity, meaning that e^x goes to -0 (the negative sign makes the limit - infinity, take e^-500)</p>
<p>therefore, its (-inf)^2/(-0) = inf^2/-0 = -inf tricky question really.</p>
<p>rw9700, you don't have to memorize them for slope fields.</p>
<p>just get a dy/dx = u differential equation. solve for whatever U is at each x and y value, and graph a line around that slope (say you have dy/dx = 2x + 2y and you are at the point (1,2), dy/dx = 2 + 2(2) = 6. therefore, at (1,2) draw a short little segment of a line that looks like it has a slope of 6)</p>
<p>drawing an entire slopefield is long as hell, and probably something they won't make us do. if you're just looking for practice though, that's the way to go.</p>
<p>so are you saying that (-∞)^2 is (-∞), not +∞?</p>
<p>no. i'm saying that e^-inf (how do you even do that infinity sign) = -0. sounds rediculous b/c 0 can't have a sign to it, but it's a limit, meaning it never actually hits 0. it's just some really small number, like -9 x 10^-10000000000000000000000000000000. but it maintains the negativity.</p>
<p>anything squared is always positive. except for i.</p>
<p>Alt+236</p>
<p>∞</p>
<p>beast, thanks.</p>
<p>btw, i'm done studying, so keep shootin questions at me until I go to bed. ∞</p>
<p>evilbooyaa, no positive number to a power can equal a negative number.</p>
<p>Good luck to all.</p>
<p>good luck :)
i'm sleeping. you guys, get some rest before the exam, or else you won't be able to concentrate.</p>