<p>For the link above, how is the answer to 5a. 12? The solution doesn't really go into depth and I'm confused. Can someone please elaborate?</p>
<p>you need to factor out a 1/12 from dp/dt to get dp/dt = (P/60)(12-P). this is in the form of kp(l-p). L is the carrying capacity which the limit as x approaches infinity will always equal. 12 is the L value in this problem. 12 is also the answer for 5 b</p>
<p>wow first logistic question i’ve seen as a free response question.</p>
<p>@Isu: I’m still confused. How did you get dp/dt = (P/60)(12-P)? I’ve been trying to factor 1/12 out but I don’t get that result.</p>
<p>Also, I’ve never seen the equation kp(l-p) before in my life, lol. My calc. teacher and my Princeton review book never covered that. Is it crucial to memorize?</p>
<p>no don’t memorize that. heres an intuitive way of thinking about it.</p>
<p>they tell you that its a logistical equation, so it isn’t hard to realize that there will be a starting point and a carrying capacity. picture the classic logistic equation in your head. the most used example is of water pollution levels after a big spill.</p>
<p>so you know that there will be two horizontal asymptotes. as t goes to negative infinity, P will approach one of the asymptotes. as t goes to positive infinity, P will approach the other.</p>
<p>now the question is, what are those asymptotes? what are the values of P as t goes to either end? </p>
<p>when we approach the limiting P values, dP/dt will approach zero. setting the differential equation equal to zero gives us P=0, and P=12. in the interval 0 to 12 dP/dt is always positive, so as t approaches infinity, P approaches 12.</p>
<p>@dchau503, divide each term in (1-(P/12)) by (1/12). So we have ((1/(1/12))-((P/12)/(1/12)), simplifying the complex fractions, (12-(P/12)(12)) = (12-P). So we have (1/12) on the outside and (1/12)(P/5)= P/60 so finally we have (P/60)(12-P).</p>
<p>Thanks for the answers guys. I also found out the the logistic equation for population is dP/dt=kP(1-P/K) from the kaplan’s book.</p>
<p>Also, for the same test link, can someone please elaborate the answer to 6b as well?</p>
<p>From the results in 6(a), we can conclude that the nth derivative of the function f evaluated at x = 0 would contain 5^n*sqrt(2)/2. There’s also an alternator of sorts; it alternates every other term, which tells us that this term is negative. (The general form of the alternator could be provided with a (-1)^[(n^2 - n)/2]. However, simply recognizing that every term that’s a multiple of 4 or one more than a multiple of 4 is positive, while also recognizing that every term that’s two more than a multiple of 4 or three more than a multiple of 4 is negative, should be enough to get you the sign for this term.) </p>
<p>The coefficient of the x^22 term of a Taylor series is found by taking the 22nd derivative of f(x) evaluated at x = 0, and dividing by 22!. Putting those elements together should get you the term -(5^22)*sqrt(2) / [2 * 22!] as provided in the solutions.</p>