<p>Question 5, (b)</p>
<p>I dont get this limit...... help please</p>
<p>This is a fundamental application of logistic growth.</p>
<p>for some logistic function y = f(t), </p>
<p>dy/dt = ky(P - y)</p>
<p>f(t) = P / (1 + C * e^(-Pkt))</p>
<p>as t -> infinity, its easy to see that e^(-Pkt) approaches 0, so lim t -> infinity of f(t) = P</p>
<p>This holds true for all logistic growth equations.</p>
<p>f’(t) = kinda complicated, but just use quotient rule. As lim t -> infinity, the e^(-Pkt) term approaches 0, so it’s basically:</p>
<p>(1 - 1)/ 1 = 0/1 = 0</p>
<p>I think TheMathProf could give a better explanation for f’(t), but I hope this helps.</p>
<p>I’d just go the long way and solve for the equations. </p>
<p>dy/(2y(3-y))= dx </p>
<p>Partial fractions.</p>
<p>A/2y + B/(3-y) = 1/(2y(3-y)</p>
<p>A(3-y) +B(2y)=1</p>
<p>A and B are constant so, suppose y = 0.</p>
<p>3A = 1 , A =1/3</p>
<p>Suppose y = 3 </p>
<p>6B = 1</p>
<p>B = 1/6. Thus</p>
<p>1/(3(y - 3)) + 1/(6(2y)) dy = dx</p>
<p>Integrate.</p>
<p>-3 ln (3-y) + (1/12) ln y = x + C</p>
<p>Times 12 to both sides.</p>
<p>-36 ln (3-y) + ln y = 12x + C(diff. c)</p>
<p>Translate -.</p>
<p>36 ln(3-y) - ln y = -12x + C
ln((3-y)^36 y^-1) = 12x + C</p>
<p>e^ both sides</p>
<p>((3-y)^36)/y = Ce^(-12x)</p>
<p>As x–> infinity, c cancels and thus you’d get 0. The only value of y that makes it zero is 3.</p>
<p>…</p>
<p>an0maly has the right way. xD</p>
<p>i did the same as u did ^, but the C value is kinda whacked. its like 2/e^24 or something…</p>
<p>and “As x–> infinity, c cancels and thus you’d get 0. The only value of y that makes it zero is 3.”<br>
Doesn’t C become C? since e^(-infinity)= 1?</p>
<p>anyway, thanks guys 
i definitely will look into that function anomaly provided</p>
<p>i just redid the problem; C, though doesn’t matter, is 2e^24.
I find it easier to go the long route, which is what Xav did, because its hard for me to memorize a formula w/o knowing wat it is.
i still don’t get the derivative of that function though…i tried deriving using the quotient rule, but i ended up getting -3…
anyone?</p>
<p>thanks~</p>
<p>Yeah, C is unimportant. As for the derivative, there’s a cute little trick. We know that the other side is going to be zero, since d/dx of e^-x is just -e^-x, thus, still 0. We’ll do quotient rule. We know that y^2 is going away, so we’ll focus on the top.</p>
<p>Here’s the trick, however: We know that, whatever we get, has to equal to zero. Thus, whatever the ho d hi - hi d ho has in common, it must be equal to zero. Lucky for us, that’s y’ . So, you don’t even have to expand it. You’ll know that y’ is zero in order for the whole thing to be zero. </p>
<p>Feel me?</p>