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Numerade Educator



Problem 16 Easy Difficulty

Von Bertalanffys equation states that the rate of growth in length of an individual fish is proportional to the difference between the current lenght $ L $ and the asymptotic length $ L_x $ (in centimeters).
(a) Write a differential equation that expresses this idea.
(b) Make a rough sketch of the graph of a solution of a typical initial-value problem for this differential equation.


(a) $\frac{d L}{d t}=k\left(L_{\infty}-L\right)$
$k$ is the proportionality constant
(b) Let initial length of a fish is $L_{0}$ .
Then the initial-value problem can be written as,
$\frac{d L}{d t}=k\left(L_{\infty}-L\right), L(0)=L_{0}$
To find the length $L$ at any time $t,$ integrate both sides,
$\int_{L_{0}}^{L(t)} \frac{d L}{L_{\infty}-L}=\int_{0}^{t} k d t$
$\left[\ln \left(L_{\infty}-L\right)\right]_{L_{0}}^{L(t)}=-k t$
$\ln \left[\frac{L_{\infty}-L(t)}{L_{\infty}-L_{0}}\right]=-k t$
$\frac{L_{\infty}-L(t)}{L_{\infty}-L_{0}}=e^{-k t}$
Solve for $L(t)$
$L(t)=L_{\infty}-\left(L_{\infty}-L_{0}\right) e^{-k t}$
From the solution it can be seen as time increases, the length increases and reaches $L_{\infty}$ as $t$ tends to infinity.


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Video Transcript

Okay, So today we're going to show you the application off the differential curation. According to the question. The rate off growth in the lands off the individuals fresh proportional to the differential between the current nous and awesome political ends l X So we Right, so here l e this wise l X is this question. So this is the question. Say no. So it's things. It's a propulsion. We don't know what's the ratio we can sink here Is the the racial rate. Okay, it's the rate. It's a coefficient you can see here, and it's a constant. Normally it's the normally speaks Luxan zero. So if if case is less than zero if case less than zero, it's not really right. The differential will be negative. So So we get some products. Mhm. So the less will be we'll beat will be different, right? So Kate must be larger than zero. So for part, for second one. So for the for the for sketch the problem, we just go Just take look this one since we knew case larger than zero And here is the only thing we need to we need to know is large or picks in zero is large or less than zero. Apparently. Must be large since you right. And when When Ellis increase, this whole thing will be a poked me to zero When? The hour. So this thing should be zero. Right? So here is the girl.