Category Archives: Product Rule

Deriving the Logistic Growth Equation

The logistic differential equation

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

where r is the growth parameter and k is the carrying capacity.

And the maximum rate of increase happens when P=\frac{k}{2}

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

    \begin{equation*}\frac{dP}{P(k-P)}=r dt{\end{equation}

    \begin{equation*}\int \frac{dP}{P(k-P)}=\int r dt{\end{equation}

I am going to separate the denominator on the left hand side

\frac{1}{P(k-P)}=\frac{A}{P}+\frac{B}{k-P}
Hence,
\frac{1}{P(k-P)}=\frac{A(k-P)+BP}{P(k-P)}
1=A(k-P)+BP
When P=0,
1=Ak\Rightarrow A=\frac{1}{k}
When P=k,
1=BK\Rightarrow B=\frac{1}{k}

So our equation is,

    \begin{equation*}\int \frac{\frac{1}{k}}{P}+\frac{\frac{1}{k}}{k-P} dP=\int r dt\end{equation}

    \begin{equation*}\frac{1}{k}\int \frac{1}{P}+\frac{1}{k-P} dP=\int r dt\end{equation}

    \begin{equation*}\int \frac{1}{P}+\frac{1}{k-P} dP=\int kr dt\end{equation}

    \begin{equation*}ln\lvert{P}\rvert-ln\lvert{k-P}\rvert=krt+c\end{equation}

    \begin{equation*}ln\lvert{\frac{P}{k-P}\rvert=krt+c\end{equation}

    \begin{equation*}\frac{P}{k-P}=e^{krt+c}\end{equation}

    \begin{equation*}\frac{P}{k-P}=e^{krt}e^{c} \end{equation}

When t=0, P=P_0,

    \begin{equation*}\frac{P_0}{k-P_0}=e^{c} \end{equation}

The equation is now

    \begin{equation*}\frac{P}{k-P}=\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P=\frac{P_0}{k-P_0}e^{krt}(k-P)\end{equation}

    \begin{equation*}P=k\frac{P_0}{k-P_0}e^{krt}-P\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P+P\frac{P_0}{k-P_0}e^{krt}=k\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P(1+\frac{P_0}{k-P_0}e^{krt})=k\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{1+\frac{P_0}{k-P_0}e^{krt}}\end{equation}

    \begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{\frac{k-P_0+P_0e^{krt}}{k-P_0}}\end{equation}

    \begin{equation*}P=\frac{kP_0e^{rkt}}{k-P_0+P_0e^{rkt}}\end{equation}

Divide by e^{rkt}

    \begin{equation*}P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}

    \begin{equation*}}\frac{dP}{dt}=rP(k-P)\Longleftrightarrow P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}

Proving the Maximum Rate of Increase Happens When P=\frac{k}{2}

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=r\frac{dP}{dt}(k-P)+rP(-\frac{dP}{dt})\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=\frac{dP}{dt}(rk-rP-rP)\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=0\end{equation}

    \begin{equation*}\frac{dP}{dt}(rk-rP-rP)=0\end{equation}

    \begin{equation*}r\frac{dP}{dt}(k-2P)=0\end{equation}

    \begin{equation*}\frac{dP}{dt}(k-2P)=0\end{equation}

    \begin{equation*}rP(k-P)(k-2P)=0\end{equation}

Hence P=k or P=\frac{k}{2}

(1)   \begin{equation*}\frac{d^3P}{dt^3}=\frac{dP^2}{dt^2}(rk-2rP)+\frac{dP}{dt}(-2\frac{dP}{dt})\end{equation*}

Substitute P=k into equation 1

    \begin{equation*}\frac{d^3P}{dt^3}=rk(k-k)(rk-2rk)(rk-2rk)-2(rk(k-k))^2=0\end{equation}

Hence, not a maximum.

Substitute P=\frac{k}{2} into equation 1

    \begin{equation*}\frac{d^3P}{dt^3}=rk(k-\frac{k}{2})(rk-2r\frac{k}{2})(rk-2r\frac{k}{2})-2(rk(k-\frac{k}{2}))^2=0\end{equation}

    \begin{equation*}\frac{d^3P}{dt^3}=-2(rk^2-\frac{rk^2}{2})^2\end{equation}

    \begin{equation*}\frac{d^3P}{dt^3}=-2\frac{r^2k^4}{4}\end{equation}

-2\frac{r^2k^4}{4}\le 0 For all values of P, r and k.

Hence maximum when P=\frac{k}{2}

We will look at a worked example in the next post.

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Filed under Differential Equations, Differentiation, Implicit, Logistic Growth, Optimisation, Product Rule, Uncategorized, Year 12 Specialist Mathematics

Deriving the Product Rule for Differentiation

In my previous post we looked at the Chain Rule for Differentiation, this post is on the Product Rule. Differentiating a function in the form y=f(x)\times g(x).

For example, y=(3x^3+2x-1)(x^4+2x^2)

Remember differentiating from first prinicpals:

f'(x)=\lim_{\limits h \to 0} \frac{f(x+h)-f(x)}{h}

y=f(x)g(x)

\frac{dy}{dx}=\lim_{\limits h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}}

\small{  \frac{dy}{dx}=\lim_{\limits h \to 0} \frac{f(x+h)g(x+h)-g(x+h)f(x)+g(x+h)f(x)-f(x)g(x)}{h}}

By subtracting and then adding g(x+h)f(x) we haven’t changed the limit, but it means we can do some factorising.

\frac{dy}{dx}=\lim_{\limits h \to 0}\frac{g(x+h)(f(x+h)-f(x))+f(x)(g(x+h)-g(x))}{h}

\small{\frac{dy}{dx}=\lim_{\limits h \to 0}g(x+h)\lim_{\limits h \to 0}\frac{f(x+h)-f(x)}{h}+\lim_{\limits h \to 0}f(x)\lim_{\limits h \to 0}\frac{g(x+h)-g(x)}{h}}

When we evaluate the limits

\frac{dy}{dx}=g(x)f'(x)+f(x)g'(x)

Example

Find the derivative of y=(3x^3+2x-1)(x^4+2x^2)

I remember the rule in words ‘derivative of the first times the second plus the derivative of the second times the first’.

y'=(9x^2+2)(x^4+2x^2)+(4x^3+4x)(3x^3+2x-1)

y'=9x^6+18x^4+2x^4+4x^2+12x^6+8x^4-4x^3+12x^4+8x^2-4x

y'=21x^6+40x^4-4x^312x^2-4x

Most exam questions have ‘don’t simplify’, so the first line of working above would be enough.

Onto the Quotient Rule.

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Filed under Calculus, Differentiation, Product Rule, Year 12 Mathematical Methods