17.3 Tangent plane approximations

To extend the notion of a linear approximation, we apply the tangent plane approximation at the point $x=a$, $y=b$:

$$\begin{equation} L(x,y) = f(a,b) + f_{x}(a,b) \cdot (x-a) + f_{y}(a,b) \cdot (y-b), \end{equation}$$

where $f_{x}$ is the partial derivative of $f(x,y)$ with respect to $x$ and $f_{y}$ is the partial derivative of $f(x,y)$ with respect to $y$. For simplicity let’s work with the equilibrium solution at $(0,0)$. If we say that $f(H,L)=.3 H - .1 HL$, we know $f(0,0)=0$. Furthermore the locally linear approximation is:

$$\begin{equation} \begin{split} f_{H} = .3 - .1L & \rightarrow f_{H}(0,0)=.3 \\ f_{L} = -.1H & \rightarrow f_{L}(0,0)=0 \end{split} \end{equation}$$

With this information we are able to take this and then write down the locally linear approximation for $f(H,L)$:

$$\begin{equation} f(H,L) \approx 0 + .3 \cdot (H-0) - 0 \cdot (L-0) = .3H \end{equation}$$

Likewise if we consider $g(H,L)= .05HL -.2L$, then we have:

$$\begin{equation} \begin{split} g_{H} &= .05L-.2L \rightarrow g_{H}(0,0)=0 \\ g_{L} &= .05H-.2 \rightarrow f_{L}(0,0)=-.2 \end{split} \end{equation}$$

With these results, our locally linear approximation for $g(H,L)$ is:

$$\begin{equation} g(H,L) \approx 0 + 0 \cdot (H-0) -.2 \cdot (L-0) = 0 - .2L. \end{equation}$$

So, at the equilibrium solution $(0,0)$, Equation (17.2) behaves like the following system of equations:

$$\begin{align} \frac{dH}{dt} &= .3H\\ \frac{dL}{dt} &= - .2L \tag{17.3} \end{align}$$

Notice how Equation (17.3) is an entirely decoupled linear system of equations that can be solved easily by separation of variables. For an initial condition $(H_{0},L_{0})$ Equation (17.3) has a solution $H(t)=H_{0}e^{.3t}$ and $L(t)=L_{0}e^{-.2t}$.