# 14.3 | Local Linearity and the Differential

Suppose $$f$$ has continuous partial derivatives. An equation of the tangent plane to the surface $$z = f(x,y)$$ at the point $$P(x_0, y_0, z_0)$$ is

$$z = f(x_0,y_0) + f_x(x_0, y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$$
\begin{aligned} z = &f(x_0,y_0) + f_x(x_0, y_0)(x-x_0) \\ &+ f_y(x_0,y_0)(y-y_0) \end{aligned}

The tangent plane at $$P$$ is also referred to as the linearization of $$f$$ and in this case we make use of the notation $$L(x,y)$$ instead of $$z$$. $$L(x,y)$$ approximates $$f$$ near $$P$$.

If $$z = f(x,y)$$, then $$f$$ is differentiable at $$(a,b)$$ if $$\Delta z$$ can be expressed in the form

$$\Delta z = f_x(a,b) \Delta x + f_y(a,b) \Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta y$$
\begin{aligned} \Delta z = & f_x(a,b) \Delta x + f_y(a,b) \Delta y \\ &+ \epsilon_1 \Delta x + \epsilon_2 \Delta y \end{aligned}

where $$\epsilon_1 \rightarrow 0$$ and $$\epsilon_2 \rightarrow 0$$ as $$( \Delta x, \Delta y) \rightarrow (0,0)$$.

If the partial derivatives $$f_x$$ and $$f_y$$ exist near $$(a,b)$$ and are continuous at $$(a,b)$$, then $$f$$ is differentiable at $$(a,b)$$.

Find an equation of the tangent plane to the surface
$$z = 3(x-1)^2 + 2(y+3)^2+7$$
at $$(2,-2,12)$$

An equation of the tangent plane to the surface $$z = 3(x-1)^2 + 2(y+3)^2+7$$ at the point $$(2,-2,12)$$ is given by

$$z = f(x_0, y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$$
\begin{aligned} z = & f(x_0, y_0) + f_x(x_0,y_0)(x-x_0) \\ & + f_y(x_0,y_0)(y-y_0) \end{aligned}

where $$x_0 = 2$$ and $$y_0 = -2$$.

The first partials of $$f(x,y)$$ are:
$$f_x = 6(x-1) \text{ and } f_y = 4(y+3)$$
and the first partials evaluated at $$(x_0,y_0)$$ are:
$$f_x(2,-2) = 6 \text{ and } f_y(2,-2) = 4$$

The function value at $$(x_0, y_0)$$ is:
$$f(2,-2) = 12$$

So an equation for the tangent plane to the surface at $$(2,-2, 12)$$ is given by
$$z = 12 +6(x-2) + 4(y+2).$$

Explain why $$f(x,y) = \frac{x}{x+y}$$ is differentiable at $$(2,1)$$. Then find the linearization $$L(x,y)$$ of the function at that point.

$$f$$ is differentiable at $$(2,1)$$ if its first partials are continuous near $$(2,1)$$. The first partials of $$f$$ are:
\begin{aligned} f_x & = \frac{y}{(x+y)^2} \\ f_y & = \frac{-x}{(x+y)^2} \end{aligned}
both of which are rational functions and hence continuous everywhere except where $$x=-y$$. So the first partials are continuous near $$(2,1)$$ and hence $$f$$ is differentiable at $$(2,1)$$.

The linearization of $$f$$ at $$(2,1)$$ is given by:

$$L(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$$
\begin{aligned} L(x,y) = & f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) \\ & + f_y(x_0,y_0)(y-y_0) \end{aligned}

where $$x_0 = 2$$ and $$y_0 = 1$$.

The first partials evaluated at $$(x_0,y_0)$$ are: $$f_x(2,1) = 1/9$$ and $$f_y(2,1) = -2/9$$. The function value at $$(x_0,y_0)$$ is: $$f(2,1) = 2/3$$.

Thus the linearization of $$f$$ at $$(2,1)$$ is:
$$L(x,y)= \frac{2}{3} + \frac{1}{9}(x-2) – \frac{2}{9}(y – 1)$$

Table 1

Determine if table 1 could represent a linear function $$z = f(x,y)$$? If not, can the given table represent an elliptic paraboloid $$z = f(x,y)$$?

Note that the general form of a plane $$z = f(x,y)$$ is given by the equation
$$z = ax + by + c$$
Hence $$z_x = a$$ and $$z_y = b$$. That is, the rate of change in the $$x$$-direction or $$y$$-direction is constant. But when $$y = -1$$, as $$x$$ increases the function values decrease by 3 and then by 1. So, $$f_x$$ is not constant, and so the table can not represent a linear function.

Note that the general form of an elliptic paraboloid $$z = f(x,y)$$ is given by the equation
$$\frac{z}{c} = \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}$$
Hence $$z_x = \frac{2c}{a^2}(x-h)$$ and $$z_y = \frac{2c}{b^2}(y-k)$$.

That is, the rate of change in the $$x$$-direction or $$y$$-direction is linear. The rate of change in the $$x$$-direction when $$y = -1$$ is given by $$z_x = -3 +2(x+2)$$. So, for example, $$z_x(-2,-1) = -3$$ and $$z_y(-1,-1) = -1$$, which is consistent with the rates of change in the $$x$$-direction in the column $$y = -1$$ of the table. This same linear function works for $$y = 0$$ and $$y = 1$$.
The rate of change in the $$y$$-direction when $$y = -2$$ is given by $$z_y = -1 +2(y+1)$$. So, for example, $$z_y(-2,-1) = -1$$ and $$z_y(-2,0) = 1$$, which is consistent with the rate of change in the $$y$$-direction in the row $$x = -2$$ of the table. This same linear function works for $$x = -1$$ and $$x=0$$.

So, the table could represent an elliptic paraboloid $$z = f(x,y)$$.