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:
$$
\begin{equation}
\begin{aligned}
f_x & = \frac{y}{(x+y)^2} \\
f_y & = \frac{-x}{(x+y)^2}
\end{aligned}
\end{equation}
$$
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:
$$
\begin{equation}
L(x,y)= \frac{2}{3} + \frac{1}{9}(x-2) – \frac{2}{9}(y – 1)
\end{equation}
$$

114TPALAE3

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 \(z = f(x,y)\).

E 14.3 Exercises