The total differential approximation works in three$-$dimensional space, too. For example, consider the paraboloid

$z=F(x,y)=\frac{x^2}{4^2}\frac{y^2}{5^2}$

at the point $3,5,\frac{25}{16}.$

The total differential approximation to $z=F(x,y)$ near $3,5$ is given by the formula

$F(x,y)$~~$\frac{25}{16}\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{x}}(3,5)(x-3)\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{y}}(3,5)(y-5)$

In fact, this gives the equation of the tangent plane to the surface at the point $3,5,\frac{25}{16}.$ Now

$\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{x}}(3,5)=$

$\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{y}}(3,5)=$

So we have the linear approximation to $F(x,y)$ near $3,5$ :

$F(x,y)$~~

Use this formula to approximate $($to $3$ decimal places$)$

$F(3\mathrm{.}1,5\mathrm{.}1)$~~

$$

Compute directly from the definition $($to $3$ decimal places$)$ the value

$F(3\mathrm{.}1,5\mathrm{.}1)=$

$$