How to find a linear approximation in two variables

If you’ve used a tangent line to estimate a function in one variable, we can do the same thing with a tangent plane in two variables. This time we need two slopes, one for a change in x and one for a change in y.

Build the linear approximation from two slopes

For a differentiable function $f$ near $(a, b)$ , start with the height $f (a, b)$ . The partial derivative $f_{x} (a, b)$ is the slope in the $x$ direction, so multiply it by the change $x - a$ . The partial derivative $f_{y} (a, b)$ is the slope in the $y$ direction, so multiply it by $y - b$ .

\begin{array}{l} L (x, y) = f (a, b) \\ + f_{x} (a, b) (x - a) \\ + f_{y} (a, b) (y - b) \end{array}

The formula gives the height $z = L (x, y)$ on the tangent plane. At the base point, both changes are zero and we get $L (a, b) = f (a, b)$ . For points close to $(a, b)$ , we use $L (x, y)$ to estimate $f (x, y)$ .

Find the height at the base point

Let $f (x, y) = 6 x^{3} - 2 x y^{2}$ and use the base point $(1, 2)$ . Plug both coordinates into the function to find the starting height.

\begin{array}{l} f (1, 2) = 6 {(1)}^{3} \\ - 2 (1) {(2)}^{2} = - 2 \end{array}

Find the slopes in the x and y directions

When we differentiate with respect to $x$ , treat $y$ as a constant. When we differentiate with respect to $y$ , treat $x$ as a constant. Then plug $(1, 2)$ into each partial derivative.

\begin{array}{l} f_{x} (x, y) = 18 x^{2} - 2 y^{2} \\ f_{x} (1, 2) = 18 {(1)}^{2} \\ - 2 {(2)}^{2} = 10 \\ f_{y} (x, y) = - 4 x y \\ f_{y} (1, 2) = - 4 (1) (2) \\ = - 8 \end{array}

Put the height and the slopes together

\begin{array}{l} L (x, y) = - 2 + 10 (x - 1) \\ - 8 (y - 2) \\ = 10 x - 8 y + 4 \end{array}

For example, at $(1.02, 1.99)$ , the $x$ coordinate increases by $0.02$ and the $y$ coordinate decreases by $0.01$ . The plane gives an estimate of $- 2 + 10 (0.02) - 8 (- 0.01) = - 1.72$ . The function itself gives about $- 1.7114$ at that point.

Learn mathKrista KingAugust 6, 2020math, learn online, online course, online math, calculus iii, calc 3, calc iii, partial derivatives, linear approximation, multivariable, multivariable functions, multivariable calculus, multivariate, multivariate equations, multivariate calculus, linear approximation equation, calculus 3

How to find a linear approximation in two variables

Build the linear approximation from two slopes

Find the height at the base point

Find the slopes in the x and y directions

Put the height and the slopes together

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