Find the linear approximation of the function f(x,y) = In(x - 4y) at (9,2) and use it to...

Question:

Find the linear approximation of the function f(x,y) = In(x - 4y) at (9,2) and use it to approximate f(9.1,1.95).

Linear Approximation

For a function of two variables z=f(x,y), points near {eq}({x_0},{y_0}) {/eq} points near the point can be approximated using the tangent plane at that point.

Hence,

{eq}L({x_0},{y_0}) = {f_x}({x_0},{y_0})(x - {x_0}) + {f_y}({x_0},{y_0})(y - {y_0}) {/eq}

Answer and Explanation:

Given, the function is

{eq}f(x,y)=ln(x-4y) {/eq}

Differentiating f(x,y) partially with respect to x and y, we get

{eq}\eqalign{ & \frac{{\delta f}}{{\delta x}} = \frac{1}{{x - 4y}}\left( 1 \right) \cr & \frac{{\delta f}}{{\delta x}} = \frac{1}{{x - 4y}}\left( { - 4} \right) \cr} {/eq}

Using linear approximation formula and taking point {eq}({x_0},{y_0}) = (9,2) {/eq}

At (9,2)

{eq}\eqalign{ & {f_x}(9,2) = \frac{1}{{9 - 4(2)}} = 1 \cr & {f_y}(9,2) = \frac{{( - 4)}}{{9 - 4(2)}} = - 4 \cr} {/eq}

Hence, linear approximation at (9,2) becomes,

{eq}\eqalign{ & L(x,y) = 1(x - 9) - 4(y - 2) \cr & L(x,y) = x - 9 - 4y + 8 \cr & L(x,y) = x - 4y - 1 \cr} {/eq}

Putting, x= 9.1 and y=1.95, we get

{eq}\eqalign{ & L(x,y) = x - 4y - 1 \cr & L = 9.1 - 4(1.95) - 1 \cr & L = 0.3 \cr} {/eq}

Hence, f(9.1,1.95) is approximately equal to 0.3


Learn more about this topic:

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Linear Approximations Using Differentials: Definition & Examples

from GRE Math: Study Guide & Test Prep

Chapter 6 / Lesson 4
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