# Use Linear Approximation to estimate the change \Delta f = f(a + \Delta x, b + \Delta y) - f(a,b)...

## Question:

Use Linear Approximation to estimate the change

{eq}\Delta f = f(a + \Delta x, b + \Delta y) - f(a,b) {/eq}

in {eq}f {/eq} when

{eq}\Delta x = -0.3, \Delta y = 0.1 {/eq},

and

{eq}f_{x}(a,b) = -4, f_{y}(a,b) = -3 {/eq}.

a. 1

b. 0.6

c. 0.8

d. 0.9

e. 0.7

## Linearization

The linearization of a function {eq}\displaystyle f(x,y) {/eq} at the point {eq}\displaystyle (x_0,y_0) {/eq} is given by

{eq}\displaystyle 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), {/eq}

where {eq}\displaystyle f_x =\frac{\partial f}{\partial x}, f_y =\frac{\partial f}{\partial y}, {/eq} are the partial derivatives of {eq}\displaystyle f(x,y), {/eq} with respect to {eq}\displaystyle x, \text{ and } y, \text{ respectively}. {/eq}

Above linearization is used in approximating he change in f, for a given change in x, {eq}\displaystyle \Delta x {/eq} and y, {eq}\displaystyle \Delta y. {/eq}

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

To estimate {eq}\displaystyle \Delta f = f(a + \Delta x, b + \Delta y) - f(a,b) {/eq} knowing {eq}\displaystyle \Delta x = -0.3, \Delta y = 0.1, f_{x}(a,b) = -4, f_{y}(a,b) = -3. {/eq}

we will use the linearization of the function, given by the tangent plane to the function at {eq}\displaystyle (a,b). {/eq}

{eq}\displaystyle \begin{align}\Delta f &= f(a + \Delta x, b + \Delta y) - f(a,b)\\ &=f_x(a,b)\Delta x+f_y(a,b)\Delta y\\ &=-4\cdot (-0.3)+(-3)\cdot 0.1\\ &=\boxed{0.9- \text{ option d)}}. \end{align} {/eq} 