Approximating function change Let z = f(x, y) =x² + y?Approximate the change in z when the independent variables change from (-1,2) to(-0.93, 1.94).

given: z=fx,y=5x2+y2 the independent variables change from (-1,2) to (-0.93,1.94)

Answered: Approximating function change Let z =…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.6: Variation

Problem 33E

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Question

Approximating function change Let z = f(x, y) =
x² + y?
Approximate the change in z when the independent variables change from (-1,2) to
(-0.93, 1.94).

Expert Solution

Step 1

given:

$z = f (x, y) = \frac{5}{x^{2} + y^{2}}$

the independent variables change from (-1,2) to (-0.93,1.94)

Step 2

$\begin{array}{rcl} \frac{\partial z}{\partial x} & = & \frac{- 5}{{(x^{2} + y^{2})}^{2}} (2 x) \\ = & \frac{- 10 x}{{(x^{2} + y^{2})}^{2}} \end{array}$

when x=-1 and y=2

$\begin{array}{rcl} \frac{\partial z}{\partial x} & = & \frac{- 10 (- 1)}{{({(- 1)}^{2} + 2^{2})}^{2}} \\ = & \frac{10}{25} \\ = & \frac{2}{5} \\ = & 0.4 \end{array}$

$\begin{array}{rcl} \frac{\partial z}{\partial y} & = & \frac{- 5}{{(x^{2} + y^{2})}^{2}} (2 y) \\ = & \frac{- 10 y}{{(x^{2} + y^{2})}^{2}} \end{array}$

when x=-1 and y=2

$\begin{array}{rcl} \frac{\partial z}{\partial y} & = & \frac{- 10 (2)}{{({(- 1)}^{2} + {(2)}^{2})}^{2}} \\ = & \frac{- 20}{25} \\ = & - \frac{4}{5} \\ = & - 0.8 \end{array}$