Question

If z = x2 − xy + 8y2 and (x, y) changes from (3, −1) to (2.96, −0.95), compare the values of Δz and dz. (Round your answers to four decimal places.) dz = Δz =

Answer 1

Answer:

dz = -1.23

$\Delta z$ = -1.2064

Step-by-step explanation:

$z=x^2− x y +8y^2$

$dz = \frac{\partial z}{\partial x} \; dx + \frac{\partial z}{\partial y} \; dy$

$\frac{\partial z}{\partial x} = 2x - y$

$\frac{\partial z}{\partial y} = -x + 16y$

From (3, −1) to (2.96, −0.95):

$\Delta x = dx = 2.96 - 3 = -0.04$

$\Delta y = dy = -0.95 - (-1) = 0.05$

$dz = [2(3) - (-1)] (-0.04) + [-(3) + 16(-1)] (0.05)$

$dz = -1.23$

$\Delta z = z(2.96,-0.95) - z(3,-1)$

$\Delta z = [2.96^2 - 2.96 (-0.95) + 8 (-0.95)^2] - [3^2 - 3 (-1) + 8 (-1)^2]$

$\Delta z = -1.2064$