Respuesta :
[tex]\text{We have been given that}\\ z = x^2 - xy + 8y^2\\ dx=\bigtriangleup x= 1.04-1=0.04\\ dy= \bigtriangleup y = -1.05+1= -0.05\\ \text{Now, we find the partial derivatives with respect to x and y}\\ \frac{\partial z}{\partial x}= 2x-y\\ \\ \frac{\partial z}{\partial y}=-x+16y\\ \\ \text{Thus, we have}\\ dz=\left ( \frac{\partial z}{\partial x} \right )dx+\left ( \frac{\partial z}{\partial y} \right )dy\\ \\ dz=(2x-y)dx+(-x+16y)dy\\ \\ dz=(2\cdot 1+1)0.04 +(-1-16)(-0.05)\\ \\ dz=0.97\\[/tex]
[tex]\\ z(1,-1)=1+1+8=10\\ z(1.04,-1.05)= 10.9936\\ \text{Thus, we have}\\ \bigtriangleup z = z(1.04,-1.05)- z(1,-1)\\ \\ \bigtriangleup z = 10.9936- 10\\ \\ \bigtriangleup z= 0.9936[/tex]
On comparing the value of δz is 0.99 and dz is 0.97.
Given
If z = x2 − xy + 8y2 and (x, y) changes from (1, −1) to (1.04, −1.05),
Differentiation;
The instantaneous rate of change of a function with respect to another quantity is called differentiation.
To compare both the function first differentiate the function on both sides with respect to x.
[tex]\rm dx=\triangle x=x_2-x_1=1.04-1=0.04\\\\dy=\triangle y =y_2-y_1=-1.05-(-1)=-1.05+1=-0.05[/tex]
Then,
[tex]\rm z = x^2 - xy + 8y^2\\\\[/tex]
[tex]\rm \dfrac{\delta z}{dx}=3x-y\\\\\dfrac{\delta z}{dy}=-x+16y[/tex]
On comparing the values;
[tex]\rm dz=\dfrac{\delta z}{\delta x} dx+\dfrac{\delta z}{\delta y} dy\\\\dz=(2x-1)dx+(-x+16y)dy\\\\dz=(2(1)+1)0.04+(-1-16)(-0.05)\\\\dz=(2+1)0.04+(-17)(-0.05)\\\\dz=3(0.04)+0.85\\\\dz=0.12+0.85\\\\dz=0.97[/tex]
And the value of z at point (1, -1) is 10 and at (1.04, -1.05) is 10.99.
Therefore,
The value of δz is;
= 10.99 - 10 = 0.99
Hence, the value of δz is 0.99 and dz is 0.97.
To know more about differentiation click the link given below.
https://brainly.com/question/7141828
