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If z=3x²+y² and (x,y) changes from (1,3) to (1.05,2.9), compare the values of Δz and dz dz=

Δz=

User Tobiash
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On comparing the value of 8z 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.

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

Then,

z = x ^ 2 - xy + 8y ^ 2 (delta*z)/(dx) = 3x - y (delta*z)/(dy) = - x + 16y On comparing the value of 8z 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.

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

Then,

z = x ^ 2 - xy + 8y ^ 2 (delta*z)/(dx) = 3x - y (delta*z)/(dy) = - x + 16y. On comparing the values;

dz = (delta*z)/(delta*x) * dx + (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

And the value of z at point (1, -1) is 10 and at (1.04, -1.05) is 10.99.

Therefore,

The value of 6z is; = 10.99 - 10 = 0.99

Hence, the value of 8z is 0.99 and dz is 0.97.
User LMG
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