Answer :
To solve the given problem, we need to compare the change in the value of the function [tex]\( z \)[/tex] and the differential change in [tex]\( z \)[/tex].
Given the function:
[tex]\[ z = x^2 - xy + 3y^2 \][/tex]
and the points:
- Initial point: [tex]\( (x_0, y_0) = (3, -1) \)[/tex]
- Final point: [tex]\( (x_1, y_1) = (2.96, -0.95) \)[/tex]
We will perform the following steps:
1. Calculate [tex]\( z \)[/tex] at the initial and final points.
2. Calculate [tex]\(\Delta z\)[/tex] (the actual change in [tex]\( z \)[/tex]).
3. Calculate the partial derivatives of [tex]\( z \)[/tex] with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex] at the initial point.
4. Calculate the differentials [tex]\( dx \)[/tex] and [tex]\( dy \)[/tex].
5. Calculate [tex]\( dz \)[/tex] (the differential change in [tex]\( z \)[/tex]).
6. Compare [tex]\(\Delta z\)[/tex] and [tex]\( dz \)[/tex].
### Step 1: Calculate [tex]\( z \)[/tex] at the initial and final points
For the initial point [tex]\( (3, -1) \)[/tex]:
[tex]\[ z(3, -1) = 3^2 - 3(-1) + 3(-1)^2 = 9 + 3 + 3 = 15 \][/tex]
For the final point [tex]\( (2.96, -0.95) \)[/tex]:
[tex]\[ z(2.96, -0.95) = (2.96)^2 - 2.96(-0.95) + 3(-0.95)^2 \][/tex]
Calculating each term separately:
[tex]\[ (2.96)^2 = 8.7616 \][/tex]
[tex]\[ - 2.96(-0.95) = 2.96 \times 0.95 = 2.812 \][/tex]
[tex]\[ 3(-0.95)^2 = 3 \times 0.9025 = 2.7075 \][/tex]
Adding these together:
[tex]\[ 8.7616 + 2.812 + 2.7075 = 14.2811 \][/tex]
So, we get:
[tex]\[ z(2.96, -0.95) = 14.2811 \][/tex]
### Step 2: Calculate [tex]\(\Delta z\)[/tex] (the actual change in [tex]\( z \)[/tex]):
[tex]\[ \Delta z = z(2.96, -0.95) - z(3, -1) = 14.2811 - 15 = -0.7189 \][/tex]
### Step 3: Calculate the partial derivatives at the initial point [tex]\( (3, -1) \)[/tex]:
[tex]\[ \frac{\partial z}{\partial x} = 2x - y \][/tex]
[tex]\[ \frac{\partial z}{\partial y} = -x + 6y \][/tex]
At [tex]\( (3, -1) \)[/tex]:
[tex]\[ \frac{\partial z}{\partial x} \bigg|_{(3,-1)} = 2(3) - (-1) = 6 + 1 = 7 \][/tex]
[tex]\[ \frac{\partial z}{\partial y} \bigg|_{(3,-1)} = -3 + 6(-1) = -3 - 6 = -9 \][/tex]
### Step 4: Calculate the differentials [tex]\( dx \)[/tex] and [tex]\( dy \)[/tex]:
[tex]\[ dx = x_1 - x_0 = 2.96 - 3 = -0.04 \][/tex]
[tex]\[ dy = y_1 - y_0 = -0.95 + 1 = 0.05 \][/tex]
### Step 5: Calculate [tex]\( dz \)[/tex] (the differential change in [tex]\( z \)[/tex]):
[tex]\[ dz = \frac{\partial z}{\partial x} \bigg|_{(3,-1)} dx + \frac{\partial z}{\partial y} \bigg|_{(3,-1)} dy \][/tex]
Substituting the values:
[tex]\[ dz = 7(-0.04) + (-9)(0.05) \][/tex]
[tex]\[ dz = -0.28 - 0.45 = -0.73 \][/tex]
### Step 6: Compare [tex]\(\Delta z\)[/tex] and [tex]\( dz \)[/tex]:
The actual change in [tex]\( z \)[/tex]:
[tex]\[ \Delta z = -0.7189 \][/tex]
The differential change in [tex]\( z \)[/tex]:
[tex]\[ dz = -0.73 \][/tex]
These values are quite close, which verifies that the differential [tex]\( dz \)[/tex] is a good approximation for the actual change [tex]\(\Delta z\)[/tex] when the changes in [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are small.
To summarize:
- [tex]\(\Delta z = -0.7189\)[/tex]
- [tex]\(dz = -0.73\)[/tex]
Both calculations show similar values indicating the accuracy of the differential approximation for small changes in the variables.
Given the function:
[tex]\[ z = x^2 - xy + 3y^2 \][/tex]
and the points:
- Initial point: [tex]\( (x_0, y_0) = (3, -1) \)[/tex]
- Final point: [tex]\( (x_1, y_1) = (2.96, -0.95) \)[/tex]
We will perform the following steps:
1. Calculate [tex]\( z \)[/tex] at the initial and final points.
2. Calculate [tex]\(\Delta z\)[/tex] (the actual change in [tex]\( z \)[/tex]).
3. Calculate the partial derivatives of [tex]\( z \)[/tex] with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex] at the initial point.
4. Calculate the differentials [tex]\( dx \)[/tex] and [tex]\( dy \)[/tex].
5. Calculate [tex]\( dz \)[/tex] (the differential change in [tex]\( z \)[/tex]).
6. Compare [tex]\(\Delta z\)[/tex] and [tex]\( dz \)[/tex].
### Step 1: Calculate [tex]\( z \)[/tex] at the initial and final points
For the initial point [tex]\( (3, -1) \)[/tex]:
[tex]\[ z(3, -1) = 3^2 - 3(-1) + 3(-1)^2 = 9 + 3 + 3 = 15 \][/tex]
For the final point [tex]\( (2.96, -0.95) \)[/tex]:
[tex]\[ z(2.96, -0.95) = (2.96)^2 - 2.96(-0.95) + 3(-0.95)^2 \][/tex]
Calculating each term separately:
[tex]\[ (2.96)^2 = 8.7616 \][/tex]
[tex]\[ - 2.96(-0.95) = 2.96 \times 0.95 = 2.812 \][/tex]
[tex]\[ 3(-0.95)^2 = 3 \times 0.9025 = 2.7075 \][/tex]
Adding these together:
[tex]\[ 8.7616 + 2.812 + 2.7075 = 14.2811 \][/tex]
So, we get:
[tex]\[ z(2.96, -0.95) = 14.2811 \][/tex]
### Step 2: Calculate [tex]\(\Delta z\)[/tex] (the actual change in [tex]\( z \)[/tex]):
[tex]\[ \Delta z = z(2.96, -0.95) - z(3, -1) = 14.2811 - 15 = -0.7189 \][/tex]
### Step 3: Calculate the partial derivatives at the initial point [tex]\( (3, -1) \)[/tex]:
[tex]\[ \frac{\partial z}{\partial x} = 2x - y \][/tex]
[tex]\[ \frac{\partial z}{\partial y} = -x + 6y \][/tex]
At [tex]\( (3, -1) \)[/tex]:
[tex]\[ \frac{\partial z}{\partial x} \bigg|_{(3,-1)} = 2(3) - (-1) = 6 + 1 = 7 \][/tex]
[tex]\[ \frac{\partial z}{\partial y} \bigg|_{(3,-1)} = -3 + 6(-1) = -3 - 6 = -9 \][/tex]
### Step 4: Calculate the differentials [tex]\( dx \)[/tex] and [tex]\( dy \)[/tex]:
[tex]\[ dx = x_1 - x_0 = 2.96 - 3 = -0.04 \][/tex]
[tex]\[ dy = y_1 - y_0 = -0.95 + 1 = 0.05 \][/tex]
### Step 5: Calculate [tex]\( dz \)[/tex] (the differential change in [tex]\( z \)[/tex]):
[tex]\[ dz = \frac{\partial z}{\partial x} \bigg|_{(3,-1)} dx + \frac{\partial z}{\partial y} \bigg|_{(3,-1)} dy \][/tex]
Substituting the values:
[tex]\[ dz = 7(-0.04) + (-9)(0.05) \][/tex]
[tex]\[ dz = -0.28 - 0.45 = -0.73 \][/tex]
### Step 6: Compare [tex]\(\Delta z\)[/tex] and [tex]\( dz \)[/tex]:
The actual change in [tex]\( z \)[/tex]:
[tex]\[ \Delta z = -0.7189 \][/tex]
The differential change in [tex]\( z \)[/tex]:
[tex]\[ dz = -0.73 \][/tex]
These values are quite close, which verifies that the differential [tex]\( dz \)[/tex] is a good approximation for the actual change [tex]\(\Delta z\)[/tex] when the changes in [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are small.
To summarize:
- [tex]\(\Delta z = -0.7189\)[/tex]
- [tex]\(dz = -0.73\)[/tex]
Both calculations show similar values indicating the accuracy of the differential approximation for small changes in the variables.
