Sure! Let's go through the solution step-by-step.
Given the function [tex]\( z = f(x, y) = x^2 + y^3 \)[/tex], we need to solve two parts:
a) Use differentials to estimate [tex]\(\Delta z\)[/tex].
b) Find [tex]\(\Delta z\)[/tex] by evaluating [tex]\( f(x + \Delta x, y + \Delta y) - f(x, y) \)[/tex].
Let's start with part (a):
### Part (a)
1. Calculate the partial derivatives of [tex]\( f \)[/tex] with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x^2 + y^3) = 2x
\][/tex]
[tex]\[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (x^2 + y^3) = 3y^2
\][/tex]
2. Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = 3 \)[/tex] into the partial derivatives:
[tex]\[
\frac{\partial f}{\partial x} \bigg|_{(5, 3)} = 2 \times 5 = 10
\][/tex]
[tex]\[
\frac{\partial f}{\partial y} \bigg|_{(5, 3)} = 3 \times 3^2 = 3 \times 9 = 27
\][/tex]
3. Use differentials to estimate [tex]\(\Delta z\)[/tex]:
[tex]\[
\Delta z \approx \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y
\][/tex]
[tex]\[
\Delta z \approx 10 \times 0.01 + 27 \times 0.03
\][/tex]
[tex]\[
\Delta z \approx 0.1 + 0.81 = 0.91
\][/tex]
So, the estimated value of [tex]\(\Delta z\)[/tex] is [tex]\(0.9100\)[/tex].
### Part (b)
1. Calculate [tex]\( f(x + \Delta x, y + \Delta y) \)[/tex]:
[tex]\[
f(x + \Delta x, y + \Delta y) = f(5 + 0.01, 3 + 0.03) = f(5.01, 3.03)
\][/tex]
[tex]\[
f(5.01, 3.03) = (5.01)^2 + (3.03)^3
\][/tex]
2. Calculate [tex]\( f(x, y) \)[/tex]:
[tex]\[
f(5, 3) = 5^2 + 3^3 = 25 + 27 = 52
\][/tex]
3. Calculate the change [tex]\(\Delta z\)[/tex]:
[tex]\[
\Delta z = f(5.01, 3.03) - f(5, 3)
\][/tex]
4. Compute the function value at the new point:
[tex]\[
(5.01)^2 = 25.1001
\][/tex]
[tex]\[
(3.03)^3 \approx 27.8181
\][/tex]
[tex]\[
f(5.01, 3.03) = 25.1001 + 27.8181 = 52.9182
\][/tex]
5. Calculate the actual [tex]\(\Delta z\)[/tex]:
[tex]\[
\Delta z = 52.9182 - 52 = 0.9182
\][/tex]
So, the actual value of [tex]\(\Delta z\)[/tex] is [tex]\(0.9182\)[/tex].
To summarize:
a) The estimated value is [tex]\(\Delta z = 0.9100\)[/tex]
b) The actual value is [tex]\(\Delta z = 0.9182\)[/tex]
These values are rounded to four decimal places as needed.
