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Chapter 9: Problem 68

The function \(z\) is defined by $$ z(x, y)=x^{2} y-3 y $$ Find \(\Delta z\) and \(\mathrm{d} z\) when \(x=4, y=3, \Delta x=-0.01\) and \(\Delta y=0.02\)

Short Answer

Expert verified
\( \Delta z = 0.079502 \) and \( \mathrm{d} z = 0.02 \).

Step by step solution

01

Substitute into the Function

First, calculate the value of the function at \( (x, y) = (4, 3) \). Substitute these values into the function: \[ z(4, 3) = 4^2 \times 3 - 3 \times 3 = 48 - 9 = 39. \]
02

Calculate \( z(x + \Delta x, y + \Delta y) \)

Calculate the value of \( z \) at the new point \( (x + \Delta x, y + \Delta y) = (4 + (-0.01), 3 + 0.02) = (3.99, 3.02) \). Substitute these into the function: \[ z(3.99, 3.02) = (3.99)^2 \times 3.02 - 3 \times 3.02. \]Compute each step:\[ (3.99)^2 = 15.9601, \]\[ 15.9601 \times 3.02 = 48.139502, \]\[ 3 \times 3.02 = 9.06, \]\[ z(3.99, 3.02) = 48.139502 - 9.06 = 39.079502. \]
03

Calculate \( \Delta z \)

\( \Delta z \) is the change in \( z \): \[ \Delta z = z(3.99, 3.02) - z(4, 3) = 39.079502 - 39 = 0.079502. \]
04

Find the Partial Derivatives

Compute the partial derivatives of \( z \) with respect to \( x \) and \( y \).\[ \frac{\partial z}{\partial x} = 2xy, \]\[ \frac{\partial z}{\partial y} = x^2 - 3. \]
05

Evaluate the Partial Derivatives

Substitute \( x = 4 \) and \( y = 3 \) into the partial derivatives:\[ \frac{\partial z}{\partial x} (4, 3) = 2 \times 4 \times 3 = 24, \]\[ \frac{\partial z}{\partial y} (4, 3) = 4^2 - 3 = 16 - 3 = 13. \]
06

Calculate the Differential \( \mathrm{d}z \)

Calculate \( \mathrm{d} z \) using the formula:\[ \mathrm{d} z = \frac{\partial z}{\partial x} \cdot \Delta x + \frac{\partial z}{\partial y} \cdot \Delta y. \]Substitute the known values:\[ \mathrm{d} z = 24 \cdot (-0.01) + 13 \cdot 0.02 = -0.24 + 0.26 = 0.02. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is basically putting specific values into a function and calculating the result. For the function \( z(x, y) = x^2 y - 3y \), we begin by finding \( z \) at a certain point. This involves substituting given values of \( x \) and \( y \) into the function.
For example, when \( x = 4 \) and \( y = 3 \), the calculation is straightforward:
  • First, calculate \( x^2 \): \( 4^2 = 16 \).
  • Multiply by \( y \): \( 16 \times 3 = 48 \).
  • Subtract \( 3 \times 3 = 9 \) from 48.
This gives us a \( z \) value of 39 at \( (x, y) = (4, 3) \). Function evaluation is an essential first step in understanding how a function behaves at different input values.
Differential Calculus
Differential calculus is all about understanding how functions change. In this context, we use partial derivatives to see how changes in \( x \) and \( y \) affect \( z \).
The process involves:
  • Calculating \( \frac{\partial z}{\partial x} = 2xy \), which shows how \( z \) changes when only \( x \) changes, while \( y \) stays constant.
  • Calculating \( \frac{\partial z}{\partial y} = x^2 - 3 \), which tells us how \( z \) changes when \( y \) changes, with \( x \) constant.
To find these derivatives at specific points, such as \( x = 4 \) and \( y = 3 \), we substitute these values into the formulas:
  • \( \frac{\partial z}{\partial x} = 2 \times 4 \times 3 = 24 \).
  • \( \frac{\partial z}{\partial y} = 4^2 - 3 = 13 \).
These calculations provide insight into how each variable contributes to the change in the function's value, which is crucial for predicting behavior under small changes.
Change in Functions
The change in functions is about observing how a small tweak in inputs affects the output. In mathematics, this involves calculating "\( \Delta z \)" and "\( \mathrm{d}z \)", which represent the change and the differential, respectively.
Starting with \( \Delta z \):
  • Compute the function at both the original point and the slightly changed point, like from \( (4, 3) \) to \( (3.99, 3.02) \).
  • Find the difference between the two: \( \Delta z = z(3.99, 3.02) - z(4, 3) \).
  • This will give a concrete change value, such as 0.079502, which indicates how much \( z \) changes due to \( \Delta x \) and \( \Delta y \).
For the differential \( \mathrm{d}z \):
  • Use partial derivatives to estimate the change: \( \mathrm{d} z = \frac{\partial z}{\partial x} \cdot \Delta x + \frac{\partial z}{\partial y} \cdot \Delta y \).
  • Substitute known values like \( \Delta x = -0.01 \) and \( \Delta y = 0.02 \) into this formula.
  • Calculate to find \( \mathrm{d}z = 0.02 \), an approximation of the change in \( z \).
Understanding these concepts shows how differential calculus helps in predicting changes, ensuring precision in mathematics and real-world applications.

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