Problem 1 - The Calculus with Analytic Geometry

Question

If \(f(x, y)=3 x^{2}+2 x y-y^{2}, \Delta x=0.03\), and \(\Delta y=-0.02\), find (a) the increment of \(f\) at \((1,4)\) and (b) the total differential of \(f\) at \((1,4)\).

Step-by-Step Solution

Answer

For (a) and (b), the increment and total differential of f at (1, 4) is approximately 0.54.

Step 1: Find the Partial Derivatives

To find the increment and total differential, first compute the partial derivatives of the function with respect to each variable. \[ \frac{\text{d}f}{\text{d}x} = \frac{\text{d}}{\text{d}x} (3x^2 + 2xy - y^2) = 6x + 2y \]\[ \frac{\text{d}f}{\text{d}y} = \frac{\text{d}}{\text{d}y} (3x^2 + 2xy - y^2) = 2x - 2y \]

Step 2: Evaluate Partial Derivatives at (1,4)

Now, evaluate these partial derivatives at the given point (1, 4). \[ \frac{\text{d}f}{\text{d}x}\bigg|_{(1,4)} = 6(1) + 2(4) = 6 + 8 = 14 \]\[ \frac{\text{d}f}{\text{d}y}\bigg|_{(1,4)} = 2(1) - 2(4) = 2 - 8 = -6 \]

Step 3: Compute the Total Differential

The total differential, \text{d}f, is given by: \[ \text{d}f = \frac{\text{d}f}{\text{d}x} \text{d}x + \frac{\text{d}f}{\text{d}y} \text{d}y \]Substitute the values obtained: \[ \text{d}f = (14) (0.03) + (-6) (-0.02) = 0.42 + 0.12 = 0.54 \]

Step 4: Compute the Increment of f

The increment of f, denoted as \(\frac{\text{ \. }}{\text{ \. }}f\), is approximated by the total differential. Thus, \[\Delta f \approx \text{d}f = 0.54\]

Key Concepts

Partial Derivatives

Partial derivatives help us find the rate at which a function changes with respect to one of its variables, while keeping the other variables constant.
To compute partial derivatives, we differentiate the function with respect to one variable at a time. For the given function \( f(x, y) = 3x^2 + 2xy - y^2 \), the steps are as follows:

• Differentiate with respect to \( x \): \ \ \[ \frac{\text{d} f}{\text{d} x} = \frac{\text{d}}{\text{d} x} (3x^2 + 2xy - y^2) = 6x + 2y \]


• Differentiate with respect to \( y \):
  
  \[ \frac{\text{d} f}{\text{d} y} = \frac{\text{d}}{\text{d} y} (3x^2 + 2xy - y^2) = 2x - 2y \]

These partial derivatives tell us how \( f \) changes as we change \( x \) and \( y \) individually. Understanding these rates is crucial for finding the total differential and increment.

Total Differential

The total differential, denoted as \( \text{d}f \), gives us a linear approximation of how much a function changes due to small changes in its variables.
The total differential of a function \( f(x, y) \) is calculated using its partial derivatives and small changes \( \text{d}x \) and \( \text{d}y \):
\[ \text{d}f = \frac{\text{d}f}{\text{d}x} \text{d}x + \frac{\text{d}f}{\text{d}y} \text{d}y \]
For our function at the point (1, 4) with given changes \( \text{d}x = 0.03 \) and \( \text{d}y = -0.02 \), we substitute the partial derivatives:
\[ \text{d}f = (14)(0.03) + (-6)(-0.02) = 0.42 + 0.12 = 0.54 \]
Here, \( 14 \) and \( -6 \) are the evaluated partial derivatives at (1, 4). The result, 0.54, is the total differential, representing an approximate adjustment in the function's value due to the small changes in both variables.

Increment

The increment of a function, denoted \( \Delta f \), measures the change in the function's value resulting from changes in its input variables.
In practical terms, when the changes \( \Delta x \) and \( \Delta y \) are small, the increment can be approximated by the total differential:
\[ \Delta f \approx \text{d}f \]
Thus, for the function \( f(x, y) = 3x^2 + 2xy - y^2 \) at the point (1, 4) with \( \Delta x = 0.03 \) and \( \Delta y = -0.02 \), the increment \( \Delta f \) is approximated by the calculated total differential 0.54:
\[ \Delta f \approx 0.54 \]
This tells us that the function's value increases by approximately 0.54 units due to the given small changes in \( x \) and \( y \). Recognizing the difference between a function's exact increment and its differential approximation is essential in use cases of calculus.