Problem 4 - The Calculus with Analytic Geometry

Question

If \(f(x, y, z)=x^{2} y+2 x y z-z^{3}, \Delta x=0.01, \Delta y=0.03\), and \(\Delta z=-0.01\), find (a) the increment of \(f\) at \((-3,0,2)\) and (b) the total differential of \(f\) at \((-3,0,2)\).

Step-by-Step Solution

Answer

The increment of \(f\) at \((-3,0,2)\) is \(0.03\). The total differential is also \(0.03\).

Step 1: Find the partial derivatives of the function

Compute the partial derivatives of the given function with respect to each variable. \[\frac{\partial f}{\partial x} = 2xy + 2yz, \quad \frac{\partial f}{\partial y} = x^2 + 2xz, \quad \frac{\partial f}{\partial z} = 2xy - 3z^2\]

Step 2: Evaluate the partial derivatives at the given point

Substitute \((-3, 0, 2)\) into the partial derivatives obtained in Step 1.\[\frac{\partial f}{\partial x} \bigg|_{(-3,0,2)} = 2(-3)(0) + 2(0)(2) = 0,\quad\frac{\partial f}{\partial y} \bigg|_{(-3,0,2)} = (-3)^2 + 2(-3)(2) = 9 - 12 = -3,\quad\frac{\partial f}{\partial z} \bigg|_{(-3,0,2)} = 2(-3)(0) - 3(2^2) = -12\]

Step 3: Use the partial derivatives to find the total differential

The total differential is given by\[df = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \frac{\partial f}{\partial z} \Delta z\]Substitute the values from Step 2 and the increments \(\Delta x=0.01\), \(\Delta y=0.03\), and \(\Delta z=-0.01\).\[df = 0 \times 0.01 + (-3) \times 0.03 + (-12) \times (-0.01) = -0.09 + 0.12 = 0.03\]

Step 4: Find the increment of f

The increment of \(f\) is the total differential calculated in Step 3. Hence, the increment is \(0.03\).

Key Concepts

total differential

The total differential is a way to approximate the change in a multivariable function based on small changes in its input variables. For a function \(f(x, y, z)\), the total differential \(df\) is given by the formula:
\[\text{df} = \frac{\text{d} f}{\text{d} x} \text{d} x + \frac{\text{d} f}{\text{d} y} \text{d} y + \frac{\text{d} f}{\text{d} z} \text{d} z \]
This formula tells us how a small change in each variable (\text{d} x, \text{d} y, \text{d} z) contributes to the overall change in the function \(f\).

increment of function

The increment of the function measures the change in the value of the function when the input variables are changed by small amounts (\text{d} x, \text{d} y, \text{d} z). In our problem, for the function \(f(x, y, z) = x^2y + 2xyz - z^3\), the increments of the variables are given as \(\Delta x = 0.01\), \(\Delta y = 0.03\), and \(\Delta z = -0.01\). To find the increment of the function, we first compute the total differential and then evaluate it at the given point to get the incremental change.

evaluating partial derivatives

Evaluating partial derivatives is essential for finding both the total differential and the increment of a function. A partial derivative measures how the function changes as one variable changes while keeping the others constant.
In our exercise, the partial derivatives of the given function \(f(x, y, z) \) with respect to \(x\), \(y\), and \(z\) were computed as follows:
\[\frac{\text{d} f}{\text{d} x} = 2xy + 2yz, \frac{\text{d} f}{\text{d} y} = x^2 + 2xz, \frac{\text{d} f}{\text{d} z} = 2xy - 3z^2 \]
Next, these partial derivatives were evaluated at the point \((-3, 0, 2)\) to get:
\[\frac{\text{d} f}{\text{d} x} = 0, \frac{\text{d} f}{\text{d} y} = -3, \frac{\text{d} f}{\text{d} z} = -12 \]
Finally, the total differential combined with the increments gives us the increment of the function, which was found to be \(0.03\).