Problem 51 - Calculus an Applied Approach

Question

Find the four second partial derivatives. Observe that the second mixed partials are equal. $$ z=x^{3}-4 y^{2} $$

Step-by-Step Solution

Answer

The second order partial derivatives are \( \frac{\partial^2 z}{\partial x^2} = 6x \), \( \frac{\partial^2 z}{\partial y^2} = -8 \), \( \frac{\partial^2 z}{\partial y \partial x} = 0 \) and \( \frac{\partial^2 z}{\partial x \partial y} = 0 \). Since the mixed partial derivatives are equal to each other, Schwarz's theorem holds for this function.

Step 1: Understanding Partial Derivatives

Partial derivatives involve finding the derivative of a function with respect to one variable, while keeping the other variables constant. This gives us the rate at which the function changes with respect to that variable. The first-order partial derivatives of the given function are found as follows:

Step 2: Finding the 1st Order Partial Derivatives

The first order partial derivative with respect to x is found by differentiating \( z = x^{3} - 4y^{2} \) with respect to x, treating y as a constant. This gives us \( \frac{\partial z}{\partial x} = 3x^{2} \). Similarly, the first order derivative with respect to y is found by differentiating the same function with respect to y, treating x as a constant. This gives us \( \frac{\partial z}{\partial y} = -8y \).

Step 3: Finding the 2nd Order Partial Derivatives

The second order partial derivatives are found by differentiating the first order partial derivatives. Differentiating \( \frac{\partial z}{\partial x} = 3x^{2} \) with respect to x gives us \( \frac{\partial^2 z}{\partial x^2} = 6x \). Differentiating \( \frac{\partial z}{\partial y} = -8y \) with respect to y gives us \( \frac{\partial^2 z}{\partial y^2} = -8 \).

Step 4: Finding the Mixed Partial Derivatives

The mixed partial derivatives are found by differentiating the first order partial derivatives with respect to the other variable. Differentiating \( \frac{\partial z}{\partial x} = 3x^{2} \) with respect to y gives us \( \frac{\partial^2 z}{\partial y \partial x} = 0 \) (because \( 3x^{2} \) does not depend on y). Differentiating \( \frac{\partial z}{\partial y} = -8y \) with respect to x gives us \( \frac{\partial^2 z}{\partial x \partial y} = 0 \) (because \( -8y \) does not depend on x).

Key Concepts

Partial Derivatives

When dealing with multivariable functions, such as the given function \( z = x^3 - 4y^2 \), we need to consider partial derivatives to understand how the function changes with respect to each variable independently. A partial derivative measures the rate of change of a function concerning one specific variable while treating all other variables as constants.

Here’s what to do:

• Differentiating with respect to \( x \), we treat \( y \) as constant. Thus, the partial derivative of \( z \) with respect to \( x \) is \( \frac{\partial z}{\partial x} = 3x^2 \).
• Differentiating with respect to \( y \), we treat \( x \) as constant, which gives us the partial derivative of \( z \) with respect to \( y \) as \( \frac{\partial z}{\partial y} = -8y \).

By obtaining these expressions, we capture the intrinsic changes occurring in the function when each variable is varied individually.

Mixed Partial Derivatives

Mixed partial derivatives are a step beyond simple partial derivatives. They represent the derivatives of a multivariable function with respect to two different variables, one after another. For the function \( z = x^3 - 4y^2 \), think of mixed partial derivatives as exploring how the change in one variable influences another after some initial change.

To find these:

• First, we differentiate \( \frac{\partial z}{\partial x} = 3x^2 \) with respect to \( y \). Since \( 3x^2 \) has no \( y \) component, the derivative is zero: \( \frac{\partial^2 z}{\partial y \partial x} = 0 \).
• Next, differentiate \( \frac{\partial z}{\partial y} = -8y \) with respect to \( x \). Again, since \( -8y \) contains no \( x \) component, this derivative is zero: \( \frac{\partial^2 z}{\partial x \partial y} = 0 \).

Both mixed partial derivatives equal zero, demonstrating symmetry: \( \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} \). This symmetry is often true due to Clairaut's theorem, given that the necessary conditions for continuity are met.

First Order Partial Derivatives

First order partial derivatives give us the immediate rate of change of a function with respect to one of the variables, while all other variables remain unchanged. These derivatives are foundational because they set the stage for tackling more complex second-order derivatives.

For the function \( z = x^3 - 4y^2 \):

• The first order partial derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = 3x^2 \). This tells us how \( z \) changes as \( x \) changes, assuming \( y \) remains constant. If we imagine moving along the surface of the function in the direction of the \( x \)-axis, this derivative describes that slope.
• Similarly, the first order partial derivative with respect to \( y \) is \( \frac{\partial z}{\partial y} = -8y \), describing how \( z \) changes with \( y \), while \( x \) stays the same.

By calculating these derivatives, we gain helpful insights into the function’s behavior at any given point, which is crucial for applications in fields like optimization and physics.