Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the gradient \(\nabla f\). $$ f(x, y)=x^{3} y-y^{3} $$

Step-by-Step Solution

Answer

The gradient \( \nabla f \) is \( (3x^2 y, x^3 - 3y^2) \).

Step 1: Identify the Function

We are given the function \( f(x, y) = x^3 y - y^3 \). Our task is to find its gradient \( abla f \). The gradient of a function in two variables, say \(f(x, y)\), is a vector of partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

Step 2: Calculate the Partial Derivative with respect to \(x\)

To find \( \frac{\partial f}{\partial x} \), we differentiate the function \( f(x, y) = x^3 y - y^3 \) with respect to \( x \), treating \( y \) as a constant. The derivative of \( x^3 y \) with respect to \( x \) is \( 3x^2 y \), and the derivative of \( -y^3 \) with respect to \( x \) is \( 0 \) since it does not contain \( x \). Hence, \( \frac{\partial f}{\partial x} = 3x^2 y \).

Step 3: Calculate the Partial Derivative with respect to \(y\)

To find \( \frac{\partial f}{\partial y} \), we differentiate the function \( f(x, y) = x^3 y - y^3 \) with respect to \( y \), treating \( x \) as a constant.The derivative of \( x^3 y \) with respect to \( y \) is \( x^3 \), and the derivative of \( -y^3 \) with respect to \( y \) is \( -3y^2 \). Therefore, \( \frac{\partial f}{\partial y} = x^3 - 3y^2 \).

Step 4: Write the Gradient Vector

The gradient \( abla f \) is the vector composed of the partial derivatives we calculated: \[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left( 3x^2 y, x^3 - 3y^2 \right) \].

Key Concepts

Partial Derivatives

Partial derivatives are like regular derivatives, but they focus on multivariable functions. When you have a function with more than one variable, such as \( f(x, y) = x^3 y - y^3 \), you can figure out how this function changes with respect to one variable while keeping the other variables constant. In our case, we calculate the partial derivative with respect to \( x \) and another with respect to \( y \). This helps us understand how small changes in \( x \) or \( y \) affect the function.

The partial derivative \( \frac{\partial f}{\partial x} \) is determined by treating \( y \) as a constant and differentiating the function concerning \( x \). Conversely, \( \frac{\partial f}{\partial y} \) is determined by treating \( x \) as a constant and differentiating with respect to \( y \). Calculating both gives you an insight into how the function behaves as each variable independently changes.

Multivariable Calculus

Multivariable calculus beautifully expands the concepts you learn in single-variable calculus into functions of several variables. Functions such as \( f(x, y) = x^3 y - y^3 \) rely on an understanding of multivariable calculus to understand their behavior over different dimensions.

Here, functions have inputs in vector form instead of scalar numbers. This change complicates things a bit as we need to consider various variables simultaneously. We make use of tools like gradients, partial derivatives, and multiple integrals to handle analysis on these functions.

• Calculating area under surfaces or evaluating rates of change in every direction calls for working within multivariable calculus.
• Understanding of certain physical phenomena like heat distribution can be modeled better with multivariable functions.

As you delve deeper, you'll see bridges to applications in physics, engineering, and beyond!

Vector Calculus

Vector calculus adds another layer by combining vector analysis with calculus. It is particularly useful when dealing with quantities that have both magnitude and direction, like gradients.

In our exercise, we explored the gradient of a multivariable function. The gradient \( abla f = \left( 3x^2 y, x^3 - 3y^2 \right) \) forms a vector. This gradient vector indicates the direction in which the function \( f \) increases most rapidly.

• The gradient acts perpendicular to level curves (or surfaces in higher dimensions), giving insight into how the function changes.
• Calculating the gradient is an essential step in optimization problems, where finding the maxima or minima of functions is crucial.

Vector calculus helps streamline these analyses, offering methods to work through more complex scenarios where directionality in two or more dimensions matters.