Problem 17 - Calculus for Biology and Medicine

Question

Find the gradient of each function. $$ f(x, y)=x^{3} y^{2} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Gradient

The gradient of a function is a vector. For a function of two variables, such as \( f(x,y) \), the gradient \( abla f(x,y) \) consists of the partial derivatives with respect to each variable.

Step 2: Differentiate with Respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant and differentiate \( f(x,y) = x^3 y^2 \) with respect to \( x \). This results in \( 3x^2 y^2 \).

Step 3: Differentiate with Respect to y

Now, find \( \frac{\partial f}{\partial y} \) by treating \( x \) as a constant and differentiating \( f(x,y) = x^3 y^2 \) with respect to \( y \). This gives \( 2x^3 y \).

Step 4: Construct the Gradient Vector

Combine the partial derivatives from Steps 2 and 3 to form the gradient vector. Therefore, \( abla f(x,y) = \left( 3x^2 y^2 , 2x^3 y \right) \).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They let us understand how a function changes as one of its variables changes, while keeping other variables constant. This is crucial when dealing with functions that have more than one variable, such as our original function, \( f(x, y) = x^3 y^2 \). Here, the function depends on both \( x \) and \( y \).

To find the partial derivative of a function with respect to a specific variable, such as \( x \) in our function, we treat all other variables as constants. This simplifies the differentiation process. In the case of \( f(x, y) = x^3 y^2 \), to compute \( \frac{\partial f}{\partial x} \), we treat \( y^2 \) as a constant. We only differentiate \( x^3 \), resulting in \( 3x^2 y^2 \).

Next, for the partial derivative with respect to \( y \), we consider \( x^3 \) as a constant while differentiating \( y^2 \). This results in \( 2x^3 y \). Understanding these basic principles of partial derivatives helps in analyzing scenarios where a function's rate of change has to be measured in multiple dimensions.

Multivariable Calculus

Multivariable calculus expands the concepts of calculus to functions of more than one variable. This kind of calculus is essential for describing and analyzing complex systems involving several independent factors. Functions can now depend on multiple inputs, such as position coordinates \( (x, y) \) or \( (x, y, z) \).

In our example, \( f(x, y) = x^3 y^2 \), the function involves two variables, making it a suitable candidate for multivariable calculus techniques. Such functions are depicted on a plane or in space, offering more detailed analysis and power than single-variable functions.

The concept of the gradient is a key element in multivariable calculus. It provides a vector that indicates the direction and rate of steepest ascent of a function. The gradient is composed of the partial derivatives of the function.

• It points in the direction where the function increases the most.
• Its magnitude tells us how steep the slope is in that direction.

Understanding these components allows one to tackle and simplify real-world problems involving surfaces and fields.

Differentiation

Differentiation in calculus is the process of finding a derivative, representing how a function changes at any given point. It allows us to understand rates of change, which is foundational in various fields such as physics, engineering, and economics. With single-variable calculus, we differentiate a function with respect to one independent variable.

In our context of multivariable functions, differentiation adapts to involve partial derivatives. Each partial derivative shows how a function changes with respect to one particular variable, while all others are held constant. This is an extension of single-variable differentiation but applied in a multidimensional space.

For the function \( f(x, y) = x^3 y^2 \), differentiation involves finding both \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). The results, \( 3x^2 y^2 \) and \( 2x^3 y \), respectively, illustrate how each variable influences the function. By combining these, we obtain the gradient vector, which summarizes these changes into a single, comprehensive picture of the function's behavior. Understanding differentiation in this multi-variable context provides a robust framework for solving complex, real-life problems.