Problem 21 - Calculus for Biology and Medicine

Question

Find the gradient of each function. $$ f(x, y)=\exp \left[\sqrt{x^{2}+y^{2}}\right] $$

Step-by-Step Solution

Answer

The gradient is \( \nabla f = \left( \frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}} \right) \cdot \exp \left( \sqrt{x^2 + y^2} \right) \).

Step 1: Understand the Problem

We need to find the gradient of the function \( f(x, y) = \exp \left[ \sqrt{x^2 + y^2} \right] \). The gradient is a vector that contains partial derivatives of the function with respect to \( x \) and \( y \).

Step 2: Find Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), apply the chain rule: \( \frac{\partial f}{\partial x} = \exp \left( \sqrt{x^2 + y^2} \right) \cdot \frac{d}{dx} \sqrt{x^2 + y^2} \). Calculate \( \frac{d}{dx} \sqrt{x^2 + y^2} = \frac{x}{\sqrt{x^2 + y^2}} \). So, \( \frac{\partial f}{\partial x} = \frac{x \cdot \exp \left( \sqrt{x^2 + y^2} \right)}{\sqrt{x^2 + y^2}} \).

Step 3: Find Partial Derivative with Respect to y

Now find \( \frac{\partial f}{\partial y} \) using the chain rule: \( \frac{\partial f}{\partial y} = \exp \left( \sqrt{x^2 + y^2} \right) \cdot \frac{d}{dy} \sqrt{x^2 + y^2} \). Here, \( \frac{d}{dy} \sqrt{x^2 + y^2} = \frac{y}{\sqrt{x^2 + y^2}} \). Thus, \( \frac{\partial f}{\partial y} = \frac{y \cdot \exp \left( \sqrt{x^2 + y^2} \right)}{\sqrt{x^2 + y^2}} \).

Step 4: Form the Gradient Vector

The gradient of the function \( f(x, y) \) is the vector of the partial derivatives found in the previous steps. Therefore, the gradient is given by:\[abla f = \left( \frac{x \cdot \exp \left( \sqrt{x^2 + y^2} \right)}{\sqrt{x^2 + y^2}}, \frac{y \cdot \exp \left( \sqrt{x^2 + y^2} \right)}{\sqrt{x^2 + y^2}} \right)\]

Key Concepts

Partial Derivatives Made Simple

In multivariable calculus, understanding partial derivatives is a key step in finding the gradient of a function. A partial derivative represents how a multivariable function changes as one variable changes, keeping other variables constant. In our example, the function is \( f(x, y) = \exp(\sqrt{x^2 + y^2}) \).
By calculating partial derivatives, we determine how the function \( f \) changes with respect to each variable, \( x \) and \( y \).

• To find \( \frac{\partial f}{\partial x} \), look at how \( f \) changes as \( x \) alone varies.
• For \( \frac{\partial f}{\partial y} \), consider changes in \( f \) with \( y \) held variable.

The result is a depiction of sensitivity along each axis, providing foundational insight into the function's behavior. Calculating each by applying relevant rules, like the chain rule, we eventually build towards forming the gradient vector.

The Chain Rule and Its Magic

The chain rule is an invaluable tool in calculus, especially when taking derivatives of composite functions. This rule provides a method to differentiate a function that is nested within another function. In the exercise, the chain rule is critical in finding our partial derivatives.
The function involves an exponential function composed with a square root, \( \exp(\sqrt{x^2 + y^2}) \). As embedded functions appear, the chain rule guides us by proposing how to differentiate the inner and outer functions systematically.

• For \( \frac{\partial f}{\partial x} \), apply the chain rule: derive \( \exp \) with respect to the square root, then the square root concerning \( x \).
• Similarly, for \( \frac{\partial f}{\partial y} \), first differentiate the entire expression and then focus inside.

The chain rule ensures each layer of complexity is managed, giving us both the tools and a structured approach to tackling derivative problems involving multistep functions.

Exploring Multivariable Calculus and Gradients

Multivariable calculus extends the concepts of calculus into higher dimensions, allowing for the exploration of functions of several variables. Here, we consider functions like \( f(x, y) \) and how they change over two variables. This exercise asked us to focus on finding the gradient of a function.
The gradient is a vector formed by the partial derivatives of a function, indicating the direction and rate of its steepest increase. It plays a crucial role in physics and engineering, optimizing functions and navigating geometry fields.

• The gradient of \( f(x, y) \) is given as a vector: \( abla f = \left( \frac{x \cdot \exp(\sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}}, \frac{y \cdot \exp(\sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}} \right) \).
• In context, it tells us how \( f \) changes most rapidly, bringing together the insights gained from partial derivatives.

Recognizing how gradients function in multivariable calculus helps solve complex problems, from finding directional flow in fluids to adjusting parameters in operations research.