Problem 7 - Calculus and its applications

Question

Find \(f_{x}, f_{y}, f_{x}(-2,1),\) and \(f_{y}(-3,-2)\). $$f(x, y)=\sqrt{x^{2}+y^{2}}$$

Step-by-Step Solution

Answer

\( f_x(x, y) = \frac{x}{\sqrt{x^2 + y^2}}, f_y(x, y) = \frac{y}{\sqrt{x^2 + y^2}}, f_x(-2,1) = -\frac{2}{\sqrt{5}}, f_y(-3,-2) = -\frac{2}{\sqrt{13}} \).

Step 1: Find partial derivative with respect to x (f_x)

To find the partial derivative of the function \( f(x, y) = \sqrt{x^2 + y^2} \) with respect to \( x \), apply the chain rule. Let \( u = x^2 + y^2 \), then \( f(x, y) = \sqrt{u} \). The derivative of \( \sqrt{u} \) with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \). Thus, the partial derivative with respect to \( x \) is \( \frac{d}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \). Here, \( \frac{du}{dx} = 2x \), so \( f_x = \frac{x}{\sqrt{x^2 + y^2}} \).

Step 2: Find partial derivative with respect to y (f_y)

Similarly, find the partial derivative of \( f(x, y) = \sqrt{x^2 + y^2} \) with respect to \( y \). Again, using the chain rule: \( \frac{du}{dy} = 2y \), so \( f_y = \frac{d}{dy}\sqrt{u} = \frac{1}{2\sqrt{u}} \cdot 2y = \frac{y}{\sqrt{x^2 + y^2}} \).

Step 3: Evaluate \( f_x(-2, 1) \)

Using the expression \( f_x = \frac{x}{\sqrt{x^2 + y^2}} \), substitute \( x = -2 \) and \( y = 1 \): \( f_x(-2, 1) = \frac{-2}{\sqrt{(-2)^2 + 1^2}} = \frac{-2}{\sqrt{4 + 1}} = \frac{-2}{\sqrt{5}} = -\frac{2}{\sqrt{5}} \).

Step 4: Evaluate \( f_y(-3, -2) \)

Using the expression \( f_y = \frac{y}{\sqrt{x^2 + y^2}} \), substitute \( x = -3 \) and \( y = -2 \): \( f_y(-3, -2) = \frac{-2}{\sqrt{(-3)^2 + (-2)^2}} = \frac{-2}{\sqrt{9 + 4}} = \frac{-2}{\sqrt{13}} = -\frac{2}{\sqrt{13}} \).

Key Concepts

Chain Rule

The chain rule is a fundamental principle in calculus used to find the derivative of composite functions. Consider you have a function composed of an inner and an outer function. In the case of the exercise, the function \(f(x, y) = \sqrt{x^2 + y^2}\) can be seen as a composite function where \(u = x^2 + y^2\) is the inner part and \(\sqrt{u}\) is the outer part.

Using the chain rule, the derivative of the outer function with respect to the inner part is multiplied with the derivative of the inner function to obtain the final result for the partial derivative.

• For \(f_x\), set \(u = x^2 + y^2\), and find \(\frac{du}{dx} = 2x\).
• The derivative of the outer function \(\sqrt{u}\) with respect to \(u\) is \(\frac{1}{2\sqrt{u}}\).
• Thus, \(f_x = \frac{1}{2\sqrt{u}} \cdot 2x = \frac{x}{\sqrt{x^2 + y^2}}\).

This method allows tackling complex functions by breaking them down into manageable parts.

Multivariable Calculus

Multivariable calculus extends principles of calculus to functions of several variables, like \(x\) and \(y\). It introduces new concepts such as partial derivatives, which involve differentiating a function with respect to one variable while holding others constant.

In the given example, \(f(x, y) = \sqrt{x^2 + y^2}\) is a function of two variables, and finding individual derivatives \(f_x\) and \(f_y\) illustrates the concept of partial differentiation.

• To find \(f_x\), treat \(y\) as a constant and differentiate with respect to \(x\).
• To find \(f_y\), treat \(x\) as a constant and differentiate with respect to \(y\).

This approach is crucial in analyzing systems where multiple factors influence outcomes. With multiple variables, calculus helps in understanding how changes in one element affect the overall system.

Function Evaluation

Function evaluation involves calculating the value of a function at specific points. This process uses the expressions derived for partial derivatives. For example, evaluating \(f_x(-2, 1)\) and \(f_y(-3, -2)\) helps us understand the behavior of the function at those particular points.

Here's how to approach it:

• For \(f_x(-2, 1)\), substitute \(x = -2\) and \(y = 1\) into \(f_x = \frac{x}{\sqrt{x^2 + y^2}}\).
• Perform the arithmetic: \(f_x(-2, 1) = \frac{-2}{\sqrt{(-2)^2 + 1^2}} = \frac{-2}{\sqrt{5}}\).
• Similarly, for \(f_y(-3, -2)\), substitute \(x = -3\) and \(y = -2\) into \(f_y = \frac{y}{\sqrt{x^2 + y^2}}\).
• Perform the arithmetic: \(f_y(-3, -2) = \frac{-2}{\sqrt{9 + 4}} = \frac{-2}{\sqrt{13}}\).

Such evaluations provide valuable insights into the function’s rate of change at certain points, often used in practical scenarios to predict behavior and influence decision-making.