Problem 21 - Calculus with Concepts in Calculus

Question

Find the first partial derivatives of \(f\) at the given point. $$ f(x, y)=\sqrt{4 x^{2}+y^{2}} ;(2,-3) $$

Step-by-Step Solution

Answer

The partial derivatives at (2,-3) are \( \frac{\partial f}{\partial x} = \frac{8}{5} \) and \( \frac{\partial f}{\partial y} = \frac{-3}{5} \).

Step 1: Understanding the Partial Derivative

To find the partial derivatives of a function with two variables, we differentiate the function with respect to each variable, treating the other variable as a constant. For our function \( f(x, y) = \sqrt{4x^2 + y^2} \), this means deriving \( \frac{\partial f}{\partial x} \) while keeping \( y \) constant, and \( \frac{\partial f}{\partial y} \) while keeping \( x \) constant.

Step 2: Finding \( \frac{\partial f}{\partial x} \)

To find \( \frac{\partial f}{\partial x} \), use the chain rule. Start by setting \( u = 4x^2 + y^2 \), so \( f(x, y) = \sqrt{u} \). Derive \( \sqrt{u} \) with respect to \( u \), getting \( \frac{1}{2\sqrt{u}} \). Then, derive \( u \) with respect to \( x \), giving \( 8x \). Thus, \( \frac{\partial f}{\partial x} = \frac{8x}{2\sqrt{4x^2 + y^2}} = \frac{4x}{\sqrt{4x^2 + y^2}} \).

Step 3: Finding \( \frac{\partial f}{\partial y} \)

To find \( \frac{\partial f}{\partial y} \), follow a similar process. Use the chain rule with the same \( u = 4x^2 + y^2 \). Derive \( \sqrt{u} \) with respect to \( u \) to get \( \frac{1}{2\sqrt{u}} \). Then, derive \( u \) with respect to \( y \), giving \( 2y \). Thus, \( \frac{\partial f}{\partial y} = \frac{2y}{2\sqrt{4x^2 + y^2}} = \frac{y}{\sqrt{4x^2 + y^2}} \).

Step 4: Substituting the Values of the Point (2,-3)

Now, substitute \( x = 2 \) and \( y = -3 \) into the partial derivatives obtained. For \( \frac{\partial f}{\partial x} \), substitute to obtain \( \frac{4(2)}{\sqrt{4(2)^2 + (-3)^2}} = \frac{8}{\sqrt{16 + 9}} = \frac{8}{5} \). For \( \frac{\partial f}{\partial y} \), substitute to get \( \frac{-3}{\sqrt{4(2)^2 + (-3)^2}} = \frac{-3}{\sqrt{25}} = \frac{-3}{5} \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus for dealing with derivatives of composed functions. It allows us to break down complex derivatives into manageable pieces. In our exercise, we're dealing with a function, \( f(x, y) = \sqrt{4x^2 + y^2} \). To use the chain rule, we first identify an inner function \( u \), such that \( u = 4x^2 + y^2 \), and an outer function, \( \sqrt{u} \).

This approach simplifies the differentiation process:

• Differentiate the outer function \( \sqrt{u} \) with respect to \( u \), yielding \( \frac{1}{2\sqrt{u}} \).
• Next, differentiate the inner function \( u \) with respect to the variable of interest (either \( x \) or \( y \)).

Combining these results using the chain rule, we effectively find the partial derivatives needed. Remember, each step in the chain rule provides a crucial link to solving the differentiation problem.

Partial Derivative with Respect to x

To find the partial derivative of a function with respect to \( x \), you hold \( y \) constant and differentiate the function as if it were a single-variable function in terms of \( x \). In our context for \( f(x, y) = \sqrt{4x^2 + y^2} \), we apply the chain rule.First, substitute \( u = 4x^2 + y^2 \), and consider \( f(x, y) \) as \( \sqrt{u} \). Differentiate \( \sqrt{u} \) with respect to \( u \) to get \( \frac{1}{2\sqrt{u}} \).

Next, differentiate \( u \) with respect to \( x \), which results in \( 8x \). Multiply these derivatives: \( \frac{8x}{2\sqrt{u}} = \frac{4x}{\sqrt{4x^2 + y^2}} \).

This expression, \( \frac{4x}{\sqrt{4x^2 + y^2}} \), is the partial derivative of \( f \) with respect to \( x \). It reflects the rate of change in \( f \) as \( x \) changes, while \( y \) stays constant.

Partial Derivative with Respect to y

Finding the partial derivative with respect to \( y \) involves differentiating while keeping \( x \) constant. Using our function \( f(x, y) = \sqrt{4x^2 + y^2} \), follow a similar process as before by using the chain rule. Here, set \( u = 4x^2 + y^2 \).

Again, differentiate the outer function \( \sqrt{u} \) with respect to \( u \), resulting in \( \frac{1}{2\sqrt{u}} \). Then focus on \( u \) with respect to \( y \), which gives \( 2y \).

Combining both derivatives yields \( \frac{2y}{2\sqrt{u}} = \frac{y}{\sqrt{4x^2 + y^2}} \).

This expression captures how \( f \) changes if \( y \) shifts, with \( x \) remaining unchanged. Each derivation leads us closer to understanding how variables interact within functions of two variables.

Substitution of Values

Once you've determined the partial derivatives for a function, the next step is often to evaluate these derivatives at a specific point to gain insights into the function's behavior at that location. For \( f(x, y) = \sqrt{4x^2 + y^2} \), we need to substitute \((x, y) = (2, -3)\) into the partial derivatives.

Substituting these values into the partial derivative with respect to \( x \):

• \( \frac{4(2)}{\sqrt{4(2)^2 + (-3)^2}} = \frac{8}{\sqrt{16 + 9}} = \frac{8}{5} \)

For the partial derivative with respect to \( y \):

• \( \frac{-3}{\sqrt{4(2)^2 + (-3)^2}} = \frac{-3}{\sqrt{25}} = \frac{-3}{5} \)

Substitution not only provides numerical values that describe the slope of the tangent plane to the surface defined by \( f \), but also helps in understanding practical applications and geometrical interpretations.

Function of Two Variables

A function of two variables, such as \( f(x, y) = \sqrt{4x^2 + y^2} \), depends on two independent variables (here, \( x \) and \( y \)). Such functions can represent complex systems like surfaces or thermal maps, reflecting how one quantity changes relative to two others.

The function values can be visualized in 3D space:

• The input pair \( (x, y) \) determines a unique point or height on the surface.
• The "form" of the function - that is, how it behaves across all input points - offers insights into the dynamic interactions between \( x \) and \( y \).

This mathematical model underpins various disciplines, including physics, economics, and engineering, offering a practical framework for studying multi-dimensional changes and optimizing real-world phenomena.