Problem 34 - Calculus with Concepts in Calculus

Question

Find \(f_{x x}, f_{y y}\), and \(f_{z z}\) (where applicable). $$ f(x, y)=\sqrt{16-9 x^{2}-4 y^{2}} $$

Step-by-Step Solution

Answer

The second partial derivatives are \(f_{xx}\) and \(f_{yy}\); \(f_{zz}\) doesn't exist as \(z\) is not a variable in the function.

Step 1: Identify Partial Derivatives

The first task involves identifying the function provided and recognizing that we need to find second partial derivatives. For the given function \(f(x, y)=\sqrt{16-9x^2-4y^2}\), observe that it depends on \(x\) and \(y\). There is no \(z\) variable, hence we cannot find \(f_{zz}\). We will only calculate \(f_{xx}\) and \(f_{yy}\).

Step 2: Compute First Partial Derivative with Respect to x

To find \(f_{xx}\), we first need to calculate \(f_x\). Start by using the chain rule. Let \(u = 16 - 9x^2 - 4y^2\), then \(f = \sqrt{u}\), making \(f_x = \frac{1}{2\sqrt{u}} \cdot (-18x)\). Simplifying, we have \(f_x = \frac{-9x}{\sqrt{16-9x^2-4y^2}}\).

Step 3: Compute Second Partial Derivative with Respect to x

Now, differentiate \(f_x = \frac{-9x}{\sqrt{16-9x^2-4y^2}}\) with respect to \(x\) again. You'll use the quotient rule: \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\). Find the derivatives: \(u' = -9\), \(v = \sqrt{16-9x^2-4y^2}\), and differentiate \(v\) with respect to \(x\) using the chain rule to get \(v' = \frac{-18x}{2\sqrt{16-9x^2-4y^2}}\). Substitute these back to find \(f_{xx}\). Simplifying, \(f_{xx} = \frac{-9(\sqrt{16-9x^2-4y^2}) + 9x(-18x)/(2\sqrt{16-9x^2-4y^2})}{16-9x^2-4y^2}\).

Step 4: Compute First Partial Derivative with Respect to y

Analogously, calculate \(f_y\) by differentiating the function \(f(x, y)\) with respect to \(y\) first: \(f_y = \frac{1}{2\sqrt{16-9x^2-4y^2}} \cdot (-8y)\), which simplifies to \(f_y = \frac{-4y}{\sqrt{16-9x^2-4y^2}}\).

Step 5: Compute Second Partial Derivative with Respect to y

Finally, compute \(f_{yy}\) by differentiating \(f_y = \frac{-4y}{\sqrt{16-9x^2-4y^2}}\) with respect to \(y\) using the quotient rule. Calculate \(u' = -4\), \(v = \sqrt{16-9x^2-4y^2}\), and \(v' = \frac{-8y}{2\sqrt{16-9x^2-4y^2}}\). Use the quotient rule to find \(f_{yy}\), then simplify \(f_{yy} = \frac{-4(\sqrt{16-9x^2-4y^2}) + 4y(8y)/(2\sqrt{16-9x^2-4y^2})}{16-9x^2-4y^2}\).

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept in multivariable calculus. They represent how a multivariable function changes as one of its variables is altered, while others are kept constant.
Given a function like \( f(x, y) = \sqrt{16-9x^2-4y^2} \), partial derivatives with respect to \( x \) (\( f_x \)) and \( y \) (\( f_y \)) provide insight into how the function changes when only one variable is varied.

• **For \( f_x \):** Think of it as slicing the function in the x-direction. It explores how the function height (output) changes if \( x \) is nudged by a tiny amount.
• **For \( f_y \)**: Here, you are slicing in the y-direction, checking how changes to \( y \) affect the height.

Understanding the pattern of change helps in predicting and modeling behaviors associated with the function in multidimensional spaces.

Chain Rule

When dealing with functions composed of other functions, such as \( f(x, y) = \sqrt{u} \) with \( u = 16 - 9x^2 - 4y^2 \), the chain rule is indispensable. It provide a way to differentiate composite functions.
To find the partial derivative of \( f \) with respect to \( x \), using chain rule extends the derivative process from simple functions to this setup:

• **Differentiate the outer function:** If we have \( f=\sqrt{u} \), its derivative with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \).
• **Differentiate the inner function:** Here, \( u = 16 - 9x^2 - 4y^2 \). So, \( \frac{du}{dx} = -18x \).
• **Combine:** Multiply the outer derivative by the inner derivative: \( \frac{-9x}{\sqrt{16-9x^2-4y^2}} \).

This rule simplifies the process of dealing with layers of functions and is vital for solving complicated calculus problems.

Quotient Rule

The quotient rule is essential when you need to differentiate a function that is a ratio of two other functions. In our context, after finding the first partial derivative using the chain rule, \( f_x = \frac{-9x}{\sqrt{16-9x^2-4y^2}} \), you again differentiate using the quotient rule to find \( f_{xx} \).
The rule is represented as:

• For a quotient \( \frac{u}{v} \), the derivative is \( \frac{u'v - uv'}{v^2} \).
• Here, with \( u = -9x \), \( u' = -9 \), \( v = \sqrt{16-9x^2-4y^2} \), and \( v' \) requiring another chain rule application, the result is plugged into the quotient formula.

The quotient rule helps manage calculations where division between functions complicates derivative solutions, ensuring the results are both precise and manageable.

Second Derivative

The second derivative helps us understand the curvature and concavity of functions, indicating how the rate of change itself is changing for a function. For \( f(x, y) = \sqrt{16-9x^2-4y^2} \), second partial derivatives \( f_{xx} \) and \( f_{yy} \) were calculated.

• **Concavity:** A positive second derivative indicates the function is concave up (like a smiling parabola), while a negative second derivative indicates concavity down (like a frowning parabola).
• **Interpretations:** These derivatives can help infer stability of points, such as determining local minima or maxima.

Second derivatives are crucial in optimization problems and in analyzing the behavior of graphs, giving deeper insights beyond just the immediate direction of change.