Problem 16 - Calculus for Biology and Medicine

Question

Show that \(f(x, y)\) is differentiable at the indicated point. \(f(x, y)=\left(x^{2}+y^{2}\right) e^{-x^{2}-y^{2}} ;(1,1)\)

Step-by-Step Solution

Answer

Yes, \(f(x, y)\) is differentiable at \((1,1)\).

Step 1: Define Differentiability for Two Variables

A function \(f(x, y)\) is differentiable at a point \((a, b)\) if it has partial derivatives at \((a, b)\), and the linear approximation \( f(x, y) \approx f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \) holds when \((x, y)\) is near \((a, b)\). A function is differentiable at \((a, b)\) if the limit \[ \lim_{(h, k) \to (0, 0)} \frac{f(a + h, b + k) - f(a, b) - f_x(a, b)h - f_y(a, b)k}{\sqrt{h^2 + k^2}} = 0 \] exists.

Step 2: Find Partial Derivatives

Calculate \(f_x(x, y)\) and \(f_y(x, y)\) using the product and chain rules. The partial derivative with respect to \(x\) is \[ f_x(x, y) = \frac{\partial}{\partial x} \left((x^2 + y^2) e^{-x^2 - y^2}\right) = 2xe^{-x^2-y^2} - 2x(x^2 + y^2)e^{-x^2 - y^2} \]Simplifying, we get, \[ f_x(x, y) = 2x e^{-x^2-y^2}(1 - x^2 - y^2) \]Similarly, for \(y\): \[ f_y(x, y) = \frac{\partial}{\partial y} \left((x^2 + y^2) e^{-x^2 - y^2}\right) = 2ye^{-x^2-y^2} - 2y(x^2 + y^2)e^{-x^2 - y^2} \]Simplifying, we get, \[ f_y(x, y) = 2y e^{-x^2-y^2}(1 - x^2 - y^2) \]

Step 3: Evaluate the Partial Derivatives at (1, 1)

Substitute \((x, y) = (1, 1)\) into \(f_x(x, y)\) and \(f_y(x, y)\): \[ f_x(1, 1) = 2(1) e^{-1 - 1}(1 - 1 - 1) = 0 \]\[ f_y(1, 1) = 2(1) e^{-1 - 1}(1 - 1 - 1) = 0 \]

Step 4: Use the Differentiability Criterion

As shown in Step 1, we need to check if the limit \[ \lim_{(h, k) \to (0, 0)} \frac{f(1 + h, 1 + k) - f(1, 1) - f_x(1, 1)h - f_y(1, 1)k}{\sqrt{h^2 + k^2}} = 0 \] exists.Since \(f_x(1, 1) = 0\) and \(f_y(1, 1) = 0\), this simplifies to \[ \lim_{(h, k) \to (0, 0)} \frac{f(1 + h, 1 + k) - f(1, 1)}{\sqrt{h^2 + k^2}} = 0 \]

Step 5: Calculate \(f(1, 1)\) and \(f(1+h, 1+k)\)

Evaluate \(f(x, y)\) at \((1, 1)\): \[ f(1, 1) = (1^2 + 1^2) e^{-1^2 - 1^2} = 2e^{-2} \]For \((1+h, 1+k)\), use Taylor expansion: \[ f(1+h, 1+k) = ((1+h)^2 + (1+k)^2)e^{-(1+h)^2 - (1+k)^2} \approx (2 + 2h + 2k)e^{-2} + \text{higher-order terms} \] Higher-order terms vanish faster than \(\sqrt{h^2 + k^2}\).

Step 6: Evaluate the Limit

Substitute the function values into the differentiability condition: \[ \lim_{(h, k) \to (0, 0)} \frac{(2 + 2h + 2k)e^{-2} - 2e^{-2}}{\sqrt{h^2 + k^2}} = \lim_{(h, k) \to (0, 0)} \frac{2he^{-2} + 2ke^{-2}}{\sqrt{h^2 + k^2}} = 0 \]Since the numerator becomes negligible when compared to \(\sqrt{h^2 + k^2}\), the limit is 0, proving differentiability.

Key Concepts

Partial Derivatives

Imagine you have a surface described by a function of two variables, like a smoothly sloping hillside. When we talk about partial derivatives, we're considering how the elevation of this surface changes as we move in specific directions, along the x-axis and the y-axis separately.

The partial derivative with respect to x, denoted as \(f_x(x, y)\), measures how the function changes as only the x variable is varied, keeping y constant. Similarly, \(f_y(x, y)\) measures changes of the function as only y varies.

• **Product Rule:** Since the given function \(f(x, y) = (x^2 + y^2)e^{-x^2-y^2}\) is a product of two expressions, we use the product rule to find its derivative.
• **Chain Rule:** This rule helps when differentiating the exponential term \(e^{-x^2-y^2}\), since it involves another layer of functions.

These rules allow us to find the rate of change for the function in each direction. Once we have these derivatives, we can evaluate them at a point, for instance at (1,1), to see how the function behaves locally as we vary x or y.

Limits and Continuity

In calculus, particularly in multivariable calculus, limits help us understand the behavior of a function as it approaches a certain point. For a function to be continuous at a point, the function must be unbroken as it nears that point.

When demonstrating differentiability, we are actually checking for a specific type of limit condition. The criterion requires that the difference between the actual function value and a linear approximation becomes negligible as we get very close to the point.

• This involves calculating a limit where both horizontal and vertical changes \((h, k)\) go to zero.
• The notation \(\lim_{(h, k) \to (0, 0)}\) denotes that we're considering the function's behavior in a multidirectional approach.

Continuity ensures that a function doesn't jump or skip in values. However, being continuous doesn't necessarily mean a function is differentiable. Differentiability implies a higher level of smoothness, akin to a road without potholes or bumps.

Multivariable Calculus

Multivariable calculus extends the principles of differentiation and integration to functions with more than one variable, such as \(f(x, y)\). Dealing with multiple variables allows for a richer analysis of how functions behave in a space rather than along a single line.

When determining differentiability, it's not only about finding derivatives but also understanding their significance in the context of a space formed by multiple dimensions.

• **Visualization:** Imagine the function as a shape in three-dimensional space, where any movement results in a change both in direction and value.
• **Concept of Multidirectional Approach:** Here, we need to ensure the linear approximation works as it should in all directions around a point, making the calculation of limits crucial.

Multivariable calculus equips us with tools to explore fields like fluid dynamics, economics, and more, through the thorough analysis of systems influenced by several variables. Each derivative and limit taken gives us glimpses into the behavior of complex surfaces, helping break down intricate real-world phenomena into understandable units.