Problem 9 - The Calculus with Analytic Geometry

Question

Given \(f(x, y)= \begin{cases}x+y-2 & \text { if } x=1 \text { or } y=1 \\ 2 & \text { if } x \neq 1 \text { and } y \neq 1\end{cases}\) Prove that \(D_{1} f(1,1)\) and \(D_{2} f(1,1)\) exist, but \(f\) is not differentiable at \((1,1)\).

Step-by-Step Solution

Answer

Partial derivatives exist; however, the limit for differentiability does not equal zero, implying the function is not differentiable at \((1,1)\).

Step 1 - Define Partial Derivatives

The partial derivatives of a function at a point give the rates of change in each direction. To find \(D_1 f(1,1)\) and \(D_2 f(1,1)\), compute the partial derivatives \(f_x(x,y)\) and \(f_y(x,y)\).

Step 2 - Compute \(D_1 f(1,1)\)

Compute the partial derivative with respect to \(x\) at \((1,1)\). For \(x = 1\), the function simplifies to \(f(x,y) = x + y - 2\). Therefore, \(f_x(x,y) = \frac{d}{dx}(x + y - 2) = 1\). Thus, \(D_1 f(1,1) = 1\).

Step 3 - Compute \(D_2 f(1,1)\)

Compute the partial derivative with respect to \(y\) at \((1,1)\). For \(y = 1\), the function simplifies to \(f(x,y) = x + y - 2\). Therefore, \(f_y(x,y) = \frac{d}{dy}(x + y - 2) = 1\). Thus, \(D_2 f(1,1) = 1\).

Step 4 - Check Differentiability

A function is differentiable at a point if it can be well-approximated by a linear function near that point. Compute the limit that defines differentiability. If \(f\) were differentiable at \((1,1)\), the limit \( \lim_{(h,k) \to (0,0)} \frac{f(1+h,1+k) - f(1,1) - D_1 f(1,1) h - D_2 f(1,1) k}{\sqrt{h^2 + k^2}} \) must be 0. Compute this limit.

Step 5 - Analyze the Limit for Differentiability

For the function given, note that when \(x eq 1\) and \(y eq 1\), \(f(1+h,1+k)=2\). Thus, \[ \frac{f(1+h,1+k) - f(1,1) - D_1 f(1,1) h - D_2 f(1,1) k}{\sqrt{h^2 + k^2}} = \frac{2 - 0 - 1h - 1k}{\sqrt{h^2 + k^2}} = \frac{2-h-k}{\sqrt{h^2+k^2}} \]. For this limit to be zero, the numerator must approach 0, but as \(h, k \rightarrow 0\), \(2 - h - k\) does not approach 0. Therefore, the limit does not equal 0, indicating that \(f\) is not differentiable at \((1,1)\).

Key Concepts

Differentiability

In calculus, differentiability defines whether a function has a derivative at each point in its domain. A function is differentiable at a point if it is locally well-approximated by a linear function near that point. This means that the function can be represented as a tangent plane in multivariable calculus. However, differentiability goes beyond just having partial derivatives at a point. It requires that these partial derivatives are continuous.

For the exercise problem, the partial derivatives at \( (1,1) \) exist, but the function is not differentiable because the limit defining differentiability does not exist. Specifically, the function must satisfy:
\[ \lim_{(h,k) \to (0,0)} \frac{f(1+h, 1+k) - f(1,1) - D_1 f(1,1) h - D_2 f(1,1) k}{\sqrt{h^2 + k^2}} = 0 \]

In this case, this limit does not go to zero, hence proving that the function is not differentiable at \( (1,1) \).

Limits

Limits are a fundamental concept in calculus and crucial for understanding continuity, derivatives, and integrals. A limit describes the value that a function approaches as the input approaches a certain point. For a function \( f(x, y) \), the limit as \( (x, y) \) approaches \( (a, b) \) is written as \(\lim_{(x,y) \to (a,b)} f(x, y)\)

In the context of the original exercise, we investigate the limit to determine the differentiability of function \( f \). Here, the limit:
\[ \lim_{(h,k) \to (0,0)} \frac{2 - h - k}{\sqrt{h^2 + k^2}} \]

does not approach zero as \( (h, k) \) approaches zero. This is evident because even though \( h \) and \( k \) get smaller, the numerator does not diminish to zero. Hence, this particular limit helps prove that \( f \) isn’t differentiable at \( (1,1) \).

Analytic Geometry

Analytic geometry, also known as coordinate geometry, involves geometric principles using a coordinate system. It allows the study of geometric figures through algebraic equations and can help visualize concepts in multivariable calculus.

In our example, the function \( f(x, y) \) behaves differently depending on whether \( x \) or \( y \) is equal to 1 or not. When visualizing this function in the \( xy \)-plane, there are distinct behaviors depending on the regions defined by \( x = 1 \) and \( y = 1 \).

This visualization aids in understanding why the function is not differentiable at \( (1,1) \). When approaching \( (1,1) \) from different directions, the function values and slopes change inconsistently, stopping the formation of a single tangent plane. This makes limits and derivatives more tangible, helping link algebraic and geometric interpretations.