Problem 17 - Active Calculus Multivariable

Question

We know that if a function of a single variable is differentiable at a point, then that function is also continuous at that point. In this exercise we determine that the same property holds for functions of two variables. A function \(f\) of the two variables \(x\) and \(y\) is continuous at a point \(\left(x_{0}, y_{0}\right)\) in its domain if $$\lim _{(x, y) \rightarrow\left(x_{0}, y_{0}\right)} f(x, y)=f\left(x_{0}, y_{0}\right)$$ or (letting \(x=x_{0}+h\) and \(y=y_{0}+k\), $$\lim _{(h, k) \rightarrow(0,0)} f\left(x_{0}+h, y+k\right)=f\left(x_{0}, y_{0}\right)$$ Show that if \(f\) is differentiable at \(\left(x_{0}, y_{0}\right),\) then \(f\) is continuous at \(\left(x_{0}, y_{0}\right) .\) (Hint: Multiply both sides of the equality that comes from differentiability by \(\lim _{(h, k) \rightarrow(0,0)} \sqrt{h^{2}+k^{2}}\).)

Step-by-Step Solution

Answer

To show that if a function \(f(x, y)\) is differentiable at \((x_0, y_0)\), then it is also continuous at \((x_0, y_0)\), we start by recalling the definitions of differentiability and continuity. We multiply both sides of the differentiability equation by \(\lim_{(h, k) \rightarrow (0,0)} \sqrt{h^2 + k^2}\). We observe a connection between the definitions and then prove continuity from differentiability using the facts that the limits of the linear terms in \(h\) and \(k\) are \(0\) and that the limit of \(R(h,k)\) as \((h, k) \rightarrow (0,0)\) is also \(0\). This leads to the conclusion that \(\lim_{(x, y) \rightarrow (x_0, y_0)} f(x, y) = f(x_0, y_0)\), which means \(f\) is continuous at \((x_0, y_0)\) if it is differentiable at \((x_0, y_0)\).

Step 1: Recall the definition of differentiability

A function \(f(x, y)\) is differentiable at a point \((x_0, y_0)\) if the partial derivatives of \(f\) with respect to \(x\) and \(y\) exist at that point and $$ f(x_0+h, y_0+k) = f(x_0, y_0) + h\frac{\partial f}{\partial x}(x_0,y_0) + k\frac{\partial f}{\partial y}(x_0,y_0) + R(h,k) $$ where \(R(h,k)\) is a function that satisfies $$ \lim_{(h, k) \rightarrow(0,0)} \frac{R(h,k)}{\sqrt{h^2+k^2}} = 0 $$

Step 2: Define continuity for a function of two variables

A function \(f(x, y)\) is continuous at a point \((x_0, y_0)\) if $$\lim _{(x, y) \rightarrow\left(x_{0}, y_{0}\right)} f(x, y)=f\left(x_{0}, y_{0}\right)$$

Step 3: Use the hint to modify differentiability definition

We multiply both sides of the differentiability equation by \(\lim_{(h, k) \rightarrow (0,0)} \sqrt{h^2 + k^2}\): $$\lim_{(h, k) \rightarrow(0,0)} \sqrt{h^2 + k^2}\left[f(x_0+h, y_0+k) - f(x_0, y_0) - h\frac{\partial f}{\partial x}(x_0,y_0) - k\frac{\partial f}{\partial y}(x_0,y_0)\right] = 0$$

Step 4: Observe a connection between definitions

We observe that the expression inside the square brackets in Step 3 is equal to \(R(h, k)\). Therefore, we get: $$\lim_{(h, k) \rightarrow(0,0)} R(h, k) \sqrt{h^2 + k^2} = 0$$

Step 5: Prove continuity from differentiability

Now that we have linked the definitions, we can prove continuity from differentiability: Let \(h=x-x_0\) and \(k=y-y_0\). Then we have $$\lim_{(x, y) \rightarrow (x_0, y_0)} f(x, y) = \lim_{(h, k) \rightarrow (0,0)} f(x_0+h, y_0+k)$$ $$= f(x_0, y_0) + \lim_{(h, k) \rightarrow (0,0)} \left[h\frac{\partial f}{\partial x}(x_0,y_0) + k\frac{\partial f}{\partial y}(x_0,y_0) + R(h, k)\right]$$ Since \(h\frac{\partial f}{\partial x}(x_0,y_0)\) and \(k\frac{\partial f}{\partial y}(x_0,y_0)\) are linear in \(h\) and \(k\) respectively, the limits of these terms as \((h, k) \rightarrow (0,0)\) are \(0\). Moreover, we have $$\lim_{(h, k) \rightarrow (0,0)} R(h, k) \sqrt{h^2 + k^2} = 0$$ $$\implies \lim_{(h, k) \rightarrow (0,0)} R(h, k) = 0$$ Combining these results, we get $$\lim_{(x, y) \rightarrow (x_0, y_0)} f(x, y) = f(x_0, y_0)$$ which proves that \(f\) is continuous at \((x_0, y_0)\) if it is differentiable at \((x_0, y_0)\).

Key Concepts

Differentiability

Differentiability is a key concept in Multivariable Calculus that extends the idea of deriving a function from single to multiple variables. When a function of two variables, say \( f(x, y) \), is said to be differentiable at a point \((x_0, y_0)\), it not only means that its partial derivatives exist at that point, but it also fits a particular approximation. This is where the concept differentiates from just having partial derivatives, since differentiability encompasses a more complete condition.

Differentiability implies that the function can be approximated by a tangent plane at that point. More technically, it means that the function's behavior can be closely estimated by a linear combination of its partial derivatives plus a remainder term \( R(h,k) \) that tends to zero faster than the distance \( \sqrt{h^2 + k^2} \) as \((h, k) \rightarrow (0, 0)\).

• A function is differentiable at \((x_0, y_0)\) if:
• Partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) exist at \((x_0, y_0)\).
• The expression for \( f(x_0 + h, y_0 + k) \) holds as:

\[f(x_0 + h, y_0 + k) = f(x_0, y_0) + h\frac{\partial f}{\partial x}(x_0,y_0) + k\frac{\partial f}{\partial y}(x_0,y_0) + R(h,k)\]where \( \lim_{(h, k) \rightarrow (0,0)} \frac{R(h,k)}{\sqrt{h^2+k^2}} = 0 \). This means the remainder term becomes negligibly small relative to the components of \( h \) and \( k \) themselves.

Essentially, differentiability at a point guarantees that locally, the function behaves predictably and linearly.

Continuity

Continuity is a foundational element of calculus and applies to multivariable functions as well. A function \(f(x, y)\) is continuous at a point \((x_0, y_0)\) if its value approaches the function's value at that point as the input, \( (x, y) \), gets closer to \( (x_0, y_0) \).

In plain terms, if you approach the point from any direction in the \(x-y\) plane, the function behaves nicely without sudden jumps or breaks.

• Mathematically, this is expressed as:

\[\lim_{(x, y) \rightarrow (x_0, y_0)} f(x, y) = f(x_0, y_0)\]This elegant simplicity has profound implications, ensuring that if a function is differentiable at a point, it must also be continuous there. This relationship underscores the well-behaved nature of differentiable functions; they do not only steadily change with no abrupt jumps but also map seamlessly in a localized manner.

Thus, by proving a function differentiable at a spot, you directly show it’s continuous at the same spot too. This beautiful mathematical connection aligns concepts in a tangible way, linking the existence of derivatives to the smooth, uninterrupted flow of a graph.

Partial Derivatives

Partial derivatives help unlock the world of functions with more than one variable, much like how regular derivatives relate to single-variable functions. They measure how a multivariable function changes as you vary just one variable while keeping others constant.

For a function \(f(x, y)\), the partial derivative with respect to \(x\) is denoted by \(\frac{\partial f}{\partial x}\), and similarly for \(y\), it is \(\frac{\partial f}{\partial y}\). These derivatives provide the slope of the tangent line along the respective axes, offering pivotal insight into the function's behavior in different directions.

• To find a partial derivative, handle the other variables as constants:
• \( \frac{\partial f}{\partial x} \) describes the rate of change of \(f\) in the \(x\) direction,
• while \( \frac{\partial f}{\partial y} \) does so in the \(y\) direction.

These partial derivatives play a crucial role in determining differentiability. Not only do they help calculate linear approximations of the function at a point, but their existence is a necessary condition for differentiability. However, merely having them is not enough to guarantee differentiability, which must also satisfy specific continuity-like conditions mentioned earlier.

In the landscape of multivariable calculus, partial derivatives offer a detailed map of the function's changing nature, making them indispensable in understanding and predicting the way functions behave over their domains.