Problem 11 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The function is differentiable at the point (1,1) since the limit of the difference over the distance approaches 0.

Step 1: Evaluate the partial derivatives

To determine differentiability at a point, we first compute the partial derivatives of the function at that point. The partial derivative of \(f\) with respect to \(x\) is \(f_x(x, y) = 1\), and the partial derivative with respect to \(y\) is \(f_y(x, y) = 2y\). Evaluating at the point \((1, 1)\), we have \(f_x(1, 1) = 1\) and \(f_y(1, 1) = 2\).

Step 2: Compute the linear approximation

The linear approximation of \(f(x, y)\) at \((1, 1)\) is given by the formula: \(L(x, y) = f(1, 1) + f_x(1, 1)(x-1) + f_y(1, 1)(y-1)\). Since \(f(1, 1) = 1 + 1^2 = 2\), the linear approximation becomes: \(L(x, y) = 2 + 1(x-1) + 2(y-1)\). Simplifying, we find: \(L(x, y) = x + 2y - 2\).

Step 3: Calculate the limit for differentiability

For differentiability, the limit of \(\frac{f(x, y) - L(x, y)}{\sqrt{(x-1)^2 + (y-1)^2}}\) as \((x, y)\) approaches \((1, 1)\) must be 0. We compute \(f(x, y) - L(x, y) = (x + y^2) - (x + 2y - 2) = y^2 - 2y + 2\). Substituting back, we check: \(\frac{y^2 - 2y + 2}{\sqrt{(x-1)^2 + (y-1)^2}}\). As \((x, y)\) approaches \((1, 1)\), this expression approaches 0, confirming differentiability.

Key Concepts

Partial Derivatives

Understanding partial derivatives is crucial when dealing with functions of multiple variables. Imagine you have a multivariable function like \( f(x, y) = x + y^2 \). Partial derivatives measure the rate at which the function changes as you vary one variable while keeping the other fixed.

• For the partial derivative with respect to \( x \), you differentiate as if \( y \) is a constant. In our example, this simplifies to \( f_x(x, y) = 1 \), because the derivative of \( x \) is 1 and the derivative of \( y^2 \) with respect to \( x \) is 0.
• With respect to \( y \), treat \( x \) as constant, resulting in \( f_y(x, y) = 2y \), since \( x \) vanishes and the derivative of \( y^2 \) is \( 2y \).

To ascertain differentiability at point \((1, 1)\), these calculations are necessary starting points. At this point, \( f_x(1, 1) = 1 \) and \( f_y(1, 1) = 2 \), giving insight into how each variable influences the function's behavior at that specific location.

Linear Approximation

Linear approximation is a technique that allows us to estimate the value of a function near a particular point using its derivatives. It's like creating a tangent line, but in a multivariable context, it becomes a tangent plane. The formula for creating this plane is: \[L(x, y) = f(1, 1) + f_x(1, 1)(x - 1) + f_y(1, 1)(y - 1) \]For the function \( f(x, y) = x + y^2 \), and point \((1, 1)\), we found earlier that \( f(1, 1) = 2 \). Therefore, substituting the values yields:\[L(x, y) = 2 + 1(x - 1) + 2(y - 1) = x + 2y - 2\]This equation represents a flat plane that closely fits the function's surface near \((1, 1)\). Linear approximation simplifies complex surfaces, helping to visualize and calculate function changes nearby.

Differentiability Limit

The concept of differentiability revolves around a limit. For a function to be differentiable at a point, the function must closely resemble its linear approximation there. Specifically, the difference between the function and its linear approximation must shrink faster than the distance between points. Mathematically, this is captured by the expression:\[\frac{f(x, y) - L(x, y)}{\sqrt{(x-1)^2 + (y-1)^2}}\]We concluded previously that:\[ f(x, y) - L(x, y) = y^2 - 2y + 2 \]This expression needs to approximate 0 as \((x, y)\) approaches \((1,1)\). Simplifying reveals it indeed approaches zero, confirming differentiability.In simple terms, as you get closer to the point under consideration, the function almost behaves like its tangent plane, affirming the smoothness at that specific point.