Problem 14 - Calculus for Biology and Medicine

Question

Show that \(f(x, y)\) is differentiable at the indicated point. \(f(x, y)=\sin (x-y) ;(1,0)\)

Step-by-Step Solution

Answer

\( f(x, y) = \sin(x-y) \) is differentiable at the point \((1, 0)\).

Step 1: Understand the Differentiability Condition

To show that a function is differentiable at a point, the function must be continuous at that point and the limit defining the derivative must exist. This means \:\( f(x, y) \) is differentiable at \((1, 0)\) if there exists a linear approximation such that \:\( f(x, y) = f(1,0) + f_x(1,0)(x-1) + f_y(1,0)(y-0) + R(x,y) \) \:where \( R(x,y) \) is the remainder term that tends to zero as \( (x,y) \to (1,0) \).

Step 2: Calculate the Partial Derivatives

Compute the partial derivative of \( f(x, y) = \sin(x-y) \) with respect to \( x \) and with respect to \( y \). \:\( f_x(x, y) = \cos(x-y) \) \:\( f_y(x, y) = -\cos(x-y) \).

Step 3: Evaluate the Partial Derivatives at the Point

Evaluate the partial derivatives at the given point \((1, 0)\): \:\( f_x(1, 0) = \cos(1-0) = \cos 1 \) \:\( f_y(1, 0) = -\cos(1-0) = -\cos 1 \).

Step 4: Linear Approximation

Write the linear approximation of the function \( f(x,y) \) at \((1,0)\): \:\( L(x, y) = f(1,0) + f_x(1,0)(x-1) + f_y(1,0)(y-0) \).\:\( f(1,0) = \sin(1-0) = \sin 1 \). \:\( L(x, y) = \sin 1 + \cos 1 \cdot (x-1) - \cos 1 \cdot y \).

Step 5: Show Remainder Term Tends to Zero

Consider the remainder \( R(x, y) = f(x, y) - L(x, y) \), and show it tends to 0 as \((x,y) \to (1,0)\).\:\( R(x, y) = \sin(x-y) - \sin 1 - \cos(1)(x-1) + \cos(1)y \). \:We need to show \( \left| R(x, y) \right| / \sqrt{(x-1)^2 + y^2} \to 0 \). Since the sine function and its derivatives are continuous, this condition is satisfied.

Key Concepts

Partial Derivatives

When dealing with functions of multiple variables, partial derivatives are essential for understanding how a function changes with respect to each variable individually. In this problem, we have the function \( f(x, y) = \sin(x-y) \). To find the partial derivatives, we need to differentiate the function with respect to each variable while holding the other constant.

• Partial derivative with respect to \( x \), denoted as \( f_x(x, y) \), is calculated by treating \( y \) as a constant. For \( f(x, y) = \sin(x-y) \), this gives us \( f_x(x, y) = \cos(x-y) \).
• Partial derivative with respect to \( y \), denoted as \( f_y(x, y) \), is found by treating \( x \) as a constant. This results in \( f_y(x, y) = -\cos(x-y) \).

By evaluating these derivatives at the point \((1,0)\), we can see how the function \( f \) behaves near this point, providing a foundational step for finding the linear approximation.

Linear Approximation

A linear approximation allows us to estimate the value of a function near a given point using a linear function. It's the tangent plane's equation at that point. For the function \(f(x, y) = \sin(x-y)\) at the point \((1,0)\), the linear approximation, \( L(x, y) \), takes the form of: \[ L(x, y) = f(1,0) + f_x(1,0)(x-1) + f_y(1,0)(y-0) \] To apply this formula:

• Evaluate \( f(1,0) \), which equals \( \sin(1) \).
• Substitute the calculated values of the partial derivatives at \((1,0)\): \( f_x(1,0) = \cos(1) \) and \( f_y(1,0) = -\cos(1) \).
• The linear approximation becomes: \( L(x, y) = \sin 1 + \cos 1 \cdot (x-1) - \cos 1 \cdot y \).

This equation is a local linearization of \( f \), providing a simpler representation of the function around the point \((1,0)\).

Remainder Term

The remainder term plays a crucial role in determining whether a function is differentiable at a point. After establishing a linear approximation, we define the remainder \( R(x, y) \) as the difference between the actual function \( f(x, y) \) and its linear approximation \( L(x, y) \). This can be expressed as: \[ R(x, y) = f(x, y) - L(x, y) \] In our problem, this becomes: \[ R(x, y) = \sin(x-y) - \sin 1 - \cos 1 \cdot (x-1) + \cos 1 \cdot y \]For differentiability at \((1,0)\), this remainder must tend to zero faster than the distance from \((x,y)\) to \((1,0)\) approaches zero, ensuring that: \[ \frac{|R(x, y)|}{\sqrt{(x-1)^2 + y^2}} \to 0 \] Since the sine function and its derivatives are continuous, the condition is satisfied, confirming that \( f \) is differentiable at the point.