Problem 26 - Calculus for Biology and Medicine

Question

Let $$f(x, y)=\sqrt{4-x^{2}-y^{2}}$$ Compute \(f_{x}(1,1)\) and \(f_{y}(1,1)\), and interpret these partial derivatives geometrically.

Step-by-Step Solution

Answer

\( f_x(1,1) = -\frac{\sqrt{2}}{2}, f_y(1,1) = -\frac{\sqrt{2}}{2} \). These represent the rates of change of \( f \) along the x- and y-axes at (1,1), respectively.

Step 1: Differentiate with respect to x

Calculate the partial derivative \( f_x(x, y) \) by differentiating \( f(x, y) = \sqrt{4 - x^2 - y^2} \) with respect to \( x \). Use the chain rule: If \( u = 4 - x^2 - y^2 \), then \( f(x, y) = \sqrt{u} = u^{1/2} \). First find the derivative of \( u^{1/2} \) with respect to \( u \), which is \( \frac{1}{2}u^{-1/2} \), and then multiply by the derivative of \( u \) with respect to \( x \), which is \(-2x\). Thus, \( f_x(x, y) = \frac{1}{2}(4 - x^2 - y^2)^{-1/2} \times (-2x) = \frac{-x}{\sqrt{4 - x^2 - y^2}} \).

Step 2: Differentiate with respect to y

Calculate the partial derivative \( f_y(x, y) \) by differentiating \( f(x, y) = \sqrt{4 - x^2 - y^2} \) with respect to \( y \). Again use the chain rule. Following the same procedure as in step 1, we multiply the outer derivative \( \frac{1}{2}(4 - x^2 - y^2)^{-1/2} \) by the inner derivative of \( -y^2 \), which gives us \(-2y\). Therefore, \( f_y(x, y) = \frac{-y}{\sqrt{4 - x^2 - y^2}} \).

Step 3: Evaluate f_x and f_y at (1, 1)

Substitute \( x = 1 \) and \( y = 1 \) into the expressions for \( f_x(x, y) \) and \( f_y(x, y) \). For \( f_x(1, 1) \), compute \( f_x(1, 1) = \frac{-1}{\sqrt{4 - 1^2 - 1^2}} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \). Similarly, for \( f_y(1, 1) \), compute \( f_y(1, 1) = \frac{-1}{\sqrt{4 - 1^2 - 1^2}} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \).

Step 4: Geometric interpretation

The partial derivatives \( f_x(1, 1) \) and \( f_y(1, 1) \) represent the rate of change of the function \( f \) in the direction of the \( x \)-axis and \( y \)-axis, respectively, when evaluated at the point (1, 1). Both values being \(-\frac{\sqrt{2}}{2}\) indicate that as we move in the positive direction along either axis from the point (1, 1), the function \( f(x, y) \) decreases at the same rate.

Key Concepts

Chain Rule in Partial Differentiation

The chain rule is a fundamental tool in calculus that helps us find derivatives when we have composite functions. In the context of partial differentiation, the chain rule allows us to differentiate a function with respect to one variable while treating another variable as constant. Consider the function given in our exercise:

• Firstly, we have \( f(x, y) = \sqrt{4 - x^2 - y^2} \), which can be rewritten in terms of a new variable \( u = 4 - x^2 - y^2 \).
• Hence, the function becomes \( f(x, y) = \sqrt{u} = u^{1/2} \).
• When we differentiate \( u^{1/2} \) with respect to \( u \), we obtain \( \frac{1}{2}u^{-1/2} \).
• We then multiply by the derivative of \( u \) with respect to \( x \), which is \(-2x\), leading to the partial derivative with respect to \( x \).

This process is analogous when taking the partial derivative with respect to \( y \). The chain rule is an elegant method that simplifies this differentiation process by breaking it down into manageable parts.

Geometric Interpretation of Partial Derivatives

The geometric interpretation of partial derivatives provides insight into how a function changes in relation to each variable. Consider the function \( f(x,y) = \sqrt{4-x^2-y^2} \). At the point (1, 1), we calculated both partial derivatives \( f_x(1,1) \) and \( f_y(1,1) \) and found them to both be \(-\frac{\sqrt{2}}{2}\). This tells us:

• The partial derivative \( f_x(1, 1) \) represents the rate at which the function \( f \) changes as we move a small distance in the direction of the \( x \)-axis, holding \( y \) constant.
• Similarly, \( f_y(1, 1) \) reveals how the function changes as we move along the \( y \)-axis, with \( x \) constant.

What's fascinating here is that both partial derivatives are equal, indicating the rate of decrease of the function is the same in both the \( x \) and the \( y \) directions at the point (1,1). This symmetry can sometimes hint at specific geometric properties of the surface represented by the function, such as level curves or contours being circular around that point.

Introduction to Partial Derivatives

Partial derivatives are a cornerstone in multivariable calculus, allowing us to understand how a function depends on multiple variables. If we have a function of multiple variables, like \( f(x, y) \), the partial derivative with respect to one variable gives us the derivative of the function while treating all other variables as constants. Here's the step-by-step insight into finding partial derivatives for our exercise:

• For \( f_x(x, y) \), differentiate with respect to \( x \) while treating \( y \) as constant. Our function transforms to \( \frac{-x}{\sqrt{4-x^2-y^2}} \).
• For \( f_y(x, y) \), differentiate with respect to \( y \), treating \( x \) as constant. This results in \( \frac{-y}{\sqrt{4-x^2-y^2}} \).
• Partial derivatives thus provide tangent lines to the graph's surface in the \( x \) and \( y \) directions.

Understanding the concept of partial derivatives is crucial as it lays the groundwork for more complex topics in calculus, such as gradients, divergence, and curl, which play pivotal roles in fields like physics and engineering.