Problem 28 - Calculus for Biology and Medicine

Question

Let $$f(x, y)=2 x^{3}-3 y x$$ Compute \(f_{x}(1,2)\) and \(f_{y}(1,2)\), and interpret these partial derivatives geometrically.

Step-by-Step Solution

Answer

\(f_x(1,2) = 0\); \(f_y(1,2) = -3\). The function is flat in the x-direction and decreases in the y-direction at (1, 2).

Step 1: Partial Derivative with Respect to x

To find the partial derivative of the function with respect to \(x\), we treat \(y\) as a constant and differentiate. Given: \[f(x, y) = 2x^3 - 3yx\]The partial derivative \(f_x\) is:\[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(2x^3 - 3yx)\]Applying the derivative rules, we get:\[\frac{\partial f}{\partial x} = 6x^2 - 3y\]Now substituting \(x = 1\) and \(y = 2\):\[f_x(1,2) = 6(1)^2 - 3(2) = 6 - 6 = 0\]

Step 2: Partial Derivative with Respect to y

To find the partial derivative with respect to \(y\), treat \(x\) as constant.Given:\[f(x, y) = 2x^3 - 3yx\]The partial derivative \(f_y\) is:\[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(2x^3 - 3yx)\]The only term involving \(y\) is \(-3yx\), so:\[\frac{\partial f}{\partial y} = -3x\]Substitute \(x = 1\) into the equation:\[f_y(1,2) = -3(1) = -3\]

Step 3: Geometrical Interpretation of Partial Derivatives

The partial derivative \(f_x(1,2) = 0\) indicates that at the point \((1, 2)\), the rate of change of the function along the x-direction is zero. This means the tangent line to the surface at this point is flat along the x-axis.The partial derivative \(f_y(1,2) = -3\) suggests that the function decreases at a rate of 3 units in the y-direction. Thus, the surface slopes downwards in the y-direction at this point.

Key Concepts

Geometrical Interpretation of Partial Derivatives

Visualizing how functions behave in a multi-dimensional space involves considering surfaces. When we calculate partial derivatives, we look at how the surface defined by a function changes as we move a little in one direction while keeping the other direction constant. In simple terms, partial derivatives tell us the slope of the curve we get if we slice through the surface along a coordinate axis.
For the function given, when we compute the partial derivative with respect to \(x\) at a specific point, it shows how steeply the surface rises or falls as we move just a tiny bit in the \(x\)-direction. Specifically in our problem, \(f_x(1,2) = 0\) shows that at point \((1,2)\), movement along the \(x\)-axis doesn't change the function's value; the surface is flat in this direction.

• Flat surface in the direction of the \(x\)-axis at point \((1, 2)\).

On the other hand, \(f_y(1,2) = -3\) means that as we move slightly in the \(y\)-direction, the function decreases at a rate of 3 units. In layman's terms, the surface tilts downwards as you head in the \(y\) direction at that point.

• Downward slope in the \(y\)-direction at point \((1, 2)\).

This geometrical interpretation makes it easier to visualize how changes in input variables impact the overall shape and behavior of the surface.

Rate of Change in Multi-Variable Functions

In mathematics, the concept of rate of change is crucial for understanding how a function's output behaves in response to varying its inputs. Particularly, in multi-variable functions, partial derivatives represent these rates of changes.
When you take the partial derivative with respect to a particular variable, you're essentially zooming in on how changing just that one variable alters the function when all other variables are kept constant. Imagine you are probing a multi-dimensional surface for inclines.
In the context of this exercise, when we calculate \(f_x(1,2) = 0\), it signifies that infinitesimal shifts in the \(x\)-value while \(y\) is fixed do not change the function's value at that spot. It's like standing on level ground, where a step forward won't change your height.

• \(f_x(1,2) = 0\) implies no change along the \(x\) direction.

Conversely, \(f_y(1,2) = -3\) illustrates a clear rate of change as you adjust \(y\), showing that for each small positive increase in \(y\), there's a corresponding decrease in \(f\)'s value, as if you're marching down a slope.

• \(f_y(1,2) = -3\) represents {change by 3 units decrease/increase in \(f\) for unit movement in \(y\).

This understanding informs us how sensitive a function is to changes in its input variables, echoing how dynamic or stable the function appears in different directions.

Differentiation Rules for Calculating Partial Derivatives

Calculating partial derivatives is an application of differentiation, but with a special focus on multi-variable functions. To compute these derivatives, we adhere to foundational differentiation rules, modifying them slightly due to the presence of several variables.
Let's delve into the differentiation rules used in calculating the partial derivatives in the given exercise:

• Product Rule: It's important when dealing with multi-variable functions, especially when variables are multiplied together. However, in this exercise, simplification is straightforward as each differentiation is carried out while considering one variable constant.
• Constant Rule: When differentiating with respect to one variable, treat the other variables as constants. This simplification outlines the difference between single and partial derivatives.

For example, to find \(\frac{\partial f}{\partial x}\) in our function, treat \(y\) as a constant, and apply normal differentiation rules to \(x\). Similarly, \(x\) is constant when differentiating with respect to \(y\).
By using these rules consistently, we simplify the process of understanding how each variable uniquely influences the function. Hence, differentiation rules act like lenses through which we can distinctly see each component's effect on the function's geometry and behavior.