Question

Find the function values. \(f(x, y)=3 x y+y^{2}\) (a) \(f(x+\Delta x, y)\) (b) \(\frac{f(x, y+\Delta y)-f(x, y)}{\Delta y}\)

Short answer

(a) The value of the function at \(f(x + \Delta x, y)\) is \(3xy + 3\Delta xy + y^2\). (b) The value of \(\[(f(x, y + \Delta y) - f(x, y))/\Delta y\]\) is \(3x + 2y + \Delta y\).

Step-by-step solution

Compute for f(x + Δx, y)

Substitute \(x + \Delta x\) for 'x' into the original function: \(f(x + \Delta x, y) = 3(x + \Delta x)y + y^2 = 3xy + 3\Delta xy + y^2\)

Compute for f(x, y + Δy)

Substitute \(y + \Delta y\) for 'y' into the original function: \(f(x, y + \Delta y) = 3x(y + \Delta y) + (y + \Delta y)^2 = 3xy + 3x\Delta y + y^2 + 2y\Delta y + (\Delta y)^2\)

Subtract f(x, y) from f(x, y + Δy)

Subtract the original function, \(f(x, y)\), from the answer in Step 2. This results in: \(\Delta f = f(x, y + \Delta y) - f(x, y) = 3x\Delta y + 2y\Delta y + (\Delta y)^2\)

Compute for (f(x, y + Δy) - f(x, y))/Δy

We divide \(\Delta f\) by \(\Delta y\) which gives the answer for part (b): \(\[(f(x, y + \Delta y) - f(x, y))/\Delta y\] = 3x + 2y + \Delta y\]

Key concepts

Partial Differentiation

Understanding partial differentiation is essential when dealing with functions that have more than one variable, much like the one in our exercise, (f(x, y) = 3xy + y^2). Unlike regular differentiation, which deals with functions of a single variable, partial differentiation involves taking the derivative of a function with respect to one variable while keeping the others constant.

In the context of our exercise, if we wanted to find the rate of change of f with respect to x, while keeping y constant, we would partially differentiate f with respect to x. The notation for this derivative is typically ∂f/∂x. Likewise, to find the rate of change with respect to y, we'd use ∂f/∂y. These derivatives represent the slopes of the tangent line to the surface defined by the function, in directions parallel to the x- and y-axis, respectively.
The step-by-step solution for computing f(x + Δx, y) and the difference quotient (f(x, y + Δy) - f(x, y))/Δy starts to approach the concept of partial derivatives as Δx and Δy become very small, thus resembling the process of finding the gradient of the function at a specific point.

Function Values Calculation

The process of function values calculation is instrumental in evaluating a function at a certain point or along a specific path in its domain. This is especially true for functions of two or more variables, like the given function f(x, y).

Calculating function values often starts with the substitution of given values into the function. In the exercise, we see this when x is replaced by (x + Δx) and when y is replaced by (y + Δy), allowing us to evaluate the function's value at a new point. This is crucial in various applied fields, such as physics and economics, where the ability to predict outcomes based on variable changes is needed.

The resulting expressions from the solution, having 3xy + 3Δxy + y^2 and 3xy + 3xΔy + y^2 + 2yΔy + (Δy)^2, give us the new function values after the specified changes in x and y have been applied. By understanding how to effectively perform these substitutions and calculations, we can predict how small changes in variables affect the overall function value.

Limit of a Function

The concept of the limit of a function plays a fundamental role in calculus, especially in understanding how a function behaves as the input approaches a particular value. Limits form the foundation on which the entire structure of calculus is built, particularly in the definition of derivatives and integrals.

In the difference quotient (f(x, y + Δy) - f(x, y))/Δy, as Δy approaches zero, we are effectively examining the limit of the difference quotient. This gives us the instantaneous rate of change of the function with respect to y, which is the essence of taking a partial derivative. The exercise touches on this concept by dividing the difference in function values by Δy, hinting at the limit process but without taking the final step of letting Δy approach zero.

Understanding limits not only enables the computation of derivatives but also allows us to explore continuous and discrete functions' behavior at points of interest, thereby deeply understanding their characteristics and how they can be applied to real-world scenarios.