Problem 14 - Applied Calculus

Question

$$\text { If } f(x, y)=x^{3}+3 y^{2}, \text { find } f(1,2), f_{x}(1,2), f_{y}(1,2)$$

Step-by-Step Solution

Answer

\( f(1, 2) = 13 \), \( f_x(1, 2) = 3 \), \( f_y(1, 2) = 12 \).

Step 1: Evaluate the function at given point

We are given the function \( f(x, y) = x^3 + 3y^2 \). To find \( f(1, 2) \), substitute \( x = 1 \) and \( y = 2 \) into the function: \( f(1, 2) = 1^3 + 3(2)^2 \). Calculate the terms: \( 1^3 = 1 \) and \( 3(2)^2 = 12 \). Add these results to get \( f(1, 2) = 1 + 12 = 13 \).

Step 2: Compute the partial derivative with respect to x

To find \( f_x(x, y) \), take the partial derivative of \( f(x, y) = x^3 + 3y^2 \) with respect to \( x \). The derivative of \( x^3 \) with respect to \( x \) is \( 3x^2 \), and the derivative of \( 3y^2 \) with respect to \( x \) is 0 because it is a constant when treating \( y \) as constant. Thus, \( f_x(x, y) = 3x^2 \). Substitute \( x = 1 \) and \( y = 2 \) to find \( f_x(1, 2) = 3(1)^2 = 3 \).

Step 3: Compute the partial derivative with respect to y

To find \( f_y(x, y) \), take the partial derivative of \( f(x, y) = x^3 + 3y^2 \) with respect to \( y \). The derivative of \( x^3 \) with respect to \( y \) is 0, and the derivative of \( 3y^2 \) with respect to \( y \) is \( 6y \). So, \( f_y(x, y) = 6y \). Substitute \( x = 1 \) and \( y = 2 \) to find \( f_y(1, 2) = 6(2) = 12 \).

Key Concepts

Understanding Multivariable Calculus

Multivariable calculus involves the study of functions with more than one input. It's like regular calculus, but we deal with more variables.
In this problem, we are working with a function, \( f(x, y) = x^3 + 3y^2 \), which has two variables: \( x \) and \( y \).

• This function can represent a surface in three-dimensional space, akin to a landscape with hills and valleys.
• Every point on this surface corresponds to specific \( x \) and \( y \) values, which determine the height \( f(x, y) \).

To deeply understand how multivariable functions behave, we calculate partial derivatives. These show how changes in one variable affect the function, while keeping the other variable constant.

Mastering Function Evaluation

Function evaluation is a core concept in calculus and involves finding the value of a function at a particular point.
This process is pretty straightforward once you know the inputs. For the function \( f(x, y) = x^3 + 3y^2 \), finding \( f(1, 2) \) requires substituting \( x = 1 \) and \( y = 2 \) into the function.

• Substitute the values: \( f(1, 2) = 1^3 + 3(2)^2 \).
• Calculate: \( 1^3 = 1 \) and \( 3(2)^2 = 12 \).
• Add the results: \( f(1, 2) = 1 + 12 = 13 \).

Evaluating functions like this helps to understand the specific behavior and output at given points.

Exploring Derivatives and Partial Derivatives

Derivatives measure the rate at which a function changes with respect to one of its variables.
Partial derivatives are like regular derivatives but are used when dealing with multivariable functions.

• With respect to \( x \): Here, we calculate \( f_x(x, y) \) while keeping \( y \) constant. For our function, the partial derivative with respect to \( x \) is \( 3x^2 \).
• With respect to \( y \): For \( f_y(x, y) \), we hold \( x \) constant. In our example, the partial derivative becomes \( 6y \).

To find the partial derivatives at the point (1, 2), we substitute the values into the derivatives:

• \( f_x(1, 2) = 3(1)^2 = 3 \)
• \( f_y(1, 2) = 6(2) = 12 \)

Understanding partial derivatives lets us analyze how small changes in \( x \) or \( y \) will affect the function's value.