Problem 5 - Calculus and its applications

Question

For \(f(x, y)=\ln x+y^{3},\) find \(f(e, 2), f\left(e^{2}, 4\right),\) and \(f\left(e^{3}, 5\right)\).

Step-by-Step Solution

Answer

9, 66, 128

Step 1: Understand the Function

We are given the function \(f(x, y) = \ln x + y^3\), which combines a natural logarithm and a cubic term. Our task is to evaluate this function at several specific points.

Step 2: Evaluate \(f(e, 2)\)

Substitute \(x = e\) and \(y = 2\) into the function. The calculation will be: \\[\ln(e) + 2^3 = 1 + 8 = 9.\] The result is \(f(e, 2) = 9\).

Step 3: Evaluate \(f(e^2, 4)\)

Substitute \(x = e^2\) and \(y = 4\) into the function. The calculation will be: \\[\ln(e^2) + 4^3 = 2 + 64 = 66.\] The result is \(f(e^2, 4) = 66\).

Step 4: Evaluate \(f(e^3, 5)\)

Substitute \(x = e^3\) and \(y = 5\) into the function. The calculation will be: \\[\ln(e^3) + 5^3 = 3 + 125 = 128.\] The result is \(f(e^3, 5) = 128\).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental concept in calculus and mathematics in general. It is the logarithm to the base \( e \), where \( e \approx 2.71828 \). The natural logarithm of a number is the power to which \( e \) must be raised to obtain that number. For instance, \( \ln(e) = 1 \) because \( e^1 = e \).

In the context of multivariable calculus, the natural logarithm often appears when dealing with functions that include exponential growth or decay. It's especially important in calculus due to its derivative properties:

• The derivative of \( \ln(x) \) is \( \frac{1}{x} \).

In our exercise, the natural logarithm is used in combination with a cubic function to form a multivariable function, \( f(x, y) = \ln x + y^3 \), making the evaluation across different points straightforward once you understand how to handle \( \ln \).

Cubic Function

A cubic function is a type of polynomial that includes a term of degree three, in this case, \( y^3 \). The general form of a cubic function can be expressed as \( ax^3 + bx^2 + cx + d \), but in our exercise, we consider only the term \( y^3 \), making it a simplified cubic expression.

Cubic functions are significant because they can model various real-world phenomena, such as volume calculations, physics problems, and complex motion dynamics. Their graphs typically have one or two turning points, and because they are polynomials, they are continuous throughout their domain.

• A key feature of the function \( y^3 \) is its symmetry around the origin, resulting in no horizontal or vertical asymptotes.
• In multivariable calculus, the combination such as \( \ln x + y^3 \) in the exercise showcases how different mathematical forms can coexist and be evaluated specifically at different points \((x,y)\).

Function Evaluation

Function evaluation in multivariable calculus involves substituting specific values into the function and finding the resulting output. This is a fundamental concept that allows us to determine how variables interact and influence each other in mathematical models.

In the given function \( f(x, y) = \ln x + y^3 \), we evaluate the function at various points. This involves inputting the given \( x \) and \( y \) values into the formula and calculating the result. For instance, evaluating \( f(e, 2) \) requires replacing \( x \) with \( e \) and \( y \) with \( 2 \):

• \( \ln(e) + 2^3 = 1 + 8 = 9 \)

Through similar steps, different points \((e^2, 4)\) and \((e^3, 5)\) are calculated, showing how each specific set of inputs yields distinct outputs:

• For \( f(e^2, 4) \): \( \ln(e^2) + 4^3 = 2 + 64 = 66 \)
• For \( f(e^3, 5) \): \( \ln(e^3) + 5^3 = 3 + 125 = 128 \)

These evaluations help illustrate how combinatorial functions are affected by both the natural logarithm and cubic term across varying inputs.