Question

If \(z=f(x, y)=3 \mathrm{e}^{x}-2 \mathrm{e}^{y}+x^{2} y^{3}\) find \(z(1,1)\)

Short answer

Answer: The value of the function at point (1,1) is \(z(1,1) = e + 1\).

Step-by-step solution

Define the function f(x, y)

The given function is: \(z = f(x, y) = 3e^x - 2e^y + x^2y^3\).

Substitute x = 1 and y = 1

Replace x with 1 and y with 1 in the function: \(z(1,1) = f(1,1) = 3e^1 - 2e^1 + (1^2)(1^3)\).

Evaluate the function

Compute the function by calculating the terms: \(z(1,1) = 3e - 2e + (1)(1) = 3e - 2e + 1\).

Simplify the function

Simplify the function by combining the terms: \(z(1,1) = e + 1\). So, the value of the function at point (1,1) is \(z(1,1) = e + 1\).

Key concepts

Partial Derivatives

In multivariable calculus, partial derivatives help us understand how a function changes as one of its variables changes, while keeping other variables constant. Essentially, they are the building blocks for many concepts in calculus.
Consider the function \(z = f(x, y) = 3e^x - 2e^y + x^2y^3\). Here, \(x\) and \(y\) are the variables. To find the partial derivative of \(z\) with respect to \(x\), we symbolically treat \(y\) as a constant and differentiate accordingly. Similarly, for \(y\), we treat \(x\) as a constant.
- **For \(\frac{\partial z}{\partial x}\)**, differentiate \(3e^x\) and \(x^2y^3\), while treating \(2e^y\) as constant: - \(\frac{\partial}{\partial x}(3e^x) = 3e^x\) - \(\frac{\partial}{\partial x}(x^2y^3) = 2xy^3\) - **For \(\frac{\partial z}{\partial y}\)**, differentiate \(-2e^y\) and \(x^2y^3\), treating \(3e^x\) as constant: - \(\frac{\partial}{\partial y}(-2e^y) = -2e^y\) - \(\frac{\partial}{\partial y}(x^2y^3) = 3x^2y^2\) These derivatives are useful for analyzing functions and understanding how changes in one variable affect overall function behavior.

Function Evaluation

Function evaluation is the process of determining the output of a function for specific input values. In our example, the task is to evaluate the function \(z = f(x, y) = 3e^x - 2e^y + x^2y^3\) at the point \((1,1)\).
This means we replace each occurrence of \(x\) and \(y\) in the expression with 1, and then compute the resulting value. Here are the steps for evaluation:

• Identify all variable terms: \(3e^x\), \(-2e^y\), and \(x^2y^3\).
• Substitute \(x = 1\) and \(y = 1\) into the function: \(z(1,1) = 3e^1 - 2e^1 + 1^2*1^3\).
• Perform the calculations: \(z(1,1) = 3e - 2e + 1\).
• Simplify the expression to find the output: \(z(1,1) = e + 1\).

Evaluating functions is a pivotal skill, allowing us to find specific values for the function based on input parameters.

Exponential Functions

Exponential functions are a crucial component in mathematics, characterized by their base raised to a power that includes a variable. Here, functions like \(3e^x\) and \(-2e^y\) appear in the given function.
The constant \(e\) is approximately 2.718, representing the base of natural logarithms.
- **Describing \(3e^x\):** - This term grows significantly as \(x\) increases due to the exponential behavior of \(e^x\). - As \(x\) becomes increasingly positive, \(3e^x\) accelerates in growth, demonstrating exponential growth characteristics. - **Understanding \(-2e^y\):** - Similar to \(3e^x\), but negative, meaning it decreases as \(y\) increases. - When plotting, this term would cause the graph to decline rapidly in negative values of \(y\), showcasing exponential decay.Exponential functions are highly applicable, found in growth models, decay processes, and they offer insight into how functions behave across vast ranges of input values.