Problem 13 - Calculus for Biology and Medicine

Question

Evaluate each function at the given point. \(h(x, t)=\exp \left[-\frac{(x-2)^{2}}{2 t}\right]\) at \((1,5)\)

Step-by-Step Solution

Answer

The function value is approximately 0.9048.

Step 1: Understand the Function

We are given the function \( h(x, t) = \exp \left[-\frac{(x-2)^{2}}{2t}\right] \). It involves the exponential function with variables \( x \) and \( t \). Our goal is to evaluate this function at the specified point \((x, t) = (1, 5)\).

Step 2: Substitute the Given Values

Substitute \( x = 1 \) and \( t = 5 \) into the function: \[ h(1, 5) = \exp \left[-\frac{(1-2)^{2}}{2 \cdot 5}\right] \].

Step 3: Simplify the Expression Inside the Exponent

Calculate the expression inside the exponent: 1. Compute \((1-2)^2 = (-1)^2 = 1\).2. Substitute this into the expression: \[ -\frac{1}{2 \cdot 5} = -\frac{1}{10} \].

Step 4: Evaluate the Exponential Expression

The simplified expression inside the exponent is \(-\frac{1}{10}\). Thus, the value of the function is: \[ h(1, 5) = \exp \left(-\frac{1}{10}\right) \]. This is approximately equal to \( e^{-0.1} \).

Step 5: Calculate Numerical Value

Using a calculator, evaluate \( e^{-0.1} \). This gives approximately \( 0.9048 \). Therefore, \( h(1, 5) \approx 0.9048 \).

Key Concepts

Substitution in Functions

Substitution is a fundamental technique in evaluating functions. It's like plugging numbers into a formula to see what outcome they create. With the function given as \( h(x, t) = \exp \left[-\frac{(x-2)^{2}}{2t}\right] \), we start by substituting the point \((x, t) = (1, 5)\). This means wherever we see an \(x\) in the function, we replace it with 1, and wherever we see a \(t\), we replace it with 5.
The goal of substitution is to transform the variable expression into a simpler, numerical form. This simplifies solving, making it more like arithmetic. After substitution, our main task is to handle these new expressions with numbers replacing the variables completely.

Exponential Expressions

Exponential expressions are functions where a constant is raised to a power that typically includes a variable. In this function, we see \(\exp\) representing the exponential function, often involving Euler’s number \(e\), known for its mathematical significance.
The expression \(\exp \left[-\frac{(1-2)^{2}}{2 \cdot 5}\right]\) translates to \(e\) raised to the power of the processed expression \(-\frac{1}{10}\). Handling exponential expressions requires understanding this base, \(e\), and how its powers grow or shrink.
Exponential expressions are crucial in many fields, modeling growth patterns, decay, and various natural phenomena. After crafting this expression by substitution, simplifying this part correctly is vital for a precise outcome.

Function Simplification

Simplification is the act of rewriting an expression in a more accessible form without changing its value. After substituting the given values into our function, we find ourselves simplifying the expression inside the exponential function \(-\frac{(1-2)^{2}}{2 \times 5}\).
This involves basic algebraic steps:

• First, calculate \((1-2)^2\): we find that \((-1)^2 = 1\).
• Next, substitute this square value into the division: \(-\frac{1}{10}\).

Through these steps, the original complex-looking function becomes simpler and more manageable. Simplifying a function is essential because it makes it easier to evaluate and interpret correctly. These streamlined versions are often easier to compute manually or using technology.

Numerical Evaluation

Once the function is simplified, it's time to compute its value numerically. Numerical evaluation involves calculating the exact value of expressions, which can often require a calculator or approximation methods, especially for transcendental numbers like \(e^{-x}\).
With our simplified expression \(\exp \left(-\frac{1}{10}\right)\), we calculate its value. This process involves evaluating \(e^{-0.1}\), which approximates to about \(0.9048\).
Numerical evaluation allows us to interpret the result in the context of real-world applications where precise values are needed for prediction, understanding phenomena, or solving equations. The final evaluated number gives concrete meaning to the initial function, showing its practical application.