Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Let \(E(t)=\left(\frac{1}{2}\right)^{t}\). a. Approximate \(E^{\prime}(0)\) using the centered difference quotient on [-0.0001,0.0001] . b. Use your value for \(E^{\prime}(0)\) and \(E^{\prime}(t)=E^{\prime}(0) E(t)\) to approximate \(E^{\prime}(-1), E^{\prime}(1),\) and \(E^{\prime}(2)\). c. Sketch the graphs of \(E(t)\) and \(E^{\prime}(t)\).

Step-by-Step Solution

Answer

a. \(E'(0) \approx -0.3465\) b. \(E'(-1) \approx -0.693\), \(E'(1) \approx -0.17325\), \(E'(2) \approx -0.086625\) c. Graphs are exponential decay for both \(E(t)\) and \(E'(t)\).

Step 1: Evaluate the Centered Difference Quotient

The centered difference quotient is given by the formula \( \frac{E(h) - E(-h)}{2h} \). To approximate \( E'(0) \), use \( h = 0.0001 \): \[\frac{E(0.0001) - E(-0.0001)}{2 \times 0.0001} = \frac{\left(\frac{1}{2}\right)^{0.0001} - \left(\frac{1}{2}\right)^{-0.0001}}{0.0002} \]Calculate this value using the small exponential approximations: \( E(0.0001) \approx 1 - 0.0001 \ln(2)/2 \) and \( E(-0.0001) \approx 1 + 0.0001 \ln(2)/2 \), providing\[E'(0) \approx \frac{-0.0001 \ln(2)}{0.0002} = -\frac{\ln(2)}{2}\]

Step 2: Calculate Value for \(E'(0)\)

Since \(E'(0) \approx -\frac{\ln(2)}{2}\), we calculate its approximate numerical value. Remember that \(\ln(2) \approx 0.693\), thus: \[E'(0) \approx -\frac{0.693}{2} = -0.3465\]

Step 3: Approximate \(E'(t)\) Values

Using the relationship \(E'(t) = E'(0) E(t)\), we find the derivative approximation at \(t = -1, 1, \) and \(2\):1. For \( E'(-1) \):\[E'(-1) = -0.3465 \left(\frac{1}{2}\right)^{-1} = -0.3465 \times 2 = -0.693\]2. For \( E'(1) \):\[E'(1) = -0.3465 \left(\frac{1}{2}\right) = -0.3465 \times 0.5 = -0.17325\]3. For \( E'(2) \):\[E'(2) = -0.3465 \left(\frac{1}{2}\right)^2 = -0.3465 \times 0.25 = -0.086625\]

Step 4: Sketch Graphs of \(E(t)\) and \(E'(t)\)

1. **Graph of \(E(t)\):** The function \(E(t) = \left(\frac{1}{2}\right)^t\) is a decaying exponential curve that starts at 1 when \(t = 0\) and approaches zero as \(t\) increases. It quickly decreases from 1 to approach zero for positive values of \(t\), and increases for negative values of \(t\).2. **Graph of \(E'(t)\):** The derivative \(E'(t) = E'(0) E(t)\) represents the slope of \(E(t)\). It will follow a similar pattern as \(E(t)\) but starts at \(E'(0) \) and becomes less steep as \(t\) increases, reflecting how the rate of change decreases as \(E(t)\) approaches zero.

Key Concepts

Exponential Functions

Exponential functions are a critical aspect in mathematics and the life sciences. They model processes where quantities grow or decay at rates proportional to their current value. In the given exercise, we have the function \(E(t) = \left(\frac{1}{2}\right)^t\), a classic example of exponential decay, where the base \(\frac{1}{2}\) signals decay. This function describes how values decrease rapidly at first and then slowly approach zero as \(t\) increases. To understand this better:

• As \(t\) becomes more positive, \(E(t)\) tends towards zero, displaying a rapid decrease.
• As \(t\) becomes more negative, the function increases, demonstrating an inverse relationship.

The concept of exponential functions is often illustrated through their behavior in graphs, where they show a smooth curve representing continuous change.

Derivative Approximation

Derivative approximation involves estimating the rate at which a function is changing at any given point. For the function \(E(t)\), the derivative \(E'(t)\) gives us the slope of the tangent line to the curve at any point \(t\). In many applications, especially in life sciences, we're interested in understanding how quickly a process unfolds or reverses. Calculating derivatives exactly might be challenging, especially for complex functions, but approximations help:

• In the exercise, the centered difference quotient was used for \(E'(0)\), providing an approximation of the instantaneous rate of change.
• This approach is particularly useful when symbolic differentiation seems complicated or impossible.

Understanding the rate of decay or growth in functions helps predict future values and trends, an essential skill in quantitative analysis.

Difference Quotient

The difference quotient is a key method used to approximate derivatives. It is defined as the ratio of the change in the function value to the change in the variable. In this exercise, the centered difference quotient was utilized to approximate \(E'(0)\). Here’s a breakdown:

• We use the formula \( \frac{E(h) - E(-h)}{2h} \) with a small \(h\) value, leading to an accurate estimate of the derivative at a point.
• This is centered because it uses values equally spaced around the point of interest, providing a balanced approximation.

This approach is particularly useful when dealing with data or situations where exact differentiation isn’t possible, offering a practical means to analyze change.

Graph Sketching

Graph sketching is an invaluable tool that helps in visualizing how functions behave. It is especially important when examining exponential and derivative functions. In this exercise, sketching the graphs of \(E(t)\) and \(E'(t)\):

• The graph of \(E(t)\), being an exponential decay function, presents a descending curve. This starts prominently at 1 and swoops toward zero as \(t\) rises.
• Meanwhile, \(E'(t)\) shows changes in the slope over time, reflecting the decreasing speed of decay.

Graph sketching helps in identifying trends and slopes graphically, making complex mathematical concepts more tangible and easier to interpret.