Problem 11 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

The following difference equations, initial data, and solutions have been scrambled. 1\. Match each solution to a correct initial condition and difference equation that it satisfies. 2\. Compute \(Q_{50}\) using the solution. 3\. Show algebraically that the solution satisfies the proposed difference equation. $$ \begin{aligned} &\text { Solutions }\\\ &\begin{array}{ll} S_{1} & P_{t}=2+4 t \\ S_{2} & P_{t}=3 \times 5^{t} \\ S_{3} & P_{t}=5 \times 3^{t} \\ S_{4} & P_{t}=2 t^{2}+6 t \\ S_{5} & P_{t}=4+t^{2} \\ S_{6} & P_{t}=1 \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Initial Data }\\\ &\begin{array}{ll} I D_{1} & P_{0}=0 \\ I D_{2} & P_{0}=1 \\ I D_{3} & P_{0}=2 \\ I D_{4} & P_{0}=3 \\ I D_{5} \quad P_{0} & =4 \\ I D_{6} \quad P_{0} & =5 \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Difference Equations }\\\ &D E_{1} \quad P_{t+1}-P_{t}=4 t+8\\\ &D E_{2} \quad P_{t+1}-P_{t}=0\\\ &D E_{3} \quad P_{t+1}-P_{t}=2 t+1\\\ &D E_{4} \quad P_{t+1}-P_{t}=2 \times P_{t}\\\ &D E_{5} \quad P_{t+1}-P_{t}=4\\\ &D E_{6} \quad P_{t+1}-P_{t}=4 \times P_{t} \end{aligned} $$

Step-by-Step Solution

Answer

Q_{50} values vary for each solution; see corresponding steps for computation.

Step 1: Match Solution S1 and Verify

For the solution \(P_{t} = 2 + 4t\), the initial condition should be \(P_0 = 2\). This matches with \(ID_3: P_0 = 2\). Checking the difference equation, we have \(P_{t+1} - P_t = (2 + 4(t+1)) - (2 + 4t) = 4\). This matches with \(DE_5: P_{t+1} - P_{t} = 4\).

Step 2: Match Solution S2 and Verify

For the solution \(P_{t} = 3 \times 5^t\), the initial condition should be \(P_0 = 3\). This matches with \(ID_4: P_0 = 3\). Checking the difference equation, we have \(P_{t+1} - P_{t} = 3 \times 5^{t+1} - 3 \times 5^t = 12 \times P_{t}\). Simplifying, it reduces to \(DE_6: P_{t+1} - P_{t} = 4 \times P_{t}\).

Step 3: Match Solution S3 and Verify

For the solution \(P_{t} = 5 \times 3^t\), the initial condition should be \(P_0 = 5\). This matches with \(ID_6: P_0 = 5\). Checking the difference equation, we have \(P_{t+1} - P_{t} = 5 \times 3^{t+1} - 5 \times 3^t = 10 \times P_t\). Simplifying, it matches with \(DE_4: P_{t+1} - P_{t} = 2 \times P_{t}\).

Step 4: Match Solution S4 and Verify

For the solution \(P_{t} = 2t^2 + 6t\), the initial condition should be \(P_0 = 0\). This matches with \(ID_1: P_0 = 0\). Checking the difference equation, \(P_{t+1} - P_t = 2(t+1)^2 + 6(t+1) - (2t^2 + 6t)\) yields \(4t + 8\). This matches with \(DE_1: P_{t+1} - P_{t} = 4t + 8\).

Step 5: Match Solution S5 and Verify

For the solution \(P_{t} = 4 + t^2\), the initial condition should be \(P_0 = 4\). This matches with \(ID_5: P_0 = 4\). Checking the difference equation, \(P_{t+1} - P_{t} = 4 + (t+1)^2 - (4 + t^2)\) gives \(2t + 1\). This matches with \(DE_3: P_{t+1} - P{t} = 2t + 1\).

Step 6: Match Solution S6 and Verify

For the solution \(P_{t} = 1\), the initial condition should be \(P_0 = 1\). This matches with \(ID_2: P_0 = 1\). Checking the difference equation, \(P_{t+1} - P_t = 0\), aligns with \(DE_2: P_{t+1} - P_{t} = 0\).

Step 7: Calculate Q_{50} for the Matched Solutions

Using each solution formula, compute:- For \(P_{t} = 2 + 4t\), \(P_{50} = 2 + 4 \times 50 = 202\).- For \(P_{t} = 3 \times 5^t\), \(P_{50}\) is computationally large.- For \(P_{t} = 5 \times 3^t\), similarly large.- For \(P_{t} = 2t^2 + 6t\), \(P_{50} = 2 \times 50^2 + 6 \times 50 = 5500\).- For \(P_{t} = 4 + t^2\), \(P_{50} = 4 + 50^2 = 2504\).- For \(P_{t} = 1\), constant \(P_{50} = 1\).

Key Concepts

Initial Conditions

In mathematics, initial conditions are pivotal when solving difference equations. They provide the value of a sequence at the start, most often when \(t = 0\). This initial value allows us to determine further values of the sequence by applying the difference equation repeatedly. For example, in the problem you're addressing, we have solutions like \(P_{t} = 2 + 4t\), where the initial condition is \(P_0 = 2\). This starting point is the basis from which all future values \(P_{t}\) are calculated. Without the correct initial condition, predicting the future values of a sequence becomes impossible.
To match an initial condition with a solution:

• Identify the expression for \(P_0\) in each provided solution.
• Compare it to the available initial data until you find a match.

Ensuring these conditions are aligned correctly is crucial to solving difference equations accurately.

Solution Verification

Solution verification involves confirming that a proposed solution satisfies the given difference equation. This step ensures that each solution not only meets the initial condition but evolves according to the specified rule of change, the difference equation.
Let's take the solution \(P_{t} = 3 \times 5^t\) and verify it against its difference equation. First, we calculate \(P_{t+1} - P_t = 3 \times 5^{t+1} - 3 \times 5^t\), which simplifies to \(4 \times P_t\). This conforms with the difference equation \(DE_6: P_{t+1} - P_{t} = 4 \times P_t\).
Verification helps to ensure that the solution:

• Accurately describes the behavior of the sequence over time.
• Consistently matches the incremental changes dictated by the difference equation.

Through this process, you confirm the validity of the matched solution and its alignment with initial conditions.

Algebraic Demonstration

Algebraic demonstration is the process of using algebraic manipulation to convincingly show that a difference equation is satisfied by a solution. This involves expanding, simplifying, and rearranging terms using known algebraic rules.
Consider the solution \(P_{t} = 5 \times 3^t\). To show that it satisfies \(DE_4: P_{t+1} - P_{t} = 2 \times P_{t}\), we proceed by calculating both sides of the equation:

• Left side: \(P_{t+1} = 5 \times 3^{t+1} = 15 \times 3^t\)
• Right side: \(P_{t} = 5 \times 3^t\), so \(2 \times P_{t} = 10 \times 3^t\)

The difference \(P_{t+1} - P_{t} = (15 - 5) \times 3^t = 10 \times 3^t\), matches the right side, confirming the solution satisfies the equation. This clear algebraic demonstration reassures us that the correct matching has been made.

Sequence Computation

Sequence computation involves using the matching solution formulas to find specific values of the sequence at given points. For instance, computing \(Q_{50}\) means finding the sequence's value when \(t = 50\). Applying the appropriate solution will yield the desired result.
Let's use \(P_{t} = 2 + 4t\) to compute \(P_{50}\):

• Substituting \(t = 50\) into the equation gives \(P_{50} = 2 + 4 \times 50 = 202\).

This calculation shows the power of sequence computation, providing us with precise numerical answers derived from the solutions paired with initial conditions.
From large numbers like \(P_{t} = 3 \times 5^t\) to constant sequences like \(P_{t} = 1\), each computation allows us to see different dynamics at play within difference equations. Exploring these provides a deeper understanding of how such sequences evolve over time.