Problem 13 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

For each equation find a numbers \(b\) and \(c\) such that \(P_{t}=b t+c\) is a solution. a. \(\quad P_{t+1}-0.9 P_{t}=t\) b. \(P_{t+1}-0.5 P_{t}=3+2 t\) c. \(P_{t+1}+0.9 P_{t}=1+2 t\) d. \(P_{t+2}-0.2 P_{t+1}-0.6 P_{t}=5+4 t\) e. \(P_{t+1}+0.9 P_{t}=3-t\) f. \(P_{t+2}+0.2 P_{t+1}-0.4 P_{t}=2-3 t\)

Step-by-Step Solution

Answer

a: \( b = 10, c = 0 \); b: \( b = 4, c = 6 \); c: \( b = \frac{2}{1.9}, c = \frac{1}{1.9} \); d: No solution; e: \( b = -\frac{1}{1.9}, c = \frac{3}{1.9} \); f: No solution.

Step 1: Understanding Linear Solutions

To solve for a linear solution of a recurrence relation, assume a form \( P_t = bt + c \), where \( b \) and \( c \) are constants that need to be determined. By substituting \( P_{t+1} \) with \( b(t+1) + c \), we can establish equations based on the given recurrence relations.

Step 2: Solving Part (a): Determine b and c

Given the equation \( P_{t+1} - 0.9P_t = t \), substitute the assumed form: \( b(t+1) + c - 0.9(bt + c) = t \). Simplify to get \((1-0.9)b + c-0.9c = t - 0.9bt\). This results in two equations, 1st comparing \( t \) coefficients: \( 0.1b = 1 \) giving \( b = 10 \); 2nd with constants: \( 0.1c = 0 \), so \( c = 0 \). Thus, \( b = 10 \) and \( c = 0 \).

Step 3: Solving Part (b): Find b and c

For \( P_{t+1} - 0.5P_t = 3 + 2t \), use the form \( P_t = bt + c \) and follow substitution: \( b(t+1) + c - 0.5(bt + c) = 3 + 2t \). Simplifying: \( (1-0.5)b = 2 \) results in \( 0.5b = 2 \) from which \( b = 4 \); constant equation: \( 0.5c = 3 \), giving \( c = 6 \). Therefore, \( b = 4 \), \( c = 6 \).

Step 4: Solving Part (c): Determine b and c

With \( P_{t+1} + 0.9P_t = 1 + 2t \), substitute \( bt + c \): \( b(t+1) + c + 0.9(bt + c) = 1 + 2t \). Simplify and match coefficients: \( (1+0.9)b = 2 \) yields \( 1.9b = 2 \), thus \( b = \frac{2}{1.9} \approx 1.053 \); \( c(1+0.9) = 1 \), gives \( c = \frac{1}{1.9} \approx 0.526 \). Therefore, \( b = \frac{2}{1.9} \), \( c = \frac{1}{1.9} \).

Step 5: Solving Part (d): Find b and c

For \( P_{t+2} - 0.2P_{t+1} - 0.6P_t = 5 + 4t \), assuming the form requires shifting: \( P_{t+1} = b(t+1) + c \), \( P_{t+2} = b(t+2) + c \), solve: \( b(t+2) + c - 0.2[b(t+1) + c] - 0.6(bt + c) = 5 + 4t \). Simplifying: \( b - 0.2b - 0.6b = 4 \) results in \( 0.2b = 4 \) thus \( b = 20 \); \( 1.4c - 0.2c - 0.6c = 5 \) gives no solution, implying none in linear form.

Step 6: Solving Part (e): Determine b and c

For \( P_{t+1} + 0.9P_t = 3 - t \), solve: \( b(t+1) + c + 0.9(bt + c) = 3 - t \). \( b(1+0.9) = -1 \) gives \( 1.9b = -1 \), which results in \( b = -\frac{1}{1.9} \approx -0.526 \); constants: \( (1+0.9)c = 3 \), gives \( c = \frac{3}{1.9} \approx 1.579 \). Thus, \( b = -0.526 \), \( c = 1.579 \).

Step 7: Solving Part (f): Find b and c

Given \( P_{t+2} + 0.2P_{t+1} - 0.4P_t = 2 - 3t \), rewrite with shifts: \( P_{t+1} = bt + c \), \( P_{t+2} = b(t+2) + c \), substitute: \( b(t+2) + c + 0.2(bt + c) - 0.4(bt + c) = 2 - 3t \). \( 1.2b - 0.4b = -3 \) simplifies to \( 0.8b = -3 \), gives \( b = -3.75 \); \( 0.8c - 0.4c = 2 \) yields no valid solution in linear form.

Key Concepts

Linear Equations in Calculus

Linear equations are foundational elements in calculus that involve constant rates of change. These equations are characterized by their simplicity and form, typically expressed as \( y = mx + b \). However, in the context of recurrence relations, we often encounter slightly more complex forms where the variable depends linearly on earlier terms.
In calculus, linear equations allow us to derive these forms that predict the future behavior of sequences or locations at specific times. Understanding the basic nature of linear equations aids in analyzing more complex calculus problems, where each step follows logically from the preceding one. This concept is crucial when studying behavior patterns over time in sequences.

• Express the equation in a linear form \( P_t = bt + c \).
• Solve for the unknown coefficients \( b \) and \( c \) that match the conditions set by the sequence.
• Use derivatives to explore changes in \( P_{t} \) over time, employing linear equations to describe these relationships effectively.

Recurrence Relation Solutions

Solving recurrence relations requires finding a formula that relates each term in a sequence to previous terms. For linear recurrence relations, the solution often comes in the form \( P_t = bt + c \), which simplifies understanding and tracking sequences over time.
Recurrence relation solutions bridge the gap between discrete sequences and continuous calculus practices. The key is to transform these relations into equations that can be solved using algebraic methods. By doing this, we can predict future terms based on past data.

• Start by assuming a general solution form, often linear like \( P_t = bt + c \).
• Substitute this general form into the original recurrence relation to set up equations for \( b \) and \( c \).
• Solve the equations for these coefficients by isolating terms and equating them to corresponding coefficients from the recurrence relation.

Understanding how to handle these solutions provides insight into dynamic systems encountered in advanced calculus.

Step-by-Step Calculus Problems

Breaking down calculus problems into step-by-step solutions is an effective way to learn complex concepts. Each step builds on the previous one, guiding you through the process of solving recurrence relations and understanding linear equations in a structured way.
By dissecting problems, you can focus on one aspect at a time, reducing complexity and enhancing comprehension. This method is particularly effective for students tackling advanced calculus, where each detail is crucial.

• Begin by understanding the problem statement and what is required.
• Identify the type of problem—whether it's about finding specific coefficients in recurrence relations or analyzing equations.
• Follow a logical sequence of steps to substitute, simplify, and solve.
• Verify the solution against the original problem to ensure accuracy.

Using a step-by-step approach lightens the learning load and fortifies your comprehension of calculus fundamentals.