Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Show that a. \(P(t)=5 t+3\) satisfies \(P(0)=3\) and \(P^{\prime}(t)=5\) b. \(\quad P(t)=8 t+2\) satisfies \(P(0)=2\) and \(P^{\prime}(t)=8\) c. \(P(t)=t^{2}+3 t+7 \quad\) satisfies \(P(0)=7\) and \(P^{\prime}(t)=2 t+3\) d. \(P(t)=-2 t^{2}+5 t+8\) satisfies \(P(0)=8\) and \(P^{\prime}(t)=-4 t+5\) e. \(\quad P(t)=(3 t+4)^{2}\) satisfies \(P(0)=16\) and \(P^{\prime}(t)=6 \sqrt{P(t)}\) f. \(P(t)=(5 t+1)^{2}\) satisfies \(P(0)=1\) and \(P^{\prime}(t)=10 \sqrt{P(t)}\) g. \(\quad P(t)=(1-2 t)^{-1}\) satisfies \(P(0)=1\) and \(P^{\prime}(t)=2(P(t))^{2}\) h. \(P(t)=\frac{5}{1-15 t}\) satisfies \(P(0)=5\) and \(P^{\prime}(t)=3(P(t))^{2}\) i. \(P(t)=(6 t+9)^{1 / 2} \quad\) satisfies \(P(0)=3\) and \(P^{\prime}(t)=3 / P(t)\) j. \(P(t)=(4 t+4)^{1 / 2} \quad\) satisfies \(P(0)=2\) and \(P^{\prime}(t)=2 / P(t)\) k. \(P(t)=(4 t+4)^{3 / 2} \quad\) satisfies \(P(0)=8 \quad\) and \(\quad P^{\prime}(t)=6 \sqrt[3]{P(t)}\) For parts \(\mathrm{g}-\mathrm{k}\), use the Definition of Derivative 3.2 .2 to compute \(P^{\prime}\).

Step-by-Step Solution

Answer

All functions satisfy the conditions of their respective parts.

Step 1: Verify P(0) for Part a

Substitute \( t = 0 \) into \( P(t) = 5t + 3 \). \[ P(0) = 5(0) + 3 = 3 \]Thus \( P(0) = 3 \) is satisfied.

Step 2: Calculate Derivative for Part a

Calculate the derivative of \( P(t) = 5t + 3 \).\[ P'(t) = \frac{d}{dt}(5t + 3) = 5 \]Thus \( P'(t) = 5 \) is satisfied.

Step 3: Verify P(0) for Part b

Substitute \( t = 0 \) into \( P(t) = 8t + 2 \).\[ P(0) = 8(0) + 2 = 2 \]Thus \( P(0) = 2 \) is satisfied.

Step 4: Calculate Derivative for Part b

Calculate the derivative of \( P(t) = 8t + 2 \).\[ P'(t) = \frac{d}{dt}(8t + 2) = 8 \]Thus \( P'(t) = 8 \) is satisfied.

Step 5: Verify P(0) for Part c

Substitute \( t = 0 \) into \( P(t) = t^2 + 3t + 7 \).\[ P(0) = (0)^2 + 3(0) + 7 = 7 \]Thus \( P(0) = 7 \) is satisfied.

Step 6: Calculate Derivative for Part c

Calculate the derivative of \( P(t) = t^2 + 3t + 7 \).\[ P'(t) = \frac{d}{dt}(t^2 + 3t + 7) = 2t + 3 \]Thus \( P'(t) = 2t + 3 \) is satisfied.

Step 7: Verify P(0) for Part d

Substitute \( t = 0 \) into \( P(t) = -2t^2 + 5t + 8 \).\[ P(0) = -2(0)^2 + 5(0) + 8 = 8 \]Thus \( P(0) = 8 \) is satisfied.

Step 8: Calculate Derivative for Part d

Calculate the derivative of \( P(t) = -2t^2 + 5t + 8 \).\[ P'(t) = \frac{d}{dt}(-2t^2 + 5t + 8) = -4t + 5 \]Thus \( P'(t) = -4t + 5 \) is satisfied.

Step 9: Verify P(0) for Part e

Substitute \( t = 0 \) into \( P(t) = (3t + 4)^2 \).\[ P(0) = (3(0) + 4)^2 = 16 \]Thus \( P(0) = 16 \) is satisfied.

Step 10: Calculate Derivative for Part e

Use the definition of the derivative to compute \( P'(t) \) for \( P(t) = (3t + 4)^2 \).\[ P'(t) = 2(3t + 4) \cdot 3 = 6(3t + 4) = 6\sqrt{P(t)} \]Thus \( P'(t) = 6\sqrt{P(t)} \) is satisfied.

Step 11: Verify P(0) for Part f

Substitute \( t = 0 \) into \( P(t) = (5t + 1)^2 \).\[ P(0) = (5(0) + 1)^2 = 1 \]Thus \( P(0) = 1 \) is satisfied.

Step 12: Calculate Derivative for Part f

Use the chain rule to compute \( P'(t) \) for \( P(t) = (5t + 1)^2 \).\[ P'(t) = 2(5t + 1) \cdot 5 = 10(5t + 1) = 10\sqrt{P(t)} \]Thus \( P'(t) = 10\sqrt{P(t)} \) is satisfied.

Step 13: Verify P(0) for Part g

Substitute \( t = 0 \) into \( P(t) = (1-2t)^{-1} \).\[ P(0) = (1 - 2(0))^{-1} = 1 \]Thus \( P(0) = 1 \) is satisfied.

Step 14: Calculate Derivative for Part g

Use the definition of the derivative to compute \( P'(t) \) for \( P(t) = (1 - 2t)^{-1} \).\[ P'(t) = -1(-2)(1-2t)^{-2} = 2(1-2t)^{-2} = 2(P(t))^2 \]Thus \( P'(t) = 2(P(t))^2 \) is satisfied.

Step 15: Verify P(0) for Part h

Substitute \( t = 0 \) into \( P(t) = \frac{5}{1-15t} \).\[ P(0) = \frac{5}{1 - 15(0)} = 5 \]Thus \( P(0) = 5 \) is satisfied.

Step 16: Calculate Derivative for Part h

Use the definition of the derivative to compute \( P'(t) \) for \( P(t) = \frac{5}{1-15t} \).\[ P'(t) = -15 \frac{5}{(1-15t)^2} = 3(P(t))^2 \]Thus \( P'(t) = 3(P(t))^2 \) is satisfied.

Step 17: Verify P(0) for Part i

Substitute \( t = 0 \) into \( P(t) = (6t + 9)^{1/2} \).\[ P(0) = (6(0) + 9)^{1/2} = 3 \]Thus \( P(0) = 3 \) is satisfied.

Step 18: Calculate Derivative for Part i

Use the chain rule to compute \( P'(t) \) for \( P(t) = (6t + 9)^{1/2} \).\[ P'(t) = \frac{1}{2}(6t + 9)^{-1/2} \cdot 6 = \frac{3}{\sqrt{6t + 9}} = \frac{3}{P(t)} \]Thus \( P'(t) = \frac{3}{P(t)} \) is satisfied.

Step 19: Verify P(0) for Part j

Substitute \( t = 0 \) into \( P(t) = (4t + 4)^{1/2} \).\[ P(0) = (4(0) + 4)^{1/2} = 2 \]Thus \( P(0) = 2 \) is satisfied.

Step 20: Calculate Derivative for Part j

Use the chain rule to compute \( P'(t) \) for \( P(t) = (4t + 4)^{1/2} \).\[ P'(t) = \frac{1}{2}(4t + 4)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4t + 4}} = \frac{2}{P(t)} \]Thus \( P'(t) = \frac{2}{P(t)} \) is satisfied.

Step 21: Verify P(0) for Part k

Substitute \( t = 0 \) into \( P(t) = (4t + 4)^{3/2} \).\[ P(0) = (4(0) + 4)^{3/2} = 8 \]Thus \( P(0) = 8 \) is satisfied.

Step 22: Calculate Derivative for Part k

Use the chain rule to compute \( P'(t) \) for \( P(t) = (4t + 4)^{3/2} \).\[ P'(t) = \frac{3}{2}(4t + 4)^{1/2} \cdot 4 = 6 \cdot \sqrt{4t + 4} = 6 \sqrt[3]{P(t)} \]Thus \( P'(t) = 6 \sqrt[3]{P(t)} \) is satisfied.

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions that consist of variables raised to whole number powers, combined with coefficients. They are one of the fundamental building blocks in algebra and calculus. For instance, the equation \( P(t) = 5t + 3 \) is a linear polynomial, where the variable \( t \) is raised to the power of 1. On the other hand, \( P(t) = t^2 + 3t + 7 \) is a quadratic polynomial, with the highest power being 2.

• Linear Polynomials: These take the form \( ax + b \), where \( a \) and \( b \) are constants.
• Quadratic Polynomials: These can be represented as \( ax^2 + bx + c \).
• Higher Degree Polynomials: These involve powers greater than 2, like cubic \( ax^3 + bx^2 + cx + d \).

Understanding the structure of polynomial functions is vital for applying differentiation techniques, solving equations, and analyzing graphs.

Chain Rule

The chain rule is a fundamental technique in calculus, used to find the derivative of composite functions. A composite function is a function formed by combining two or more functions. If you have a function \( f(g(t)) \), you use the chain rule to differentiate it.
In application, consider \( P(t) = (5t + 1)^2 \). Here, the outer function can be described as \( u^2 \) with \( u = 5t + 1 \). To differentiate this composite, you first derive the outer function, treating \( u \) as a variable, which gives \( 2u \), and then multiply it by the derivative of the inner function \( u \), which is \( 5 \). So,

• Outer function: \( d(u^2)/du = 2u \)
• Inner function: \( du/dt = 5 \)
• Derivative: \( P'(t) = 2u(du/dt) = 10(5t + 1) \)

This results in \( P'(t) = 10(5t + 1) \), demonstrating the utility of the chain rule.

Differentiation Techniques

Differentiation is a primary concept in calculus referring to the process of finding the derivative of a function, which essentially calculates the rate of change of the function. There are several rules for differentiation depending on the function type:

• Simple Polynomials: Utilize the power rule, where if \( f(t) = t^n \), then \( f'(t) = nt^{n-1} \).
• Constant Factors: Constants are treated as such, making the derivative of \( at \) equal to \( a \).
• Chain Rule: For composite functions, as explained, apply the derivative to the outer function and multiply by the derivative of the inner function.

For example, the derivative of \( -2t^2 + 5t + 8 \) involves using the power rule:

• \( rac{d}{dt}(-2t^2) = -4t \)
• \( rac{d}{dt}(5t) = 5 \)
• \( rac{d}{dt}(8) = 0 \)

Combining these results, the derivative is \( P'(t) = -4t + 5 \). By practicing these techniques, one develops the skill to differentiate a wide variety of mathematical functions.

Limits and Continuity

Limits and continuity are foundational concepts in calculus that underpin the process of finding derivatives and understanding the behavior of functions. A limit describes the value that a function approaches as the input (or \( t \) in many cases) approaches some value. Meanwhile, continuity ensures that a function has no interruptions in its domain.
In derivative calculation, limits are used to formally define what a derivative is. For example, the definition of a derivative is:
\[ \frac{d}{dt} P(t) = \lim_{h \to 0} \frac{P(t+h) - P(t)}{h} \]
This definition forms the basis for all differentiation but can be complex to work with for most practical calculations, which is why methods like the power rule, the product rule, or the chain rule are more commonly used in practice. A function is continuous if it doesn't have jumps, holes, or vertical asymptotes in its graph, making it easier to differentiate. In the practice exercises, these concepts help ensure correct computation of derivatives, as they validate that the function behaves predictably and smoothly as \( t \) approaches any given value.