Problem 3

Question

Suppose a mold colony is growing in a nutrient solution and that on day zero the area was $0.5 \mathrm{~cm}^{2}$ and for every time, $t \geq 0$, the instantaneous rate of growth of the area of the colony is $2 t \mathrm{~cm}^{2}$ per day. Let $P(t)$ be the colony area at time $t$. a. Argue that for every time, $t, P^{\prime}(t)=2 t$. b. Show that for $Q(t)=t^{2}, Q^{\prime}(t)=2 t$. Then $P^{\prime}(t)=Q^{\prime}(t)$. c. From the Parallel Graph Theorem, it follows that there is a constant $C$ such that $P(t)=Q(t)+C$ d. Use $P(0)=0.5$ to evaluate $C$. e. What is the area of the mold colony on day $8 ?$

Step-by-Step Solution

Verified

Answer

The area of the mold colony on day 8 is 64.5 cm².

1Step 1: Understand the given information

The growth of the mold colony is given by the rate of change $P'(t) = 2t$. Initially, the area of the colony is $P(0) = 0.5\, \text{cm}^2$.

2Step 2: Confirm the Statement a

We are given that the instantaneous rate of growth $P'(t) = 2t$. Since this is directly provided, we accept that $P'(t) = 2t$ without further need to prove.

3Step 3: Show Statement b

For $Q(t) = t^2$, let's find its derivative: $Q'(t) = \frac{d}{dt}(t^2) = 2t$. This confirms that $Q'(t) = 2t$. Thus, $P'(t) = Q'(t)$.

4Step 4: Use the Parallel Graph Theorem for Statement c

Since $P'(t) = Q'(t)$, according to the Parallel Graph Theorem, $P(t) = Q(t) + C$ for some constant $C$.

5Step 5: Evaluate Constant C using Initial Condition

At $t = 0$, $P(0) = 0.5$. Substitute into the equation: $0.5 = 0^2 + C$, giving $C = 0.5$. Thus, $P(t) = t^2 + 0.5$.

6Step 6: Calculate Area on Day 8

To find the area of the mold colony on day 8, substitute $t = 8$ into the equation: $P(8) = 8^2 + 0.5 = 64 + 0.5 = 64.5\, \text{cm}^2$.

Key Concepts

Differential EquationsInstantaneous Rate of ChangeInitial Value Problem

Differential Equations

Differential equations are a central topic in calculus, and they are equations that involve a function and its derivatives. The main goal is to find a function that satisfies the given equation. In this specific exercise, we deal with a simple differential equation where the derivative of a function, denoted by $P'(t)$, equals $2t$. This equation tells us how the area of the mold colony changes over time.

Working with differential equations often involves these steps:

Identify the equation and understand what it represents.
Recognize any initial conditions provided, which help determine the specific solution.
Solve the differential equation by finding a general solution.
Use initial conditions to find the particular solution that fits all given circumstances.

By understanding these steps, students can grasp how to approach and solve differential equations with more confidence.

Instantaneous Rate of Change

The instantaneous rate of change is a key concept in calculus that describes how a quantity changes at a specific moment in time. In this exercise, it's defined by the derivative $P'(t) = 2t$. Essentially, this tells us how fast the mold colony is growing at any given time $t$.

The concept can be broken down into two main points:

It is represented mathematically by the derivative of a function.
Provides insight into the trend of change at a specific point rather than over an interval.

Understanding the instantaneous rate of change helps students apply calculus to real-world problems, like predicting how populations such as the mold colony in this scenario, evolve over time. The derivative calculated in this context reflects the slope of the tangent line to the curve at any point $t$, offering precise details about growth behavior at that moment.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation accompanied by a condition that specifies the value of the unknown function at a certain point. This exercise illustrates an IVP with the given that $P(0) = 0.5$. This initial condition is crucial for finding the particular solution to the differential equation $P'(t) = 2t$.

To solve an IVP effectively, you should:

Identify the differential equation and the initial condition, which sets the specific starting value for the solution.
Solve the differential equation to obtain a general solution, in this case, $P(t) = t^2 + C$.
Apply the initial condition to determine the constant $C$. For this problem, substituting $t = 0$ and applying $P(0) = 0.5$ helps us find $C = 0.5$.

This approach ensures that the solution not only satisfies the differential equation but also matches the specific conditions at the starting point. Understanding IVPs allows students to find solutions relevant to particular scenarios, reflecting real-world applications where initial conditions have significant impacts.

Problem 3

Problem 4

Other exercises in this chapter

Problem 3

Compute $\int\left(1+t^{4}\right)^{3} d t$.

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Problem 3

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Problem 4

Which of the two indefinite integrals $$ \int \frac{1}{1+t^{2}} t d t \quad \text { or } \int \frac{1}{1+t^{2}} d t $$ is $\ln \left(1+t^{2}\right)^{0.5}+C ?$

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Problem 4

Suppose the rate of glucose production in a corn plant is proportional to sunlight intensity and can be approximated by $$ R(t)=K(t+7)^{2}(t-7)^{2}=K\left(t^{4}

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