Problem 10
Question
Suppose that the amount of phosphorus in a lake at time \(t\), denoted by \(P(t)\), follows the equation $$\frac{d P}{d t}=3 t+1 \quad \text { with } P(0)=0$$ Find the amount of phosphorus at time \(t=10\).
Step-by-Step Solution
Verified Answer
The amount of phosphorus at \(t=10\) is 160 units.
1Step 1: Understanding the Problem
We are given an ordinary differential equation (ODE) describing the rate of change of phosphorus in a lake. Our goal is to find the amount of phosphorus, denoted as \(P(t)\), at \(t=10\). We have an initial condition \(P(0) = 0\).
2Step 2: Solving the Differential Equation
The differential equation is \(\frac{dP}{dt} = 3t + 1\). To find \(P(t)\), we need to integrate both sides with respect to \(t\). The integration of \(3t + 1\) with respect to \(t\) is \(\frac{3}{2}t^2 + t + C\), where \(C\) is the constant of integration.
3Step 3: Using the Initial Condition
We apply the initial condition \(P(0) = 0\) to find \(C\). Substituting \(t = 0\) into the integrated equation, we have \(P(0) = \frac{3}{2} \cdot 0^2 + 0 + C = 0\). Solving for \(C\), we find that \(C = 0\).
4Step 4: Forming the Particular Solution
With \(C = 0\) determined, the particular solution to the differential equation becomes \(P(t) = \frac{3}{2}t^2 + t\). This equation represents the amount of phosphorus in the lake at any time \(t\).
5Step 5: Calculating the Amount at \(t=10\)
Substitute \(t = 10\) into the particular solution: \(P(10) = \frac{3}{2}(10)^2 + 10 = \frac{3}{2} \cdot 100 + 10 = 150 + 10 = 160\). So, the amount of phosphorus at \(t = 10\) is 160 units.
Key Concepts
Ordinary Differential EquationsInitial Value ProblemIntegration
Ordinary Differential Equations
Ordinary differential equations, or ODEs, are mathematical equations that relate some function with its derivatives. They play a crucial role in modeling and analyzing real-world phenomena.
In an ODE, we find the function that satisfies a given rate of change, expressed through derivatives. For instance, outcomes like population growth, movement of particles, or changes in chemical concentrations can often be described using ODEs.
In this exercise, we encounter an ODE of the form \(\frac{dP}{dt} = 3t + 1\). It implies that the rate at which phosphorus changes in the lake is dependent on both time \(t\) and a constant. The term \(3t + 1\) represents this relationship.
In an ODE, we find the function that satisfies a given rate of change, expressed through derivatives. For instance, outcomes like population growth, movement of particles, or changes in chemical concentrations can often be described using ODEs.
In this exercise, we encounter an ODE of the form \(\frac{dP}{dt} = 3t + 1\). It implies that the rate at which phosphorus changes in the lake is dependent on both time \(t\) and a constant. The term \(3t + 1\) represents this relationship.
- \(\frac{dP}{dt}\): Derivative of phosphorus with respect to time. This is the rate of change.
- \(3t + 1\): Shows what affects the rate of change - here, it's linear in time.
Initial Value Problem
An initial value problem (IVP) gives us a starting point to solve a differential equation fully. It is essentially an ODE combined with a specified value at a certain point.
The initial condition helps determine the specific solution among the infinite possibilities offered by the integration process.
In this context, we know the amount of phosphorus at time \(t=0\) as given by \(P(0) = 0\). This is a crucial step because when we integrate the equation to solve for \(P(t)\), we introduce an unknown constant, \(C\).
The initial condition helps determine the specific solution among the infinite possibilities offered by the integration process.
In this context, we know the amount of phosphorus at time \(t=0\) as given by \(P(0) = 0\). This is a crucial step because when we integrate the equation to solve for \(P(t)\), we introduce an unknown constant, \(C\).
- The initial condition helps us eliminate this unknown by providing a means to solve for \(C\).
- Here, substituting \(t=0\) into the integrated form allows us to establish that \(C=0\).
Integration
Integration is the mathematical process of finding a function given its rate of change, or derivative. It is the reverse operation of differentiation.
When solving differential equations like the one in this exercise, integration helps build a function that models the accumulation of quantities over time.
In this situation, to solve \(\frac{dP}{dt} = 3t + 1\), we need to integrate \(3t + 1\) with respect to \(t\). This involves finding a function \(P(t)\) such that its derivative gives back \(3t + 1\).
When solving differential equations like the one in this exercise, integration helps build a function that models the accumulation of quantities over time.
In this situation, to solve \(\frac{dP}{dt} = 3t + 1\), we need to integrate \(3t + 1\) with respect to \(t\). This involves finding a function \(P(t)\) such that its derivative gives back \(3t + 1\).
- The integral of \(3t\) is \(\frac{3}{2}t^2\), and the integral of \(1\) is simply \(t\).
- By integrating, we arrive at \(P(t) = \frac{3}{2}t^2 + t + C\), where \(C\) is a constant.
Other exercises in this chapter
Problem 9
Suppose that the volume \(V(t)\) of a cell at time \(t\) changes according to $$\frac{d V}{d t}=1+\cos t \quad \text { with } V(0)=5$$ Find \(V(t)\).
View solution Problem 10
Suppose that a fish population evolves according to a logistic equation and that fish are harvested at a rate proportional to the population size. If \(N(t)\) d
View solution Problem 11
Solve the given autonomous differential equations. \(\frac{d y}{d x}=3 y\), where \(y_{0}=2\) for \(x_{0}=0\)
View solution Problem 12
(Adapted from Crawley, 1997) Denote plant biomass by \(V\), and herbivore number by \(N .\) The plant-herbivore interaction is modeled as $$ \begin{array}{l} \f
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