Problem 7
Question
Solve starting from \(y_{0}=10\). At what time does \(y\) increase to 100 or drop to \(1 ?\) $$ \frac{d y}{d t}=e^{4 t} $$
Step-by-Step Solution
Verified Answer
\( y = 100 \) when \( t = \frac{\ln(361)}{4} \). \( y \) won't drop to 1.
1Step 1: Write Down the Differential Equation
We are given the differential equation \( \frac{d y}{d t} = e^{4t} \). This defines how \( y \) depends on time \( t \). The task is to determine when \( y = 100 \) and when \( y = 1 \) given that initially \( y_0 = 10 \).
2Step 2: Integrate the Differential Equation
To find \( y \) as a function of \( t \), we need to integrate the equation \( \frac{d y}{d t} = e^{4t} \). Integrating both sides with respect to \( t \), we have:\[ y(t) = \int e^{4t} \, dt = \frac{1}{4} e^{4t} + C \]where \( C \) is the constant of integration.
3Step 3: Solve for the Constant of Integration
We use the initial condition \( y(0) = 10 \) to solve for \( C \). Substituting into the equation:\[ 10 = \frac{1}{4} e^{4 \times 0} + C = \frac{1}{4} + C \]Solving for \( C \), we find:\[ C = 10 - \frac{1}{4} = \frac{40}{4} - \frac{1}{4} = \frac{39}{4} \]
4Step 4: Construct the General Solution for \( y(t) \)
With \( C = \frac{39}{4} \), the general solution for \( y(t) \) is:\[ y(t) = \frac{1}{4} e^{4t} + \frac{39}{4} \]
5Step 5: Solve for \( t \) when \( y = 100 \)
We set \( y(t) = 100 \) and solve for \( t \):\[ 100 = \frac{1}{4} e^{4t} + \frac{39}{4} \]Subtract \( \frac{39}{4} \) from both sides:\[ 100 - \frac{39}{4} = \frac{1}{4} e^{4t} \]Converting 100 to a fraction:\[ \frac{400}{4} - \frac{39}{4} = \frac{361}{4} = \frac{1}{4} e^{4t} \]Multiply by 4:\[ 361 = e^{4t} \]Take the natural logarithm on both sides:\[ \ln(361) = 4t \]Solving for \( t \):\[ t = \frac{\ln(361)}{4} \]
6Step 6: Solve for \( t \) when \( y = 1 \)
We set \( y(t) = 1 \) and solve for \( t \):\[ 1 = \frac{1}{4} e^{4t} + \frac{39}{4} \]Subtract \( \frac{39}{4} \) from both sides:\[ 1 - \frac{39}{4} = \frac{1}{4} e^{4t} \]Converting 1 to a fraction:\[ \frac{4}{4} - \frac{39}{4} = \frac{-35}{4} = \frac{1}{4} e^{4t} \]Multiply by 4:\[ -35 = e^{4t} \]Since \( e^{4t} \) is always positive, it's impossible for \( y \) to drop to 1 with this model.
Key Concepts
IntegrationInitial ConditionsExponential FunctionsConstant of Integration
Integration
Integration is a fundamental concept in calculus that is used to determine the accumulation of quantities, such as area under a curve or the total variation of a function over an interval. In this exercise, we are given a differential equation of the form \( \frac{d y}{d t} = e^{4t} \). To find the function \( y(t) \), we need to integrate both sides with respect to time \( t \).
This process effectively reverses the differentiation, allowing us to determine the function \( y(t) \) that describes how variable \( y \) changes over time.
For this integration, the antiderivative of \( e^{4t} \) with respect to \( t \) is \( \frac{1}{4} e^{4t} \), plus a constant of integration \( C \). This constant \( C \) will be determined by taking into account additional conditions, such as initial conditions.
This process effectively reverses the differentiation, allowing us to determine the function \( y(t) \) that describes how variable \( y \) changes over time.
For this integration, the antiderivative of \( e^{4t} \) with respect to \( t \) is \( \frac{1}{4} e^{4t} \), plus a constant of integration \( C \). This constant \( C \) will be determined by taking into account additional conditions, such as initial conditions.
Initial Conditions
Initial conditions help to uniquely determine the specific solution for a differential equation among an infinite number of possible ones. In the context of this problem, we start with \( y_0 = 10 \), meaning the value of \( y \) at time \( t = 0 \) is 10.
Using this initial condition gives a point through which the solution curve must pass, allowing us to solve for the constant of integration \( C \).
For instance, we substitute \( t = 0 \) and \( y = 10 \) into our integrated equation \( y(t) = \frac{1}{4} e^{4t} + C \), which leads to finding that \( C = \frac{39}{4} \).
This makes our solution tailored to the specific scenario of this exercise.
Using this initial condition gives a point through which the solution curve must pass, allowing us to solve for the constant of integration \( C \).
For instance, we substitute \( t = 0 \) and \( y = 10 \) into our integrated equation \( y(t) = \frac{1}{4} e^{4t} + C \), which leads to finding that \( C = \frac{39}{4} \).
This makes our solution tailored to the specific scenario of this exercise.
Exponential Functions
Exponential functions are a type of mathematical function of the form \( e^{kt} \), where \( e \) is the base of the natural logarithm, and \( k \) is a constant that dictates the rate of growth or decay.
In this exercise, the equation \( d t \) represents an exponential rate of growth for \( y \).
This function increases rapidly because the value of \( e^{4t} \) grows as \( t \) becomes larger.
Understanding that exponential functions can only take positive values helps us recognize that the solution \( e^{4t} = -35 \) we encountered for \( y = 1 \) is impossible.
The exponential growth indicated by our differential equation implies that \( y \) can increase significantly over time but never decrease to the level of 1 under the given conditions.
In this exercise, the equation \( d t \) represents an exponential rate of growth for \( y \).
This function increases rapidly because the value of \( e^{4t} \) grows as \( t \) becomes larger.
Understanding that exponential functions can only take positive values helps us recognize that the solution \( e^{4t} = -35 \) we encountered for \( y = 1 \) is impossible.
The exponential growth indicated by our differential equation implies that \( y \) can increase significantly over time but never decrease to the level of 1 under the given conditions.
Constant of Integration
The constant of integration \( C \) represents an arbitrary constant that arises when computing the integral of a function.
Since integration is the reverse process of differentiation, it inherently lacks information about the original vertical position of the function.
This is why, without any conditions, there are infinite possible functions that can be differentiated into the same result.
Adding initial conditions, such as \( y_0 = 10 \), allows us to solve for a specific \( C \).
In this problem, substituting \( y(0) = 10 \) into \( y(t) = \frac{1}{4} e^{4t} + C \) enables us to calculate \( C \) as \( \frac{39}{4} \).
This helps build the complete function that matches initial conditions and satisfies the differential equation.
Since integration is the reverse process of differentiation, it inherently lacks information about the original vertical position of the function.
This is why, without any conditions, there are infinite possible functions that can be differentiated into the same result.
Adding initial conditions, such as \( y_0 = 10 \), allows us to solve for a specific \( C \).
In this problem, substituting \( y(0) = 10 \) into \( y(t) = \frac{1}{4} e^{4t} + C \) enables us to calculate \( C \) as \( \frac{39}{4} \).
This helps build the complete function that matches initial conditions and satisfies the differential equation.
Other exercises in this chapter
Problem 6
Solve the following equations for \(x\) : (a) \(\log _{10}\left(10^{x}\right)=7\) (b) \(\log 4 x-\log 4=\log 3\) (c) \(\log _{x} 10=2\) (d) \(\log _{2}(1 / x)=2
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With the slow method multiply out the three terms of \(\left(1-\frac{1}{2}\right)^{2}\) and the five terms of \(\left(1-\frac{1}{4}\right)^{4}\). What are the f
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Find the derivative \(d y / d x\) in \(1-10\). $$ y=\ln (\sin x) $$
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Find the derivatives of the functions $$ (2 / 3)^{x} $$
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