Problem 361 - Calculus Volume 1

Question

For the following exercises, use \(y=y_{0} e^{k t}\). If \(y=100\) at \(t=4\) and \(y=10\) at \(t=8\), when does \(y=1 ?\)

Step-by-Step Solution

Answer

The value of \( t \) when \( y = 1 \) is 12.

Step 1: Identify Initial Conditions

We have two pieces of information: when \( t = 4 \), \( y = 100 \), and when \( t = 8 \), \( y = 10 \). We need to use these conditions to find \( y_0 \) and \( k \).

Step 2: Use First Condition to Formulate Equation

Using the equation \( y = y_0 e^{kt} \) with the condition \( t = 4 \), \( y = 100 \), we get:\[ 100 = y_0 e^{4k} \].

Step 3: Use Second Condition to Formulate Equation

Using the condition \( t = 8 \), \( y = 10 \), the equation becomes:\[ 10 = y_0 e^{8k} \].

Step 4: Solve the System of Equations for \( k \)

Divide the second equation by the first equation to eliminate \( y_0 \):\[ \frac{10}{100} = \frac{y_0 e^{8k}}{y_0 e^{4k}} \]\[ \frac{1}{10} = e^{4k} \]Take the natural logarithm on both sides:\[ \ln\left(\frac{1}{10}\right) = 4k \].Thus, solve for \( k \):\[ k = \frac{\ln\left(\frac{1}{10}\right)}{4} \].

Step 5: Apply Known Decisions to Solve for \( y_0 \)

Substitute \( k \) back into the first equation to solve for \( y_0 \):\[ 100 = y_0 e^{4 * \frac{\ln\left(\frac{1}{10}\right)}{4}} \]. Since \( e^{\ln\left(\frac{1}{10}\right)} = \frac{1}{10} \),\[ 100 = y_0 * \frac{1}{10} \].Hence, \( y_0 = 1000 \).

Step 6: Solve for \( t \) When \( y = 1 \)

Now we have \( y_0 = 1000 \) and \( k = \frac{\ln\left(\frac{1}{10}\right)}{4} \). Use the equation \( 1 = 1000 e^{kt} \) to find \( t \):\[ 1 = 1000 e^{kt} \],\[ \frac{1}{1000} = e^{kt} \].Taking the natural logarithm on both sides:\[ \ln\left(\frac{1}{1000}\right) = kt \].Substitute \( k = \frac{\ln\left(\frac{1}{10}\right)}{4} \) into the equation:\[ \frac{\ln(\frac{1}{1000})}{\ln(\frac{1}{10})/4} = t \].

Step 7: Calculate Final Value for \( t \)

Calculate both logarithms. Since \( \ln(\frac{1}{10}) = -\ln(10) \) and \( \ln(\frac{1}{1000}) = -3\ln(10) \), substitute these into the equation:\[ t = \frac{-3\ln(10)}{\frac{-\ln(10)}{4}} = 12 \].

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific known conditions, known as initial conditions. These initial conditions specify the value of the function at a certain point, allowing us to find a unique solution.

In the context of exponential growth and decay, we often deal with equations of the form:

• \( y = y_{0} e^{kt} \)
• \( y_0 \) is the initial amount when \( t = 0 \)
• \( k \) is the growth (or decay) rate

To solve such problems, we need to find out the initial value \( y_0 \) and the rate \( k \). Let's consider a problem where you are told the amount at different times. By using these values, you can form equations and solve for \( y_0 \) and \( k \). This allows us to describe how the process evolves over time with a precise mathematical function.
The given problem provides two points: \( y=100 \) at \( t=4 \), and \( y=10 \) at \( t=8 \). We can use these data points to set up equations that help determine the specific exponential model, making the initial value problem solvable.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). It is a powerful tool in solving equations involving exponential functions because it allows us to "bring down" the exponent, making the equation linear.

For example, if you have an equation like \( e^k = a \), you can take the natural logarithm of both sides to obtain:

• \( \ln(e^k) = \ln(a) \)
• This simplifies to \( k = \ln(a) \), thanks to the property of logarithms that \( \ln(e^x) = x \)

In our case, finding \( k \) required using \( \ln \) to handle the division of exponential terms. By taking the natural logarithm of both sides, we could solve for \( k \) straightforwardly:
- \( \ln\left(\frac{1}{10}\right) = 4k \)The natural log turns multiplicative processes into additive ones, which is why it's so useful in exponential equations.

System of Equations

A system of equations is a set of equations with multiple variables that you solve together, finding a common solution that satisfies all the equations in the set.

In our exponential growth and decay problem, we dealt with the system of equations from the initial conditions:

• \( 100 = y_0 e^{4k} \)
• \( 10 = y_0 e^{8k} \)

The goal was to find values for \( y_0 \) and \( k \) such that both conditions are satisfied.

To simplify the process, we used division to eliminate one variable, \( y_0 \), leaving us with a single equation that could be solved for \( k \).By solving this single equation, we could back-substitute to find the value of \( y_0 \). This strategic manipulation of a system of equations is key in multi-variable problems.

Solving Exponential Equations

Exponential equations are those where variables appear as exponents. Solving these equations often requires transformation techniques, such as using logarithms, to simplify and solve for the variable.

In our problem, we needed to find \( k \) and \( y_0 \). After simplifying our equations using division, we ended up with:

• \( e^{4k} = \frac{1}{10} \)
• \( k = \frac{\ln\left(\frac{1}{10}\right)}{4} \)

After obtaining \( k \), we solved for \( y_0 \) using back-substitution: \( y_0 = 1000 \).

Finally, to find when \( y = 1 \), we set up the equation and used: - \( \ln(\frac{1}{1000}) = kt \) - Solving gives \( t = \frac{-3\ln(10)}{-\ln(10)/4} = 12 \)
This methodical approach to solving exponential equations ensures accurate results by handling all parts of the equation systematically.