dy =ry ln dt ´K =), subject to the initial Solve the Gompertz equation condition y(0) = yo- Hint: You may wish to let u = ln K. (). For r= 0.73 per year, K = 80.6 × 10° kg, = 0.3, use the Gompertz model to find the Yo K time 7 at which y(7) = 0.9K. NOTE: Enter an exact answer. T=
dy =ry ln dt ´K =), subject to the initial Solve the Gompertz equation condition y(0) = yo- Hint: You may wish to let u = ln K. (). For r= 0.73 per year, K = 80.6 × 10° kg, = 0.3, use the Gompertz model to find the Yo K time 7 at which y(7) = 0.9K. NOTE: Enter an exact answer. T=
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![**Solve the Gompertz equation \(\frac{dy}{dt} = r \, y \, \ln\left(\frac{K}{y}\right)\), subject to the initial condition \(y(0) = y_0\).**
**Hint:** You may wish to let \(u = \ln\left(\frac{y}{K}\right)\). For \( r = 0.73 \) per year,
**Given:**
- \( K = 80.6 \times 10^6 \) kg,
- \(\frac{y_0}{K} = 0.3\),
use the Gompertz model to find the time \(\tau\) at which \(y(\tau) = 0.9K\).
**NOTE:** Enter an exact answer.
\[
\tau = \_\_\_\_\_\_
\]
In the context provided:
- **Gompertz Equation:** The differential equation given is \(\frac{dy}{dt} = r \, y \, \ln\left(\frac{K}{y}\right)\), which is a form used in modeling growth processes where \( r \) is the growth rate, \( y \) is the current size, and \( K \) is the carrying capacity.
- **Initial Condition:** \(y(0) = y_0\) signifies that at time \(t = 0\), the initial size is \( y_0 \).
- **Hint Provided:** Introducing the substitution \( u = \ln\left(\frac{y}{K}\right) \) may simplify the differential equation for easier solving.
- **Parameters:** With \( r = 0.73 \) per year and \( K = 80.6 \times 10^6 \) kg, \( \frac{y_0}{K} = 0.3 \).
- **Objective:** Determine the time \(\tau\) at which \(y(\tau) = 0.9K\), which involves solving the model with the given parameters and initial conditions.
When formulating the response on an educational website, ensure all steps are carefully broken down to enable the learner to understand the thought process and solution path.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F56945fae-aeca-4f60-8c3d-a266c334f238%2F0d45753b-3f49-4a53-98d3-59cbfa95e700%2Fp9xvr0s_processed.png&w=3840&q=75)
Transcribed Image Text:**Solve the Gompertz equation \(\frac{dy}{dt} = r \, y \, \ln\left(\frac{K}{y}\right)\), subject to the initial condition \(y(0) = y_0\).**
**Hint:** You may wish to let \(u = \ln\left(\frac{y}{K}\right)\). For \( r = 0.73 \) per year,
**Given:**
- \( K = 80.6 \times 10^6 \) kg,
- \(\frac{y_0}{K} = 0.3\),
use the Gompertz model to find the time \(\tau\) at which \(y(\tau) = 0.9K\).
**NOTE:** Enter an exact answer.
\[
\tau = \_\_\_\_\_\_
\]
In the context provided:
- **Gompertz Equation:** The differential equation given is \(\frac{dy}{dt} = r \, y \, \ln\left(\frac{K}{y}\right)\), which is a form used in modeling growth processes where \( r \) is the growth rate, \( y \) is the current size, and \( K \) is the carrying capacity.
- **Initial Condition:** \(y(0) = y_0\) signifies that at time \(t = 0\), the initial size is \( y_0 \).
- **Hint Provided:** Introducing the substitution \( u = \ln\left(\frac{y}{K}\right) \) may simplify the differential equation for easier solving.
- **Parameters:** With \( r = 0.73 \) per year and \( K = 80.6 \times 10^6 \) kg, \( \frac{y_0}{K} = 0.3 \).
- **Objective:** Determine the time \(\tau\) at which \(y(\tau) = 0.9K\), which involves solving the model with the given parameters and initial conditions.
When formulating the response on an educational website, ensure all steps are carefully broken down to enable the learner to understand the thought process and solution path.
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