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Modeling yeast populations (cont.) Verify that \(N(t)=\frac{42 e^{0.55 t}}{209.8+0.2 e^{0.55 t}}\) is an approximate solution of the differential equation $$\frac{d N}{d t}=(0.55-0.0026 N) N$$

Short Answer

Expert verified
The function \(N(t)\) is an approximate solution to the equation.

Step by step solution

01

Calculate dN/dt using N(t)

First, we need to differentiate the given function \(N(t) = \frac{42 e^{0.55 t}}{209.8 + 0.2 e^{0.55 t}}\) with respect to \(t\). We apply the quotient rule which states that for a function \(\frac{u}{v}\), the derivative \(\frac{d}{dt}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}\). Here, let \(u = 42 e^{0.55 t}\) and \(v = 209.8 + 0.2 e^{0.55 t}\). Then \(u' = 42 \cdot 0.55 e^{0.55 t}\) and \(v' = 0.2 \cdot 0.55 e^{0.55 t}\). Substitute these into the formula to get \(\frac{dN}{dt}\).
02

Express dN/dt in terms of N

From Step 1, we would have derived \(\frac{dN}{dt}\) from the differentiation. Now express \(N(t)\) using the original function. Substitute this expression back into \(\frac{dN}{dt}\) to write \(\frac{dN}{dt}\) as a function of \(N\), simplifying where possible to get the expression similar to \((0.55 - 0.0026N)N\).
03

Compare dN/dt with the differential equation

Now, compare the expression obtained from Step 2 with the right-hand side of the differential equation \((0.55 - 0.0026N)N\). They should be approximately equal indicating that the given function \(N(t)\) is approximately a solution to the differential equation.
04

Conclusion

Since the expression obtained for \(\frac{dN}{dt}\) matches the differential equation, we have verified that \(N(t)=\frac{42 e^{0.55 t}}{209.8+0.2 e^{0.55 t}}\) is indeed an approximate solution for the differential equation \(\frac{dN}{dt}=(0.55 - 0.0026N)N\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Yeast Population Modeling
Yeast population modeling helps us understand how yeast grows over time. In biology, yeast is often used in experiments because it multiplies quickly and is predictable. By modeling its population, scientists can predict how it changes under different conditions.
To model a yeast population, we use mathematical equations called differential equations. These equations describe how the rate of change of a population depends on the current population size. In this exercise, we have a differential equation \(\frac{dN}{dt}=(0.55 - 0.0026N)N\), which calculates how fast the number of yeast cells changes over time.
This model is important in various applications, like brewing, baking, and biofuel production. By understanding the growth pattern, processes can be optimized for better yields.
  • Yeast grows quickly and predictably, making it ideal for mathematical modeling.
  • We use differential equations to describe the population changes.
  • This knowledge aids in industries like brewing and baking.
Quotient Rule in Calculus
In calculus, we often find the derivative of a function to understand how it changes. When dealing with fractions of formulas, like in this yeast problem, we use the quotient rule. This rule helps find the derivative of a function given as one function divided by another.
The quotient rule states: for \( y = \frac{u}{v} \), \[ \frac{dy}{dt} = \frac{u'v - uv'}{v^2} \], where \(u\) is the numerator and \(v\) is the denominator. This formula requires knowing the derivatives of both \(u\) and \(v\).
In our exercise, \(u = 42 e^{0.55 t}\) and \(v = 209.8 + 0.2 e^{0.55 t}\). We needed to find \(u'\) and \(v'\) to apply the quotient rule successfully. Using this method, we can calculate the rate of yeast population change over time.
  • The quotient rule assists in differentiating fractions.
  • Requires derivatives of both numerator and denominator functions.
  • Essential for solving equations involving rates of change.
Approximate Solutions
In mathematics, exact solutions are sometimes difficult to obtain. Approximations help us find solutions that are close enough to the actual value, making them practical for real-world situations.
When modeling populations like yeast, exact calculations usually involve complex equations. By using approximate solutions, we simplify the math while maintaining useful accuracy. In this exercise, we verify that an approximate solution of the differential equation fits well.
Approximations simplify complex models by providing a close estimate. In the exercise, the function \(N(t) = \frac{42 e^{0.55 t}}{209.8 + 0.2 e^{0.55 t}}\) is an approximate solution to the equation. This serves as an effective model for describing how the yeast population grows.
  • Exact solutions can be impractical; approximations provide an alternative.
  • They simplify complex equations into more manageable formulas.
  • In yeast modeling, approximations are still useful for predicting population changes.
Biomathematics
Biomathematics is the intersection of biology and mathematics, providing a framework to solve biological problems using mathematical principles. This field allows for the construction of models to represent biological systems, like yeast populations.
By using biomathematics, scientists can predict biological phenomena, optimize industrial processes, and develop new technologies. The yeast population problem demonstrates how a mathematical model can describe a biological process.
Through differential equations, we model the interaction between growth rates and current population sizes in yeast. Biomathematics offers powerful tools to apply math to biological queries, opening doors to advanced research and application.
  • Biomathematics bridges biology with mathematics.
  • Models biological processes such as yeast growth.
  • Enables enhanced understanding and innovation in biological research.

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Most popular questions from this chapter

A system of differential equations is given. (a) Use a phase plane analysis to determine the values of the constant \(a\) for which the sole equilibrium of the differential equations is locally stable. (b) Obtain an expression for each equilibrium (it may be a function of the constant \(a\) ). \(x^{\prime}=a y^{2}-x+1, \quad y^{\prime}=2(1-y)\)

According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass \(m\) that has been projected vertically upward from the earth's surface is $$F=\frac{m g R^{2}}{(x+R)^{2}}$$ where \(x=x(t)\) is the object's distance above the surface at time \(t, R\) is the earth's radius, and \(g\) is the acceleration due to gravity. Also, by Newton's Second Law, \(F=m a=m(d v / d t)\) and so $$v_{0}=\sqrt{\frac{2 g R h}{R+h}}$$ [Hint: By the Chain Rule, \(m(d v / d t)=m v(d v / d x) . ]\) $$\begin{array}{l}{\text { (b) Calculate } v_{c}=\lim _{h \rightarrow \infty} v_{0 .} \text { This limit is called the }} \\ {\text { escape velocity for the earth. }} \\ {\text { (c) Use } R=3960 \text { mi and } g=32 \mathrm{ft} / \mathrm{s}^{2} \text { to calculate } v_{e} \text { in }} \\ {\text { feet per second and in miles per second. }}\end{array}$$

(a) Use a computer algebra system to draw a direction field for the differential equation. Get a printout and use it to sketch some solution curves without solving the differential equation. (b) Solve the differential equation. (c) Use the CAS to draw several members of the family of solutions obtained in part (b). Compare with the curves from part (a). \(y ^ { \prime } = y ^ { 2 }\)

The logistic differential equation Suppose that the per capita growth rate of a population of size \(N\) declines linearly from a value of \(r\) when \(N=0\) to a value of 0 when \(N=K\) Show that the differential equation for \(N\) is $$\frac{d N}{d t}=r\left(1-\frac{N}{K}\right) N$$

Use Euler's method with step size 0.1 to estimate \(y(0.5),\) where \(y(x)\) is the solution of the initial-value problem \(y^{\prime}=y+x y, y(0)=1\)

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