# Exponential Growth and Differential Equation

• MHB
• hongyuan94
In summary: C$as the term for the growth rate, what would we subtract to account for culling?You have a linear growth rate, but we need a growth rate that is in part proportional to the population itself and minus the cull... so if we have$kC$as the term for the growth rate, what would we subtract to account for culling?The culling would be a constant value, so we would subtract that from the growth term. Therefore, our ODE would be:dC/dt = kC - 20And our initial value is C(0) = 500 * (11/10)^5. Does this make sense?The culling would be a constant hongyuan94 a) On January 1 2000, the park estimated that they had 500 deer on their land. Two years later, they estimated that there were 550 deer on the land. Assume that the number of deer was changing exponentially, i.e. P(t)=ae^bt where P is the number of deer at year t, and a and b are parameters.Find the values of a and b. b) On January 1, 2005, the park management determined that the deer population is growing too quickly. So they decided to cull 20 deer from the herd every year thereafter (starting from January 1, 2006). Assume that the growth rate of the deer population is the same as in (a). Determine the differential equation that governs this situation, and solve it. Assume t=0 for the year 2005. c) On January 1, 2010, the park management felt that the deer population is still rampant. They decided to remove deer so that the instantaneous rate of decrease is 10% of the deer population. Determine the differential equation that governs this situation, and solve it. Assume t=0 for the year 2010. Need help with b and c, only able to do part (a), I got a = 500, b = 0.4476 For part a), we are told that: $$\displaystyle P(t)=ae^{bt}$$ where $$\displaystyle P(0)=500$$ and $$\displaystyle P(2)=550$$ Hence: $$\displaystyle ae^{b\cdot0}=a=500$$ And so: $$\displaystyle 500e^{b\cdot2}=550\implies e^{2b}=\frac{11}{10}\implies b=\frac{1}{2}\ln\left(\frac{11}{10}\right)\approx0.0476550899021624$$ So, you may now express the solution as: $$\displaystyle P(t)=500\left(\frac{11}{10}\right)^{\frac{t}{2}}$$ Now, for part b) we are going to need to construct the appropriate IVP, which consists of an ODE and an initial value, which will be the population of the deer at the beginning of 2005. Can you state the ODE governing the deer population given the growth rate and culling that will be taking place once a year? MarkFL said: For part a), we are told that: $$\displaystyle P(t)=ae^{bt}$$ where $$\displaystyle P(0)=500$$ and $$\displaystyle P(2)=550$$ Hence: $$\displaystyle ae^{b\cdot0}=a=500$$ And so: $$\displaystyle 500e^{b\cdot2}=550\implies e^{2b}=\frac{11}{10}\implies b=\frac{1}{2}\ln\left(\frac{11}{10}\right)\approx0.0476550899021624$$ So, you may now express the solution as: $$\displaystyle P(t)=500\left(\frac{11}{10}\right)^{\frac{t}{2}}$$ Now, for part b) we are going to need to construct the appropriate IVP, which consists of an ODE and an initial value, which will be the population of the deer at the beginning of 2005. Can you state the ODE governing the deer population given the growth rate and culling that will be taking place once a year? Part (b) we use the formula from (a) $$\displaystyle P(t)=500\left(\frac{11}{10}\right)^{\frac{t}{2}}$$, for year 2005 t=0, so P(0) = 500, next is year 2006 where 20 deer are culled every year, so we use the term -20t then substitute in the formula we have P(1) = 500 e^-20 No,$P(t)$governs the population from 2000-2005, so on January 1, 2005 the population is $$\displaystyle P(5)=500\left(\frac{11}{10}\right)^{\frac{5}{2}}$$. Let's call$C(t)$the population once culling begins. We must also assume that culling is continuous rather than discrete as actually given (otherwise we will have a difference equation rather than a differential equation). So, we know$C(0)=P(5)$is our initial value. Now you need to construct the ODE, of the form: $$\displaystyle \d{C}{t}=f(t)$$ Now$f$will have two terms, one that represent the natural growth and one that represents the culling. The growth rate is proportional to$C$while the culling rate is a constant and must be negative since deer are being taken from the population. Can you express the ODE? MarkFL said: No,$P(t)$governs the population from 2000-2005, so on January 1, 2005 the population is $$\displaystyle P(5)=500\left(\frac{11}{10}\right)^{\frac{5}{2}}$$. Let's call$C(t)$the population once culling begins. We must also assume that culling is continuous rather than discrete as actually given (otherwise we will have a difference equation rather than a differential equation). So, we know$C(0)=P(5)$is our initial value. Now you need to construct the ODE, of the form: $$\displaystyle \d{C}{t}=f(t)$$ Now$f$will have two terms, one that represent the natural growth and one that represents the culling. The growth rate is proportional to$C$while the culling rate is a constant and must be negative since deer are being taken from the population. Can you express the ODE? take from part (a) b is the growth rate, so dC/dt = bt-20 ? hongyuan94 said: take from part (a) b is the growth rate, so dC/dt = bt-20 ? No, your first term is not proportional to$C$...:) MarkFL said: No, your first term is not proportional to$C\$...:)
sorry, I can't really understand it. What means proportional to C here? I thought that b is the growth rate so b multiply with t would be the growth then minus 20 would be the after-culling growth rate.

hongyuan94 said:
sorry, I can't really understand it. What means proportional to C here? I thought that b is the growth rate so b multiply with t would be the growth then minus 20 would be the after-culling growth rate.

You have a linear growth rate, but we need a growth rate that is in part proportional to the population itself and minus the cull rate.

## What is exponential growth?

Exponential growth is a type of growth or increase in which the quantity grows at a rate proportional to its current value. This means that the larger the quantity, the faster it will grow.

## What is a differential equation?

A differential equation is an equation that relates a function with one or more of its derivatives. It is used to model many real-world phenomena, including exponential growth.

## What are the applications of exponential growth?

Exponential growth is used to model various natural and social phenomena, such as population growth, compound interest, and radioactive decay. It is also commonly used in economics, biology, and physics.

## How is exponential growth different from linear growth?

In exponential growth, the quantity grows at a rate proportional to its current value, resulting in a curve that increases rapidly over time. In contrast, linear growth is when the quantity grows at a constant rate, resulting in a straight line on a graph.

## How can differential equations be solved?

Differential equations can be solved analytically by finding an explicit formula for the function, or numerically by using computational methods. Some common techniques for solving differential equations include separation of variables, variation of parameters, and the method of undetermined coefficients.

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