Calc 2 Exponential Growth(differential equation)

In summary, studying exponential growth in Calc 2 allows one to understand how a quantity changes over time when it grows at a rate proportional to its current value. This concept is used in various fields such as finance, population growth, and radioactive decay. Exponential growth is represented in a differential equation as dy/dt = ky, where dy/dt is the rate of change of the quantity y with respect to time, and k is the growth rate constant. It is the opposite process of exponential decay, which is represented by dy/dt = -ky. The solution to a differential equation for exponential growth is y = Ce^(kt), where C is the initial value of the quantity and k is the growth rate constant. Real-life examples of
  • #1
IWANTHOTDOG
1
0

Homework Statement



A tissue culture grows until it has an area of 9 cm^2. Let A(t) be the
area of the tissue at time t. A model of the growth rate is that:
A'(t) = k*(sqrt(A(t)) * (9-A(t))

a. Without solving the equation, show that the maximum rate of growth
occurs at any time when A(t) = 3 cm^2.

b. Assume that k = 6. Find the solution corresponding to A(0) = 1 and
sketch its graph.

c. Do the same for A(0) = 4.



Homework Equations


y(t) = b + e^kt ?


The Attempt at a Solution



For part A, I took the derivative and set that equal to 0, which gave me back 3.

For part B, i took it as a separable differential equation and got
(1/3)(ln((9-A)/(3-sqrt(A))^2)) = kt + C
but i have no idea how to solve that for A

Thanks for any help
 
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  • #2
IWANTHOTDOG said:
t
(1/3)(ln((9-A)/(3-sqrt(A))^2)) = kt + C
but i have no idea how to solve that for A

Thanks for any help

Well,

[tex]\ln\left(\frac{9-A}{(3-\sqrt{A})^2}\right)=3kt+3C\implies \frac{9-A}{(3-\sqrt{A})^2}=Be^{3kt}[/tex]

where [itex]B\equiv e^{3C}[/itex]

From there, multiply both sides by [itex](3-\sqrt{A})^2[/itex], expand everything out and collect terms in powers of [itex]A[/itex]...you will be left with an equation that is quadratic in terms of [itex]\sqrt{A}[/itex], and I'm sure you know how to solve quadratic equations.
 

1. What is the purpose of studying exponential growth in Calc 2?

The purpose of studying exponential growth in Calc 2 is to understand how a quantity changes over time when it grows at a rate proportional to its current value. This concept has numerous applications in fields such as finance, population growth, and radioactive decay.

2. How is exponential growth represented in a differential equation?

Exponential growth is represented in a differential equation by the function dy/dt = ky, where dy/dt is the rate of change of the quantity y with respect to time, and k is the growth rate constant. This equation describes the relationship between the rate of change and the current value of the quantity.

3. What is the difference between exponential growth and exponential decay?

Exponential growth and decay are opposite processes. Exponential growth occurs when a quantity increases at a constant rate, while exponential decay occurs when a quantity decreases at a constant rate. In a differential equation, exponential decay is represented by dy/dt = -ky, where the negative sign indicates the decrease in the quantity over time.

4. How is the solution to a differential equation for exponential growth determined?

The solution to a differential equation for exponential growth can be determined using the general solution y = Ce^(kt), where C is the initial value of the quantity and k is the growth rate constant. This solution can then be used to find the specific value of the quantity at a given time or to graph the growth function.

5. What are some real-life examples of exponential growth?

Some real-life examples of exponential growth include population growth, compound interest in investments, and the spread of infectious diseases. These situations involve a quantity increasing at a constant rate, resulting in a rapid growth over time.

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