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To what exponential growth rate per hour does a growth rate of \(100 \%\) per day correspond?

Short Answer

Expert verified
The exponential growth rate per hour is approximately 2.89%.

Step by step solution

01

Understanding the Problem

We need to find the exponential growth rate per hour that corresponds to a 100% growth rate per day. This means that we need to convert the daily growth rate to an hourly growth rate for exponential growth.
02

Exponential Growth Formula

Exponential growth can be described by the formula \[ A(t) = A_0 e^{rt} \]where \( A(t) \) is the amount at time \( t \), \( A_0 \) is the initial amount, \( r \) is the growth rate, and \( t \) is the time. A 100% growth rate means \( A(1) = 2A_0 \) in 24 hours.
03

Setting Up the Equation for Daily Growth

We set up the equation for growth over 24 hours:\[ 2A_0 = A_0 e^{24r} \]By dividing both sides by \( A_0 \), we simplify to:\[ 2 = e^{24r} \]
04

Solve for Daily Growth Rate \( r \)

To find \( r \), take the natural logarithm of both sides:\[ \ln(2) = 24r \]Thus, the daily growth rate \( r \) is:\[ r = \frac{\ln(2)}{24} \]
05

Convert Daily to Hourly Growth Rate

To find the hourly growth rate, we set up a new equation for one hour:\[ e^{r_{hour}} = e^{\frac{\ln(2)}{24}} \]Where \( r_{hour} \) is the hourly growth rate. Simplifying, we get:\[ r_{hour} = \frac{\ln(2)}{24} \]
06

Calculate the Result

Calculate the value:\[ r_{hour} = \frac{0.6931}{24} \approx 0.0289 \]Thus, the exponential growth rate per hour is approximately 0.0289 or 2.89%.

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Key Concepts

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

Hourly Growth Rate
When dealing with any form of exponential growth, understanding how it translates into different time frames is crucial. The hourly growth rate helps us see the rate of increase per hour. Imagine you're observing a fast-growing plant or a rapidly increasing bank balance. You’re interested in knowing how much it grows every single hour instead of just each day or year.

An hourly growth rate gives you a fine-tuned snapshot, making it easier to understand fluctuations within any shorter time span. To find this rate, you often start with a daily or yearly rate. The process typically involves converting these rates using formulas, such as the one used to find the daily rate from an annual rate, already familiar from different growth models.

In the context of exponential growth, each hourly rate is a small step in the larger picture. A 100% per day growth rate translates to knowing how much exactly increases each hour – in this case, approximately 2.89% per hour. This helps in detailed planning and forecasting in finance, biology, and many other areas.
Natural Logarithm
The natural logarithm plays a pivotal role in calculating growth rates in exponential processes. If you are new to logarithms, think of them as the reverse operations of exponentiation. With exponential growth models, using the natural logarithm, denoted as \( \ln(x) \), helps to solve for variables that are stuck in an exponent.

Essentially, the natural log lets you break down the complexity of exponentials into manageable arithmetic. For example, when you encounter an equation where you need to solve for time or rate, as in the expression \( \ln(2) = 24r \), using the natural log helps isolate \( r \). The natural log is specifically ideal when dealing with growth rates because of the base of the exponential constant \( e \), approximately 2.718, inherent in natural growth processes.

It's important in many fields like physics, biology, and economics, especially when analyzing continuous growth. Utilizing \( \ln(x) \) empowers you to translate complex exponential relationships into simpler linear forms.
Growth Rate Conversion
Sometimes, it's crucial to convert growth rates from one time frame to another to make meaningful comparisons or predictions. This task, referred to as growth rate conversion, helps you understand how a daily rate might look on an hourly scale or vice versa.

The process usually involves basic math operations like division or multiplication, facilitated further by logarithmic properties. For instance, converting a daily exponential growth rate of 100% to an hourly rate involves understanding the concept of exponential compounding. It divides the natural log of 2 by 24 to get the hourly equivalent, which is approximately 0.0289.

This conversion is essential when planning resources, schedules, or even experiments. By converting these rates, you get a precise understanding of how frequently changes occur, aligned with decision-making periods. It makes the relationship between different time-perspective growths tangible and understandable, empowering you to act decisively.

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Most popular questions from this chapter

Describe the differences in the graphs of \(f(x)=3^{x}\) and \(g(x)=x^{3}\)

The tortoise population, \(P(t),\) in a square mile of the Mojave Desert after \(t\) years can be approximated by the logistic equation $$ P(t)=\frac{3000}{20+130 e^{-0.214 t}} $$ a) Find the tortoise population after \(0 \mathrm{yr}, 5 \mathrm{yr}, 15 \mathrm{yr},\) and 25 yr. b) Find the rate of change in the population, \(P^{\prime}(t)\). c) Find the rate of change in the population after \(0 \mathrm{yr}\) \(5 \mathrm{yr}, 15, \mathrm{yr},\) and \(25 \mathrm{yr}\) d) What is the limiting value (see Exercise 42 ) for the population of tortoises in a square mile of the Mojave Desert?

A lake is stocked with 400 rainbow trout. The size of the lake, the availability of food, and the number of other fish restrict population growth to a limiting value of 2500 trout. The population of trout in the lake after time \(t\), in months, is approximated by $$ P(t)=\frac{2500}{1+5.25 e^{-0.32 t}} $$ a) Find the population after 0 months, 1 month, 5 months, 10 months, 15 months, and 20 months. b) Find the rate of change, \(P^{\prime}(t)\). c) Sketch a graph of the function.

Differentiate. $$ y=\frac{e^{x}}{x^{2}+1} $$

The Rule of 70. The relationship between doubling time \(T\) and growth rate \(k\) is the basis of the Rule of \(70 .\) Since $$ T=\frac{\ln 2}{k}=\frac{0.693147}{k}=\frac{69.3147}{100 k} \approx \frac{70}{100 k} $$ we can estimate the length of time needed for a quantity to double by dividing the growth rate \(k\) (expressed as a percentage) into \(70 .\) Estimate the time needed for an amount of money to double, if the interest rate is \(7 \%,\) compounded.

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