Question

A yeast grows at a rate proportional to its present size. If the original amount doubles in two hours, in how many hours will it triple?

Short answer

Answer: Approximately 3.19 hours.

Step-by-step solution

Understand the exponential growth formula

The exponential growth formula is represented by the equation: P(t) = P0 * e^(k * t) Where: - P(t) is the final amount of yeast after time t - P0 is the initial amount of yeast - e is the base of natural logarithm (approximately 2.718) - k is the growth constant - t is the time in hours

Apply the given information to the exponential growth formula

We are given that the original amount doubles in two hours. This means when t = 2, P(t) = 2 * P0. Plug these values into the formula and solve for k: 2 * P0 = P0 * e^(k * 2)

Solve for the growth constant k

Divide both sides of the equation by P0 and then take the natural logarithm of both sides to isolate k: 2 = e^(2 * k) ln(2) = ln(e^(2 * k)) ln(2) = 2 * k Now, divide by 2 to find the value of k: k = ln(2) / 2

Set up the equation to find when the original amount triples

We are asked to find how many hours it takes for the yeast to triple its original size. Let t3 be the number of hours it takes the amount to triple. Set up the equation for this situation with P(t) = 3 * P0: 3 * P0 = P0 * e^(k * t3)

Solve for t3, the time it takes for the original amount to triple

Divide both sides by P0 and substitute the value of k from step 3 into the equation, then take the natural logarithm of both sides: 3 = e^((ln(2) / 2) * t3) ln(3) = ln(e^((ln(2) / 2) * t3)) ln(3) = (ln(2) / 2) * t3 Finally, isolate t3 by dividing by (ln(2) / 2): t3 = ln(3) / (ln(2)/2)

Calculate the value of t3

Evaluate the expression on the right side of the equation to find the value of t3, which represents the time it takes for the original amount to triple t3 = ln(3) / (ln(2) / 2) ≈ 3.19 hours So, it will take approximately 3.19 hours for the original amount of yeast to triple.

Key concepts

Exponential Growth Formula

Exponentially growing situations occur frequently in science and finance, where a quantity increases at a rate that is proportional to its current value. The exponential growth formula that captures this situation is \( P(t) = P_0 \cdot e^{kt} \), where:

• \( P(t) \) is the amount at time \( t \).
• \( P_0 \) is the initial amount.
• \( e \) is the base of the natural logarithm, an irrational number approximately equal to 2.71828.
• \( k \) is the growth rate constant.
• \( t \) is the time elapsed.

To use this formula effectively, first, determine the initial value and growth rate. The growth rate is often derived from given conditions, such as in the problem where the yeast's quantity doubles over a certain time period. After calculating the growth rate, we can make predictions about future values at different times by plugging values into the formula.

Natural Logarithm

The natural logarithm is an operation that allows us to 'reverse' the effect of exponentiation with base \( e \). If we have \( e^{x} = y \), then the natural logarithm of \( y \), denoted as \( \ln(y) \), equals \( x \).

\(\ln(e^x) = x\)

It is a critical tool in solving equations involving the exponential growth formula because it allows us to isolate the growth rate \( k \) or time \( t \) when either is unknown. For instance, in our problem, we used the natural logarithm to determine the growth rate \( k \) when the yeast doubled in size after a certain period.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In terms of growth problems, differential equations can express how the rate of change of a quantity is proportional to the quantity itself. For exponential growth, the differential equation would be \( \frac{dP}{dt} = kP \).

From Exponential Models to Differential Equations

When the rate of growth is proportional to the current amount, that proportionality constant \( k \) is what we found using the natural log in the earlier steps. This relationship forms the cornerstone of understanding how changes occur over time in various systems from physics to finance. Solving differential equations can tell us how quantities change dynamically under certain rules, or 'laws' of growth or decay.

Integral Calculus

Integral calculus is the branch of calculus that deals with finding the quantity where the rate of change is known. It helps in solving for a function given its derivative -- essentially the inverse process of differentiation.

Integrating Growth Rates

Integral calculus doesn't appear directly in our original problem, but it is inherently part of the background process. If we had the differential equation and needed to find the exponential function that models our yeast growth, we'd use integral calculus. It translates the concept of accumulation - adding up tiny increments to find a total. In terms of our growth formula, integrating the rate of growth \( kP \) with respect to time would lead us back to the exponential function that describes the amount of yeast over time.