Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


d. If the rate constant of an exponential growth function is increased, its doubling time is decreased.

Verified step by step guidance

1. Recall the general form of an exponential growth function: \(N(t) = N_0 e^{kt}\), where \(k\) is the rate constant and \(k > 0\) for growth.
2. Understand that the doubling time \(T\) is the time it takes for the quantity to double, so \(N(T) = 2N_0\).
3. Set up the equation for doubling time: \$2N_0 = N_0 e^{kT}\(, which simplifies to \)2 = e^{kT}$.
4. Take the natural logarithm of both sides to solve for \(T\): \(\ln(2) = kT\), so \(T = \frac{\ln(2)}{k}\).
5. Analyze the relationship: since \(\ln(2)\) is constant, if the rate constant \(k\) increases, the denominator increases, making \(T\) smaller. Therefore, increasing the rate constant decreases the doubling time.

Key Concepts

Exponential Growth Function

An exponential growth function models quantities that increase at a rate proportional to their current value, typically expressed as f(t) = f_0 * e^(kt), where k > 0 is the growth rate constant. This function describes processes like population growth or compound interest.

Rate Constant and Its Effect

The rate constant k in an exponential growth function determines how quickly the quantity grows. A larger k means the function grows faster, causing the value to increase more rapidly over time compared to a smaller k.

Doubling Time in Exponential Growth

Doubling time is the time required for an exponentially growing quantity to double in size. It is calculated as T = ln(2)/k, showing that as the rate constant k increases, the doubling time decreases, meaning the quantity doubles faster.