Math Problem Statement

A population changes according to the formula P(t) = 5.4^t. Which of the following statements are always true? (i) P'(t) = lim(h→0) (5.4^h(5.4^t − 1))/h (ii) P'(1) is less than the average rate of change of P between 0 and 1. (iii) The average rate of change between 0 and 1 is 5.4. Options: (A) (i) only (B) (i) and (ii) only (C) (i) and (iii) only (D) none of them (E) (ii) and (iii) only (F) all of them (G) (iii) only (H) (ii) only

Solution

We are given a problem about population change modeled by the function P(t)=5.4tP(t) = 5.4^t. The task is to determine which of the given statements are always true. Let's examine each statement.

Statement (i):

P(t)=limh05.4h(5.4t1)hP'(t) = \lim_{h \to 0} \frac{5.4^h(5.4^t - 1)}{h} This is the definition of the derivative of P(t)P(t) at time tt, using the limit form. To confirm this, we need to find P(t)P'(t).

Given P(t)=5.4tP(t) = 5.4^t, we apply the general rule for differentiating exponential functions: P(t)=ddt(5.4t)=5.4tln(5.4)P'(t) = \frac{d}{dt} \left( 5.4^t \right) = 5.4^t \ln(5.4) Thus, the derivative of P(t)P(t) is P(t)=5.4tln(5.4)P'(t) = 5.4^t \ln(5.4). Therefore, the given limit expression is consistent with this derivative formula. So, (i) is true.


Statement (ii):

P(1) is less than the average rate of change of P between 0 and 1.P'(1) \text{ is less than the average rate of change of } P \text{ between 0 and 1.} First, we compute P(1)P'(1). Using the formula from statement (i): P(1)=5.41ln(5.4)=5.4ln(5.4)P'(1) = 5.4^1 \ln(5.4) = 5.4 \ln(5.4) Now, the average rate of change of P(t)P(t) between t=0t = 0 and t=1t = 1 is: Average rate of change=P(1)P(0)10=P(1)P(0)\text{Average rate of change} = \frac{P(1) - P(0)}{1 - 0} = P(1) - P(0) Since P(1)=5.41=5.4P(1) = 5.4^1 = 5.4 and P(0)=5.40=1P(0) = 5.4^0 = 1, the average rate of change is: 5.41=4.45.4 - 1 = 4.4 Now, we compare P(1)=5.4ln(5.4)P'(1) = 5.4 \ln(5.4) with the average rate of change, 4.44.4.

The value of ln(5.4)\ln(5.4) is approximately 1.6861.686, so: P(1)=5.4×1.6869.11P'(1) = 5.4 \times 1.686 \approx 9.11 Since 9.119.11 is greater than 4.44.4, statement (ii) is false because P(1)P'(1) is not less than the average rate of change.


Statement (iii):

The average rate of change between 0 and 1 is 5.4.\text{The average rate of change between 0 and 1 is 5.4.} We already calculated the average rate of change as 4.44.4, so (iii) is false.


Conclusion:

From the analysis above, only statement (i) is true. Therefore, the correct answer is:

A\boxed{A}

Would you like any further clarification or details on these steps?


Related Questions:

  1. How is the derivative of an exponential function with a base other than ee calculated?
  2. Why is the average rate of change of a function between two points not the same as the instantaneous rate of change?
  3. Can you explain how the natural logarithm ln\ln plays a role in differentiating exponential functions?
  4. What is the significance of the limit definition of the derivative in calculus?
  5. How can we interpret the derivative of a population growth function in real-world terms?

Tip:

For any exponential function ata^t, the derivative is atln(a)a^t \ln(a), where ln(a)\ln(a) is the natural logarithm of the base.

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Math Problem Analysis

Mathematical Concepts

Calculus
Exponential Functions
Rates of Change
Derivatives

Formulas

P'(t) = lim(h→0) (5.4^h(5.4^t − 1))/h
P'(t) = 5.4^t * ln(5.4)
Average rate of change = (P(1) − P(0))/(1 − 0)

Theorems

Derivative of exponential functions
Limit definition of the derivative
Average rate of change

Suitable Grade Level

College Calculus