Math Problem Statement

giải bài tập

Trong thí nghiệm y học, người ta cấy 1000 vi khuẩn vào môi trường dinh dưỡng. Bằng thực nghiệm, người ta xác định số lượng vi khuẩn thay đổi theo thời gian bởi công thức N(t) = 1000 + 100t / (100 + t^2) (con). Trong đó t là thời gian tính bằng giây. Tính số lượng vi khuẩn lớn nhất kể từ khi thực hiện cấy vi khuẩn vào môi trường dinh dưỡng.

Solution

To solve this problem, let's analyze and understand the given information.

We are provided with a formula describing the number of bacteria N(t)N(t) over time tt:

N(t)=1000+100t100+t2N(t) = 1000 + \frac{100t}{100 + t^2}

where:

  • tt is time in seconds,
  • N(t)N(t) represents the number of bacteria at time tt.

Task

The question asks for the maximum number of bacteria after the initial seeding of 1000 bacteria.

Solution Approach

To find the maximum value of N(t)N(t), we need to:

  1. Take the derivative of N(t)N(t) with respect to tt to find N(t)N'(t).
  2. Set N(t)=0N'(t) = 0 to find critical points, which are potential maximum points.
  3. Evaluate N(t)N(t) at the critical points to find the maximum number of bacteria.

Step 1: Derivative of N(t)N(t)

The function N(t)N(t) can be differentiated using the quotient rule for the term 100t100+t2\frac{100t}{100 + t^2}.

  1. Differentiate N(t)N(t) to find N(t)N'(t).
  2. Solve for tt when N(t)=0N'(t) = 0.

Step 2: Find the Maximum N(t)N(t)

After finding the critical points, substitute them back into N(t)N(t) to determine the maximum number of bacteria.

Let me calculate this for you.The critical points for tt are t=10t = -10 and t=10t = 10. However, since time tt represents seconds after the experiment starts, we only consider positive values of tt.

Evaluating N(t)N(t) at t=10t = 10: N(10)=1005N(10) = 1005

Thus, the maximum number of bacteria after the initial seeding is 1005 bacteria.

Would you like further details on the steps taken, or have any questions?

Here are five related questions to deepen understanding:

  1. How would the maximum number change if the initial number of bacteria was different?
  2. How does the rate of growth of the bacteria change as time progresses?
  3. What does the second derivative of N(t)N(t) tell us about the concavity and nature of the critical point?
  4. How would the solution change if tt was measured in minutes instead of seconds?
  5. What is the significance of the function's asymptotic behavior as tt \to \infty?

Tip: For any function describing growth, finding critical points and evaluating limits at large values helps understand long-term trends.

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Math Problem Analysis

Mathematical Concepts

Calculus
Optimization
Derivatives

Formulas

N(t) = 1000 + (100t) / (100 + t^2)

Theorems

First Derivative Test for Maximum

Suitable Grade Level

Grades 11-12