Math Problem Statement

  1. The number N of bacteria in a culture after ‘t’ days is modeled by N=100 (1−(4/(t3+1)^2)) , find the rate of change of N with respect to ‘t’ when t = 0 and 1.

Solution

We are tasked with finding the rate of change of NN with respect to tt for the function N(t)=100(14(t3+1)2)N(t) = 100 \left( 1 - \frac{4}{(t^3 + 1)^2} \right).

Step 1: Differentiate N(t)N(t) with respect to tt

We need to compute the derivative dNdt\frac{dN}{dt}.

Given: N(t)=100(14(t3+1)2)N(t) = 100 \left( 1 - \frac{4}{(t^3 + 1)^2} \right)

To differentiate this, we apply the chain rule. Let's first find the derivative of the function inside the parentheses:

N(t)=100(1u(t)),where u(t)=4(t3+1)2N(t) = 100 \left( 1 - u(t) \right), \quad \text{where} \ u(t) = \frac{4}{(t^3 + 1)^2}

Now, let's compute dudt\frac{du}{dt}.

Step 2: Differentiate u(t)=4(t3+1)2u(t) = \frac{4}{(t^3 + 1)^2}

Using the quotient rule or chain rule, we have:

u(t)=4(t3+1)2u(t) = 4(t^3 + 1)^{-2}

Differentiating u(t)u(t) with respect to tt using the chain rule:

dudt=8(t3+1)3ddt(t3+1)\frac{du}{dt} = -8(t^3 + 1)^{-3} \cdot \frac{d}{dt}(t^3 + 1)

The derivative of t3+1t^3 + 1 is 3t23t^2, so:

dudt=8(t3+1)33t2=24t2(t3+1)3\frac{du}{dt} = -8(t^3 + 1)^{-3} \cdot 3t^2 = -24t^2(t^3 + 1)^{-3}

Step 3: Differentiate N(t)N(t)

Now, using dNdt=100(0dudt)\frac{dN}{dt} = 100 \left( 0 - \frac{du}{dt} \right):

dNdt=100(24t2(t3+1)3)\frac{dN}{dt} = -100 \cdot \left( -24t^2(t^3 + 1)^{-3} \right)

Thus:

dNdt=2400t2(t3+1)3\frac{dN}{dt} = 2400t^2(t^3 + 1)^{-3}

Step 4: Evaluate at t=0t = 0 and t=1t = 1

  1. At t=0t = 0: dNdtt=0=240002(03+1)3=0\frac{dN}{dt}\bigg|_{t=0} = 2400 \cdot 0^2 \cdot (0^3 + 1)^{-3} = 0

So, the rate of change at t=0t = 0 is 00.

  1. At t=1t = 1: dNdtt=1=240012(13+1)3=24001(1+1)3=240018=300\frac{dN}{dt}\bigg|_{t=1} = 2400 \cdot 1^2 \cdot (1^3 + 1)^{-3} = 2400 \cdot 1 \cdot (1 + 1)^{-3} = 2400 \cdot \frac{1}{8} = 300

So, the rate of change at t=1t = 1 is 300300.

Final Answer:

  • The rate of change of NN with respect to tt at t=0t = 0 is 00.
  • The rate of change of NN with respect to tt at t=1t = 1 is 300300.

Would you like further details or have any questions?


  1. How does the chain rule apply when differentiating complex functions?
  2. What is the quotient rule and how can it be used in this context?
  3. How does the value of tt affect the rate of change in the bacteria population?
  4. What happens to the rate of change as tt becomes very large?
  5. How does the shape of the curve for N(t)N(t) relate to the bacteria's growth?

Tip: Always apply the chain rule carefully when differentiating nested functions to avoid errors in the differentiation process.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Rate of Change

Formulas

N(t) = 100 (1 - (4/(t^3 + 1)^2))
dN/dt = 2400t^2 (t^3 + 1)^{-3}

Theorems

Chain Rule
Quotient Rule

Suitable Grade Level

Undergraduate Calculus