Math Problem Statement

A population of bacteria is introduced into a culture. The number of bacteria P can be modeled P = 500(4 + 5t/(40 + t^2))

where t is the time (in hours).

Find P'.

Solution

The number of bacteria PP is modeled by the equation:

P(t)=500(4+5t40+t2)P(t) = 500 \left( 4 + \frac{5t}{40 + t^2} \right)

We are tasked with finding P(t)P'(t), the derivative of P(t)P(t) with respect to tt.

Step 1: Differentiate the constant multiplier

The constant factor 500500 can be factored out of the differentiation, so we first focus on differentiating the inner function:

P(t)=500(4+5t40+t2)P(t) = 500 \left( 4 + \frac{5t}{40 + t^2} \right)

Let g(t)=4+5t40+t2g(t) = 4 + \frac{5t}{40 + t^2}. Therefore:

P(t)=500g(t)P'(t) = 500 \cdot g'(t)

Step 2: Differentiate the inner function g(t)=4+5t40+t2g(t) = 4 + \frac{5t}{40 + t^2}

First, observe that the constant term 44 has a derivative of zero:

ddt[4]=0\frac{d}{dt} [4] = 0

Now, we differentiate the second term 5t40+t2\frac{5t}{40 + t^2} using the quotient rule. The quotient rule states:

ddt(u(t)v(t))=u(t)v(t)u(t)v(t)[v(t)]2\frac{d}{dt} \left( \frac{u(t)}{v(t)} \right) = \frac{u'(t) v(t) - u(t) v'(t)}{[v(t)]^2}

For the term 5t40+t2\frac{5t}{40 + t^2}:

  • u(t)=5tu(t) = 5t, so u(t)=5u'(t) = 5
  • v(t)=40+t2v(t) = 40 + t^2, so v(t)=2tv'(t) = 2t

Now apply the quotient rule:

ddt(5t40+t2)=5(40+t2)5t(2t)(40+t2)2\frac{d}{dt} \left( \frac{5t}{40 + t^2} \right) = \frac{5(40 + t^2) - 5t(2t)}{(40 + t^2)^2}

Simplifying the numerator:

=5(40+t2)10t2(40+t2)2=200+5t210t2(40+t2)2=2005t2(40+t2)2= \frac{5(40 + t^2) - 10t^2}{(40 + t^2)^2} = \frac{200 + 5t^2 - 10t^2}{(40 + t^2)^2} = \frac{200 - 5t^2}{(40 + t^2)^2}

Thus, the derivative of g(t)g(t) is:

g(t)=2005t2(40+t2)2g'(t) = \frac{200 - 5t^2}{(40 + t^2)^2}

Step 3: Multiply by 500

Finally, multiply g(t)g'(t) by 500 to get P(t)P'(t):

P(t)=5002005t2(40+t2)2P'(t) = 500 \cdot \frac{200 - 5t^2}{(40 + t^2)^2}

Simplifying:

P(t)=1000002500t2(40+t2)2P'(t) = \frac{100000 - 2500t^2}{(40 + t^2)^2}

Final Answer:

P(t)=1000002500t2(40+t2)2P'(t) = \frac{100000 - 2500t^2}{(40 + t^2)^2}

Let me know if you'd like further details, or if you have any questions!


Related Questions:

  1. How can the derivative help in analyzing population growth over time?
  2. What is the meaning of the quotient rule in this context?
  3. Can this population growth model be used for long-term predictions? Why or why not?
  4. How would we find the time at which the bacteria population is increasing most rapidly?
  5. What other applications could this type of growth model have?

Tip: When dealing with fractions, the quotient rule is essential for finding derivatives correctly!

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Quotient Rule
Population Growth Modeling

Formulas

P(t) = 500(4 + 5t / (40 + t^2))
Quotient Rule: d/dt (u/v) = (u'v - uv')/v^2

Theorems

Quotient Rule

Suitable Grade Level

Grades 11-12 (or Calculus I students)