2. Bacterial Culture Problem: When you grow a
culture of bacteria in a petri dish, the area of
the culture is a function of the number of
bacteria present. Suppose that the area of the
culture, A(t), measured in square millimeters,
is given by this function of time, t, measured
in hours:
A(t) = 9(1.1¹)
a. Find the area at times t=0, 1=5, and t = 10.
Is the area changing at an increasing rate or
at a decreasing rate? How do the values of
A(0), 4(5), and A(10) allow you to answer
this question?
b. Assume that the bacteria culture is circular.
Use the results of part a to find the radius
of the culture at these three times. Is the
radius changing at an increasing rate or at a
decreasing rate?
c. The radius of the culture is a function of the
area. Write an equation for the composite
function R(A(t)).

2 Bacterial Culture Problem When you grow a culture of bacteria in a petri dish the area of the culture is a function of the number of bacteria present Suppose class=

Respuesta :

The area is changing at an increasing rate because the values of A(0), A(5), and A(10) increases as t increases

Find the area at times t = 0, t = 5, and t = 10.

The function is given as:

A(t) = 9 * (1.1)^t

Substitute t = 0, t = 5, and t = 10.

A(0) = 9 * (1.1)^0 = 9

A(5) = 9 * (1.1)^5 = 14.5

A(10) = 9 * (1.1)^10 = 23.3

The area is changing at an increasing rate because the values of A(0), A(5), and A(10) increases as t increases

Use the results of part a to find the radius of the culture at these three times

The area of a circle is

A = πr^2

Make r the subject

r = √(A/π)

So, we have:

r = √(9/π) =  1.7

r = √(14.5/π) =  2.1

r = √(23.3/π) =  2.7

The radius is changing at an increasing rate

Write an equation for the composite function R(A(t)).

In (b), we have:

r = √(A/π)

Express as a function

r(A(t)) = √(A/π)

Hence, the equation for the composite function is R(A(t)) = √(A/π)

The restriction for the function R * A

We have:

r(A(t)) = √(A/π)

A(t) = 9 * (1.1)^t

This gives

r(A(t)) = √(9 * (1.1)^t/π)

The radius is 30.

So, we have:

√(9 * (1.1)^t/π) = 30

Square both sides

(9 * (1.1)^t/π)= 900

Divide by 9

(1.1)^t/π= 100

Multiply by π

(1.1)^t= 314

Take the logarithms of both sides

t * log(1.1) = log(314)

Solve for t

t = 60

Hence, the restriction on the domain is 0 <= t <= 60

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