2. Bacterial Culture Problem: When you grow a
1 answer:
The area is changing at an increasing rate because the values of A(0), A(5), and A(10) increases as t increases
<h3>Find the area at times t = 0, t = 5, and t = 10.</h3>The complete question is added as an attachment
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
<h3>Use the results of part a to find the radius of the culture at these three times</h3>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
<h3>Write an equation for the composite function R(A(t)).</h3>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)
<h3>The restriction for the function R * A</h3>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
Read more about composite functions at:
#SPJ1
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