To solve for the function [tex]\( A(t) \)[/tex] that represents the area of the expanding circle in [tex]\( t \)[/tex] seconds, we need to follow these steps:
1. Identify the radius function [tex]\( r(t) \)[/tex]:
Given: [tex]\( r(t) = 1 + 4t \)[/tex]
2. Recall the formula for the area of a circle:
The area [tex]\( A \)[/tex] of a circle is given by [tex]\( A = \pi r^2 \)[/tex].
3. Substitute the radius function [tex]\( r(t) \)[/tex] into the area formula:
[tex]\( A(t) = \pi [r(t)]^2 \)[/tex]
Hence, [tex]\( A(t) = \pi (1 + 4t)^2 \)[/tex]
4. Expand the squared term [tex]\( (1 + 4t)^2 \)[/tex]:
[tex]\[
(1 + 4t)^2 = (1 + 4t)(1 + 4t)
\][/tex]
Using the distributive property or binomial expansion:
[tex]\[
(1 + 4t)^2 = 1^2 + 2(1)(4t) + (4t)^2 = 1 + 8t + 16t^2
\][/tex]
5. Substitute the expanded form back into the area formula:
[tex]\[
A(t) = \pi (1 + 8t + 16t^2)
\][/tex]
6. Distribute π across each term:
[tex]\[
A(t) = \pi + 8\pi t + 16\pi t^2
\][/tex]
So, the function [tex]\( A(t) \)[/tex] representing the area of the expanding circle in [tex]\( t \)[/tex] seconds is:
[tex]\[
A(t) = \pi + 8\pi t + 16\pi t^2
\][/tex]
Now, we match this result with the given options:
[tex]\[
A. \ A(t) = \pi\left(16 t^2 + 8 t + 1\right)
\][/tex]
[tex]\[
B. \ A(t) = \pi\left(16 t^2 + 4 t + 1\right)
\][/tex]
[tex]\[
C. \ A(t) = 16 \pi t^2 + 1
\][/tex]
[tex]\[
D. \ A(t) = 4 \pi t^2 + 1
\][/tex]
We can see that option A matches our derived function:
[tex]\[
A(t) = \pi (1 + 8t + 16t^2) = \pi\left(16 t^2 + 8 t + 1\right)
\][/tex]
Therefore, the correct answer is:
A. [tex]\( A(t) = \pi\left(16 t^2 + 8 t + 1\right) \)[/tex]
