To solve the problem of finding the function [tex]\( A(t) \)[/tex] that represents the area of the expanding circle in [tex]\( t \)[/tex] seconds, we start with the given radius function:
[tex]\[ r(t) = 1 + 4t \][/tex]
The area [tex]\( A \)[/tex] of a circle is given by the formula:
[tex]\[ A = \pi r^2 \][/tex]
To find [tex]\( A(t) \)[/tex], we need to express the area in terms of [tex]\( t \)[/tex]. Substituting the radius function [tex]\( r(t) \)[/tex] into the area formula, we get:
[tex]\[ A(t) = \pi \left( r(t) \right)^2 \][/tex]
Next, we substitute [tex]\( r(t) \)[/tex] with [tex]\( 1 + 4t \)[/tex]:
[tex]\[ A(t) = \pi \left( 1 + 4t \right)^2 \][/tex]
To find the explicit form of [tex]\( A(t) \)[/tex], we need to expand the squared term:
[tex]\[ \left( 1 + 4t \right)^2 = (1 + 4t)(1 + 4t) \][/tex]
Using the distributive property (FOIL method):
[tex]\[ (1 + 4t)(1 + 4t) = 1 \cdot 1 + 1 \cdot 4t + 4t \cdot 1 + 4t \cdot 4t \][/tex]
[tex]\[ = 1 + 4t + 4t + 16t^2 \][/tex]
[tex]\[ = 1 + 8t + 16t^2 \][/tex]
So, we have:
[tex]\[ (1 + 4t)^2 = 1 + 8t + 16t^2 \][/tex]
Therefore, the area function [tex]\( A(t) \)[/tex] becomes:
[tex]\[ A(t) = \pi (1 + 8t + 16t^2) \][/tex]
Simplifying further, we get:
[tex]\[ A(t) = \pi \left( 16t^2 + 8t + 1 \right) \][/tex]
Thus, the function [tex]\( A(t) \)[/tex] that represents the area of the expanding circle in [tex]\( t \)[/tex] seconds is:
[tex]\[ \boxed{A(t) = \pi \left( 16t^2 + 8t + 1 \right)} \][/tex]
So, the correct answer is:
A. [tex]\(\quad A(t)=\pi\left(16 t^2+8 t+1\right)\)[/tex]
