A car is leaking oil. The oil that has leaked spreads in a circular layer. The area of the circular layer of oil is given by A(r)=πr2

Question

. The radius of the circular layer of oil is increasing every hour, which is given by the function r(t)=0.5t3
. What function can be used to determine the area of the circular layer of oil after t hours? What is the domain?(1 point)
Responses

(A∘r)(t)=0.5πt6
and the domain is (−∞, ∞)
.
open paren cap A circ r close paren times t is equal to 0 point 5 pi t to the 6th power and the domain is open paren negative infinity comma infinity close paren .

(A∘r)(t)=0.25πt6
and the domain is t≥0
.
open paren cap A circ r close paren times t is equal to 0 point 2 5 pi t to the 6th power and the domain is t is greater than or equal to 0 .

(A∘r)(t)=0.5π3t6
and the domain is t≥0
.
open paren cap A circ r close paren times t is equal to 0 point 5 pi cubed t to the 6th power and the domain is t is greater than or equal to 0 .

(A∘r)(t)=0.5π3t6
and the domain is (−∞, ∞)
.

Answer 1

To find the function that can be used to determine the area of the circular layer of oil after \( t \) hours, we will substitute the radius function \( r(t) = 0.5t^3 \) into the area function \( A(r) = \pi r^2 \).

Starting with the area:

\[ A(r) = \pi r^2 \]

Substituting \( r(t) \):

\[ A(r(t)) = A(0.5t^3) = \pi (0.5t^3)^2 \]

Calculating \((0.5t^3)^2\):

\[ (0.5t^3)^2 = 0.25t^6 \]

So, we have:

\[ A(r(t)) = \pi \cdot 0.25t^6 = 0.25\pi t^6 \]

Thus, the function that determines the area of the circular layer of oil after \( t \) hours is:

\[ (A\circ r)(t) = 0.25\pi t^6 \]

Now, considering the domain: Since \( t \) represents time and cannot be negative in this context, the domain is \( t \geq 0 \).

Therefore, the correct response is:

\((A\circ r)(t) = 0.25\pi t^6\) and the domain is \( t \geq 0 \).