Question
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
. 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 (−∞, ∞)
.
. 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 (−∞, ∞)
.
Answers
GPT-4o mini
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 \( t \geq 0 \).
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 \( t \geq 0 \).