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 (−∞, ∞). 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 open paren negative infinity comma infinity close paren .(A∘r)(t)=0.5π3t6 and the domain is t≥0.
Answers
GPT-4o mini
To determine the area of the circular layer of oil after \( t \) hours, we need to use the area function \( A(r) = \pi r^2 \) and substitute the radius function \( r(t) = 0.5t^3 \) into it.
1. **Substituting \( r(t) \) into \( A(r) \)**:
\[
A(r(t)) = A(0.5t^3) = \pi(0.5t^3)^2
\]
2. **Calculating \( A(0.5t^3) \)**:
\[
A(0.5t^3) = \pi(0.25t^6) = 0.25\pi t^6
\]
So the function that gives the area of the circular layer of oil after \( t \) hours is:
\[
A(r(t)) = 0.25\pi t^6
\]
3. **Finding the domain**:
- The function \( r(t) = 0.5t^3 \) is defined for all real numbers \( t \), but since \( t \) represents time, it must be non-negative. Thus, the domain is \( t \geq 0 \).
Therefore, the correct function to determine the area after \( t \) hours is:
\[
(A\circ r)(t) = 0.25\pi t^6
\]
and the domain is:
\[
t \geq 0.
\]
The corresponding response is:
\[
(A \circ r)(t) = 0.25\pi t^6 \quad \text{and the domain is} \quad t \geq 0.
\]
1. **Substituting \( r(t) \) into \( A(r) \)**:
\[
A(r(t)) = A(0.5t^3) = \pi(0.5t^3)^2
\]
2. **Calculating \( A(0.5t^3) \)**:
\[
A(0.5t^3) = \pi(0.25t^6) = 0.25\pi t^6
\]
So the function that gives the area of the circular layer of oil after \( t \) hours is:
\[
A(r(t)) = 0.25\pi t^6
\]
3. **Finding the domain**:
- The function \( r(t) = 0.5t^3 \) is defined for all real numbers \( t \), but since \( t \) represents time, it must be non-negative. Thus, the domain is \( t \geq 0 \).
Therefore, the correct function to determine the area after \( t \) hours is:
\[
(A\circ r)(t) = 0.25\pi t^6
\]
and the domain is:
\[
t \geq 0.
\]
The corresponding response is:
\[
(A \circ r)(t) = 0.25\pi t^6 \quad \text{and the domain is} \quad t \geq 0.
\]