Respuesta :
Part A: A[r(t)] = π(2t)^2 = 4πt^2
Part B: A(5) = 4 * 3.14 * 5^2 = 314
Part B: A(5) = 4 * 3.14 * 5^2 = 314
Answer:
The area of the circle of spilled sauce as a function of time is [tex]A(r(t)) = 4 \pi t^2[/tex]. The area of spilled sauce after 5 minutes is 314.
Step-by-step explanation:
Consider the provided statement.
The sauce flow can be expressed with the function [tex]r(t) = 2t[/tex].
Where t represents time in minutes and r represents how far the sauce is spreading.
The spilled sauce is creating a circular pattern and the area of the pattern can be expressed as: [tex]A(r) = \pi r^2[/tex].
Part A: Find the area of the circle of spilled sauce as a function of time, or A[r(t)].
Substitute [tex]r = 2t[/tex] in [tex]A(r) = \pi r^2[/tex].
[tex]A(r(t)) = \pi (2t)^2[/tex]
[tex]A(r(t)) = 4 \pi t^2[/tex]
Hence, the area of the circle of spilled sauce as a function of time is [tex]A(r(t)) = 4 \pi t^2[/tex].
Part B: How large is the area of spilled sauce after 5 minutes?
Substitute t = 5 in [tex]A(r(t)) = 4 \pi t^2[/tex].
[tex]A(5) = 4 \pi (5)^2[/tex]
[tex]A(5) = 100 \pi [/tex]
Use π = 3.14 in above equation.
[tex]A(5) = 100 \times 3.14[/tex]
[tex]A(5) = 314[/tex]
Hence, the area of spilled sauce after 5 minutes is 314.