Respuesta :
A. It's a composite function, so basically, wherever you see a p, replace it with 5t, because we are given that information. So, your answer is:
[tex]A[p(t)] = 5t \pi 2=10t \pi [/tex]
B. Let's use the function we created, and just plug in 2 for t:
[tex]A[p(2)] = 10(2) \pi [/tex]
[tex]A[p(2)] = 62.83[/tex]
So, your answer is (approximately) 62.83 units².
[tex]A[p(t)] = 5t \pi 2=10t \pi [/tex]
B. Let's use the function we created, and just plug in 2 for t:
[tex]A[p(2)] = 10(2) \pi [/tex]
[tex]A[p(2)] = 62.83[/tex]
So, your answer is (approximately) 62.83 units².
Answer:
The area of the circle of spilled paint as a function of time is [tex]A(p(t)) = 25 \pi t^2[/tex]. The area of spilled paint after 2 minutes is 314.
Step-by-step explanation:
Consider the provided statement.
The paint flow can be expressed with the function [tex]p(t) = 5t[/tex].
Where t represents time in minutes and p represents how far the paint is spreading.
The flowing paint is creating a circular pattern on the tile. The area of the pattern can be expressed as: [tex]A(p) = \pi p^2[/tex].
Part A: Find the area of the circle of spilled paint as a function of time, or A[p(t)].
Substitute [tex]p = 5t[/tex] in [tex]A(p) = \pi p^2[/tex].
[tex]A(p(t)) = \pi (5t)^2[/tex]
[tex]A(p(t)) = 25 \pi t^2[/tex]
Hence, the area of the circle of spilled paint as a function of time is [tex]A(p(t)) = 25 \pi t^2[/tex].
Part B: How large is the area of spilled paint after 2 minutes?
Substitute t = 2 in [tex]A(p(t)) = 25 \pi t^2[/tex].
[tex]A(2) = 25 \pi (2)^2[/tex]
[tex]A(2) = 100 \pi [/tex]
Use π = 3.14 in above equation.
[tex]A(2) = 100 \times 3.14[/tex]
[tex]A(2) = 314[/tex]
Hence, the area of spilled paint after 2 minutes is 314.