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Sagot :
To write a composite function to find the area of the circular oil spill at time [tex]\(t\)[/tex], we need to follow a few steps, using the given functions for radius and area.
1. Understand the functions:
- The radius of the oil spill as a function of time [tex]\(t\)[/tex] is given by:
[tex]\[ r(t) = 0.5 + 2t \][/tex]
- The area of a circle as a function of its radius [tex]\(r\)[/tex] is given by:
[tex]\[ A(r) = \pi r^2 \][/tex]
2. Substitute [tex]\(r(t)\)[/tex] into [tex]\(A(r)\)[/tex] to form a composite function:
- To determine the area of the oil spill as a function of time, we need to substitute the radius function [tex]\(r(t)\)[/tex] into the area function [tex]\(A(r)\)[/tex].
- The composite function [tex]\(A[r(t)]\)[/tex] represents the area of the oil spill at any time [tex]\(t\)[/tex]:
[tex]\[ A[r(t)] = A(0.5 + 2t) \][/tex]
3. Calculate the composite function [tex]\(A[r(t)]\)[/tex]:
- Substitute [tex]\(r(t) = 0.5 + 2t\)[/tex] into the area function [tex]\(A(r)\)[/tex]:
[tex]\[ A(0.5 + 2t) = \pi (0.5 + 2t)^2 \][/tex]
4. Simplify the expression (optional, to understand the function better):
- We can expand the expression inside the parentheses to simplify it further:
[tex]\[ A(0.5 + 2t) = \pi (0.5 + 2t)^2 = \pi (0.25 + 2(0.5) \cdot 2t + (2t)^2) = \pi (0.25 + 2t + 4t^2) \][/tex]
- So, the area of the oil spill as a function of time is:
[tex]\[ A[r(t)] = \pi (4t^2 + 2t + 0.25) \][/tex]
Therefore, given the composite function [tex]\(A[r(t)] = \pi (0.5 + 2t)^2\)[/tex], we can calculate the area of the oil spill at any time [tex]\(t\)[/tex].
For example, to find the area at [tex]\(t = 1\)[/tex] hour:
[tex]\[ r(1) = 0.5 + 2 \cdot 1 = 2.5 \text{ miles} \][/tex]
[tex]\[ A(2.5) = \pi \cdot (2.5)^2 = \pi \cdot 6.25 \approx 19.635 \text{ square miles} \][/tex]
Thus, at [tex]\(t = 1\)[/tex] hour, the radius of the oil spill is [tex]\(2.5\)[/tex] miles and the area of the oil spill is approximately [tex]\(19.635\)[/tex] square miles.
1. Understand the functions:
- The radius of the oil spill as a function of time [tex]\(t\)[/tex] is given by:
[tex]\[ r(t) = 0.5 + 2t \][/tex]
- The area of a circle as a function of its radius [tex]\(r\)[/tex] is given by:
[tex]\[ A(r) = \pi r^2 \][/tex]
2. Substitute [tex]\(r(t)\)[/tex] into [tex]\(A(r)\)[/tex] to form a composite function:
- To determine the area of the oil spill as a function of time, we need to substitute the radius function [tex]\(r(t)\)[/tex] into the area function [tex]\(A(r)\)[/tex].
- The composite function [tex]\(A[r(t)]\)[/tex] represents the area of the oil spill at any time [tex]\(t\)[/tex]:
[tex]\[ A[r(t)] = A(0.5 + 2t) \][/tex]
3. Calculate the composite function [tex]\(A[r(t)]\)[/tex]:
- Substitute [tex]\(r(t) = 0.5 + 2t\)[/tex] into the area function [tex]\(A(r)\)[/tex]:
[tex]\[ A(0.5 + 2t) = \pi (0.5 + 2t)^2 \][/tex]
4. Simplify the expression (optional, to understand the function better):
- We can expand the expression inside the parentheses to simplify it further:
[tex]\[ A(0.5 + 2t) = \pi (0.5 + 2t)^2 = \pi (0.25 + 2(0.5) \cdot 2t + (2t)^2) = \pi (0.25 + 2t + 4t^2) \][/tex]
- So, the area of the oil spill as a function of time is:
[tex]\[ A[r(t)] = \pi (4t^2 + 2t + 0.25) \][/tex]
Therefore, given the composite function [tex]\(A[r(t)] = \pi (0.5 + 2t)^2\)[/tex], we can calculate the area of the oil spill at any time [tex]\(t\)[/tex].
For example, to find the area at [tex]\(t = 1\)[/tex] hour:
[tex]\[ r(1) = 0.5 + 2 \cdot 1 = 2.5 \text{ miles} \][/tex]
[tex]\[ A(2.5) = \pi \cdot (2.5)^2 = \pi \cdot 6.25 \approx 19.635 \text{ square miles} \][/tex]
Thus, at [tex]\(t = 1\)[/tex] hour, the radius of the oil spill is [tex]\(2.5\)[/tex] miles and the area of the oil spill is approximately [tex]\(19.635\)[/tex] square miles.
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