1. suppose that an offshore oil rig is leaking and that the oil forms a circular region whose radius (r)…

1. suppose that an offshore oil rig is leaking and that the oil forms a circular region whose radius (r) increases over the first 12 hours according to the function (r = f(t)=\frac{t}{2t + 4},0leq tleq12) where (r) is measured in miles, (t) is in hours, and (t = 0) corresponds to the instant that the leak begins. (a) using function - notation, the area of a circle of radius (r) is given by (a(r)=pi r^{2}). find ((acirc f)(t)) and interpret the result. (b) show that ((acirc f)(0)=0) and that ((acirc f)(6)=\frac{9pi}{64}).

1. suppose that an offshore oil rig is leaking and that the oil forms a circular region whose radius (r) increases over the first 12 hours according to the function (r = f(t)=\frac{t}{2t + 4},0leq tleq12) where (r) is measured in miles, (t) is in hours, and (t = 0) corresponds to the instant that the leak begins. (a) using function - notation, the area of a circle of radius (r) is given by (a(r)=pi r^{2}). find ((acirc f)(t)) and interpret the result. (b) show that ((acirc f)(0)=0) and that ((acirc f)(6)=\frac{9pi}{64}).

Answer

Explanation:

Step1: Recall the definition of function - composition

The composition ((A\circ f)(t)=A(f(t))). Given (A(r)=\pi r^{2}) and (r = f(t)=\frac{t}{2t + 4}), we substitute (r=f(t)) into (A(r)).

Step2: Substitute (r = f(t)) into (A(r))

((A\circ f)(t)=A(f(t))=\pi\left(\frac{t}{2t + 4}\right)^{2}=\frac{\pi t^{2}}{(2t + 4)^{2}},0\leq t\leq12). This function represents the area of the circular - oil spill as a function of time (t) (in hours) since the leak began.

Step3: Calculate ((A\circ f)(0))

Substitute (t = 0) into ((A\circ f)(t)): ((A\circ f)(0)=\frac{\pi(0)^{2}}{(2(0)+4)^{2}} = 0).

Step4: Calculate ((A\circ f)(6))

Substitute (t = 6) into ((A\circ f)(t)): [ \begin{align*} (A\circ f)(6)&=\frac{\pi(6)^{2}}{(2\times6 + 4)^{2}}\ &=\frac{36\pi}{(12 + 4)^{2}}\ &=\frac{36\pi}{16^{2}}\ &=\frac{36\pi}{256}\ &=\frac{9\pi}{64} \end{align*} ]

Answer:

((A\circ f)(t)=\frac{\pi t^{2}}{(2t + 4)^{2}},0\leq t\leq12); ((A\circ f)(0) = 0); ((A\circ f)(6)=\frac{9\pi}{64})