Environmental Science The spread of a contaminant is increasing in a circularpattern on the surface of a lake. The radius of the contaminant can be modeledby r(t) = 5.25Vr, where r is the radius in meters and + time in hours since contamination. Recall,the formula for the area of circle, A(r) = Tr?a) Using the terms: function, inputs, and outputs, explain the meaning of the mathematicalexpression, (A or) (36).b)in the context of the situation, interpret The meaning of a function (A o r)(t).c) Use A(+(1) = 7(5.25VT) to find the size of the contaminated area after 36 hours.d)Find when the size of the contaminated area is 6250 square meters.

means that the input of the function is 36 and the output is the value that we get when we input 36 onto the function r which is put on the function A.

(c)

Let's find the composite function

[tex](A\circ r)=\pi(5.25\sqrt[]{t})^2[/tex]

Now, we find the value of (A o r) (36):

[tex]\begin{gathered} (A\circ r)(36)=\pi(5.25\sqrt[]{36})^2 \\ =\pi(5.25(6))^2 \\ =3117.25 \end{gathered}[/tex]

So the size of contamination is approximately 3117.25 sq. m.

(d)

We want the time (t) when the size of the contamination would be 6250. Let's find it:

[tex]\begin{gathered} (A\circ r)=\pi(5.25\sqrt[]{t})^2 \\ 6250=\pi(5.25\sqrt[]{t})^2 \\ \frac{6250}{\pi}=(5.25\sqrt[]{t})^2 \\ 5.25\sqrt[]{t}=\sqrt[]{\frac{6250}{\pi}} \\ \sqrt[]{t}=\frac{\sqrt[]{\frac{6250}{\pi}}}{5.25} \\ t=(\frac{\sqrt[]{\frac{6250}{\pi}}}{5.25})^2 \\ t=(\frac{\frac{\sqrt[]{6250}}{\sqrt[]{\pi}}}{5.25})^2 \\ t=\frac{(\frac{\sqrt[]{6250}}{\sqrt[]{\pi}})^2}{(5.25)^2} \\ t=\frac{\frac{6250}{\pi}}{27.5625} \\ t\approx72.18 \end{gathered}[/tex]Answer - About 72.18 hours