Question

Airborne chemicals will disperse from their release point in a circular pattern. Suppose that a train crash results in the release of chlorine gas into the atmosphere. After t minutes, the radius of the circular area containing the gas plume is given by the function \[r{\text{ }} = {\text{ }}f\left( t \right){\text{ }} = {\text{ }}0.17t\] . The area of the gas plume as a function of the radius is.
(a) Evaluate \[g\left( {f\left( {30} \right)} \right)\] . What are its units? Explain what this expression means in the context of this problem.
(b) Evaluate${f^{^ - 1}}(4)$. What are its units? Explain what this expression means in the context of this problem.
(c) Evaluate ${g^{^ - 1}}(100).$What are its units? Explain what this expression means in the context of this problem.
I think for a I got and the units is minutes but I'm not sure if this is correct. For B and C how would you turn the equation and solve for the values?

Answer 1

Hint: Use properties of functions and their inverse according to question
According to the questions, we are going to use the properties of function and their inverse application and after we simplify the given expression, we can apply the function value and evaluate them at the given points such that we get the required answer. For some of the given functions that have combined functions, we solve the inner part first and use the value obtained to solve the outer function.

Complete step-by-step answer:
(a)Start with
 \[f(t) = 0.17t\]
And evaluate it at \[t = 30\] :
Then we get
 \[
  f(30) = 0.17(30) \\
  f(30) = 5.1m \\
\]
Then evaluate \[g(r) = \pi {r^2}\] at \[r = 5.1m\]
 \[
  g(5.1m) = \pi {(5.1m)^2} \\
  g(5.1m) = 26.01\pi \;{m^2} \;
\]

(b) A way to find the inverse to a function (in an inverse exists) is:
Start with the function:
 \[f(t) = 0.17t\]
Substitute \[{f^{ - 1}}(r)\;\] wherever you see \[t\] .
 \[f({f^{ - 1}}(r)) = 0.17{f^{ - 1}}(r)\]
Use a property that all inverses and their function must be equal to $r$
 \[{f^{ - 1}}(r) = \dfrac{r}{{0.17}}\]
At $r = 4$
 \[
  {f^{ - 1}}(4m) = \dfrac{{4m}}{{0.17}} \\
  {f^{ - 1}}(4) \approx 25.2s \;
\]

(c)
Start with \[g(r)\]
 \[g(r) = \pi {r^2}\]
Substitute \[{g^{ - 1}}(A)\;\] everywhere you see and \[\;r\] :
 \[g({g^{ - 1}}(A)) = \pi {({g^{ - 1}}(A))^2}\]
We know that
 \[A = \pi {({g^{ - 1}}(A))^2}\]
Solve for \[{g^{ - 1}}(A)\] :
We get
at \[A = 100{m^2}\]
 \[{g^{ - 1}}(100 = \sqrt {\dfrac{{100{m^2}}}{\pi }} \]
 \[{g^{ - 1}}(100\;{m^2}) = \dfrac{{10}}{{\surd \pi }}\;m\]

Note: While solving for functions combined, we have to make sure we solve the inner function first and then the outer function, and also, we have to substitute the correct form of the function which is given rather than mismatching them.