Question

I'm confused on how to find the inverse function and solve?

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 = f(t) = 0.17t. The area of the gas plume as a function of the radius is A = g(r) = πr².

(a) Evaluate g(f(30)). 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 g(f(30))=π(30)²=900π and the units is minutes but I'm not sure if this is correct. For B and C how would you turn the equation A = g(r) = πr² and solve for the values?

Answer 1

(a) A way to evaluate `g(f(30)` is:

Start with:

`f(t) = 0.17t`

And evaluate it at `t = 30`:

`f(30) = 0.17(30)`

`f(30) = 5.1 " m"`

Then evaluate `g(r) = pir^2` at `r = 5.1 " m"`

`g(5.1" m") = pi(5.1" m")^2`

`g(5.1" m") = 26.01pi" 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)` everywhere you see a t:

`f(f^-1(r)) = 0.17f^-1(r)`

Use a property that all inverses and their function must have `f(f^-1(r)) = r`:

`r = 0.17f^-1(r)`

Solve for `f^-1(r)`:

`f^-1(r) = r/0.17`

Now, evaluate `f^-1(r)` at `r = 4" m"`:

`f^-1(4" m") = (4" m")/0.17`

`f^-1(4) ~~ 25.2" s"`

I knew that the inverse function accepted, as its argument, a radius (in meters) and returned a value of time (in seconds), because the function accepts, as is argument, at time (in seconds) and returns a value of a radius (in meters). All inverses "undo" what the original function does.

(c) Find `g^-1(A)` using the same method that I used for `f^-1(r)`:

Start with `g(r)`

`g(r) = pir^2`

Substitute `g^-1(A)` everywhere you see and r:

`g(g^-1(A)) = pi(g^-1(A))^2`

Use the same property `g(g^-1(A)) = A`:

`A = pi(g^-1(A))^2`

Solve for `g^-1(A)`:

`A/pi = (g^-1(A))^2`

`(g^-1(A))^2= A/pi`

`g^-1(A)= +-sqrt(A/pi)`

All radii must be positive, therefore, we discard the `+-`

`g^-1(A)= sqrt(A/pi)`

Evaluate at `A = 100" m"^2`

`g^-1(100" m"^2) = sqrt((100" m"^2)/pi)`

`g^-1(100" m"^2) = 10/sqrt(pi)" m"`

I knew that the units using the same logic as in part (b).