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?

1 Answer
Feb 23, 2018

(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).