Question

Can someone help out with the question below?

Answer 1

`P'(16) = 75/32`

Explanation:

The idea here is that you can find the rate of change of the pollution with respect to time by taking the first derivative of your function `P(t)`.

So this is pretty much an exercise in finding the derivative of the function

`P(t) = (t^(1/4) + 3)^3`

which can be calculated by using the chain rule and the power rule. Keep in mind that you have

`color(blue)(ul(color(black)(d/(dt)[u(t)^n] = n * u(t)^(n-1) * d/(dt)[u(t)])))`

In your case,

`{(u(t) = t^(1/4) + 3), (n = 3) :}`

This means that the derivate of `P(t)` will be

`overbrace(d/d(dt)[P(t)])^(color(blue)(=P^(')(t))) = 3 * (t^(1/4) + 3)^2 * d/(dt)(t^(1/4) + 3)`

`P^'(t) = 3 * (t^(1/4) + 3)^2 * 1/4 * t^((1/4 - 1))`

`P^'(t) = 3/4 * t^(-3/4) * (t^(1/4) + 3)^2`

Now all you have to do is plug in `16` for `t` to find

`P^'(t) = 3/4 * 16^(-3/4) * (16^(1/4) + 3)^2`

Since you know that

`16 = 2^4`

you can rewrite the equation as

`P^'(16) = 3/4 * 2^[[4 * (-3/4)]] * [2^((4 * 1/4)) + 3]^2`

`P^'(16) = 3/4 * 2^(-3) * (2 + 3)^2`

`P^'(16) = 3/4 * 1/8 * 25`

`P^'(16) = 75/32`

And there you have it -- the rate at which the pollution changes after `16` years.