# Can someone help out with the question below?

##### 1 Answer

#### 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

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

#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

#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