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