Question

On a particular day, the amount of untreated water coming into the plant can be modeled by f(t) = 100 + 30cos(t/6) where t is in hours since midnight and f(t) represents thousands of gallons of water. The amount of treated water at any given time, t, can be modeled by g(t) = 30e^cos(/2)a) Define a new function, a(t), that would represent the amount of untreated water inside the plant, at any given time, t.b) Find a′ (t).c) Determine the critical values of this function over the interval [0, 24).d) Determine whether the critical values represent local maximums or minimums.e) Determine the maximum and minimum amount of untreated water in the plant for the day.

Answer 1

Explanation

The amount of untreated water coming into a plant can be modeled by the following function:

$f(t)=100+30\cos\frac{t}{6}$

The amount of water treated by that same plant is given by:

$g(t)=30e^{\cos\frac{t}{2}}$

In part (a) we must define the function a(t) that represents the amount of untreated water inside the plant, at any given time, t. This quantity is equal to the amount of untreated water entering the plant less the amount of water treated by it at the same time. In summary we have:

$a(t)=f(t)-g(t)=100+30\cos\frac{t}{6}-30e^{\cos\frac{t}{2}}$

In part (b) we have to find its derivative. In order to do that let's remember how to derivate a cosine and an exponential expression:

$\begin{gathered} h(t)=\cos(k(t))\Rightarrow h^{\prime}(t)=-\sin(k(t))\cdot k^{\prime}(t) \\ b(t)=e^{c(t)}\Rightarrow b^{\prime}(t)=e^{c(t)}\cdot c^{\prime}(t) \end{gathered}$

Using these properties we can find a'(t):

$\begin{gathered} a^{\prime}(t)=f^{\prime}(t)-g^{\prime}(t)=(100+30\cos\frac{t}{6})^{\prime}-(30e^{\cos\frac{t}{2}})^{\prime} \\ a^{\prime}(t)=-\frac{30}{6}\sin\frac{t}{6}-30e^{\cos\frac{t}{2}}\cdot(-\frac{1}{2}\sin\frac{t}{2}) \\ a^{\prime}(t)=-5\sin\frac{t}{6}+15e^{\cos\frac{t}{2}}\sin\frac{t}{2} \end{gathered}$

In part (c) we must find the critical points of the function over [0,24). The critical values are t-values where the derivative of the function is not defined or where it is 0. Then we must solve a'(t)=0:

$a^{\prime}(t)=0=-5\sin(\frac{t}{6})+15e^{\cos(t\/2)}\sin(\frac{t}{2})$