Pharmacy exam 2019-12-16

Degrees: Pharmacy and Biotechnology
Date: Dic 16, 2019

Question 1

A lagoon contaminated with nitrates contains 1000 tons of nitrates dissolved in 6 millions of cubic meters of water. To decontaminate the lagoon, we start to introduce pure water into the lagoon at a rate of 100000 cubic meters per day, and we take out the same amount of contaminated water. Assuming that the concentration of nitrates remains uniform in the lagoon, what amount of nitrates will be in the lagoon after two weeks? If the maximum concentration of nitrates to consider a water not contaminated is $0.1$ kg/m $^{3}$ , when will the lagoon be decontaminated?

Solution

Let

n (t)

the amount of nitrates in the lagoon at time

t

.
Differential equation:

n^{'} = - n / 60

.
Solution:

n (t) = 10^{6} e^{- t / 60}

n (14) = 791889.6

kg.
The lagoon will be decontaminated after

30.6495

days.

Question 2

The temperature $T$ of a chemical reaction depends on the concentrations of two substances $x$ and $y$ according to the function $T (x, y) = - x^{3} + 4 x^{2} y - 3 y^{2}$ .

If the concentration of $x$ and $y$ are 2 gr/dl and 1 gr/dl respectively, how must the two concentrations be changed to increase the temperature the maximum? How is the variation of the temperature if we change the two concentration in that direction?
How must the two concentrations be changed to increase the temperature at a rate of 10 ºC (gr/dl) $^{- 1}$ ?

Solution

$x$ and $y$ must be changed along the direction of the gradient $\nabla T (2, 1) = (4, 10)$ . Along this direction the rate of change of the temperature is $| \nabla T (2, 1) | = 10.77$ ºC (gr/dl) $^{- 1}$ .
$x$ and $y$ must be changed along the direction of the unit vector $(0, 1)$ , that is $x$ must be keep constant.

Question 3

It is known the concentration in blood of the active ingredient of a drug $t$ hours after applying the drug is given by the function $c (t) = t^{2} e^{- t / 2}$ mg/ml.

Compute the maximum value for the concentration of the active ingredient and give the time when the maximum is reached.
Study the concavity and compute the inflection points of the concentration.

Solution

The maximum is reached at $t = 4$ hours and $c (4) = 16 e^{- 2}$ mg/dl.
There are two inflection points at $t = 1.1716$ and $t = 6.8284$ .
The function is concave up in $(- \infty, 1.1716) \cup (6.8284, \infty)$ and concave down in $(1.1716, 6.8284)$ .

Exam

Last updated on Sep 15, 2020