Pharmacy exam 2016-01-10

Degrees: Pharmacy and Biotechnology
Date: Jan 10, 2017

Question 1

The rate of growth of certain bacteria population is the square root of the number of bacteria in the population. How much will the population have increased after 1 hour from the beginning of the growth? How long will it take until the population is four times the population at the beginning?

Solution

Naming $x$ to the number of bacteria and $t$ to time, $x(t)=(\frac{t}{2}+C)^2$.
The number of bacteria has increased $\frac{1}{4}+C$ after 1 hour from the beginning.
The number of bacteria is four times the population at the beginning at time $t=2C$.

Question 2

The temperature of a chemical process depends on the amounts $x$ and $y$ of two substances, according to the function $T(x,y)=4x^3+y^3-3xy$. Determine the local extrema and saddle points of the temperature function (recall that the amounts $x$ and $y$ cannot be negative).

Solution

$T$ has a saddle point at $(0,0)$ and a local minimum at $(\frac{\sqrt[3]{4}}{4},\frac{\sqrt[3]{2}}{2})$.

Question 3

An ecological model explains the number of individuals in a population through the function $$f(x,y)=\dfrac{e^t}{x},$$ where $t$ is the time and $x$ the number of predators in the area. Give an approximation of the number of individuals at $t=0.1$ and $x=0.9$ using the second order Taylor polynomial of function at point $(1,0)$.

Solution

Second order Taylor polynomial of $f$ at point $(1,0)$: $P^2_{f,(1,0)}(x,y)=3-3x+2t+x^2+\frac{t^2}{2}-xt$.
$P^2_{f,(1,0)}(0.9,0.1)=1.225$.

Question 4

The position of a moving object in space is given by the function $f(t)=(e^{t/2}, \sin^2(t), \sqrt[3]{1-t})$.

Compute the velocity and acceleration vectors at time $t=0$.
Remark: velocity is the variation of space with respect to time, and acceleration is the variation of velocity with respect to time.
Compute an equation of the plane normal to the trajectory at time $t=0$.

Solution

$f’(t)=(\frac{e^{t/2}}{2},2\sin t \cos t, \frac{-(1-t)^{-2/3}}{3})$ and $f’(0)=(\frac{1}{2},0,-\frac{1}{3})$. $f’’(t)=(\frac{e^{t/2}}{4},2(\cos^2 t-\sin^2 t), \frac{-2(1-t)^{-5/3}}{9})$ and $f’’(0)=(\frac{1}{4},2,-\frac{2}{9})$.
Normal plane to the trajectory at time $t=0$: $3x-2z=1$.

Exam

Last updated on Sep 15, 2020