sábado, 25 de octubre de 2014

Relative velocity: Boat problems

Problem 1. 

A river  flows due East at a speed of 1.3 metres per second. A girl in a rowing boat, who can row at 0.4 metres per second in still water, starts from a point on the South bank and steers due North. The boat is also blown by a wind with speed 0.6 metres per second from a direction of N20ºE.

Figure 1: The red arrows represent the velocities of the boat (b), wind (w) and flow (r).

  1. Find the resultant velocity of the boat and its magnitude.
  2. If the river has a constant width of 10 metres, how long does it take the girl to cross the river, and how far upstream or downstream has she then travelled?


Problem 2. 

A river  flows due West at a speed of 2.5 metres per second and has a constant width of 1 km. You want to cross the river from point A (South) to a point B (North) directly opposite with a motor boat that can manage to a speed of  5 metres per second.

  1. If you head out pointing your boat at an angle of 90 degrees to the bank. How long does it take to get from point A to point B?
  2. After crossing the river you realised that it took  longer than expected. In what direction should you point you motor boat in order to reduce the time to cross the river? How long will it take you to get from point A to point B? Is it a better time?


Applet GeoGebra

The following applet shows a representation of the problem 2, considering that the boat starts from a point A. It also shows the velocities (vectors) and their magnitudes (speeds) of the boat and current.


  1. Move the sliders to change the magnitude and direction of vectors.
  2. Click the 'Start' button to activate the motion of the boat.
  3. Click the 'Reset' button to put back the boat to its original position.
  4. You can also change the width of the river. Chose a number between 5 and 1000.
  5. All velocities can be considered either as metres per second or km per second. 


Open this applet in an external window: Relative Velocity: Boat Problem

domingo, 18 de mayo de 2014

Graphical representation of the domain and range of real functions

A function specifies a rule by which an input is converted to a unique output.

More precisely:

A function $f$ is a rule that assigns to each element $x$ in a set $D$ exactly one element, called $f(x)$, in a set $E$.

The domain of a function is the set of all possible $x$ values that can be used as inputs, and the range is the set of all possible $f(x)$ values that arise as outputs.

It's helpful to think of a function as a machine (see Figure 1). If $x$ is in the domain of the function then when enters the machine, it is accepted as an input and the machine produces an output according to the rule of the function. Thus we can think of the domain as the set of all possible inputs $x$ and the range as the set of all possible outputs  $f(x)$.

Figure 1.
The most common method for visualising a function is its graph; which consists of all points $(x,y)$ in the coordinate plane such  that  $y=f(x)$ and $x$ is in the domain of $f$.

The graph of a function $f$ gives us a useful picture of the behaviour of the function. Since the $y$-coordinate of any point $(x,y)$ on the graph is $y=f(x)$, we can read the value of $f(x)$ from the graph as being the height of the graph above the point $x$ (see Figure 2).

Figure 2

The graph of $f$ also allows us to picture the domain of $f$ on the $x$-axis and its range on the $y$-axis as in Figure 3.

Figure 3
For the domain of the function we need to ask: What is the set of all the valid inputs $x$?

Meanwhile, for the range of a the function we need to ask: What is the set of all the valid outputs $f(x)$?

In the next applet you can see a graphic representation of the domain and range of functions. In this case, the green horizontal line represents the domain and the salmon vertical line represents the range. The function is represented with the dotted curve.

Type your function and see how the domain and range change. Selecting the asymptotes will show you particular cases where the function you typed is whether defined or not, in particular values of $x$.



Some particular cases: x^2 for $x^2$, exp(x) for $e^x$, abs(x) for $|x|$, 1/(x^2+1) for $\frac{1}{x^2+1}$, ln(x) for $\ln x$ and sqrt(x) for $\sqrt{x}$.

Follow the next link for opening the applet in an external window:

Derivación de la fórmula para calcular las raíces de una cuadrática

En varios sitios de internet se puede encontrar la derivación de la fórmula
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
para resolver la ecuación cuadrática $ax^2+bx+c=0$.

Por ejemplo:

http://www.elosiodelosantos.com/sergiman/div/formula_cuadratica.htm

http://www.disfrutalasmatematicas.com/algebra/cuadratica-ecuacion-derivacion.html

http://www.montereyinstitute.org/courses/Algebra1/COURSE_TEXT_RESOURCE/U10_L1_T3_text_final_es.html

El siguiente procedimiento me parece menos rebuscado:

                  Sean $a, b$ y $c$ números reales con $a\neq 0$. Consideremos la ecuación
$$ax^2+bx+c=0$$
                  Entonces tenemos
$$ax^2+bx=-c$$
                  Multiplicamos por $4a$ en ambos lados
$$4a^2x^2+4abx=-4ac$$
                  Ahora sumamos en ambos lados $b^2$
$$4a^2x^2+4abx+b^2=-4ac+b^2$$
                  Lo anterior lo podemos escribir como
$$(2ax+b)^2=b^2-4ac$$
                  Entonces, aplicando raíz cuadrada resulta
$$2ax+b=\pm \sqrt{b^2-4ac}$$
                  Despejando $x$ obtenemos la expresión:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Al parecer este procedimiento data del siglo IX y se suele atribuir un matemático de la India llamado Sridhara Acharya.

Referencias

- O'Connor, J. & Robertson, E. F. Sridhara. The McTutor History of Mathematics Archive, University of St Andrews. Disponible en: http://www-history.mcs.st-andrews.ac.uk/Biographies/Sridhara.html