Extraneous solutions to radical equations

Extraneous solutions - a solution of a simplified  version of an equation that does not satisfy the original equation. Referred from  http://www.mathwords.com/e/extraneous_solution.htm


    Extraneous solutions to radical equations: Extraneous Solutions to Radical Equations

Dependent and independent variables

For p=5q

p is dependent on what happens to q so p is the dependent variable.


    Dependent and independent variables exercise: the basics: Here we have a problem that asks us to identify which variables are dependent and independent. Hint: independent variables are not influenced and remain unchanged by the other variable.

How many people can a blimp carry

Density = Mass /Volume so volume is proportional to mass.




    How many people can a blimp carry:

Surface Area

The problems for this topic may be quite hard to get through.

You may need to draw the object or the net of it.

 Write down the dimensions one by one if you are having difficulty, calculating the area first, then the number of sides.

Read the question carefully!

Here is one of the odd objects you may encounter
-tetrahedron

 


 

Triangle inequality theorem

The triangle inequality theorem states that any side of a triangle is always shorter than the sum of the other two sides.

By the same theorem, the third side must be also larger than the difference between the other two sides.

Recognizing functions (example 1)

If y has more than one value, then y is not a function of x.


    Recognizing functions (example 1):

Limits

From Khan Academy 

Evaluate limx→11−x√1−x using algebraic methods.

If we try substitution, we obtain the indeterminate form 

. 

The first solution requires rationalizing the numerator  

limx→11−x√1−x=limx→11−x√1−x⋅1+x√1+x√=

 The rest is not too difficult. Sometimes, have to factorise expressions like 25-x.
This becomes (5-x^0.5)(5+x^0.5). The above case can be done with this method.
1-x becomes (1-x^0.5)(1+x^0.5)

Asymptotes of hyperbola

To derive the equation, find  it in terms of y. The constant term is removed.

I use a formula like the one below to solve the problem in Khan Academy https://www.khanacademy.org/math/precalculus/hyperbolic_trig_topic/hyperbolic_trig_intro/e/equation_of_a_hyperbola. 
 y1 is moved to the other side giving a  value that has an opposite sign. Please take note- this is not a fixed formula but one that I derive on my own to solve the problem faster. 

y=±(b/a)(x+x1)-y1

The fixed ones are as follows:-   y = ±(b/a)(x - h) + k for a hyperbola that opens horizontally and y = ±(a/b)(x - h) + k for a hyperbola that opens vertically. For more info, see

http://www.purplemath.com/modules/hyperbola.htm

http://laurashears.info/math122/unit4/conics/whyBoverAisAsymp.htm

Parabola intuition 2

Concavity can help to determine what a parabola looks like. See http://jwlchin.blogspot.com/2014/08/recognizing-concavity.html

Substitute a value of x so that the R.H.S of the equation is equal to 0 for the vertical parabola.  

The  distance between the directrix or focus to the vertex =1/(4 * a).  This formula is actually for finding the focus. (focus to vertex)

These hints can help to solve the problem 
https://www.khanacademy.org/math/algebra2/conics_precalc/parabolas_precalc/e/parabola_intuition_2

Parabola focus and directrix

From http://www.mathopenref.com/parabolafd.html
http://www.softschools.com/math/calculus/writing_the_equation_of_parabolas/

One of the ways to define a parabola -

 the locus of all points that are equidistant from a given point (focus) and a given line (directrix).

To write the equation of a parabola....

1. Determine which pattern to use (based on whether it is horizontal or vertical)

2. Substitute in h and k ,which is the vertex of the parabola.

3. Choose a coordinate to substitute in and solve for a.

4. Write your final equation with a, h, and k.

Remember the patterns: 

To derive, see http://www.mathopenref.com/parabolafdderive.html


   
    Using the focus and directrix to find the equation of a parabola:

Foci (focus points) of an ellipse

From  http://www.mathopenref.com/ellipsefoci.html

where F  is the distance from each focus to the center (see figure below) j  is the semi-major axis (major radius) n  is the semi-minor axis (minor radius) By definition, a+b always equals the major axis length QP, no matter where R is. So a+b equals OP+OQ. So b must equal OP. (And a equals OQ).

General Equation of an Ellipse

Read about this from http://www.mathopenref.com/coordgeneralellipse.html
which I think is also a good reference for mathematics



where
a is the radius along the x-axis
b is the radius along the y-axis
h, k are the x,y coordinates of the ellipse's center.

Equation of a circle in non-factored form

This can be done by completing the square in x and y. The formulas for this can be obtained from http://www.analyzemath.com/Calculators/Cen_Rad_Cal.html


For equation
x ² + y ² + a x + b y = c

 the center has coordinates (-a/2 , -b/2) and radius r such that r ² = c + (a/2) ² + (b/2) ²

    Completing the square to write equation in standard form of a circle: 

Recognizing concavity

Update:

The second derivative, is more than 0 wherever the function is concaved up. It is less than 0 when it is concaved down.

Below is an image from http://clas.sa.ucsb.edu/staff/lee/inflection%20pt%20x%5E3.gif
that helps to explain it.

 This can be quite hard to understand so I assume
Concaved up - slope is getting more positive
Concaved down - slope is  getting less positive.

    Recognizing concavity exercise:





To understand more on this, here is another video about concavity


    Concavity, concave upwards and concave downwards intervals: