Instantaneous Rate of Change of a Trigonometric Function

Now, everyone knows the basic method to find the instantaneous rate of change of an equation ; that is, to find the difference between 2 points nearest to the point required a rate of change, divided by the difference in x-values of the respective y-values chosen. SIMPLE. What about trigonometric functions? Well, I have to say that is JUST THE SAME. Like 100% the same.

There is only a few things you need to note:

- Given a trigonometric graph with/out transformation, the turning points of the graph give an instantaneous rate of ZERO!!!
- Given a plain trigonometric function, and asked to find the instantaneous rate when a < x < b, graph the function and use the Tangent Operation to find the instantaneous rate of change  simply subsititute the a and b into the function to find the y-values. Then use the traditional way to find the instantaneous rate of change.
- It really doesnt matter how complicated or how simple the function is, because your only need two distinct points to find the instanatneous rate of change

Take a look at this video to understand some detailed examples of this topic, if you are still unsure or how to find instantaneous rate of change. Remember to practise!

If you come across a quadratic trigonometric equation........hmmmm

Quadratic trigonometric equations are just like any other quadratic fequations. Dont see sinx / cosx / tanx as a BIG thing...if you look carefully, replace the trigonometric identities with a simple x, you will see the nostalgicity of it ; it will look like a regular quadratic equation.

there are generally 4 methods you can use to solve a quadratic trigonometric equation. stay calm and look carefully =)

method 1. common factor (factor out the common trigonometric identity and equate them to zero to find the acute angle,x)

method 2. trinomial factor (factorise the whole equation to get something like (kx + a)(jx + b). equate each polynomial to zero to find the acute angle, x)

method 3. replace 'complex' trigonometric identities with simpler ones (using the trigonometric formulas. then factor them, and again, equate it to zero to find the acute angle, x)

method 4. quadratic formula (use the quadratic formula to determine the roots of an unfactorable trigonometric equation. or simply use the GC)

Sorry to be repetitive, but seriously, it's nothing much of a big deal. Take it as a quadratic equation you are required to solve.

For example,

tanxcos²x = tanx    (uh-ohhhhhhh...what now?!)

Remember the identity cos²x = 1 – sin²x  ? Use it!

Substitute it into the equation and you will get tanx (cos²x – 1)Factor out (cos²x – 1) so you can find the acute angle of x.
See...Told you it isnt THAT tough   =D

Solving Trigonometric Equations

Solving trigonometric equations is all about finding the x value in the equation, or what may be also known as the angle in the equation.
First things first, to be able to solve trigonometric equations, you must first master your special angles and must know it by heart. refer to previous posts if you need kick-start reviews on special angles

Step by Step:
1. Simplify the equation so the left hand side has only the trigonometric identity and right hand side has only the integers
2. Note the positive/negative sign on the right hand side. Then determine which quadrant the angle is in (with reference to the trigonometric identity on the left hand side)
3. Find the value/s of x  (note that the value of x determined from the equation is an acute angle. Thus, the possible values of x must be determined by adding or substracting angles to the acute angle, depending on which quadrant the angle is in)

e.g.
sin x = - s.r3 / 2  ;   0  <  x  <  2pi

Since we know that the acute angle of x is pi / 3 (use special angles to determine the angle in pink), and now after sketching the graph, we know that sin x lies in the 3rd and 4th quadrant (because in the 3rd and 4th quadrant, sin is a negative ; this matches with the answer determined). Hence, there are 2 values of x which lie in the 3rd and 4th quadrant (obeying the restriction that x's maximum is 2pi)

x = pi / 3 + pi     >>>     because to find an angle in the 3rd quadrant is pi + acute angle
   = 4pi / 3

AND

x = 2pi - pi / 3
   = 5pi / 3

Solving trigonometric identities is really about practising by doing more exercises and familiarizing yourself with the special angles and methods to solve it.

Definitely not a one-time-go math topic.

Check this video out. It'll probably help you understand the above much, much better.
Good Luck!

Sinusoidal Functions of the form f(x) = a sin/cos [k (x-d)] + c

SINUSOIDALLLLLL! (big bombastic word here) ; but it is not self-explanatory. the things under it is not as bombastic as it sounds. hehehehehe...

As usual, when you get a function  f(x) = a sin/cos [k (x-d)] + c,

|a| = amplitude of the graph (refer to the previous 2 posts for a clear definition of an amplitude)
k = horizontal stretch / compression
d = horizontal shift
c = vertical shift

The only thing that's gonna be new here is the method to find the graph's period, given its equation. Using the formula
2π / k = Period
remember, k is the horizontal stretch / compression you can determine by the equation given. Hence, substitute k into the formula, and you will be able to find the period of your trigonometric graph.

Example,
Transform a sine function such that g(x) has amplitude 4, period π, phase shift π/6rad and 2 units up.

Using the formula given above,
the period of the function = 2π/k = π. Therefore, k is 2.
Hence, g(x) = 4 sin [2 (x + π/6)] + 2

I found this exercise quite beneficial especially if ur new to these. After mastering a few questions, you're set and ready to go! (of course, only for this subtopic...hehehe) EXERCISES!

Sayonara!

Graphs of Reciprocal Trigonometric Functions

Now, the R-E-C-I-P-R-O-C-A-L!!! you really cant deny that mathematicians from the past have genuinely creative and curious minds. =]

This part of math has nothing new to learn, but only new things to remember...
Since trigonometric function graphs are periodic, they are continuous and will definitely have x-intercepts.

To remember:

1. The x-intercepts of a trigonometric function become the vertical asymptotes of the respective trigonnometric function's reciprocal.

2. The reciprocal of a trigonometric function IS NOT THE SAME as the inverse of the trigonometric function.

3. Practise, practise, practise with the graphing calculator!

Examples:

If y = sin x is graphed, its x-intercept lies at π.
If y = 1/sin x is graphed, its x-intercept of  π becomes its vertical asymptote. Hence, y = 1/sin x has a vertical asymptote at x =  π

The reciprocal of a trigonometric function is 1 divided by the trigonometric function. However, Trigo^-1  is asking for the angle that has the trigonometric ratio equal to x (the inverse).

For further help of this, this video can probably help you understand a little better as it provides detailed examples. Good luckie!!!~

Graphs of Sine, Cosine and Tangent Functions

FIRST OF ALL, you must remember that trigonometric functions like sine, cosine and tangent are periodic. In other words, they are continuous with the same pattern  =D  think of eternal~
I would say this part of math is fairly easy to understand. But it's just the matter of understanding and knowing it well without memorizing.

Remember how polynomial functions could have their transformations? Like the vertical stretch, horizontal shift, etc? Trigonometric functions can too. But there are 'special ways' to understand their transformation. Not that much of a difference, its just the way to find the value of each transformation.

Before that, note there are 3 graphs at the top. Now, these 3 graphs are all trigonometric graphs over ONE CYCLE.

What is a PERIOD?
A period is basically the x value at the end of a cycle (from the beginning of the first cycle to the end of the first cycle). In this case, the period for sinx graph is 2pi, for cosx graph is 2pi, and pi for the tanx graph. It means how 'long' it takes for one cycle to be completed.

What is an AMPLITUDE?
An amplitude is like the 'height' of the graph from the x axis. Ask yourself "How high and how low does the graph go, in terms of the y values?" and you will get the amplitude of the graphs. In this case, the amplitude for the sinx graph is 1unit, for the cosx graph is 1unit, and 1 unit as well for the tanx graph.
To refresh,

y = a sin k (x - d) + c, where

a = vertical stretch / vertical compression
k = horizontal stretch / horizontal compression
d = horizontal shift (left if +, and right if -)
c = vertical shift (upwards or downwards)

Normally, the questions asked in this topic will ask for a new equation based on the information they provide you with, such as the vertical shift, horizontal compression and more. All you need to do is just to fill in the value they provide at the correct space in the general equation. For example,

A sine function has a vertical stretch of 2.
Hence, the equation of the function is  y = 2sinx

However, sometimes they give you the period of the graph and expect you to find its horizontal stretch/compression. Then comes the formula  

Period = 2pi / k, where k is the horizontal stretch/compression.

For example,
Find the value of k if the period is 2pi
Hence, 2pi = 2pi / k   >>>   k = 4.
y = sin4x

Simple as that! If you need more help, Click Here for a YouTube Tutorial

Bye!

By the way. . .

After learning about speedometers and secants yesterday, I became mathcrazee while driving my way home.

I noted the mileage on my dashboard and the time I left college. Drove home as usual and noted my end mileage and the time I arrived home. Just for the fun of it, I calculated my average rate of speed throughout my journey. 

I subtracted the difference in my beginning and end mileage and divided it by the time I reached home. As a result, I found out that I was driving at 80km/h. 

Ok I know...sounds crazy. But why not apply these stuffs to real life?? Wouldn't it be much easier to remember or understand it this way, instead of blindly memorizing from the book?

Try it yourself. It really works! 

Now...it's the Slope of Tangents and Instantaneous Rate of Change

In the previous post, I explained all about slopes of a secant on a graph. Just refreshing, secant line is a line connecting any 2 points on a graph. Whereas, a tangent line is a line passing through only 1 point on a graph.

The point on the graph is where the instantaneous rate of change takes place ; that is, the
change in y ÷change in x without an interval
Unlike the slope of a secant line, we have 2 reference points to determine its slope. How about a tangent line where there is only ONE point to refer to?

Steps:
1. Draw a tangent line through the point of reference ; a line cutting through only 1 point on the graph. NOTHING MORE THAN THAT ONE POINT!

2.  As you can see, the tangent cuts 1 point on the graph. Now, take any 2 points from the yellow straight line, a.k.a the tangent line, and find its slope, just the way you found the slope for a secant line.
3. When you have found your slope of the tangent line, this is called the instantaneous rate of change at the red point when x = ...

Note : You have to know the secant line well enough first to be able to master the tangent line questions.

See you!

Slopes of Secants and Average Rate of Change

Average rates of change are like a mixture of Physics and a little basic maths. It's basically the y-values divided by its x-values.  In other words, it is like the gradient/slope of any graph with respect to its x-values.

Let's just say...

A circular wave is formed on the surface of a basin filled with water. The table below shows the radius of the circular wave during the first 10s.

Time, t (s)	Radius, r (m)
0	0
1	2
2	4
3	6
4	8
5	10
6	12
7	14
8	16
9	18
10	20

If a question asks, determine the average rate of change of the radius, this means,

∆ y ÷ ∆ x = yx^-1

in this case, the specific formula would be... 

∆ r ÷ ∆ t = rt^-1

then,

(20 - 0) m ÷ (10 - 0) s = 2ms^-1

Note:

- Secant is the name for the line on a graph which connects ANY TWO points on a graph ; be it a quadratic graph, cubic graph or what not.

- Remember to have the x value written with a power of negative 1 (the small -1 on the top of it) or the answer will have no meaning

This part of math is just about UNDERSTANDING its formula, NOT memorizing it, but spending time to PRACTISE on it. I've found a pretty useful website that gives you examples and its solutions as well as some necessary explanation. Check it out Here!

Until next time!~

From the Unsensible to the "Ohhhh this makes sense!!!"

It's finally clearer now. Well, "Equations and Graphs of Polynomial Functions" isn't as difficult after all. Not until you differentiatet the terms properly. Phew, now I don't need to look like a blur girl who never attended math class before. =D

So! Two terms, that both make the world turn differently are...

1. Even Degree

As I've mentioned in my previous posts, even degree is a degree with an even number (2,4,6,8,...). Thus, a function with an even degree can be x to the power of 'something'.

WHEREAS *drum rolls*

2. Even Function (the graph looks like a mirror reflection across the y-axis)

- has a LINE SYMMETRY at x = 0

- ALL the powers of x's in the function are EVEN powers

- f(x) = f(-x)

  since any negative number to the power of an even power results in its original function, therefore,    

  when the function is EQUAL to the negative of its function, the function is even.

AND...

3. Odd Function (the graph looks like a mirror reflection across the y-axis)

- has a POINT SYMMETRY at (0,0)

- ALL the powers of x's in the function are ODD powers

- f(-x) ≠ - f(x)

   since any negative number to the power of an odd power results in a totally different function,

   therefore, when the function is NOT EQUAL to the negative of its function, the function is odd.

Warm Note : A function with a mix of odd and even powers is a 'neither odd nor even function'

Interesting way to remember:

Think of the white as the even function, the black as the odd function. Now these 2 functions are distinctly different from 1 another. A neither odd nor even function would be, the grey line seperating the black and white. =) simply simple!

Happy Maths!

Equations & Graphs of a Polynomial Function (going mumbo jumbo)

Oh my goddddddddddddddddddddddddddddddddddddddd!!! This isn't as easy as I expected! Especially with the textbook answers all different from Darren's. Sh*t.....what's going onnn...?? The more I look at those unanswered questions, the more I go mumbo jumbo up in my head. I REALLY do get the concept. But what is up with those weird questions answers? =(

Nevermind. Let's get the concet straight first!

Given the polynomial function

y = x² + 1

   = (x + 1)(x – 1)

Solely from an equation, this is the information you need to spot in order to sketch the function's graph without using a graphing calculator.

i) The sum of exponents of all factors (the equation has 2 factors with an exponent of 1. Thus, The sum of exponents of the equation is 2. note: each bracket is a factor.)

ii) The sign in front of the product of all x's (x x x = x² . Therefore, we can see that the sign in front of the product of all x's in the function is a positive sign)

iii) The x-intercept and y-intercept of the function (substitute y = 0 to get x-intercept and x = 0 to get y-intercept)

iv) The positive/negative sign of f(x) (substitute an x value to determine if f(x) is positive or negative at that particular x point)

 

Similarly, from a given graph, you will be able to identify and state the

i) Degree = 2 ; because it has n x-intercept, n - 1 local maximum/minimum point and its graph extends from quadrant 2 to quadrant 1. This information is enough to show that the graph has an even degree of 2 =D

ii) Sign of leading coefficient = Positive ; because the graph extends from quadrant 2 to quadrant 1

iii) x-intercept = -1, 1

      y-intercept = -1

iv)

Intervals	x < -1	-1 < x < 1	x > 1
Sign of f(x)	+	-	+

The new phrase of the day!

...has a zero at x = ? with order of ?...

Ok Ok...calm down. Read more, you'll understand better.

Let's say... y = -2 (x – 3)²

Expand it,    x - 3 = 0

                    x = 3

Hence, the function has a zero at x = 3 with the order of 2

Not too difficult right? The term "with the order of" is basically just the highest power of the function

Ps: I found this site where this person posted a quartic function question and there were many comments posted there. I think the comments, explanations and answers really help. Click Here

Alright then...not-too-happy-math! Till I solve more questions!

Characteristics of a Polynomial Function

Did you know???

- Negative leading coefficients result in a negative finite difference. Similarly, a positive leading coefficient results in a positive finite difference.

   i.e. y = -2x³ Since -2 is the leading coefficient and is a negative number, hence, the constant finite 

         difference for this polynomial function will also be a negative constant

- The number of finite differences is the same as the degree of the polynomial function.

   i.e. y = -2x³ Since the degree of the polynomial function is 3, hence, this polynomial function has 3

         finite difference / hence, this polynomial function reaches its constant at the 3rd finite difference

- Simple??? I say very this is pretty straight forward. What makes it simpler is the existence of a formula for finding finite differences without using a graphing calculator!

                                      Constant Finite Difference = Leading Coefficient x Degree!

C.F.D = a x n!

For example,

A polynomial function has 4 finite difference and its constant is -48

Therefore, n = 4   >>>   Remember? The number of finite differences = The degree of a P. function

Using the formula,

C.F.D = a x n!

    -48 = a x (4 x 3 x 2 x 1)

       a = -2

Don't believe me? Try more questions Here

At first, I didn't quite understand how an even/odd degree polynomial function affected the maximum/minimum points of its graph. But after i read through this table, everything seemed to make MUCH more sense! Hoorayyy!!!

Description	Odd Degree (n)		Even Degree (2n)
Leading Coefficient	+	-	+	-
General Shape	Refer to picture A	A reflection in the x-axis of its original shape	A reflection in the x-axis of its original shape	Refer to picture B
Domain	{ x ϵ R }	{ x ϵ R }	{x ϵ R }	{ x ϵ R }
Range	{ y ϵ R }	{ y ϵ R }	{ y ϵ R | y has a restriction }	{ y ϵ R | y has a restriction }
End Behavior	Extends from quadrant 3 to quadrant 1	Extends from quadrant 2 to quadrant 4	Extends from quadrant 2 to quadrant 1	Extends from quadrant 3 to quadrant 4
Number of absolute maximum / minimum points	0	0	1	1
Total number of local maximum / minimum points	Only a maximum of n-1	Only a maximum of n-1	Only a maximum of n-1	Only a maximum of n-1
Total number of x-intercepts	Maximum n, minimum 1	Maximum n, minimum 1	Maximum n, minimum 0	Maximum n, minimum 0

Note :

1) Both ends of the graph with an odd degree faces an opposite direction (1 north, 1 south). Whereas, the ends of the graph with an even degree faces the same direction (both north / both south).

2) ONLY the even degree has :

   - restrictions for its range (y values)

   - ONE absolute maximum / minimum point (ABSOLUTE basically means MOST. So, absolute

     maximum/ minimum point is the point on the graph with even degree that has the highest or lowest

     peak)

3) ALL the types of polynomial graphs can have a maximum number of local maximum / minimum

    points of n-1. Nothing more than that!!!

*** By the way, an absolute maximum / minimum point need not be a turning point on a graph. It can also be an end of the graph which has not restrictions***

I hope this piece of information helped you well! It cleared my doubts, totally!

See you soon!