Trig graphs (2 Viewers)

[hello-there]

Can some1 show me with working how you would graph y=20sin(theta/2) for
0 < θ < pie

 

[Gussy Booo]

Hi there!

Well, you look at the function, and you have to recognise that its different from our regular y=sin(-).
We have a 20 at the front. and a x/2.

Now, I'm going to try and explain this, in a way in which we can relate to a graph.
So we'll start by figuring out how the 20 changes the curve.

so y=sinx. From previous years, we learnt that, in order to find the Y VALUE at a specific X VALUE, we simply sub in that x value into the RHS (considering the eqn y=sinx), and we are hence given a number which represents the y value.

Now, look at this --> 20 TIMES sin(x). We can see here, that whatever value, is given by sinx, it will be multiplied by 20 !. Therefore, our y value will be 20 times the original graph.

Hence we can conclude that, the maximum amplitude of the new equation is 20 and -20. (Y axis has 20--->-20).

We will now observe [x/2]. So lets consider: y=sin(x/2). We know that the x value is proportional the period of the curve.
So lets make x the subject.

y=sinx/2
x=2.arcsin(y) arcsin is SINE INVERSE.

As we can see, any y value we put in, will give a larger x value due to the multiplication of 2, hence making it wider. Therefore, y=20sin(x/2), will have a period of 4pi (2 TIMES its original period which is 2pi). Furthermore we are drawing a sine curve with an amplitude of 20, in the space given of 0-->4pi

But, we must also look at the restriction : 0<(-)< pi
By inspecting the restriction, and looking at the period calculated...we can conclude that we have a portion of a sine curve of amplitude +-20 on the Cartesian plane.

Can I also say that - when x = pi, y = 0. Furthermore, the restriction only permits a portion of y=20sin(-), until it first INTERSECTS the x-axis.

 

[tacogym27101990]

Can some1 show me with working how you would graph y=20sin(theta/2) for
0 < θ < pie

starts at zero, has an amplitude of 20 and has a period of T=2pi/(1/2)=2pi.
so at pi it reaches maximum amplitude for the first time.

 

[hello-there]

Hi there!

Well, you look at the function, and you have to recognise that its different from our regular y=sin(-).
We have a 20 at the front. and a x/2.

Now, I'm going to try and explain this, in a way in which we can relate to a graph.
So we'll start by figuring out how the 20 changes the curve.

so y=sinx. From previous years, we learnt that, in order to find the Y VALUE at a specific X VALUE, we simply sub in that x value into the RHS (considering the eqn y=sinx), and we are hence given a number which represents the y value.

Now, look at this --> 20 TIMES sin(x). We can see here, that whatever value, is given by sinx, it will be multiplied by 20 !. Therefore, our y value will be 20 times the original graph.

Hence we can conclude that, the maximum aplitude of the new equation is 20 and -20. (Y axis has 20--->-20).

We will now observe [x/2]. What this basically does, is divde its NORMAL period by 2. One period of a sinx curve is 0-->2pi. Hence, if we divide 2pi by 2, we get pi. Therefore, y=20sin(x/2), will have a period of pi. That is --> we are drawing a REGULAR sine curve with an amplitude of 20, in the space given of 0-->pi

Now, we must also look at the restriction : 0<(-)< pi
By inspecting the restriction, and looking at the period calculated...we can conclude that we have 1 sine curve of amplitude +-20 on the cartesian plane.

thanx alot man butttt...... how come when you input x=0 in the equation you get y=0 thus the amplitude at x=0 is not 20 but 0 which does not obide with what you just said.
So if i were to use a table of values i would get a diff graph how strange?

 

[annabackwards]

y=asin(bx)+c

a = amplitude (max displacement from rest)
Period = 2Pi/b (Time it takes for once complete curve to to be sketched)
c = translates the graph up or down

Gussy Booo explained the rest very well

thanx alot man butttt...... how come when you input x=0 in the equation you get y=0 thus the amplitude at x=0 is not 20 but 0 which does not obide with what you just said.
So if i were to use a table of values i would get a diff graph how strange?

aplitutde = MAXIMUM displacement. So you must look at the maximum displacement from rest (the axis of symmetry for the graph).

 

[hello-there]

Ohh so it does start at 0?
The answer has it starting at y=10 when x=0

 

[Gussy Booo]

thanx alot man butttt...... how come when you input x=0 in the equation you get y=0 thus the amplitude at x=0 is not 20 but 0 which does not obide with what you just said.
So if i were to use a table of values i would get a diff graph how strange?

Hmmm, perhaps you misunderstood me? Or maybe i just explained that part poorly. In terms of maths..it is correct

Ummm ok. so. When x=0
y=20 TIMES sin(0)
y=0?
Does it mathematically make sense?

 

[Gussy Booo]

Ohh so it does start at 0?
The answer has it starting at y=10 when x=0

y = 10 :| when x = 0 :|.....
that doesn't make sense..


y=20sin(x/2)
sub in x=0..what do you get ?
y=0.

Maths doesnt lie.

 

[hello-there]

The has these values for the graph
x= 0
y =10


x=90(degrees, POI)
y=10root2

x=180(degrees)
y= 20

 

[hello-there]

Yea i thought so, the answer is bugged unless it is for a diff question which is y=10sec(theta/2)

I would attach the doc for the whole question but it keeps failing

 

[Gussy Booo]

The has these values for the graph
x= 0
y =10


x=90(degrees, POI)
y=10root2

x=180(degrees)
y= 20

Ok, im lost lol.
You've given me this equation: y=20sin(theta/2)

So, if we sub, x=0...

y=20sin(0/2)
y=0?
So..wheres the problem =/

 

[Gussy Booo]

Yea i thought so, the answer is bugged unless it is for a diff question which is y=10sec(theta/2)

I would attach the doc for the whole question but it keeps failing

The values you have given me, EXACTLY MATCH: y=10sec(theta/2)
So, you are looking at the wrong question .

However, make sure you understand the concepts that I explained in my previous posts (y)

 

[hello-there]

The values you have given me, EXACTLY MATCH: y=10sec(theta/2)
So, you are looking at the wrong question .

However, make sure you understand the concepts that I explained in my previous posts (y)

yep thanx alot