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Trigonometric graphs (1 Viewer)

tennille

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Are there any easier ways of drawing a sketch of x = t + 2sint without substituting numbers like pi and all?
 

gordo

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draw the sin curve then shift it up 't' units

the 2 in fronnt means the amplitude will be 2

the period is 2 pi on t (the t after sin)
 

mojako

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Tennille said:
Are there any easier ways of drawing a sketch of x = t + 2sint without substituting numbers like pi and all?
period is 2pi !!
draw line x=t in dotted line
then plot pi/4, pi/2, 3pi/4, and connect them
then u get an idea of what it looks like.
then just follow the pattern.
its a kind of sine curve along the line x=t

alternatively, draw x=t and x=2sint,
and add them... use a few points as a model (such as pi/4, pi/2, 3pi/4 mentioned earlier)

NOTE: if I interpret the question correctly :p
 
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mojako

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oh, wait.. I dont think it technically has a "period".
but it has pattern... well, u know what I mean.
 

mojako

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Li0n said:
x<sup>2</sup> + sin<sup>2</sup>x is one scarey parabola...
its NOT a parabola.
by definition, a parabola is... the locus of point P(x,y) such that the following 2 distances are equal:
1. the perpendicular distance to a fixed line (directrix) from P
2. the distance to a fixed point (focus) from P
this creates a shape which is distinct from that quoted graph.
all parabolas are similar (like similar triangles etc), only enlarged or made smaller, and rotated around.
the enlargement factor is in the leading coefficient.
and the above curve certainly is not similar to any known parabola :p
[/end of rant]
 
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JamiL

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gordo said:
draw the sin curve then shift it up 't' units

the 2 in fronnt means the amplitude will be 2

the period is 2 pi on t (the t after sin)
no very very wrong....
it will look more like a stepps but curvey a lil, its a very hard graph 2 draw... very very hard. co t is a variuble it is always increasing in the value as the graph is drawn.. ps wots the domain
 

JamiL

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mojako said:
period is 2pi !!
draw line x=t in dotted line
then plot pi/4, pi/2, 3pi/4, and connect them
then u get an idea of what it looks like.
then just follow the pattern.
its a kind of sine curve along the line x=t

alternatively, draw x=t and x=2sint,
and add them... use a few points as a model (such as pi/4, pi/2, 3pi/4 mentioned earlier)

NOTE: if I interpret the question correctly :p
no that is wrong, u did the same thing you have to differenciate the function
and find all staionary points in the domain, u will find they are all inflection points
 

withoutaface

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mojako said:
its NOT a parabola.
by definition, a parabola is... the locus of point P(x,y) such that the following 2 distances are equal:
1. the perpendicular distance to a fixed line (directrix) from P
2. the distance to a fixed point (focus) from P
this creates a shape which is distinct from that quoted graph.
all parabolas are similar (like similar triangles etc), only enlarged or made smaller, and rotated around.
the enlargement factor is in the leading coefficient.
and the above curve certainly is not similar to any known parabola :p
[/end of rant]
Or more simply, a quadratic equation of degree 2
:)
 

mojako

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Li0n said:
that is NOT a parabola!!
look closely..
the bottom end is a bit steep / looks like straight line.
parabolas are all "similar", only enlarged by some constant factor.

the red one in the pic is mr parabola
the blue one is mr non-parabola
ok?(tm)

(tm) ok?(tm) is a trademark of its respective owner
 

mojako

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JamiL said:
no that is wrong, u did the same thing you have to differenciate the function
and find all staionary points in the domain, u will find they are all inflection points
here is the curve
I give y=x for comparison.

note: its not a sine curve along y=x.. its a kind of sine curve...
the crests & troughs are not in the middle of the U-sphaped thing.

But yea u need to differentiate to find stat pt... I cant think of any other approach.
 
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