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Function For Distance Between 2 Parametric Equations

parametrics equations distance casio prizm 9860

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#1 Sumaxi

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Posted 13 April 2015 - 11:19 AM

( Can a Mod move this to the correct board? -- I threw it in here by accident since it sounded about right )

 

Hiya,

I'm in a classroom full where everyone uses Ti calculators and no one can help me with my casio, one of the equations that everyone has is:

 

y4 = sqrt( (xt1(x) - xt2(x))^2 + (yt1(x)-yt2(x))^2 )

 

where y4 is in y= and the x1t xt2 yt1 yt2 are in parametric.

 

I'm getting a syntax error and I'm pretty sure that's because i'm trying to use parametric variables in a y= function, HOWEVER THAT'S EXACTLY WHAT EVERYONE ELSE IS DOING.

 

So please, help me out, I don't know what else to do. I haven't found any good parametric tutorials on casio and I don't know how to use that equation to find the distances between the 2 parametric equations.


Edited by Sumaxi, 13 April 2015 - 11:23 AM.


#2 pan.gejt

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Posted 14 April 2015 - 07:18 AM

Try direct substitution, see the example

 

x1(t) = 1,5*T                      x2(T) = -10+3*T

y1(T) = 0,5*T                    y2(T) = 15-1,5*T

 

So you can rewrite these equations as (T is X now):

y = sqrt((1,5*x-(-10+3*x))^2+(0,5*x-(15-1,5*x))^2)

 

which corresponds to

y = sqrt(6,25*x^2-90*x+325)


Edited by pan.gejt, 14 April 2015 - 07:27 AM.


#3 Sumaxi

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Posted 14 April 2015 - 11:11 AM

Try direct substitution, see the example

 

x1(t) = 1,5*T                      x2(T) = -10+3*T

y1(T) = 0,5*T                    y2(T) = 15-1,5*T

 

So you can rewrite these equations as (T is X now):

y = sqrt((1,5*x-(-10+3*x))^2+(0,5*x-(15-1,5*x))^2)

 

which corresponds to

y = sqrt(6,25*x^2-90*x+325)

 

direct substitution works, however if I use something in a y= from a yt1= or xt1=, the variables yt1 and xt1 will give me a error. I think the solution is some sort of converter from t to x but I don't know how to do that.

 

Thanks a ton for the reply







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