Eliminate the parameter

Question

1) x=3-2t

    y=2+3t

 

 

2) x=sqrt(t)

     y=1-t

 

 

 

3) x=4+2cos(theta)

    y=1+2sin(theta)

 

 

4) x=sec(theta)

    y=cos(theta)

 

 

 

 

5) x=e^2t

    y=e^t

 

 

                          can someone help me with just one problem so i can do the rest?

                            thank you :)

Answer 1

1)  x=3-2t  --->  t= (3-x)/2   

     Replace t in the second equation   y= 2+3((3-x)/2) = 2+ 9/2-3x/2 = -3x/2 + 13/2 = -1/2(3x-13)

     y=-1/2(3x-13), where x is any real number

2)  Replace t=1-y from the second equation, into the first eq.

     x= sqrt(1-y), where 1-y >=0, Y<=1 and x>=0

     x^2 = 1-y  --->  y= 1- x^2, where x>=0

3)  cos(theta) =(x-4)/2  --->  theta = arccos((x-4)/2), where 0 <= (x-4)/2 <= pi  --->  4<= x <=2pi+4

     y= 1+2sin(arccos((x-4)/2), where 4<= x <=2pi+4

4)  theta= arcsec(x), 

     Y=cos(arcsec(x))  where x<=-1 or x>=1

5) x=e^(2t)  --->  ln x= ln e^(2t) = 2t ln(e) --->  t= (ln x)/2, where x>0

    y=e^((ln x)/2) =e^ln (sqrt(x)) = sqrt (x), x>o  

or easier way

    x= e^(2t) = (e^t)^2 = y^2 , where x>0

   or  y= sqrt(x), where x>0 

  

    

Answer 2

Daisy,

          The idea is to solve one of the parametric equations for t in terms of x or y and then substitute that in the other equation giving a relationship between x and y.

 

1. t=(y-2)/3 and x=3-2((y-2)/3)=3-(2/3)y+4/3=(9/3)+(4/3)-(2/3)y=(13/3)-(2/3)y

 

         so (2/3)y+x=(13/3)

 

The other easy one is 5. x=y² by inspection or ln(y)=t and ln(x) =2t so ln(x)=2ln(y) or x=y^2.

 

Jim

 

 

Answer 3

1) x=3-2t

y=2+3t

 

Set one equation equal to t.Then plug what t equals into the other equation. You are trying to get rid of the t and only be left with x and y.

 

X=3-2t

 

x-3= -2t

divide -2 by both sides

 

-0.5x+1.5=t

 

plug t into the y equation

 

y=2+3t

 

y=2+ 3(-0.5x+1.5)

 

y=2+ (-1.5x+4.5)

 

y=2-1.5x+4.5

 

y= -1.5x+6.5