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 :)
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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

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=y^{2} by inspection or ln(y)=t and ln(x) =2t so ln(x)=2ln(y) or x=y^{2.}

Jim

Elana G. | A Gator Can Chomp Away Your Accounting FearsA Gator Can Chomp Away Your Accounting F...

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

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