Elimintate the parameter problem

[nameVoid]

elimintate the parameter
1.
x=lnt
y=t^(1/2)
e^x=t
y=e^(x/2)

2.
x=2t^3
y=1+4t-t^2
for slope = 1

dC/dt=(4-2t)/6t^2=(2-t)/3t^2=1
3t^2+t-2=0
t=-1
t=2/3
A(-2,-4)
B(16/27,29/9)

 

[Office_Shredder]

1 looks good. I'm not sure what's going on with 2... what do you mean by for slope=1? And then the whole second half?

 

[HallsofIvy]

elimintate the parameter
1.
x=lnt
y=t^(1/2)
e^x=t
y=e^(x/2)

2.
x=2t^3
y=1+4t-t^2
for slope = 1

dC/dt=(4-2t)/6t^2=(2-t)/3t^2=1
3t^2+t-2=0
t=-1
t=2/3
A(-2,-4)
B(16/27,29/9)

1) is correct. For 2) you have A, B, and C without defining them. I think what you are saying is that you want to find points where the slope of the graph given by the parametric equations is 1.
You "dC/dt" is simply dy/dx= (dy/dt)/(dx/dt)=(2-t)/3t^2=1 and then gives 2- t= 3t^2 so 3t^2+ t- 2= 0. Yes, t= -1 and t= 2/3. Then x(-1)= -2, y(-1)= -4 and x(2/3)= 16/27, y(2/3)= 29/9 so the two points are (-2, -4) and (16/27, 29/9).

I think your teacher would prefer that your working be not so cryptic!