Cant work this out, why is it happening,

[devanlevin]

cant work this out, why is it happening, please help!

y(t)=22t-4t^{2}
x(t)=30t-t^{3}

find x(y) or y(x)
what i did was say

y-22t+4t^{2}=0

t=\frac{22+\sqrt{484-16y}}{8}=\frac{11+\sqrt{121-4y}}{4}

x=30*\frac{11+\sqrt{121-4y}}{4}-(\frac{11+\sqrt{121-4y}}{4})^{3}


which works for every value for y(t) i plug in except for t=2s
where i get, from the original equations, x=52 y=28
but from my equation when y=28, x=62.125
why is this happening?

 

[Mark44]

Did you remember to cube the 4 in the denominator when you substituted for t^3? x=30IOW, the second term in your value for x should be:
\frac{(11+\sqrt{121-4y})^3}{4^3}

Also, you are omitting the \pm in your value for t, but this is a more minor problem.

 

[HallsofIvy]

y(t)=22t-4t^{2}
x(t)=30t-t^{3}

find x(y) or y(x)
what i did was say

y-22t+4t^{2}=0

t=\frac{22+\sqrt{484-16y}}{8}=\frac{11+\sqrt{121-4y}}{4}

The equation 4t^2- 22t+ y= 0 has two solutions! Choosing the positive root means that you are assuming that t> 11/4= 2.75 and your formula will be correct only for t> 2.75

x=30*\frac{11+\sqrt{121-4y}}{4}-(\frac{11+\sqrt{121-4y}}{4})^{3}


which works for every value for y(t) i plug in except for t=2s
where i get, from the original equations, x=52 y=28
but from my equation when y=28, x=62.125
why is this happening?

 

[devanlevin]

thanks, that works perfectly, but why do you say specifically 11/4, because of 11/4 in the fraction or because when the sqrt=0, t=11/4

 

[Mark44]

Because your value for t is 11/4 + a positive quantity (i.e., sqrt(121 - 4y)/4 ). By omitting the other solution, you are losing sight of the fact that there is a solution for t < 11/4.

 

[devanlevin]

thanks