1. Nov 14, 2008

devanlevin

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?

2. Nov 14, 2008

Staff: Mentor

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.

3. Nov 14, 2008

HallsofIvy

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

4. Nov 16, 2008

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

5. Nov 16, 2008