- #1

- 145

- 9

**Say you're given x = 3t^2 + 6t - 4.**

If you treat

*x*as an unknown variable and movie it over to the other side, you can use completing the square to solve for

*t*. This eventually gets you: t = 1/3*sqrt(3(x+7)) - 1

Then you differentiate the initial function to obtain the velocity, substitute the calculated time in and thus obtain velocity in terms of displacement. This method ultimately got me v = 2*sqrt(3x+21), which works and is completely fine.

However, when I try the alternative way it doesn't work. If we go back to the beginning and find both velocity and acceleration, we get x = 3t^2 + 6t - 4, v = 6t + 6 and a = 6. To obtain dv/dx, we need to multiple the dv/dt by dt/dx (with the dt on the numerator and denominator canceling out). We know dv/dt, as that's just another way of expressing the acceleration, which is 6. Now this is the part which I think is messing up; I contend that dt/dx would should just be the inverse of dx/dt, which we know to be 6t + 6. This should mean that dt/dx should just equal 1/(6t + 6). Thus, dv/dx = dv/dt * dt/dx = 6*(1/(6t + 6)) = 1/(t+1). To obtain velocity in terms of displacement, we then would need to differentiate dv/dx, which should leave us with v(x) = ln(|t + 1|) + c. At t = 0, we know velocity = 6. Therefore, 6 = ln(|1|) + c, c = 6. v(x) = ln(|t+1|) + 6.

However, this equation is clearly very different from the one obtained in the first method and doesn't work. Can someone please tell me how to fix the second method?