Why isn't my rearranging correct? How would you correctly rearrange this to find t?
We have the equation $\displaystyle 90t^2=t^3$ but we don't know the value of $\displaystyle t$ yet (since we're looking for it...) hence we cannot divide the equation by $\displaystyle t^2$ without considering two cases :
- First case : $\displaystyle t\neq0$. We can safely divide the equation by $\displaystyle t^2$ (since it does not equal 0) and it gives us $\displaystyle \boxed{t=90}$. That's one of the solutions.
- Second case : $\displaystyle t=0$. For $\displaystyle t=0$, the equation is true since $\displaystyle 90\times 0^2=0^3$ hence $\displaystyle \boxed{0}$ is the other solution.
Is it clearer ?
OH you're right! I had a derivative already, so I ignored the fact that I needed to find the derivative of THAT to get the maximum.
I still don't see how we got the minimums though. Unless the minimums happened to be the minimums/maximums of the original formula as well, I guess they were.
Thanks very much for your help
Just a query...I differentiated to find the maximum, and came out with t=0 or t=60. Is this right? t shouldn't be 0.
I differentiated to:
1/100(180t-3t^2)
Then, put that equal to 0, and rearranged:
0=1.8t-0.03t^2
0=t(1.8-0.03t)
So, due to the law "If A*B=0, either A or B=0", I decided that t=0 or (1.8-0.03t)=0. This gives t=0 or t=60.
Am I wrong?
No, that's it. The thing is that the points such that $\displaystyle 0=\frac{\mathrm{d}^2V}{\mathrm{d}t}$ can be local maximums or local minimums.
To show that $\displaystyle t=60$ is a local maximum, you can, for example, show that $\displaystyle \frac{\mathrm{d}V}{\mathrm{d}t}$ increases on $\displaystyle (0,60)$ and decreases on $\displaystyle (60,\infty)$. ($\displaystyle \implies \frac{\mathrm{d}^2V}{\mathrm{d}t^2}>0$ on $\displaystyle (0,60)$ and $\displaystyle \frac{\mathrm{d}^2V}{\mathrm{d}t^2}<0$ on $\displaystyle (60,\infty)$)
The same idea applies to show that $\displaystyle t=0$ is a local minimum : simply show that the derivative is negative for $\displaystyle t<0$ and positive for $\displaystyle 0<t<60$.