# Rearranging

• Jul 13th 2008, 11:07 PM
Naur
Rearranging
Why isn't my rearranging correct? How would you correctly rearrange this to find t?
• Jul 13th 2008, 11:59 PM
flyingsquirrel
Hi :)
Quote:

Originally Posted by Naur
Why isn't my rearranging correct? How would you correctly rearrange this to find t?

The only mistake you've done is that you've divided by $t$ without checking that $t\neq 0$. Can you correct this ?
• Jul 14th 2008, 03:26 AM
Naur
Hello there :D
I'm not quite sure what you mean by that.
• Jul 14th 2008, 03:48 AM
flyingsquirrel
Quote:

Originally Posted by Naur
I'm not quite sure what you mean by that.

We have the equation $90t^2=t^3$ but we don't know the value of $t$ yet (since we're looking for it...) hence we cannot divide the equation by $t^2$ without considering two cases :

• First case : $t\neq0$. We can safely divide the equation by $t^2$ (since it does not equal 0) and it gives us $\boxed{t=90}$. That's one of the solutions.
• Second case : $t=0$. For $t=0$, the equation is true since $90\times 0^2=0^3$ hence $\boxed{0}$ is the other solution.

Is it clearer ?
• Jul 14th 2008, 05:16 AM
kalagota
or simply, from $90t^2 - t^3 =0$, you can factor out $t^2$ so that it becomes $t^2(90-t)=0$..
therefore, from there, either $t^2=0$ or $90-t=0$..
• Jul 15th 2008, 03:57 AM
Naur
Oh dear.

Okay, what we had to do, is use the formula at the start, which is DV/Dt, and find the maximum.
They drew a graph, to find maximum = 60.
I put DV/Dt = 0, but only obtained the two minimum values, 90 and 0.
• Jul 15th 2008, 04:28 AM
flyingsquirrel
Quote:

Originally Posted by Naur
Okay, what we had to do, is use the formula at the start, which is DV/Dt, and find the maximum.
They drew a graph, to find maximum = 60.
I put DV/Dt = 0, but only obtained the two minimum values, 90 and 0.

If you're looking for the maximum of $\frac{\mathrm{d}V}{\mathrm{d}t}$ then you've to compute its derivative $\frac{\mathrm{d}^2V}{\mathrm{d}t^2}$ and to solve $\frac{\mathrm{d}^2V}{\mathrm{d}t^2}=0$ for $t$.
• Jul 15th 2008, 11:49 PM
Naur
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 :D

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?
• Jul 16th 2008, 01:18 AM
flyingsquirrel
Quote:

Originally Posted by Naur
[...]
This gives t=0 or t=60.
Am I wrong?

No, that's it. The thing is that the points such that $0=\frac{\mathrm{d}^2V}{\mathrm{d}t}$ can be local maximums or local minimums.

To show that $t=60$ is a local maximum, you can, for example, show that $\frac{\mathrm{d}V}{\mathrm{d}t}$ increases on $(0,60)$ and decreases on $(60,\infty)$. ( $\implies \frac{\mathrm{d}^2V}{\mathrm{d}t^2}>0$ on $(0,60)$ and $\frac{\mathrm{d}^2V}{\mathrm{d}t^2}<0$ on $(60,\infty)$)

The same idea applies to show that $t=0$ is a local minimum : simply show that the derivative is negative for $t<0$ and positive for $0.
• Jul 17th 2008, 12:24 AM
Naur
Oh, I should've guessed that. You mean, differentiating finds the maximum and the minimum (which I obviously knew), so 0 is the minimum.
Heh, easy. :P

Thanks again. :D