Rearranging

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July 13th 2008, 11:07 PM
Naur

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Rearranging

Why isn't my rearranging correct? How would you correctly rearrange this to find t?
July 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 ?
July 14th 2008, 03:26 AM
Naur

Hello there :D
I'm not quite sure what you mean by that.
July 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 but we don't know the value of yet (since we're looking for it...) hence we cannot divide the equation by 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 ?
July 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 $90-t=0$ ..
July 15th 2008, 03:57 AM
Naur

Oh dear.
See, the answer is 60.

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.
July 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 $t$ .
July 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?
July 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 $0<t<60$ .
July 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