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

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- Jul 13th 2008, 11:07 PMNaurRearranging
Why isn't my rearranging correct? How would you correctly rearrange this to find t?

- Jul 13th 2008, 11:59 PMflyingsquirrel
- Jul 14th 2008, 03:26 AMNaur
Hello there :D

I'm not quite sure what you mean by that. - Jul 14th 2008, 03:48 AMflyingsquirrel
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 ? - 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.
- Jul 14th 2008, 05:16 AMkalagota
or simply, from $\displaystyle 90t^2 - t^3 =0$, you can factor out $\displaystyle t^2$ so that it becomes $\displaystyle t^2(90-t)=0$..

therefore, from there, either $\displaystyle t^2=0$ or $\displaystyle 90-t=0$.. - Jul 15th 2008, 03:57 AMNaur
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. - Jul 15th 2008, 04:28 AMflyingsquirrel
- Jul 15th 2008, 11:49 PMNaur
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 AMflyingsquirrel
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 $\displaystyle 0<t<60$. - Jul 17th 2008, 12:24 AMNaur
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