1. ## Rearranging

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

2. Hi
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 ?

3. Hello there
I'm not quite sure what you mean by that.

4. 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 ?

5. 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 $90-t=0$..

6. 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.

7. 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 $t$.

8. 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?

9. 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.

10. 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.