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

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- July 14th 2008, 12:07 AMNaurRearranging
Why isn't my rearranging correct? How would you correctly rearrange this to find t?

- July 14th 2008, 12:59 AMflyingsquirrel
- July 14th 2008, 04:26 AMNaur
Hello there :D

I'm not quite sure what you mean by that. - July 14th 2008, 04:48 AMflyingsquirrel
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 : . We can safely divide the equation by (since it does not equal 0) and it gives us . That's one of the solutions.
- Second case : . For , the equation is true since hence is the other solution.

Is it clearer ? - First case : . We can safely divide the equation by (since it does not equal 0) and it gives us . That's one of the solutions.
- July 14th 2008, 06:16 AMkalagota
or simply, from , you can factor out so that it becomes ..

therefore, from there, either or .. - July 15th 2008, 04:57 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. - July 15th 2008, 05:28 AMflyingsquirrel
- July 16th 2008, 12:49 AMNaur
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, 02:18 AMflyingsquirrel
No, that's it. The thing is that the points such that can be local maximums or local minimums.

To show that is a local maximum, you can, for example, show that increases on and decreases on . ( on and on )

The same idea applies to show that is a local minimum : simply show that the derivative is negative for and positive for . - July 17th 2008, 01:24 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