More complicated quotients

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July 13th 2008, 11:07 PM
Naur

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More complicated quotients

See picture.
July 14th 2008, 12:25 AM
flyingsquirrel

Hi (again :D)

Both the numerator and the denominator of the fraction have been multiplied by $\sqrt{3t^2+4}$ . How could you've seen this ? Take a look at the denominator : it was $3t^2+4$ and it becomes $(3t^2+4)^{\frac32}=(3t^2+4)\sqrt{3t^2+4}$ hence it has been multiplied by $\sqrt{3t^2+4}$ . So that the new fraction equals the old one, one has to multiply the numerator of the old one by $\sqrt{3t^2+4}$ too. It can be checked that this was done by computing

$\begin{aligned}\underbrace{\left[3\sqrt{3t^2+4}-3t\times\frac{3t}{\sqrt{3t^2+4}}\right]}_{\text{old numerator}}\sqrt{3t^2+4} &= 3\sqrt{3t^2+4}^2-9t^2\frac{\sqrt{3t^2+4}}{\sqrt{3t^2+4}}\\ &= 3(3t^2+4)-9t^2\\ &= \text{new numerator}\end{aligned}$

Thus what happened is that the whole fraction has been multiplied by $\frac{\sqrt{3t^2+4}}{\sqrt{3t^2+4}}$ .
July 14th 2008, 03:28 AM
Naur

Hello again :D
Oh...So you're saying I should've applied the old rule; "What ever you do to the bottom you have to do to the top"?
You have to times the bottom by the brackets, you have top times the top as well?
I don't know how that didn't become apparent in other questions. I really really don't.
July 14th 2008, 03:52 AM
flyingsquirrel

Quote:

Originally Posted by Naur

Hello again :D
Oh...So you're saying I should've applied the old rule; "What ever you do to the bottom you have to do to the top"?
You have to times the bottom by the brackets, you have top times the top as well?

Yes, that's it !
July 14th 2008, 04:14 AM
ticbol

Quote:

Originally Posted by Naur

See picture.

I like your "Apart from the fact that they wrote it in the most confusing way possible..." (Happy)

Here is one way how they arrived at their answer. The old "fraction way".

dV/dt ={[3(3t^2 +4)^(1/2)] -[(3t)[3t/ (3t^2 +4)^(1/2)]} / {3t^2 +4}

Combine the numerator into only one fraction.
The common denominator is (3t^2 +4)^(1/2).

dV/dt ={[3(3t^2 +4)^(1/2)] -[(3t)[3t/ (3t^2 +4)^(1/2)]} / {3t^2 +4} ---we continue from here.

dV/dt = {[3(3t^2 +4)(1/2))^2 -9t^2] /(3t^2 +4)^(1/2)} / {3t^2 +4}
dV/dt = {[3(3t^2 +4) -9t^2] /(3t^2 +4)^(1/2)} / {3t^2 +4}

Hey, the exponent (1/2) disappeared in the numerator of the original numerator!

Continuing...

dV/dt = {[3(3t^2 +4) -9t^2] /(3t^2 +4)^(1/2)} * {1 /(3t^2 +4)}
dV/dt = {3(3t^2 +4) -9t^2} /[(3t^2 +4)^(1 +1/2)}
dV/dt = {3(3t^2 +4) -9t^2} /[(3t^2 +4)^(3/2)}

That is it.
July 15th 2008, 11:20 PM
Naur

Quote:

Originally Posted by flyingsquirrel

Hi (again :D)

Both the numerator and the denominator of the fraction have been multiplied by $\sqrt{3t^2+4}$ . How could you've seen this ? Take a look at the denominator : it was $3t^2+4$ and it becomes $(3t^2+4)^{\frac32}=(3t^2+4)\sqrt{3t^2+4}$ hence it has been multiplied by $\sqrt{3t^2+4}$ . So that the new fraction equals the old one, one has to multiply the numerator of the old one by $\sqrt{3t^2+4}$ too. It can be checked that this was done by computing

Okay. I almost understand that. But why did they multiply it all by that? What's the point in multiplying an entire fraction by something? How would I know to? Is it only because there's no other way of doing it?

ticbol...I'm not sure what you did here...

Quote:

dV/dt = {[3(3t^2 +4)(1/2))^2 -9t^2] /(3t^2 +4)^(1/2)} / {3t^2 +4}
July 16th 2008, 12:07 AM
flyingsquirrel

Quote:

Originally Posted by Naur

Okay. I almost understand that. But why did they multiply it all by that? What's the point in multiplying an entire fraction by something? How would I know to?

They multiplied the fraction by that in order to get rid of the square roots which appear in the numerator : $3{\color{red}\sqrt{\color{black}3t^2 +4}} -3t\times\frac{3t}{\color{red} \sqrt{\color{black}3t^2 +4}}$ . This can be done by multiplying the numerator by $\sqrt{3t^2+4}$ . In order not to modify the value of the fraction, one also has to multiply the denominator by the same number : that's why the fraction has been multiplied by $\frac{\sqrt{3t^2+4}}{\sqrt{3t^2+4}}$

Quote:

Is it only because there's no other way of doing it?

You can do it as ticbol did : ${3\sqrt{3t^2 +4}} -3t\times\frac{3t}{ \sqrt{3t^2 +4}}= \frac{{3\sqrt{3t^2 +4}^2} -3t\times 3t}{ \sqrt{3t^2 +4}}$ hence ...
July 16th 2008, 12:45 AM
Naur

Ohh, so we did it your way because it's easier, though it can be done the other way. I was going to ask before, why bother multiplying numerator and denominator by the same number, but I get it now. It's because the value of the sum doesn't change, but it's easier to manipulate.

Thanks very much. Such a great feeling when you understand something you never thought you'd be able to :D

Could you just pop over to my other thread from before? I updated the last post with an edit, and I worry you'll miss it :D