Isolate T

[mrpmorris]

Hi all

I have the following equasion

(cT)^2 = D^2 + (vT)^2

I'd like to isolate T.

It's been a while so I was wondering if someone could help me with the steps? It's not homework (if that matters?), I am just trying to follow a book in which they present the answer

T^2 = D^2 / (c^2 - v^2)

but then instruct me to trust that the answer is correct. However I am not that trusting ;-)

Thanks!

 

[galactus]

\(\displaystyle (cT)^{2}=D^{2}+(vT)^{2}\)

\(\displaystyle c^{2}T^{2}=D^{2}+v^{2}T^{2}\)

\(\displaystyle c^{2}T^{2}-v^{2}T^{2}=D^{2}\)

Factor out T^2:

\(\displaystyle T^{2}(c^{2}-v^{2})=D^{2}\)

\(\displaystyle T^{2}=\frac{D^{2}}{c^{2}-v^{2}}\)

Square root of both sides:

\(\displaystyle T=\pm\sqrt{\frac{D^{2}}{c^{2}-v^{2}}}\)

 

[mrpmorris]

Thank you very much!

This step
c[sup:iknetnw3]2[/sup:iknetnw3]T[sup:iknetnw3]2[/sup:iknetnw3] - v[sup:iknetnw3]2[/sup:iknetnw3]T[sup:iknetnw3]2[/sup:iknetnw3] = D[sup:iknetnw3]2[/sup:iknetnw3]

What was the reasoning behind the decision to first isolate D[sup:iknetnw3]2[/sup:iknetnw3]? I can see how it led you to the answer, but my instinct was to go directly towards isolating T. What I am missing is that little spark of light that makes you realise that you have to get to T by first isolating D?

 

[galactus]

What was the reasoning behind the decision to first isolate D[sup:10itf0dj]2[/sup:10itf0dj]?

Because we have to isolate T. Thus, we get D on the other side. There are other ways to go about it. There always is in algebra, and this is a rather basic operation.

 

[chrisr]

\(\displaystyle D^2\ is\ the\ \underline{only}\ term\ not\ containing\ T.\)

\(\displaystyle Hence,\ T\ is\ a\ factor\ of\ all\ others.\)
\(\displaystyle However,\ you\ may\ think\ of\ it\ as\ \underline{bringing\ all\ the\ T\ terms\ together}.\)

 

[mrpmorris]

\(\displaystyle D^2\ is\ the\ \underline{only}\ term\ not\ containing\ T.\)

\(\displaystyle Hence,\ T\ is\ a\ factor\ of\ all\ others.\)
\(\displaystyle However,\ you\ may\ think\ of\ it\ as\ \underline{bringing\ all\ the\ T\ terms\ together}.\)

Ah, thanks both!