# Is my book wrong?

• June 2nd 2008, 07:24 PM
zellster87
Is my book wrong?
This simplification came directly from my book. I think it is wrong, but I may just not understand it fully. Here it is:

$T'(t) = \\-4t(2t^2 + 1)^-2 <2t,2t^2,1> +\\ (2t^2 +1)^-1 <2,4t,0> \\ = (2t^2 + 1)^-2 <-8t^2 + 4t^2 + 2, \\-8t^3 + 8t^3 + 4t, -4t> \\ = 2(2t^2 + 1)^-2 <1-2t^2, 2t, -2t>$

I don't understand how they factored out the (2t^2 + 1)^-2. Is it wrong, or am I just missing something? Thanks.
• June 2nd 2008, 07:30 PM
mr fantastic
Quote:

Originally Posted by zellster87

This simplification came directly from my book. I think it is wrong, but I may just not understand it fully. Here it is:

$T'(t) = \\-4t(2t^2 + 1)^-2 <2t, 2t^2, 1> +\\ (2t^2 +1)^-1 <2, 4t, 0> \\ = (2t^2 + 1)^-2 <-8t^2 + 4t^2 + 2, \\-8t^3 + 8t^3 + 4t, -4t> \\ = 2(2t^2 + 1)^-2 <1-2t^2, 2t, -2t>$

I don't understand how they factored out the (2t^2 + 1)^-2. Is it wrong, or am I just missing something? Thanks.

Quote:

Originally Posted by zellster and edited by Mr F
This simplification came directly from my book. I think it is wrong, but I may just not understand it fully. Here it is:

$T'(t) = \\-4t(2t^2 + 1)^{-2} <2t, \, 2t^2, \, 1> ~ + ~ (2t^2 +1)^{-1} <2, \, 4t, \, 0> \\$

$= (2t^2 + 1)^{-2} <-8t^2 + 4t^2 + 2, \, -8t^3 + 8t^3 + 4t, \, -4t> \\$

$= 2(2t^2 + 1)^{-2} <1-2t^2, \, 2t, \, -2t>$

I don't understand how they factored out the $(2t^2 + 1)^{-2}$. Is it wrong, or am I just missing something? Thanks.

Now we can read the expression!
• June 2nd 2008, 07:44 PM
mr fantastic
Quote:

Originally Posted by zellster87
This simplification came directly from my book. I think it is wrong, but I may just not understand it fully. Here it is:

$T'(t) = \\-4t(2t^2 + 1)^-2 <2t,2t^2,1> +\\ (2t^2 +1)^-1 <2,4t,0> \\ = (2t^2 + 1)^-2 <-8t^2 + 4t^2 + 2, \\-8t^3 + 8t^3 + 4t, -4t> \\ = 2(2t^2 + 1)^-2 <1-2t^2, 2t, -2t>$

I don't understand how they factored out the (2t^2 + 1)^-2. Is it wrong, or am I just missing something? Thanks.

The expression is correct. The line in red should make this clear:

$T'(t) = -4t(2t^2 + 1)^{-2} ~ <2t, 2t^2, 1> ~ + ~ (2t^2 +1)^{-1} <2, 4t, 0>$

${\color{red}= (2t^2 + 1)^{-2} ~ <-8t^2, -8t^3, -4t> ~ + ~ (2t^2 +1)^{-2} <2(2t^2 + 1), 4t(2t^2 + 1), 0> }$

$= (2t^2 + 1)^{-2} ~ <-8t^2 + 4t^2 + 2, -8t^3 + 8t^3 + 4t, -4t>$

$= 2(2t^2 + 1)^{-2} ~ <1-2t^2, 2t, -2t>$
• June 2nd 2008, 07:53 PM
zellster87
Ahh I see. Thanks Mr. fantastic.